AOD-1:

Single Correct Answer Type Questions

1. Equation of tangent to curve x = a cos3t, y = a sin3t at 't' is -

(A) x sect – y cosec t = a (B) x sec t + y cosec t= a
(C) x cosec t + y cosec t = a (D) None of these
Ans: B
dx dy
Sol. = – 3 a cos2tsin t = 3a sin2t cost
dt dt
dy
= – tan t  y– a sin3t= – tan t (x– acos3t)
dx
 y cost + x sin t – a sin3 t cos t–a sin t –cos3 t= 0
 x tan t+ y cos t – a t sin t cos t = 0
 x sec t + y cosec t = a

2. The equation of the normal to the curve y = x + sin x cos x at x = /2 is -

(A) x = 2 (B) x =  (C) x +  = 0 (D) 2x = 
Ans: D
  
Sol. at x = ; y = +0=
2 2 2
 
 point =  , 
2 2
 dy  2 2
     = 1 + cos x – sin x
dx
  , 
2 2
=1–1=0

 normal x = x1 =
2
 2x  

3. The point on the curve y = x2 – 3x + 2 where tangent is perpendicular to y = x is -

(A) (0, 2) (B) (1, 0) (C) (–1, 6) (D) (2, –2)
Ans: B
dy
Sol. = 2x – 3 & slope of given line = 1
dx
 line  tangent
 dy 
  × 1 = – 1 2x – 3 = – 1
 dx 
 x = 1, y = 0
 point (1, 0)

4. The tangent to the curve x = a cos 2 cos , y = a cos 2 sin  at the point corresponding to 
= /6 is
(A) parallel to the x-axis (B) parallel to the y-axis
(C) parallel to line y = x (D) None of these
Ans: A
dx a cos  sin 2 (cos 2 sin   cos  sin 2) a sin 3
Sol. = –a cos 2 sin  + = –a =
d cos 2 cos 2 cos 2
 dy sin  sin 2 a cos 3
=a cos 2 cos  –a =
d cos 2 cos 2
dy dy
Hence = – cot 3 =0
dx dx  /6
So, the tangent to the curve at  = /6 is parallel to the x- axis

5. If tangent at any point of the curve y = x3 + x2 + x + 5 makes acute angle with x-axis, then -
(A) 0 << 3 (B) – 3 << 3 (C) || < 1 (D)  (0, 1)
Ans: B
Sol. acute angle  1st quadrant
tan > 0
dy
> 0  3x2 + 2x + 1 > 0
dx
Possible for D < 0  42 – 12 < 0
2 – 3 < 0  ( – 3 ) ( + 3 ) < 0
  3 3

6. If the tangent to the curve 2y3 = ax2 + x3 at the point (a, a) cuts off intercepts  and  on the
coordinate axes, where 2 + 2 = 61 then the value of | a | is -
(A) 16 (B) 28 (C) 30 (D) 31
Ans: C
dy 2ax  3x 2
Sol. The slope of the tangent is =
dx 6y 2
and the value of this slope at (a, a) is 5/6. Therefore, the equation
5 x y
y–a= (x – a)  + = 1,
6 a / 5 a/6
represents the tangent. Thus the x-intercept  is –a/5, and the y-intercept  is a/6. From
2 2 a2 a2
 +  = 61, we now get  = 61
25 36
a2 = 25 × 36 a = ± 30.

7. The points of contact of the vertical tangents to x = 2 – 3 sin , y = 3 + 2 cos  are

(A) (2, 5), (2, 1) (B) (–1, 3), (5, 3) (C) (2, 5), (5, 3) (D) (–1, 3), (2, 1)
Ans: B
dx  3
Sol. For vertical tangents = 0 so, we have–3cos  = 0  = or . Corresponding to these
d 2 2

values of , we havex = 2 – 3 sin = –1,
2

y = 3 + 2 cos = 3;
2
3
x = 2 – 3 sin = 2 + 3 = 5,
2
3
y = 3+ 2 cos =3
2
Thus the required points are (–1, 3), (5, 3).
  x
8. All the points on the curve y2 = 4a  x  a sin  at which the tangents are parallel to the axis of
 a
x, lie on a
(A) circle (B) parabola (C) line (D) None of these
Ans: B
dy x 1
Sol. 2y = 4a[1 + a cos . ]
dx a a
dy
 = 0 1 + cos x/a = 0 x/a = (2n + 1)
dx
x = (2n + 1)aFor these value’s of x, sin x/a = 0
 all points lie on parabola y2 = 4ax

9. The curve y = ax3 + bx2 + cx + 8 touches x- axis at P(–2, 0) and cuts the y-axis at a point Q
where its gradient is 3. The values of a, b, c are respectively
1 3 1 1 7
(A) – , – , 3 (B) 3, – , – 4 (C) – , – , 2 (D) None of these
2 4 2 2 4
Ans: D
dy
Sol. = 3ax2 + 2bx + c
dx
Since the curve touches x- axis at (–2, 0) so
dy
= 0  12a – 4b + c = 0... (i)
dx ( 2,0)
The curve cut the y- axis at (0, 8) so
dy
=3c=3
dx (0,8)
Also the curve passes through (–2, 0) so
0 = –8a + 4b –2c + 8  –8a + 4b –2 = 0 (ii)
1
Solving (i) and (ii) a = – ,b=0
4

10. If the relation between subnormal SN and subtangent ST at any point S on the curve
by2 = (x + a)3 is p(SN) = q(ST)2, then p/q =
a 8a 8b 8b
(A) (B) (C) (D)
27b 27b 27a 27
Ans: D
Sol. by2 = (x + a)3
differentiating both sides
dy
2by = 3(x + a)2.1
dx
dy 3 (x  a) 2
=
dx 2 by
dy 3 (x  a) 2
length of subnormal = SN = y =
dx 2 b
dx 2by 2
and length of subtangent = ST = y =
dy 3(x  a) 2
p (ST) 2
= (given)
q (SN)
 (2by 2 ) 2 .2b
=
{3(x  a)2 }2 .3(x  a)2
8b {(x  a)3 }2
= {using by2 = (x + a)3
27 (x  a) 6

8b
=
27
p 8b
=
q 27

11. The angle between the tangents at any point P and the line joining P to origin O, where P is a
point on the curve ln(x2 + y2) = c tan–1 y/x, c is a constant, is
(A) constant (B) varies at tan–1 (x)
–1
(C) varies as tan (y) (D) None of these
Ans: A
Sol. P(x, y) be a point on the curveln (x2 + y2) = c tan–1 y/x
differentiating both side with respect to x
2x  2yy ' c(xy ' y) 2x  cy
 2 y'= = m1
(x  y )
2 2
x y 2
cx  2y
slope of OP = y/x = m2
2x  cy y

m1  m 2 cx  2y x
So tan  = = = 2/c
1  m1m 2 2xy  cy 2
1 2
cx  2xy
–1
 = tan (2/c) which is independent of x and y

12. The angle of intersection of curves

y = [|sin x| + |cos x|] and x2 + y2 = 5 where [·] denotes the greatest integer function is
1 
(A) tan–1 (2) (B) tan–1   (C) tan–1 ( 2) (D)
2 2
Ans: A
Sol. We know that 1  | sin x| + |cos x|  2
So that [|sin x| + |cos x|] will be constant function y = 1
No
x2 + y2 = 5
P Q
(–2, + 1) (2, 1)
y=1

Now intersection point P and Q are (–2, 1) and (2, 1)

Slope of line y = 1 is zero and slope of tangent at P and Q are (–2) and (2) respectively
Thus the angle of intersection is tan–1 (2)

13. If the curves y2 = 6x, 9x2 + by2 = 16, cut each other at right angles then the value of b is -
(A) 2 (B) 4 (C) 9/2 (D) None of these
Ans: C
Sol. The intersection of the two curves is given by 9x2 + 6bx – 16 = 0 (i)
dy 3
Differentiating y2 = 6x, we have = .
dx y
dy 9x
Differentiating 9x2 + by2 = 16, we have =– .
dx by
3  9x 
For curves to intersect at right angles, we must have at the points of intersection   =–
y  by 
1  27x = by2. Thus we must have
9x2 + by2 = 16  9x2 + 27x – 16 = 0 (ii)
(i) and (ii) must be identical so 27 = 6b  b = 9/2.

14. If tangent to the curve y2 = x3 at point (m2, m3) is also a normal to the curve at point (M2, M3),
then mM is equal to -
(A) – 1/9 (B) – 2/9 (C) – 1/3 (D) – 4/9
Ans: D
dy
Sol. y2 = x3 differentiating  2y. = 3x2
dx
2 4
 dy  3x 3m 3
  = = 3
= m
 dx (m2 ,m3 ) 2y 2m 2
–1  2y  2M3 2
Slope of normal = =–  2 = – =–
 dy   3x  3M 4
3M
  2 3
 dx  (M ,M )
3 2 4
 m= –  mM  –
2 3M 9

x2
15. If the normal of y = f (x) at (0, 0) is given byy – x = 0,then Lim ,
x 0 f (x 2 )  20f (9x 2 )  2 f (99x 2 )
is equal to
1 1 1
(A) (B) (C) (D) D.N.E.
19 19 2
Ans: B
Sol. Equation of normaly = x
Slope of normal at (0, 0) is = 1
1
=1  f ' (0) = – 1
f ' (0)
x2 0 
Lim 2 2 2  form 
x  0 f ( x )  20 f (9 x )  2 f (99 x )  0 
2x
= Lim
x  0 2 x f ' ( x )  360x f ' (9 x 2 )  396x f ' (99x 2 )
2

1
= Lim
x  0 f ' ( x )  180 f ' (9 x 2 )  198 f ' (99x 2 )
2

1
=
f ' (0)  180 f ' (0)  198 f ' (0)
1 1
= = Ans
(1  180  198) f ' (0) 19
Multiple Answer Type Questions

16. If the line y = 1 – 4x touches the curve y = x3 + ax2 – bx + cat(0, 1)and also touches the curvey
x2
= – px + 3at()then
2
(A) b + c = 5. (B) sum of all possible value ofp is 8.
(C) sum of all possible values ofis0. (D) sum of all possible values ofis 2.
Ans: A, B, C, D
Sol. (0, 1) lies on first curvec = 1
dy
Also, = – 4– b = – 4b = 4
dx
 b+c=5
x2
Also 1 – 4x = – px + 3 x2 + (8 – 2p)x + 4 = 0
2
D = 4 (4 – p)2 – 16 = 0
(4 – p) = ± 2
if p = 6x = 2
p = 2 x = –2
point of contact with second curve(2, –7) on (–2, 9)

17. If tangent drawn to the curve f(x) = x3 – 9x – 1atP x 0 , f ( x 0 )  meets the curve again atQ, mA
denotes the slope of the tangent at A andmOB denotes the slope of the line joining'O' origin and
a point B on the curve, then
(A) mQ – 4mp = 27 (B) mQ – 4mp = 9
m OP m OP 1
(C) = 2, where x0 = 1 (D) = , wherex0 = 1
m OQ m OQ 2
Ans: A, C
Sol. f(x) = x3 – 9x – 1
f (x0) = 3x02 – 9
2 x13  9x1  1  x 30  9x 0  1
3x0 – 9 =
x1  x 0
3x0 – 9 = (x1 + x1x0 + x02) – 9
2 2
 x12 + x1x0 – 2x02 = 0
 (x1 + 2x0) (x1 – x0) = 0  x1 = –2x0
3
x  9x1  1
Now,mP = 3x02 – 9 mOQ = 1
x1
(A) mQ = 4mP = 3x1 – 9 – (3x0 – 9)=3(–2x0)2 – 9 – 4(3x02 – 9) = 27
2 2

m op x 30  9x 0  1 1 9 1  9 ( 2 )
 = = =2
m OQ  x1  9x1  1 
3   8  18  1  9
x 0   1


x   2 
 1 

18. Equation of a tangent to the curve x3 + 3x2y – y3 + x2 – y2 = 0 at the origin is

(A) x – y = 0 (B) x + y = 0 (C) x – 2y = 0 (D) 2x + y = 0
Ans: A, B
Sol. Equate to zero the lowest degree term

19. For the curve y = bex/a

(A) the subtangent is of constant length
 (B) the subnormal is of constant length
(C) the subnormal varies as the square of ordinate
(D) the subtangent varies as the radius vector
Ans: A, C
Sol. y = bex/a
consider a point (x1, y1) on the curvey1 = bex1 /a …(1)
y = bex/a
 dy  b
  = e x1 /a
 dx (x1 ,y1 ) a
 dx  a
length of subtangent = y1   = bex1 /a x1 /a  a(constant)
 dy  x1 ,y1  be

 dy  b 1 1
= be x1 /a e x1 /a =  be x1 /a   y12
2
length of subnormal = y1  
 dx  x1 ,y1  a a a

20. The abscissa of the point on the curve x y = a + x, the tangent at which cuts off equal
intersects from the co-ordinate axes is : ( a > 0)
a a
(A) (B)  (C) a 2 (D)  a 2
2 2
Ans: A, B
a2 dy a2
Sol. xy = a2 + x2 + 2axy = + x + 2a =  2 + 1= 1
x dx x
a
 2x2 = a2 x = ± ]
2

x
21. If tangents are drawn from a point P (2, 0) to the curve 1 y 2 = which meet the curve at A
3
and B, then
 5
(A) Acute angle between the tangents is tan 1  

 2 
(B) AB = 5
5 5
(C) Area of PAB =
4
(D) PAB is equilateral triangle.
Ans: A, B, C
2 y  dy  1
Sol.  =
2 1  y 2  dx  3

dy 1  y2
=
dx 3y
 dy 1 k2
=
dx ( h ,k ) 3k

9 5 
 , 
2 2 
 
A
(h, k)

P
(2, 0)
B
9  5
 , 
2 2 
 
k0 1 k2 h h2
 =  1 k 2 =  1 + k2 =
h2 3k 3 9
 h2 
3   1
3k 2  9  h h2  9
= 1 k 2   =  =h
h2 h2 3 h2
9 5
h2 – 9 = h2 – 2hh = ; k = ± .
2 2
 1 
 2·   
 1  5  = tan 1  5  .
(A) Angle between tangents  = 2 tan 1  = tan 1  
 5  1   2 
 1 5 
 
(B) AB = 5
1 5 5 5
(C) Area of PAB =   5 = sq. units.
2 2 4
(D) PAB is not equilateral.

 3u  u 3 
22. If a curvey = f(x)is parametrically represented asy = tan–1   , x = sin–1  2u  , then
2  1 u2 
 1  3u 
(A) equation of tangent toy = f (x)atu = 0is 3x – 2y = 0.
(B) equation of tangent toy = f (x)atu = 3 is3x + 2y = .
(C) equation of tangent toy = f (x)atu = 0is 2x – 3y = 0.
(D) equation of tangent toy = f (x) atu = 3 is2x – 3y = .
Ans: A, B
      
Sol. u = tan ,   ,    
 2 2   6
y = tan–1 (tan 3),x = sin–1 (sin 2)
y = 3 ,x = 2in vicinity ofu = 0
y = 3 – ,x =  – 2in vicinity ofu = 3
 dy 3 dy 3
 = , =
dx u 0 2 dx u 3 2
3
T( 0, 0) ,y = x3x – 2y = 0
2

Ifu = 3 ,x = ,y = 0
3
3 
 T , 0  : y = x  
3 2  3
3x + 2y = 

1 1
23. The values of a for which y  ax 2  ax  , x  ay 2  ay  touch each other are
24 24
2 3 13  601 13  601
(A) (B) (C) (D)
3 2 12 12
Ans: A, B, C, D
Sol. Point of contact will lie an y  x (let say (, )) and slope of tangent can be  1 and
1
  a  2  a  eliminating  we get
24
2 2
 1  a   1  a  1
   a  a
 2a   2a  24
2 3 13  601
On solving we get a  , ,
3 2 12

Numerical Answer Type Questions

7
24. The tangent to y = ax2 + bx + at (1, 2) is parallel to the normal at the point (–2, 2) on the
2
curve y = x2 + 6x + 10.Find the value of a – 2b.
Ans: 6
dy
Sol. Slope of tangent at (–2, 2) on curve y = x2 + 6x + 10 is = 2.
dx
7 1
So slope of tangent at (1, 2) of curve y = ax2+bx+ is 
2 2
1 1
[2ax + b]x=1 =   2a + b =  ....(i)
2 2
7
 point (1, 2) also lies on y = ax2 + bx +
2
7 3
so2 = a + b + a+b=  .....(ii)
2 2
5
on solving (i) and (ii) a = 1 and b = 
2

25. If x + my = 1 touches the curve (ax)n + (by)n = 1.

n n
   n 1  m  n 1
If      = p, then value of p is ..........
a b
Ans: 1
Sol. Tangent at P(x1, y1) is a n .x1n 1.x  b n .y1n 1.y =1 .... (i)
 but as givesx + my = 1 ....(ii)
 m
Hence eq. (i) & (ii) represent coincident lines, so n n 1 = n n 1 = 1
a x1 b y1
 m
so x1n 1 = n ; y1n 1 = n
a b
 m
 (ax1)n–1 = ; by1 =
a b
as (x1, y1) lies on the curve so (ax1)n + (by1)n = 1
n n
   n 1  m  n 1
    = 1.
a b

x2 y2
26. Let  be the angle in radians between + = 1 and the circle x2 + y2 = 12 at their points
36 4
k k2
of intersection. If  = tan–1 , then find the value of .
2 3 4
Ans: 4
Sol. The points of intersection,  y = ± 3 and x = ± 3
consider the point P(3, 3 )
Equation of the tangent at P to the circle is
3x + 3 y = 12
x 3
Equation of the tangent at P to the ellipse is + y=1
12 4
if  is angle between these tangents, then
2
tan =
3

27. The tangent to the curvey = x3at point M meets the curve again at N. Find the ratio of gradient
to the curve at N to the gradient at M.
Ans: 4
 dy  2
Sol.    3x1
 dx  M
 Tangent at M isy – x13  3x12 (x – x1)
This tangent passes throughN(x2, x 32 )
 x 32  x13 = 3x12 (x2 – x1)
y y=x
3

N(x2y2)

M(x1,y1)

 (x2 – x1)2 (x2 + 2x1) = 0x2 = –2x1

  dy 
  2
 dx  N 3x 2
 = 2 =4
 dy  3x1
 
 dx M

28. Let tangents be drawn to the curve y2 – 4(x + y) = 3 sin  + 4 cos  – 15,  Rfrom the origin
when the vertex of the curve is at the maximum distance from the origin. If slopes of these
1
tangents are m1 and m2 then find the value of .
m1m 2
Ans: 4
Sol. y2 – 4 (x + y) = 3 sin  + 4 cos  – 15
y2 – 4y + 4 = 4x + 3 sin  + 4 cos  – 11
 11  3 sin   4 cos  
(y – 2)2 = 4  x  
 4 
2
(y – 2) = 4(x – )
11  3 sin   4 cos  3 
=   , 4 
4 2 
Value of  for which vertex (, 2) situated at maximum distance from the origin is 4.
 (y – 2)2 = 4(x – 4)
Now, equation of tangent of slope m is
1
y – 2 = m(x – 4) +
m
which passes through the (0, 0)
1 2  4  16
 –2 = –4m + 4m2 – 2m – 1 = 0m =
m 8
2  2 5 1 5
 =
8 4
1 1
 m1m2 =  =4
4 m1m 2

29. If the tangent at P(a, b) to the curve x3 + y3 = c3 meet the curve again in Q(a1, b1), then find the
value of
a b 
–  1 1
a b
Ans: 1
a2
Sol. Slope of tangent at (a, b) = – 2
b
b1  b b1  b a 2
slope of PQ = ; =
a1  a a1  a b 2
a3 – a13 = b13 – b3
 (a – a1) (a2 + a12 + aa1) = (b1– b) (b22 + b2 + bb1)
 b  b  a 2  a12  aa1 a 2  a12  aa1 a 2
–  1 =
 2 so = 2
 a1  a  b  b1  bb1 b  b1  bb1 b
2 2 2

a b
on solving 1 + 1 = –1.
a b
 1
30. The chord of the parabola y = a2x2 + 5ax  4 touches the curve y = at the point x = 2 and
1 x
is bisected by that point. Find 'a'.
Ans: 1
Sol. y = – ax2x2 + 5ax – 4
equation of tangent to
1
y= at (2, – 1) is y + 1 = (x + 2)
1 x
isx – y = 3

(x1 , y1 ) (2, – 1)
(x2 , y2 )

1
y + 1 = (x – 2) y=
1–x
dy = 1
dx (2,1) (1 – x)2
this line cuts the parabola at 2 distinct point.
x – 3 = – a2x2 + 5ax – 4
a2x2 + (1 + 5a)x + 1 = 0 ....(1)
x1  x 2
now =2  x1 + x2 = 4;
2
1 5a
2
=– 44a2 – 5a + 1 = 04a2 – 4a – a + 1 = 0 4a(a – 1) – (a – 1) = 0
a
 a = 1ora = 1/4
but roots of (1) must be realthereforeD > 0
i.e. (1 – 5a)2 – 4a2> 0
21a2 – 10a + 1 > 0
21a2 – 7a – 3a + 1 > 0
7a(3a – 1) – 1 (3a – 1) > 0

1 1 1
7 4 3
1 1
a> ora <
3 7
1
so a  hence a = 1 Ans.
4

Passage Type Questions

Paragraph for question nos. 31 to 32
 x 2  10x  8 ; x  2
 2
Given a continuous function y = f(x) = ax  bx  c ; 2  x  0, a  0
 x 2  2x ; x0

If a line L touches the graph of y = f(x) at three points then

31. The gradient of 'L' is equal to:

(A) 16 (B) 2 (C) 4 (D) 6
Ans: A, C
Sol. Let y = mx + c is the required line on solvingy = mx + c and
y = x2 + 10x + 8  x2 + (10 –m)x + 8 – c = 0
and on solving y = mx + c & y = x2 + 2x x2 + x(2 –m) – c = 0
 D = 0 of both quadratic so m = 4 and c = –1

32. The value of (a + b + c) is equal to:

(A) 5 2 (B) 49 (C) 6 (D) 7
Ans: B, D
Sol. So y = 4x – 1 also touches y = ax2 + bx + c c = 0; so (b – 4)2 –4a = 0 …(1)
f (–2) = –8 = 4a –2b + c ....(2)
on solving (1) and (2) and (3) a = 1, b = 6, c = 0
f(0) = 0 = c …(3)

Paragraph for question nos. 33 to 34

Two functionsf (x)andg (x)are connected by the relation f ( x )  x 2  1  g ( x )  ln 2 | x |

=0 x  R – {0}and f (0) = g(0) = 1.

33. Number of solutions of the equationf(x) – g(x) = 0, is

(A) 2 (B) 3 (C) 5 (D) 1
Ans: C

1
34. y-intercept made by the tangent drawn to the curve y = g(x)atx = – ,is
2
(A) 1 – ln 2 (B) 1 + ln 2 (C) – ln 2 – 1 (D) – 1 + ln 2

Ans: B
Sol. (33-34)
f ( x )  x 2  1  g ( x )  ln 2 | x | = 0

f(x) = – x2 + 1 g(x) = ln 2 | x | g(x) = ln | x |

From the graph of y = f(x) and y = g(x) it is clear that both curves intersect at 5 points.
g(x) = ln | x | = –ln (–x) –1 < x < 0
1  1 
at x=– y = ln 2    , ln 2 
2  2 
1 1
g'(x) = – ( 1)  
x x
 1 
g'    = 2
 2
1
Equation of tangent at x = 
2
y

1
g(x)

O x
–1 f(x)
  1
y – ln2 = 2  x   2x – y + 1 + ln2 = 0
 2
 y – intercept  y = 1 + ln2

35. Match the column:

Column-I Column-II
 sin x  cos x  (P) domain is R
(A) f (x) = sin–1  
 2 
2  (Q) range contain only
(B) g (x) = sin–1  tan 1 x 
  one integer
(R) odd function
(C) h (x) = tan–1  2 (2 tan 1 x  sin 1 x  cot 1 x  cos1 x ) 
 
–1 3
(D) j(x) = tan (x + x) (S) no vertical tangent

Ans: (A) P, Q, S;(B) P, R, S;(C) Q, R, S;(D) P, R, S

 sin x  cos x  –1  2 sin x  (  4)  
Sol. (A) We knowsin–1   = sin   lie between – 1 , 1 for all
 2  2  2 2
 
xR
domain is x  R
1
sin–1x is increasing function and max. value of given function is sin–1 which is /4
2
 
and minimum value sin–1  1 2 which is – /4, so range contain only one integer which
is zero
(Q) is correct
   
f (x) = sin–1  sin x    not odd (R) is not correct
  3 
vertical tangent exist for sin–1x is x = ± 1 but this is not possible so (S) is also correct.
2 
(B) g (x) = sin–1  tan 1 x 
 
2
– 1 < tan 1 x < 1 for all x  R  (P) is correct

 range is (– /2, /2)  (Q) is wrong
g(x) is odd  (R) is correct
and no vertical tangent  (S) is also correct
 2
 
(C) h (x) = tan–1  2 tan 1 x  cot 1 x  (sin 1 x  cos 1 x )  
 
2     2 
= tan–1   tan 1 x    = tan–1  tan 1 x 
  2 2   
It is easy to analysedomain is [–1, 1]
1 2 1
–  tan–1x  forx  [–1, 1]
2  2
1
 tan–1  1  h(x) contain only one integer which is zero
2
h (x) is odd
no vertical tangent.
 (D) domain is R (cubic polynomial)
range is (– /2, /2)
x3 + x is odd
 tan–1(x3 + x) is also odd
no vertical tangent

36. Match the column:

Column – I Column - II
(A) If position of the tangent at any point on the curve x  at , y  at3 4 (P) 1
between the axes is divided by the abscissa of the point of contact in the
ratio m : n externally, then |n + m| is equal to (m and n are co-prime)

(B) The area of triangle formed by normal at the point (1, 0) on thecurve (Q) 1/2
x  esin y with axes is
(C) If the angle between curves x 2 y  1and y  e 21 x  at the point (1, 1) is (R) 7
 then tan  is equal to
(D) x (S) 3
The length of sub-tangent at any point on thecurve y  be 3 is equal to

(T) 0

Ans: A - R ;B -Q ;C -T ; D - S
dy 4t
Sol. (A) 
dx 3
4t
Tangent is y  at 4 
3
 x  at 3 

at 3 at 4
x-intercept = ,y-intercept = 
4 3
 at 3   at 4 
the point of intersection of tangent with the axes are  , 0  and  0,  
 4   3 
m 3
  som = 3, n = 4 andm + n = 7
n 4
dx
(B)  esin y cos y , so slope of normal = –1
dy
equation of normal is x + y = 1
1
Area =
2
1 dy 2
(C) y  2 :   3 : slope of tangent = –2
x dx x
dy
y  e 2 2 x :  e 2  2x  2  : slope of tangent = –2
dx
 tan   0
x
y be 3
(D) Length of sub-tangent   3
y x
1
be  
3

3
AOD-2:
Single Correct Answer Type Questions

1. If f(x) = xex(1 – x), then f(x)

 1 
(A) Increasing on   , 1 (B) Decreasing on R
 2 
 1 
(C) Increasing on R (D) Decreasing on   ,1
 2 
Ans: A
Sol. f '(x) = ex(1 – x) + xex(1 – x) (1 – 2x)
f '(x) = ex(1 – x) (1 + x – 2x2)
f '(x) = – ex(1–x) (2x + 1) (x – 1)
wavy curve Method sign of f '(x)
– –
–1/2 + 1
 1   1
f(x) is increasing in  ,1 and f(x) is decreasing in   ,   [1, )
 2   2

x x
2. If f(x) = and g(x) = where 0 < x  1, then in this interval
sin x tan x
(A) both f(x) and g(x) are increasing functions
(B) both f(x) and g(x) are decreasing functions
(C) f(x) is an increasing function
(D) g(x) is an increasing function
Ans: C
sin x  x cos x tan x  x sec 2 x
Sol. f(x) = , g(x) =
sin 2 x tan 2 x
Let u(x) = sin x – x cos x, so that u(x) = x sin x > 0 for 0 < x 1.
So u(x) >u(0) = 0.So f(x) > 0 for 0 < x  1.
Hence f increasing on (0, 1].
Let v(x) = tan x –x sec2 x, so that v(x) = –2x sec2 x tan x < 0 for 0 < x  1.
Thus v(x) < v (0), i.e., g(x) < 0 for 0 < x 1. So g decreases on (0, 1]

3. The function f(x) = sin4 x + cos 4 x increasing if -

(A) 0 < x </8 (B) /4 < x < 3/8 (C) 3/8 < x < 5/8 (D) 5/8 < x < 3/4
Ans. B
Sol. Here,ƒ(x) = sin4 x + cos4 x
ƒ(x) = 4 sin3x . cos x + 4 cos3 x (– sin x)
ƒ(x) = 4 sin x cos x (sin2 x – cos2 x)
ƒ(x) = 2 (sin 2x) (– cos 2x)
ƒ(x) = – sin 4x
Now, ƒ(x)  0 if sin 4x  0
 4x  2
 /4  x /2
Therefore, (B) is the solution.

4. f(x) = 1 + x log (x  x 2  1) – 1  x 2 where x is any real number then f(x) is

(A) increasing in (–) (B) increasing in (–, 0)
(C) increasing in (0, ) (D) None of these
Ans: C
Sol. f(x) = 1 + x log (x  x 2  1) – 1  x 2
x  2x  2x
f '(x) = 1  × 1   + log (x  x  1) .1 –
2

x  x 1  2 x 1 
2 2
2 1 x2
x x
=1+ + log (x  x 2  1) –
x 1
2
1 x2
= 1 + log (x  x 2  1)
When x > 0 f '(x) is > 0  f(x) is increasing

5. Function f(x) = | x –2 1 | is monotonic decreasing in -

x
(A) (– , ) (B) (0, 1) (C) (2, ) (D) (– , 1)  (2, )
Ans: D
1 – x
 x 2 , x  1
Sol. f(x) = 
 x –1 , x  1
 x 2
 1 2
 x 2 – x 3 , x  1
f (x) = 
 2 – 1 , x 1
 x 3 x 2
f(x) is decreasing f (x) < 0
x – 2
 x 3  0, x  1
  x < 1, x > 2
 2 – x  0, x  1
 x 3

a sin x  b cos x
6. If ƒ(x) = decreases for all x if -
c sin x  d cos x
(A) ad – bc< 0 (B) ad – bc> 0 (C) ab – cd > 0 (D) ab – cd < 0
Ans: A
(a cos x  b sin x) (c sin x  d cos x)  (a sin x  b cos x)(c cos x – d sin x)
Sol. ƒ(x) =
(c sin x  d cos x) 2
ad(cos 2 x  sin 2 x)  bc(cos 2 x  sin 2 x) ad  bc
= =
(c sin x  d cos x) 2
(c sin x  d cos x) 2
ƒ decreases for all x if and only if ƒ(x) < 0 for all x i.e., ad – bc< 0.

7. Let ƒ be a differentiable function on R and h(x) = ƒ(x) – (ƒ(x))2 + (ƒ(x))3 for all x  R. Then-
(A) h increases whenever ƒ decreases
(B) h decreases whenever ƒ increases
(C) h increases or decreases according as f increases or decreases
(D) nothing can be claimed in general
Ans: C
Sol. h(x) = ƒ(x) – 2ƒ(x) ƒ(x) + 3(ƒ(x))2 ƒ(x) = ƒ (x) [1 – 2ƒ(x) + 3(ƒ(x))2]
Since 1 – 2ƒ(x) + 3(ƒ(x))2
  1 2
2

= 3  ƒ(x)     > 0 for all x so h(x) > 0 or < 0 according as ƒ(x) > 0 or < 0. Hence h
 3  9 
increases or decreases according as ƒ increases or decreases.

8. f(x) = e2x – (a + 1)ex + 2x is monotonic increasing for all xR is

(A) (3, 4) (B) (– , 0) (C) (– , 3] (D) (3, )
Ans: C
Sol. f (x) = 2e2x + (a + 1)ex + 2  0  xR
i.e. 2 e x  x  – (a + 1)  0 xR
1
 e 
 4 – (a + 1)  0ora  3

9. Function f(x) = x3 + 6x2 + (9 + 2k) x + 1 is increasing function if

(A) k  3/2 (B) k > 3/2 (C) k < 3/2 (D) k 3/2
Ans: A
Sol. f  (x) = 3x2 + 12x + (9 + 2k)  0
144 – 4 × 3 × (9 + 2k)  0
12 – 9 – 2k  0
k 3/2

10. Let f(x) = 2x3+ ax2 + bx – 3cos2x is an increasing function for all x  R, then
(A) a2 – 6b – 18 > 0 (B) a2 – 6b + 18 < 0 (C) a2 – 3b – 6 < 0 (D) a > 0, b > 0
Ans: B
Sol. f(x) = 2x3 + ax2 + bx – 3cos2x
f '(x) = 6x2 + 2ax + b + 3 sin 2x > 0
6x2 + 2ax + b – 3 > 0 as sin 2x  – 1
 4a2 – 4b(b – 3) < 0  a2 – 6b + 18 < 0

 a4  5
11. The set of all values of a for which f (x)    1 x  3x  log 5, a  1 , decreases for all
 1  a 
real x is
 3  21   5  27 
(A)  4 ,   (1, ) (B)  3 ,   (1, )
 2   2 
(C) (, 4) (D) none of these
Ans: A
 a4 
Sol. f ( x )    1 5 x 4  3
 1 a 
For f (x) to be a decreasing function for all real x,f (x)  0xR
a4  a4 
  1  0 If  1  0 , then f ( x )  0 for large x 
1 a  1 a 
 a  4  1  a , 4  a  1
 
 a  4  1  a , a  1
(i) a  4  1  a a  4  1  a true for all a> 1 …(i)
(ii) a  4  1  a , 4 a< 1
 a  4  1  a 2  2a
 3  21
 a 2  3a  3  0 and4 a< 1  a 
2
3  21
or a and  4  a  1
2
3  21
 4  a  …(ii)
2
From (i) and (ii) all possible values of a are given by
3  21
4  a  or1 <a<
2

 
12. Let f sin x  0 and f   sin x   0  x   0,  and g  x   f  sin x   f  cos x  , then g x  is
 2
decreasing in
       
(A)  ,  (B)  0,  (C)  0,  (D)  , 
4 2  4  2 6 2
Ans: B
Sol. g  x   f   sin x  .cos x  f   cos x  .sin x
 g  x   f   cos x  .sin x  cos 2 x .f   sin x   f   cos x  .sin 2 x  f   cos x  .cos x  0
 x  0,  2
 g  x  is increasing in 0,  2 .
  
Also g   4   0  g  x   0  x  ,  and g  x  0  x  0,  4
4 2

13. If ƒ(x) > 0 and ƒ(1) = 0 such that g(x) = ƒ (cot2 x + 2 cot x + 2), where 0 < x < then the
interval in which g(x) is decreasing is -
   3   3 
(A) (0, ) (B)  ,   (C)  ,   (D)  0, 
2   4   4 
Ans: D
Sol. Here, g(x) = ƒ(cot2 x + 2 cot x + 2)
g(x) = ƒ(cot2 x + 2 cot x + 2) ·{–2 cot x cosec2 x – 2cosec2 x}
for g(x) to be decreasing, g(x) < 0
ƒ{(cot x + 1)2 + 1} · (–2 cosec2 x) (cot x + 1) < 0
ƒ{(cot x + 1)2 + 1} · (cot x + 1) > 0 … (i)
{as ƒ(x) > 0  ƒ(x) is increasing, then
 3   3 
ƒ(cot x + 1)2 + 1} > ƒ(1) = 0 x   0,    ,  
 4   4 
Thus, equation (i) holds, if cot x + 1 > 0
 3 
cot x > –1  x   0, 
 4 
Hence, (D) is the correct answer.

14. In [0, 2], let g(x) =f (x) + f (2 x) and f (x) < 0, then g(x)
(A) increases in [0, 2]
(B) decreases in [0, 2]
(C) decreases in [0, 1) and increases in (1, 2]
(D) increases in [0, 1) and decreases in (1, 2]
Ans: D
Sol. Given, g(x) = f(x) + f(2 x), 0 x 2 …(i)
f (x) < 0, 0 x 2 …(ii)
g(x) = f (x) f  (2 x) …(iii)
From (ii),f (x) is a decreasing function.
Sign of 2(x  1)

0 ve 1 +ve 2
In [0, 1), x< 2 xf (x) >f (2 x)
 from (iii), g(x) > 0 and hence g(x) increases in [0, 1).
In (1, 2], x> 2 xf (x) <f  (2 x)
 From (iii), g(x) < 0 and hence g(x) decreases in (1, 2].

x  2 
15. If (x) = 3f   + f(3– x2)  x (–3, 4) wheref"(x) > 0  x  (–3, 4), then (x) is
 3 
3   3
(A) increasing in  , 4  (B) decreasing in  3,  
2   2
 3   3
(C) increasing in   ,0  (D) decreasing in  0, 
 2   2
Ans: A
 x 2  2x x  2 
Sol. '(x) = 3f '  

– 2x f '(3 – x2)= 2x [f '   – f '(3 – x2)]
  3
3  3 
x2
'(x) = 0  = 3 – x2 or x = ± 3/2
3
f "(x) > 0  f '(x) is increasing
+ + + +++
–3 –3/2 O 3/2 4

Multiple Answer Type Questions

x  p2 pq pr
16. If p, q, r be real, thenf(x) = pq x  q2 qr
pr qr x  r2
(A) increases in interval (0, )
 2(p 2  q 2  r 2 ) 
(B) decreases in interval  , 0
 3 
 2(p 2  q 2  r 2 ) 
(C) increases in interval  , 
 3 
(D) None of these
Ans: A, B, C
1 0 0 x  p 2 pq pr x  p2 pq pr
Sol. f(x) = pq x  q 2 qr + 0 1 0 + pq x  q2 qr
pr qr xr 2
pr qr x  r 2 0 0 1
2 2 2 2
= 3x + 2(p + q + r ) x
17. Let f  x   x 2 e x and g  x   c . Let x0  R such that f  x0   g  x0  have exactly two
solutions. Then which of the following is / are correct.
(A) g  x0   4e 2
(B) f  x   g  x  is a monotonic function
(C) f  x  is decreasing in  1,  1 / 2 
(D) area bounded by y  f  x  , y  f  x0  and x  0 is more than 1 (unit)2
Ans: A, C
Sol. f '( x)  ( x 2  2 x)e x So graph of y = f(x) is like this

1
18. Letf (x) = x  xe x
, x > 0 then which of the following is(are) correct?
(A) f (x) is increasing on (0, ) (B) f (x) is decreasing on (0, )
(C) Lim f x   1 (D) f (x) is non monotonic on (0, )
x 
Ans: A, C
1 e 1 / x 1  1
Sol. f ' (x) =1 – e –1/x – x e–1/x. 2 = 1 – e –1/x – = 1 / x  e1 / x  1  
x x e  x
1  1 1 1  1
= 1 / x 1     .......... ..   1  > 0  x  (0, )
e  x 2x 2 3! x 2  x 
 f (x)on (0, )
 1 
 e 1 / x  2 
Also Lim x (1 – e–1/x)= Lim
1 e1 / x

= Lim  x  = e0 = 1.
x  x 1/ x x  1
x2

1
19. For the function f  x   x cos , x  1 ,
x
(A) for at least one x in the interval [1, ), f  x  2   f  x   2
(B) lim f   x   1
x 

(C) for all x in the interval [1, ), f  x  2   f  x   2

(D) f   x  is strictly decreasing in the interval [1, )
Ans: B, C, D
1 1 1 1 1 1
Sol. f   x   cos  x sin   2  cos  sin
x xx x x x
lim f x   1
x 
  1  1  1 1 1  1  1  1 1 1 1 1 1
f   x    sin    2   2 sin  cos    2   2 sin    2 sin  3 cos
 x  x  x x x  x  x  x x x x x x
1 1
  3 cos  0,  [1, )
x x
f   x   lim f   x   1
x 

f  x  2  f  x 
f x   1 f  x  2   f  x   2
x 2x

20. Letf : R+ Ris a strictly decreasing function for allx R+such thatf (k2 – 2k) > f (3k – 4), then
k can be
5 10 11 21
(A) (B) (C) (D)
2 3 4 5
Ans: A, B, C
Sol. Given: k2 – 2k > 0 k(k – 2) > 0
+ +
0 – 2
 k  (–, 0)  (2, ) ……(i)
4 4 
 3k – 4 > 0 k > k   ,   ……(ii)
3 3 
 f is a decreasing function
 k2 – 2k < 3k – 4  k2 – 5k + 4 < 0
 1<k<4 ……(iii)
Common value of kfrom (i), (ii) and (iii)is
2<k<4
Correct value of k  (2, 4). So option A, B, C contains (2, 4).

 2 1
x  2x sin , if x  0
21. Consider the functiong(x) =  x
0, if x  0
which of the following hold(s) true?
(A) g(x)is continuous anddifferentiable atx = 0
(B) g'(x)is not continuous atx = 0.
(C) g'(0)exists.
(D) g(x)is strictly increasing atx = 0.
Ans: A, B, C, D
 2 1

 h  2h sin   0
g(0  h )  g(0) x 
Sol. We haveg'(0+) = Lim = Lim    g ' ( 0)  1
h0 h h0 h 

|||ly g '(0–) = 1
  2 1  1  1
1  2  x cos  2   2 x sin  if x  0
g '(x) =   xx  x
1 if x  0

Lim g ' ( x ) is oscillatory g ' (x) is not continuous at x = 0.
x0
Although g '(0) = 1 but nothing definite can be said about the validity of the inequalities.
g(0 + h) > g (0)andg (0 – h) < g (0) (D) is correct.
22. Letg : R  R be a differentiable function such thatg(x) g'(x) < 0 for allx  R. Then which of
the following is (are) correct?
(A) The possible graph ofg(x)may be
y

x
O

(B) The possible graph ofg(x)may be

y
x
O

(C) The function| g(x) |is always decreasing function.

(D) lim f   x   0
x 

Ans: A, B, C, D
Sol. g(x) andg'(x) must have opposite sign.Now, check the option.

23. Iff(x)andg(x)are two positive and increasing functions, then which of the following is NOT
always true?
(A) f x g ( x ) is increasing.
(B) if f x g ( x ) is decreasing, then f(x) < 1.
(C) if f x g ( x ) is increasing, then f(x) > 1.
(D) if f(x) > 1, then f x g ( x ) is increasing.
Ans: A, B, C
Sol. Let y = f ( x ) g ( x )
dy  f ' (x ) 
 = f(x)g(x) g( x )  g' ( x ) log f ( x )
dx  f (x) 
g(x))
f(x) , g(x), f(x), f(x),f '(x)andg'(x) are positive,
dy
but log f(x)can be negative, which can cause < 0.
dx
Hence, statement A is false.
dy
Iff(x) < 1log f(x) < 0, which does not necessarily make < 0, hence, statement B is false.C
dx
is false but reverse of C is true.

Numerical Answer Type Questions

24. If the range ofall real values of bfor which the function
f(x) = (b2 – 3b + 2) (cos2 x – sin2 x)+(b – 1) x+sin 2
does not possess any critical points onRis (b1, b2),then find the value of(b1 + b2).
Ans: 4
Sol. We havef (x) =(b2 – 3b + 2) (cos2x – sin2x) + (b – 1) x + sin 2
 f '(x)=(b – 1) (b – 2) (– 2 sin 2x) + (b – 1)
Now, f '(x)  0 for every x  R,
so (b – 1) 1  2(b  2) sin 2x  0  x  R
 b1
 1  3   5
Also, >1  b   , 2    2, 
2(b  2)  2   2
Now, when b = 2, f(x) = x + sin 2f '(x) = 1 ( 0).
3 5 3 5
Hence, b  ,   b1 = and b2 =
2 2 2 2
3 5 8
 (b1 + b2) =  = =4
2 2 2

x3 x2
25. If all the real values ofm for which the function f (x) =  ( m  3)  m x  2013 is strictly
3 2
increasing inx  [0, )is [0, k],then find the value ofk.
Ans: 9
x3 x2
Sol. f (x) =  ( m  3)  mx  2013
3 2
 
 f ' (x) = x 2  ( m  3) x  m 0,x [0, )
Case-I:WhenD  0  m  [1, 9] .... (I)
Case-II: WhenD  0  m  [– , 1] [9, ) .... (i)
b
 0 (m – 3)  0  m 3 ..... (ii)
2a
and f ' (0)  0  m0 ...... (iii)
 (1)  (ii)  (iii)
 m  [0, 1] .......(II)
So, finally(I)  (II)
 m  [0, 9][0, k]
 k=9

 
26. If the functionf(x) = 2 cot x + (2a + 1) ln | cosec x | + (2 – a)xis strictly decreasing in  0, 
 2
then range of'a'is[m, ), find the value ofm.
Ans: 0
Sol. Let f(x) = 2 cot x + (2a + 1) ln | cosec x | + (2 – a)x
f '(x) = – 2 cosec2 x – (2a + 1)cot x + (2 – a) = – 2 cot2x – (2a + 1)cot x – a
 
 f ' (x) = (cot x + a)(– 2 cot x – 1)  0in  0, 
 2
 
 cot x + a  0in  0, 
 2
Hence a0

 x2  x 1
27. If the function k(x) = loge  2  is strictly decreasing in x    t , t  , then find the

 x  x 1  7 7
greatest integral value of t.
Ans: 7
2( x  1)( x  1)
Sol. Here, k ' (x) = 2
( x  x  1)( x 2  x  1)
Now, k'(x) < 0 gives
(x – 1)(x + 1) < 0
–1 < x < 1
  t = 7.

3 1 2x
28. Letf (x) = x3 – ax2 + (3a – 10)x + 6andg(x) = + sin–1 , then the number of integral
2  1 x2
values of'a'for whichfis decreasing on the setA = {g(x) : x  R}(range set of the functiong).
Sol. 6
Clearly,g(x)  [1, 2]
f '(x)  0  x  [1, 2]
f '(x) = 3x2 – 2ax + 3a – 10  0  x  [1, 2]
f '(1) = 3 + a – 10  0, f '(2) = 12 – a – 10  0
a  7, a  2
 a  [2, 7]
Therefore 6 integral values ofa.]

29. Let f (x), x  0, be a non-negative continuous function. If f (x) cos x  f (x) sin x,  x  0, then
 5 
find f   .
 3
Ans: 0
Sol. Given f (x)  0  x  0 … (1)
f (x) cos x– f (x) sin x  0
 (f (x) cos x) 0 … (2)
Let G (x) = f (x) cos x.
G (x)  0
 G (x) is a decreasing function. (from (2))
π  5π   5π 
 G    G   G  0
 2  3   3 
 5π 
 f 0 ……(3)
 3 
 5π 
From statement (1) and (3), f   = 0.
 3 

d
30. If H(x0) = 0 for some x = x0 and H(x)  2cxH(x) for all x  x0, where c > 0, then number of
dx
values of x, (x > x0) so that H(x) is equals to zero is…………………
Ans: 0
d
Sol. Given that H(x)  2cxH(x)
dx
d
 H(x)  2cxH(x)  0
dx
d 2
 (H(x) e  cx )  0
dx
2
 H(x)e  cx is an increasing function.
2
But H (x0) = 0 and ecx is always positive.
 H (x) > 0 for all x > x0
 H (x) cannot be zero for any x > x0.
Passage Type Questions

Paragraph for question nos. 31 to 32

x2
Consider the function f (x) = 2
x 1
31. The interval in which f is increasing is
(A) (–1, 1) (B) (– , – 1)(–1, 0)
(C) (– , ) – {–1, 1} (C) (0, 1)  (1, )
Ans: B

32. If f is defined from R – {–1, 1}  R thenfis

(A) injective but not surjective (B) surjective but not injective
(C) injective as well as surjective (D) neither injective nor surjective.
Ans: D
Sol. (For 31 and 32)
x2
y= 2 ;not defined at x = ± 1
x 1
1 2x
=1+ 2 ; y'=– 2
x 1 ( x  1) 2
dy
=0  x = 0(point of maxima)
dx
y

1
O x
–1 1

as x  1+, y ; x  1– ,y  – 
|||ly x  – 1+,y  –  ; x  – 1–,y 
x2
The graph of y = 2 is as shown
x 1
verify all alternativels from the graph.

Paragraph for question nos. 33 to 34

Consider the cubic f (x) = 8x3 + 4ax2 + 2bx + awhere a, b  R.

33. For a = 1 if y = f (x) is strictly increasing x  R then maximum range of values of b is

 1 1  1 
(A)   ,  (B)  ,   (C)  ,   (D) (–, )
 3 3  3 
Ans: C
Sol. a=1
f (x) = 8x3 + 4x2 + 2bx + 1
f ' (x) = 24x2 + 8x + 2b = 2(12x2 + 4x + b)
for increasing function, f ' (x) 0  x  R
1
 D 0  16 – 48b  0  b  (C)
3
34. For b = 1, if y = f (x) is non monotonic then the sum of all the integral values ofa  [1, 100], is
(A) 4950 (B) 5049 (C) 5050 (D) 5047
Ans: B
Sol. ifb = 1
f (x) = 8x3 + 4ax2 + 2x + a
f ' (x) = 24x2 + 8ax + 2or2(12x2 + 4ax + 1)
for non-monotonic f ' (x) = 0 must have distinct roots
hence D > 0 i.e. 16a2 – 48 > 0
 a2> 3;  a > 3 ora < – 3
 a  2, 3, 4, .......
sum = 5050 – 1 = 5049Ans.

Matching Type Questions

35. Match the columns:

Column-I Column-II
(A) 10 (P) ( , 0)
y is decreasing on
4x  9x 2  6x
3

(B) y  2x 2  n x is increasing on (Q) (0, 1 / 2)

(C) y  x  2 sin x decreasing on (R) (2,  )
(D) y  x 2e  x decreasing on (S) (0, 2)

Ans: A  P, Q, R ; B  R ; C  Q ; D  P, R
10
Sol. (A) y  = 3 2 2
(12x2 – 18x + 6)
(4 x  9x  6x )
1 4x 2  1
(B) y  = 4x – = where x > 0
x x
1
(C) y  = 1 – 2 cos x then y < 0 for cos x >
2
(D) y  = x2 (– e–x) + 2x e–x = e–x x [2 – x]

x3  4 x3  4
36. Letf (x) = for all x  R – {1}andh(x) = for allx  R – {–1}. Match the entries
( x  1)3 ( x  1)3
in Column-I with Column-II.
Column-I Column-II
(A) Number of distinct real roots of equationf(x) = k (P) 0
wherek  [4, )can be

(B) Number of distinct real roots of equationh(x) = k (Q) 1

wherek  [1, ) can be

(C) Number of distinct real roots of equationf(x) = k (R) 2

wherek  [0, 1] can be

(D) Number of distinct real roots of equationh(x) = k (S) 3

wherek  (0, 1) can be (T) 4
Ans: AQ, R; BQ, R, S; CQ, R, S; DQ, R, S
Sol. fis increasing in[–2, 2]
fis decreasing in R – (– 2, 2)
his increasing in R – (–2, 2)
his decreasing in[–2, 2]
andf (–x) = h(x)
f(x)

(2, 4)

. (0, 4)
y=1

(–2, 4/9)

–2 –1 0 1 2

x=1
AOD-3:
Single Correct Answer Type Questions

 b  2x , if x  1
1. Let f : R R be defined by f(x) =  . If f has a local minimum atx = – 1,then
 2x  3 , if x  1
a possible value ofbis equal to
y
+ 3
2x
y=
y=
b– 1
2x (–1, 1)
x
x=–1 O (0, 0)

1
(A) 0 (B) (C) 1 (D) – 1
2
Ans: D
Sol. For f(x) to have local minimum at x = – 1
Lim f (– 1 + h)  f (– 1)
h 0

1b+2
b–1

2. Let P(x)  a 0  a1x 2  a 2 x 4  ......  a n x 2n be a polynomial in a real variable x with

0 <a0<a1<a2<…..<an. The function P(x) has
(A) neither a maximum nor a minimum (B) only one maximum
(C) only one minimum (D) none of these
Ans: C
Sol. We have P(x)  2x(a1  2a 2 x 2  ........  na n x 2n 2 ) = 0 for P(x) to be maximum or minimum.
This gives x = 0, since other factor cannot be zero.
Now P(x)  2a1  12a 2 x 2  30a 3 x 4  .......  2n(2n  1)a n x 2n 2
 P(x)  2a1  0 at x = 0
Hence P(x) has only one minimum at x = 0.

10x12
3. Maximum value of the expression is equal to
x 24  2x12  3x16  3x 8  1
(A) 1 (B) 2 (C) 10 (D) not defined
Ans: A
10 10 10
Sol. f(x) = = 
3 1 3 1 3
x12  2  3x 4   12 x12  3x 4  4  12  2  x 4  1   2
x 4
x x x  
 x4 
3
 2   x 4 
1 1  10
 x4 +  + 2  10  f(x)  =1
x4  x4  10

tan  x  
6 
4. The minimum value of the function f(x) = is
tan x
(A) 0 (B) 1/2 (C) 1 (D) 3
Ans: D
Sol. f(x) has a period equal to & can take values (, )  3 is the local minimum value.
 =
 
2 sin x  6 cos x sin 2 x  6  sin
=
  
6
2 sin x cos x  6  
sin 2 x  6  sin   
6
1
=1+

sin 2 x  6  sin 6 
 
is minimum if2x + =
6 2
x =  ymin=1 + 2 = 3]
6

5. Difference between the greatest and the least values of the function
f (x) = x(ln x – 2) on [1, e2] is
(A) 2 (B) e (C) e2 (D) 1
Ans: B
Sol. y = x (ln x – 2)
1
y' = x   + (ln x – 2) = ln x – 1
x
dy
= ln x – 1 = 0  x=e
dx
now f (1) = – 2
f (e) = – e (least)
2
f (e ) = 0 (greatest)

 difference = 0 – (–e) = e Ans. 1 e e2

2x 2
1
6. The maximum value of   is -
x
(A) 1 (B) e (C) e1/e (D) None
Ans: C
Sol. y is maximum when log y = z is max.
1
z = 2x2 log = – 2x2 log x
x
dz
= –2[x + 2x log x] = 0 = –2x(1 + 2 log x) = 0
dx
1 1
log x = – orx = e–1/2 =
2 e
2
d z  2
= – 2  (1  2 log x).1  x. 
2
dx  x
= –2 [0 + 2] = –4 = –ive
2x 2
1
 y =  = ( e ) 2.(1/e)  e1/e .
x

x  
7. f x  , x   0,  , then
1  x tan x  2
(A) f  x  has exactly one point of minima (B) f  x  has exactly one point of maxima
   
(C) f  x  is increasing in  0,  (D) f  x  is decreasing in  0, 
2   2
Ans: B y
1  x 2 sec 2 x sec 2 x  cos x  x  cos x  x 
Sol. f x   y=x
1  x tan x  1  x tan x 
2 2

clearly f  x0   0 
x0
and f   x   0  x  0, x 0  O
x
/2
f   x   0  x  x 0 ,  2  y = cos x

n2
8. {a1, a2, ....., a4, ......} is a progression where an = .The largest term of this progression
n 3 200
is
(A) a6 (B) a7 (C) a8 (D) none of these
Ans: B
x2 d y x ( 400  x 3 )
Sol. Lety= 3 ; =
x  200 d x ( x 3  200) 2
Nowifx> (400)1/3,y is decreasingand
ifx< (400)1/3,y is increasinghence
yis greatest atx = (400)1/3.
ButxNhence practical maxima occurs at
49 64
x = 7orx = 8;a7 = ;a8 =
543 712

9. If the function f (x) = ax e–bx has a local maximum at the point (2, 10) then
(A) a = 5; b = 0 (B) a = 5e, b = 1/2 (C) a = 5e2, b = 1 (D) none
Ans: B
Sol. f (2) = 10, hence 2ae–2b = 10  ae–2b = 5 ....(1)
–bx –bx
f ' (x) = a [e – bx e ] = 0
f ' (2) = 0
a(e–2b – 2be–2b) = 0
ae–2b (1 – 2b) = 0 b = 1/2ora = 0(rejected)
from (1) ifb = 1/2; a = 5e
 a = 5eandb = 1/2 Q.

10. Let f (x) = ax2 + bx + 8, where a  0. If f has a local minimum of 6 at x = 2 then

1
(A) a = and b = – 2 (B) a = 2 and b = – 8
2
1
(C) a = – and b = 2 (D) It is impossible to determine a and b.
2

 
11. Letf(x) = sin3x +  sin2x ,   x  , then the intervals in which  should lie in order that f(x)
2 2
has exactly one minimum and one maximum is
 3 3   3   3 
(A)  ,  (B)  , 0    0, 
 2 2  2   2
 3 
 , 0    0, 
3
(C)  , 0    0,   (D)
 2   2
Ans: B
Sol. f(x)= sin3x +  sin2x
f(x) = 3 sin2x .cosx + 2sinxcosx
f(x) = 0sinxcosx (3 sinx + 2) = 0
2
sinx = 0orcosx = 0 orsinx = −
3
sinx = 0x = 0
  
cosx = rejected ,   x 
2 2 2
2
sinx = −
3
2  3 3
provided− 1 < − < 1    , 
3  2 2
at  = 0,f(x) = sinx x = 0
 3   3
    , 0    0,  .
 2   2

12. The maximum area of the rectangle whose sides pass through the angular points of a given the
rectangle is of sides a and b is -
(A) (1/2) (ab)2 (B) (1/2) (a + b) (C) (1/2) (a + b)2 (D) None of these
Ans: C
Sol. Let ABCD be the given rectangle of sides a and b and EFGH be any rectangle, whose sides
pass through A, B, C, D
A = area EFGH = (b sin  + a cos ) (a sin  + b cos )= ab + (a2 + b2) sin  cos 
G
2 2
dA/d = (a + b ) cos 2 so dA/d = 0 D C
F
 = /4 a

d2A b
d 2A
 2 = – 2 (a2 + b2) sin 2, so <0 a
d d 2  /4 H B
A 
Hence Amax = (1/2) (a + b)2.
E

13. The ratio of the altitude of the cone of greatest volume which can be inscribed in a given
sphere to the diameter of the sphere is -
(A) 2/3 (B) 1/2 (C) 4/5 (D) 1/3
Ans: A
Sol. Let h be the height of the cone and r be its radius.
 h = CL = CO + OL = a + OL C
 OL = h – a
r = LA = (OA2 – OL2) a
h
or r = {a2 – (h – a)2} = 2ah  h 2 O
1 1 1 r
V = r2h = (2ah – h2) h=  (2ah2 – h3) B L
A
3 3 3
dV/dh = (/3) (4ah – 3h2) = 0
 h = 0 or 4a/3
h = 0 is rejected  h = 4a/3 = (2/3) (2a)
2
h= (diameter).
3
14. The set of all values of the parameters a for which the points of minimum of the function
x2  x  2
y = 1 + a2x – x3 satisfy the inequality 2  0 is -
x  5x  6
(A) an empty set (B) ( 3 3, 2 3)
(C) (2 3, 3 3) (D) ( 3 3, 2 3)  (2 3, 3 3)
Ans: C
dy d2 y
Sol. = a2 – 3x2 = 0  x = ± a/ 3 . Since = –6x so y is minimum for x = –a/ 3 .
dx dx 2
x2  x  2
Since x2 + x + 2 > 0 for all x so for 2 0, we must have x2 + 5x + 6 < 0.
x  5x  6
If x = –a/ 3 , we have a2/3 – 5a/ 3 + 6 < 0 i.e. a2 – 5 3 a + 18 < 0
 (a – 2 3 ) (a – 3 3 ) < 0 i.e. a  (2 3 , 3 3 ).


15. The function f(x) = (4sin2 x – 1)n (x2 – x + 1), n N, has a local minimum at x = , then -
6
(A) n can be any even number (B) n can be any odd number
(C) n can be odd prime number (D) n can be any natural number
Ans: A
Sol. f(x) = (4sin2 x – 1)n (x2 – x + 1)
sincex2 – x + 1 > 0  x  R
f(/6) = 0
f(/6)+ = lim (4sin2x – 1)n (x2 – x + 1) =  0+

x
6

f(+/6) = lim (4sin2 x – 1)n (x2 – x + 1)


x
6

=  (0)–n (a positive value)

f(–/6) > 0 if n is an even no.

16. The function f (x)  | px  q |  r | x | ,x (, ), where p> 0, q> 0, r> 0 assumes its minimum
value only at one point if
(A) pq (B) rq (C) rp (D) p = q = r
Ans: C
q q
Sol. f (x)   px  q  rx , x  0   px  q  rx , 0  x   px  q  rx ,  x
p p
Y Y Y
y = f (x)
y = f (x) y = f (x)
q q q

X X X
O q/p O q/p O q/p

Thus, f has two points of minimum if r = p. In case pr, then x = 0 is point of minimum if
r>r and x = q/p is point of minimum if r<p.

Multiple Answer Type Questions

 3x  12x  1 , –1 x  2
2
17. If f (x)=  . Then -
37  x , 2 x 3
(A) f (x) is increasing on [–1, 2] (B) f (x) is continuous on [–1, 3]
(C) f (2) doesn't exist (D) f (x) has the maximum value at x = 2
Ans: A, B, C, D
Sol. For x  [–1, 2), f (x) = 6x + 12 > 0 on (–1, 2), so f increases on [–1, 2]. f is trivially
continuous on [–1, 3] except possibly x = 2. At point x = 2, f(2) = 35 and
lim lim
x 2 f (x)= x 2 (37 – x) = 35,
lim lim
x 2 f(x)= x 2 (3x2 + 12x – 1)= 3 × 4 + 12 × 2 – 1 = 35.
Thus, f is continuous at x = 2 as well. Now
lim f ( 2  h )  f ( 2) lim 37  ( 2  h )  35
f  (2+) = h 0  = h 0  =–1
h h
Similarly, f (2 –) = 24  f  (2+)
As f increases on [–1, 2] and decreases on [2, 3]
so f has a maximum value at x = 2.

18. f(x) is cubic polynomial which has local maximum at x =  1. If f(2) = 18, f(1) = 1 and f(x)
has local minima at x = 0, then
(A) the distance between (1, 2) and (a, f(a)), where x = a is the point of local minima is 2 5
(B) f(x) is increasing for x  [1, 2 5 ]
(C) f(x) has local minima at x = 1
(D) the value of f(0) =5
Ans: B, C
1
Sol. The required polynomial which satisfy the condition is f(x) =
4
(19x3 57x + 34) 1 1
f(x) has local maximum at x = 1 and local minimum at x = 1
Hence f(x) is increasing for x  1, 2 5  .
Hence (B) and (C) are correct.

19. If f (x) = |4 x – x2- 3| when x  [0 , 4] then

(A) x = 0 is a global maximum (B) x = 4 is a global maximum
(C) x = 2 is a local maximum (D) x = 1 and 3 are global minimum
Ans: A, B, C, D y
Sol. Clearly, x = 1, 3 are the points of global minimum (values
being equal) and x = 0, 4 are the points of global maximum 3
y = |x -4x+3| 2

(values being equal) and x = 2 is a local maximum.Hence

1
(A), (B), (C) and (D) are correct.
O x
1 2 3 4
Fig. 26 – Graph of y = |x2-4x+3|

 x 3  x 2  10x ;  1  x  0

20. 26. Let f (x) =  sin x ; 0  x   / 2 then f (x) has
 1  cos x ;  / 2  x  

(A) local maxima at x = p/2 (B) local minima atx = /2
(C) absolute maxima at x =  (D) absolute maxima at x = /2
Ans: A, C
Sol.
 Y

y=x3+x2–10x
y = sinx
(/2, 1) y = 1 + cos x

X
x= –1 (0,0) 1  (,0)
2
Graph of f(x)

Clearly, from above graph, f(x) has local maximum at x = and absolute maximum at x = –
2
1.

21. If f(x) = tan–1 x – (1/2) log x. Then

(A) the greatest value of f(x) on [1/ 3 , 3 ] is /6+ (1/4) log 3
(B) the least value of f(x) on [1/ 3 , 3 ] is /3 – (1/4) log 3
(C) f(x) decreases on (0, )
(D) f(x) increases on (–, 0)
Ans: A, B, C
Sol. The domain of f(x) is (0, ). For x > 0,
1 1 2x  (1  x 2 ) (1  x)2
f (x) = – = =–
1  x 2 2x (1  x 2 ) 2x (1  x 2 ) 2x
Thus f(x) < 0, i.e., f(x) decreases on (0, ). Also f(x) = 0 if x = 1 and f(1) = /4, f(1/ 3 )
= /6 + (1/4) log 3, f( 3 ) = /3 –(1/4) log 3, thus the greatest value is /6 + log 3 and the least
value is /3 – (1/4) log 3.

x3 x 2 x4 11x 2
22. If f(x) =   x  2 and g(x) =  2x 
3
 6x . Then f(g(x)) will have local
3 2 4 2
minimum at
(A) x = 1 (B) x = 2 (C) x = 3 (D) x = 4
Ans: A, C
Sol. Let h(x) = f(g(x))  h '( x)  f '  g ( x)  .g '( x)
As f '( x)  x 2  x  1  f '  g ( x)   0  x  R
Now g '( x )  ( x  1)( x  2)( x  3) so function have minimum at x = 1 & 3

x3  4
23. Letf(x) = then which of the following is(are) correct.
(x  1)3
(A) Lim f ( x )  1 .
x  
(B) x =–2is the point of local minimum.
4 
(C) The equationf(x) = cwill have three real and distinct roots forc   , 1  1, 4  .
9 
(D) The line y = 1 intersects the graph of f(x) at exactly one point.
Ans: B, C
Sol.
 +
y B(2,4)

y=1

 4
A  2, 
 9 x=1
x
–¥ –2 O x=2 + ¥
(0,0)
–¥
3
( x  4)
Graph off(x) =
( x  1)3
Now, verify alternatives.

   
24. Letf (x) = cos2x · etan x, x   ,  then
 2 2

(A) f '(x) has a point of local minima at x = .
4
 
(B) f '(x) has a point of local maxima in  , 0 .
 4 
 
(C) f '(x) has exactly two points of local maxima / minima in  , .
 2 2
 
(D) f "(x) = 0 has no root in  , .
 2 2
Ans: A, B, C
Sol. f(x) = cos2x · etanx
   
f '(x) = etanx.(1 – sin 2x)  0 x   ,  f (x)on  , 
 2 2  2 2
f "(x) = etan x · (–2 cos 2x) + etan x · sec2 x · (1 – sin 2x)
 2(tan 2 x  1) 
f "(x) = etan x  2
 (sec 2
x  2 tan x ) 

 1  tan x 
= f (x)·(tan x – 1) (tan3 x – tan2 x + 3 tan x + 1)
3 2
Now, letg (x) = tan x – tan x + 3 tan x + 1
 
 g'(x) = (3 tan2 x – 2 tan x + 3) sec2 x > 0 x   , .
 2 2
So, g(x)is increasing function.

Also,g   g(0) < 0
 4 
 
So, g(x) = 0 has exactly one root in  , ,
 2 2
  
where one root is and other root lies in  , 0 .
4  4 
  
So, f "(x) = 0has exactly two roots in  , .
 2 2

Atx = ,f '(x) has a local minima.
4

Numerical Answer Type Questions

x2  2
25. Number of local maxima in the function f  x   is……….
x2  4
Ans: 1
x 2  2 dy 4x dy dy
Sol. For y  ,    0 forx< 0 and  0 for x> 0. Then x = 0 is the
 
2 2
x  4 dx 2
x 4 dx dx

1
point of local maxima for y. Now y x  0  (positive). Thus x = 0 is also the point of local
2
x2  2
maxima for y  2 .
x 4

ax  b
26. If the function f  x   has a local maxima at 2,  1 , then value of a  b is……
 x  1 x  4 
Ans: 1
2a  b
Sol. Clearly f  2   1  1   2a  b  2
 2  1 2  4 
4a  5b  2bx  ax 2
Now f   x   , f   2  0
 2  1 2  4 
  x  2  x  2 
a = 1b = 0 f   x  
 x  1 x  4 
2

Clearly for x  2, f  x   0 and for x  2, f  x   0 . Thus x = 2 is indeed the point of local

maxima for y  f x  .

27. If f (x)  a log e | x |  bx 2  x has extremums at x = 1 and x = 3, then value of 4a  8b is……

Ans: 4
Sol. Around x = 1, 3 we have | x | = x
 f (x) = aloge x + bx2 + x
a
 f (x)   2bx  1
x
From the question, f(1) = 0, f(3) = 0
a
 a + 2b + 1 = 0,  6b  1  0 .
3

Min f (t) : 0  t  x ; 0  x  1
28. If f(x) = 4x3 x2 2x + 1and g(x) =[ then find value of
3x ; 1 x  2
 1  3  5 
2 g    g    g   .
 4  4  4 
Ans: 5
Sol. f (x) = 12x2 – 2x – 2
= 2[6x2 – x – 1] = 2(3x + 1)(2x – 1)
D I
 1
f (x) if 0  x 
2

1 1
Hence g(x) =  f   if  x  1 –1/3 0 1/2 1
 2 2

 3  x if 1  x  2


1 1 a
29. If absolute maximum value of the function f (x) =  is ,where a & b are
| x  4 | 1 | x  8 | 1 b
co-prime then find number of divisors of a  b
Ans: 4
1 1
Sol. Say g (x) =  = g (x + 12)
| x  4 | 1 | x  8 | 1
Now for x  (–, –8) both g (x) and g (x + 12) are increasing hence maximum value can’t
occur in this interval.
Similarly for x  (4, ) both g (x) and g (x + 12) are decreasing hence maximum value can’t
occur in this interval.
So, now for all values of x  (–8, 4)
1 1 14
f (x) =  
x  9 5  x (5  x)(x  9)
which will have maximum either at x = 4 or x = –8 or at minima of (5 – x)(x + 9). Also minima
of (5 – x)(x + 9) does not exist.
1 14
 f (4) = 1 + = = f (–8).
13 13
14
So, absolute maximum is .
13

 x 3  x 2  10x  5, x 1
30. Let f  x    . If possible real values of b such that f (x) has the
 2
2x  log 2 b  2 , x  1
 
greatest value at x = 1 is   a , b  c , d  then c  d  a  b is……….
Ans: 4
Sol. for x  1; f (x) = 3x2 –2x + 10
Discriminant of f (x) = 0 = –56 < 0 and coefficient of x2> 0
Hence f (x) > 0 for all x  1
Hence f (x) is an increasing function for (1, 5)
x1 –2x+log2(b2–2)
For x > 1; f (x) = –2
 f (x) is a decreasing function for x > 1
(1, 0)
f (x) will have the greatest value at x = 1 if (0, –5)
lim f  x   f (1)  lim f 1  h   f (1)
x 1 h 0 x3 + x2+10x –5
 –2 + log2(b2 –2)  5  log2 (b2 –2)  7
 b2 –2  27 b2 130
Again b2 –2 > 0 for log2 (b2 –2) to be defined
  2 < b2 130
 b  [  130 , –2)  ( 2 , 130 ]

x ( x  1) (x  2) ; 0  x  n
31. Let f (x) =  .If n N then find the least value of n for which f(x)
 sin (x ); n  x  2n
has more points of minima than maxima in[0,2n].
Ans: 3
Sol. Forn = 1,graphoff (x)
y
1
O 2 x

Forn = 2,graphoff (x)

y

1 3
O 2 4 x

Forn = 3, graph off (x)

y

1 3 5
x
O 2 4 6

Passage Type Questions

Paragraph for question nos. 32 to 33

Lety = g(x)and y = P(x)be two polynomial functions of degree two and four respectively such
that
(i) P(–x) = P(x) and P(x)  0  x  R
(ii) P(x)has two local minimum atx = x1, x2such that|x1 – x2| = 2.
(iii) The line y = 1touches the curvey = P(x) at point of local maximum and encloses an area
8 2
of sq. units with it
15
P( x )  g ( x )  g (  x )
(iv) Lim = g'(–1).
x 0 x2

32. Minimum value of P(x) equals

1 1
(A) 2 (B) 1 (C) (D)
2 4
Ans: C

33. If g(x) has maximum value at x = –1 then product of its roots is

1 1
(A) (B) –1 (C) (D) 1
2 2
Ans: B
Sol. (32 &33)
Clearly,x1 = –x2 and P'(x) = –P'(–x)
 x1 = 1,x2 = –1 P'(0) = 0
2
 P'(x) = A(x – 1)x
 x4 x2 
P(x) = A    + c.alsoP(0) = 1c = 1
 4 2 
 x4 x2 
Now, y = 1 =    + 1x = ± 2 , 0
 4 2 

y=1

 2 2

2   x4 x2   8 2
  1   A 4  2   1 dx = 15 A = 2
 2    
 x4 x2 
 P(x) = 2    + 1
 4 2 
1
 Minimum value =P(1) =
2

Paragraph for question nos. 34 to 35

The function f(x)= ax 3  bx 2  cx  d has its non-zero local minimum and local maximum
values at – 2 and 2 respectively. Given 'a' is root of the equation x2 – x – 6=0.

34. The value of (a + b + c) is equal to

(A) 16 (B) 18 (C) 20 (D) 22
Ans: D

35. The roots of the equation ax2 + bx + c = 0

(A) are opposite in sign (B) are imaginary
(C) are both positive (D) are both negative.
Ans: A
Sol. (34 & 35)
(i) Since minimum occurs before maximum, soa < 0.
Also, a is root ofx2 – x – 6 = 0  a =– 2.
Let g(x) = ax3 + bx2 + cx + d  g(x) =– 2x3 + bx2 + cx + d
So, g'(x) = – 6x2 + 2bx + c=– 6 (x + 2) (x – 2)b = 0, c = 24.
 a + b + c =– 2 + 0 + 24 = 22
(ii) Clearly, the equationax2 + bx + c = 0, is– 2x2 + 0·x + 24 = 0  x = ± 2 3 ,so roots of
above equation are opposite in sign.
Matching Type Questions

36. Match the column:

Column-I Column-II
(A)  x 1 (P) 5
Domain of function f(x) = n  1    is (p, q), then p + q is equal
 1 | x | 2 

to
(B) sin 2x   (Q) 1
Maximum value of f(x) =
 
in the interval 0, 2  is
sin  x  
 4
(C) Let f(x) = x3 + ax + b with a  b and suppose the tangent lines to the (R) 0
graph of f at x = a and x = b have the same gradient. Then the value of
f(1) is equal to
(D) If f is a differentiable function forall real x and f (x)  5,  xR. If f(2) = (S) –1
0 and f(5) = 15 then f(3)is equal to
Ans: A  R;B  Q;C  Q;D  P
1 1
Sol. (A)  2>1+|x|
1 | x | 2
 | x | < 1  –1 < x < 1
p+q=0

2 sin 2x  (sin x  cos x) 2  1 

(B) f(x) =  2 
sin x  cos x  sin x  cos x 
 1 
= 2 (sin x  cos x)  
 sin x  cos x 
 1 
fmax(x) = 2  2  =2–1=1
 2
(C) f (a) = f (b)  3a2 + a = 3b2 + a a = – b
So f(1) = 1 + a + b = 1
f (5)  f (2) 15  0
(D) as  =5
52 3
So average change is 5 and f (x)  5
 f (x) = 5 x  [2, 5)  f(x) = 5x + c
f(2) = 0  10 + c = 0  c = – 10  f(3) = 15 – 10 = 5

37. Column I contains function and Column II contains behavior of function in their domain. Entry
of column I are to be matched with one or more than one entries of column II.

Column-I Column-II
(A) f (x) = 5 x  5 (P) monotonic
(B) ln x (Q) non monotonic
g (x) =
x
(C) h (x) = x sgn x, where sgn x denotes signum function of x. (R) possesses extremum point
(D) tan 1 x, if x  0 (S) possesses critical point
k (x) = 2 , if x  0
cot 1 x , if x  0
(T) possesses inflection point

Ans: A P, S, T;B Q, R, S, T;C Q, R, S; D Q, R, S

Sol. (A) We have f(x) = 5 x + 5Domain off (x) = R
1
Also,f '(x) = 4 5 > 0  x  R0. y
5x
(0, 5)
f(x) is strictly increasing on R.
Clearly, f '(0) does not exist, although x = 0 is in the domain of
function. O
x
So, x = 0 is critical point of f(x).
Also, f "(x) changes sign about x = 0
x = 0 is point of inflection of f(x) = 5 x + 5.
Graph of f(x) = 5 x  5
lnx
(B) Given, g(x) =
x
Domain of g(x) = R+ y
 1
1  lnx  e, 
 e
g'(x) =
x2
Now verify alternatives.
x
O x =e
 x ; x0

(C) We have h(x) = x sgn x 0 ; x  0 = | x | x  R.
lnx
x ; x0 Graph of g(x) =
 x
y

y =– x y=x
x
O
Graph of h(x) = x sgn x

 tan 1 x ; x0

(D) We havek(x) = 2 ; x0 .
 1
cot x ; x0
y

¯ y = cot–1x
(0, 2)
y =  /2
¯ y = tan–1 x
x
O

Graph of k(x)
AOD-4:
Single Correct Answer Type Questions

1. Consider the function for x [–2, 3]

x 3  2x 2  5x  6
 if x  1
x 1
f(x) =  then:

  6 if x  1
(A) f is discontinuous at x = 1  Rolle's theorem is not applicable in [–2, 3]
(B) f(–2)  f(3)  Rolle's theorem is not applicable in [–2, 3]
(C) f is not derivable in (–2, 3)  Rolle's theorem is not applicable
(D) Rolles' theorem is applicable as f satisfies all the conditions and C of Rolle's theorem is
1/2
Ans: D
Sol. x3 – 2x2 – 5x + 6 = (x – 1) (x2 – x – 6)

 2  x 3 if x  1
2. Consider the function f (x) =  , then
3x if x  1
(A) f is continuous on [–1, 2] but is not differentiable on (–1, 2)
(B) Mean value theorem is not applicable for the function on [–1, 2]
(C) Mean value theorem is applicable on [–1, 2] and the value of c = 1
5
(D) Mean value theorem is applicable on [–1, 2] and the value of c is ±
3
Ans: D
Sol. Obviously f is continuous and differentiable in [–1, 2] ; f (–1) = 1andf (2) = 6
y
6

x
–2 –1 –1 1 2
3
f ( 2)  f ( 1) 6  1 5
 f ' (c) = = =
2 1 3 3
5 5 5 5
3c2 = c2 = c = or–
3 9 3 3

tan [ x]
2/3  , x 0
1   x
3. Given : f (x) = 4    x  g (x) = 
2   1 , x 0
h (x) = {x} k (x) = 5log 2 ( x  3)
then in[0, 1], Lagranges Mean Value Theorem is NOT applicable to
(A) f, g, h (B) h, k (C) f, g (D) g, h, k
where [x] and {x} denotes the greatest integer and fraction part function.
Ans: A
1
Sol. f is not differentiable at x =
2
g is not continuous in [0, 1] at x = 0 & 1
h is not continuous in [0, 1] at x = 1
k (x) = ( x  3)ln 2 5 = (x + 3)p where 2 < p < 3

 1 
4. If the function f(x) = ax3 + bx2 + 11x – 6 satisfies the Rolle's theorem in [1, 3] and f '  2  
 3
= 0, then the values of a and b are respectively
(A) – 6, 1 (B) – 2, 1 (C) 1, – 2 (D) 1, – 6
Ans: D
Sol. By Rolle's Theorem
f(1) = f(3)
a + b + 11 – 6 = 27a + 9b + 33 – 6
13a + 4b + 11 = 0 .......(1)
Now, f '(x) = 3ax2 + 2bx + 11
 1 
So, f '  2  = 0
 3
2
 1   1 
3a  2    2b  2   + 11 = 0 [Using equation (1)]
 3  3
6a + b = 0 .......(2)
On solving (1) and (2), we get a = 1, b = – 6.

2
5. Let f '(x) = e x and f (0) = 10. If A <f (1) < B can be concluded from the Mean Value Theorem,
then the largest value of(A – B)equals
(A) e (B) 1 – e (C) e – 1 (D) 1 + e
Ans: B
Sol. ApplyingLMVT in [0, 1] to the function y = f(x), we get
f (1)  f (0)
f ' (c) = ,for some c  (0, 1)
1 0
2 f (1)  f (0) 2
 ec   f (1) – 10 = e x for some c  (0, 1) but 1 << e in (0, 1)
1
1 < f (1) – 10 < e  11 < f (1) < 10 + e
 A = 11,B = 10 + e  A–B=1–e

6. If the function f(x) = x3 – ax2 + 2x satisfies the conditions of LMVT over the interval[0, 2] and
1
the tangent to the curve y = f(x) at x = is parallel to the chord that joins the points of
2
intersection of the curve with ordinates at x = 0 and x = 2, then the value of a is
9 11 13 15
(A) (B) (C) (D)
4 4 4 4
Ans: C
Sol. Given, f(x) = x3 – ax2 + 2x,x  [0, 2]
Now, f ' (x) = 3x2 – 2ax + 2
f(2) = (2)3 – 4a + 4 = 12 – 4aandf(0) = 0
 Using LMVT, we have
  1  f ( 2)  f ( 0) 3 (12  4a )  0
f '     a  2 
2 20 4 2

11 11 13
  a = 6 – 2aa = 6 –  a = .
4 4 4

7. If f(x) is a twice differentiable function and given that f(1) = 1, f(2) = 4, f(3) = 9, then
(A) f(x) = 2, for  x  (1, 3) (B) f(x) = f(x) = 5 for some x (2, 3)
(C) f(x) = 3,  x  (2, 3) (D) f(x) = 2, for some x  (1, 3)
Ans: D
Sol. Let g(x) = f(x) – x2.
We have g(1) = 0, g(2) = 0, g(3) = 0.
Hence by Rolle’s theorem g(x) = 0for some c  (1, 2)
and g(x) = 0 for some d  (2, 3).
Again, by Rolle’s theorem g(x) = 0 at some x (c, d)
 f(x) = 2 for some values x  (1, 3).

8. Which one of the following functions Rolle’s theorem is applicable?

sin x
x , 0 x  1  x ,  x  0
 
(A) f(x) =  on [0, 1] (B) f(x) =  on [–, 0]
0 , x 1 
0 , x 0
x 3  2 x 2  5x  6
 if x  1, on [2,3]
x2  x  6  x  1
(C) f(x) = on [–2,3] (D) f(x)= 
x 1 
6 if x  1
Ans: D
Sol. (A) discontinuous at x =1not applicable
(B) f (x) is not continuous at x =0 hence (2) is incorrect.
(C) discontinuity at x = 1not applicable
(D) Notice that x3 – 2x2 – 5x + 6 = (x–1) (x2 –x –6). Hence, f(x) = x2 – x – 6 if andf (1)= –6
f is continuous at x = 1. So f(x) = x2 –x – 6 throughout the interval[–2,3].
Also, note that f(–2) = f(3) = 0. Hence, Rolle’s theorem applies. f (x) = 2x –1.
Settingf '(x)= 0 , we obtain x = 1/2 which lies between –2 and 3.

9. Let a, b, c, d are non-zero real numbers such that 6a + 4b + 3c + 3d = 0, then the equation
ax3 + bx2 + cx + d = 0 has
(A) at-least one root in [ –2, 0] (B) at-least one root in [ 0, 2]
(C) at least two roots in [ –2, 2] (D) no root in [ –2, 2]
Ans: B
Sol. We haveax3 + bx2 + cx + d = 0
ax 4 bx 3 cx 2
Let f(x) = + + + dx + e
4 3 2
f(0) = e
8b (12a  8b  6c  6d ) 2
f(2) = 4a + + 2c + 2d + e = + e = (6a + 4b + 3c + 3d)+ e = 0 + e
3 3 3
 f (2) = e
 By Rolle’s theorem, there exist at least one value of x  (0, 2) such that f ' (x) = 0
 The equation ax3 + bx2 + cx + d = 0 has at least one real root in [0, 2]
 
10. Consider f (x) = | 1 – x |1  x  2 and g (x) = f (x) + b sin x, 1 < x < 2 then which of the
2
following is correct?
3
(A) Rolles theorem is applicable to both f, g and b =
2
1
(B) LMVT is not applicable to f and Rolles theorem if applicable to g with b =
2
(C) LMVT is applicable to f and Rolles theorem is applicable to g with b = 1
(D) Rolles theorem is not applicable to both f, g for any real b.
Ans: C
Sol. f (x) = x – 1, 1  x  2

g (x) = x – 1 + b sin x,1  x  2
2
f (1) = 0 ;f (2) = 1Rolle's theorem is not applicable to ' f ' but LMVT is applicable to f.
(x – 1is continuous and differentiable in [1, 2] and (1, 2) respectively)
Now g (1) = b ; g (2) = 1and

Function x – 1, sin x are both continuous in [1, 2] and (1, 2)
2
 For Rolle's theorem to be applicable to g.
We must have b = 1

2012 2011
 1

11. If P(x) = (2013)x – (2012)x – 16x + 8, then P(x) = 0for x  0, 8  has
2011

 
(A) exactly one real root. (B) no real root.
(C) at least one and at most two real roots. (D) at least two real roots.
Ans: D
Sol. F(x) =  P ( x ) dx = x2013 – x2012 – 8x2 + 8x + C, where C is constant of integration.
F(x) = x(x – 1) (x2011 – 8) + C
F (0) = F(1) = F (81/2011) = CF'(x) = 0 has at least two real roots.(Using Rolle's Theorem)
 1

Note that P(x) = 0 has exactly two real roots in x  0, 8 2011 
 

 
cos 2x, 0  x  4    
12. If f (x) =  , is continuous in  0,  and differentiable in 0, 
 (sin x  cos x),   x    2  2
 4 2
where  R, then
 
(A) f (x) is non-monotonic function in  0,  .
 2
 
(B) there exist more than onec  0,  such that f ' (c) = – 2.
 2
 
(C) there exist at least onec  0,  such that f '(c) =

2 2  1
.

 2 

 
(D) there cannot exist anyc  0,  such that f '(c) =

2 2  1 
 2 
Ans: C

Sol. Since, f (x) is differentiable at x = ,so = – 2 .
4
 
 cos 2x ; 0  x
 f (x) =  4
 
 2 (sin x  cos x ) ;  x
 4 2
 
Clearlyf ' (x) < 0 x   0,  .
 2
Hence,f (x) is decreasing in .
   
ApplyingL.M.V.T.tof (x)in  0,  , there exist some c   0,  such that
 2  2
 
f    f (0)
f ' (c) =  
2

=
2

 
2 1 .
0
2

Also, f '   = – 2  f ' (x) = – 2
4
 
for unique value of c  0, 
 2

13. Let f (1) = – 2 and f ' (x)  4.2 for 1  x  6. The smallest possible value of f (6), is
(A) 9 (B) 12 (C) 15 (D) 19
Ans: D
Sol. Using LMVT some c  (1, 6)s.t.
f (6)  f (1) f (6)  2
f ' (c) = =  4.2
5 5
f (6) + 2  21
f (6)  19 Ans.

14. If y = f (x) is twice differentiable function such that f (a) = f (b) = 0, and f (x) > 0 x  (a, b),
then
(A) f ''(c) < 0 for some c  (a, b) (B) f '' (c) > 0 c  (a, b)
(C) f (c) = 0 for some c  (a, b) (D) none
Ans: A
 a  b
Sol. Applying LMVT over f (x) for x  a ,
 2 
ab
f   f (a )
 2  2 ab  ab
f ' (c1) = = · f  ,c1  a, 
ba  (b  a )  2   2 
 
 2 
ab
2f  
 2  ab 
|||ly f ' (c2) = – ,c2  ,b 
ba  2 
Applying LMVT over y = f ' (x) in [c1, c2]
 4 ab
 ·f  
f ' (c 2 )  f ' (c1 ) (b  a )  2 
f ''(x) = = < 0, wherex (c1, c2)
c 2  c1 c 2  c1

d
15. Given f ' (1) = 1 and f (2x )   f ' ( x )  x > 0. If f ' (x) is differentiable then there exists a
dx
number c  (2, 4) such that f '' (c) equals
(A) – 1/4 (B) – 1/8 (C) 1/4 (D) 1/8
Ans: B
Sol. f ' (1) = 1; 2 · f ' (2x) = f ' (x)
f ' (1) 1
put x = 1, f ' (2) = =
2 2
1 1
and f ' (4) = f ' (2) =
2 4
applying LMVT for y = f '(x) is [2, 4]
1 1

f ' ( 4)  f ' ( 2) 4 2 1
f '' (c) = = = – Ans.
2 2 8

Multiple Answer Type Questions

16. Which of the following functions fail to satisfy the condition of Rolle's theorem on the interval
[– 1, 1],where [x] denotes the greatest integer less or equal to x and {x} denotes the fractional
part of x respectively.
 tan x
 , x0
(A) f (x)=| x |[ x ] (B) f ( x )   x
 0, x0
(C) f (x)={ x }+ { – x } (D) f (x)=| x | – | sin x |
Ans: A, B, C
 1, x  1

Sol. (A) f x    0 , 0  x  1  not differentiable at x = 0 in (– 1, 1)
 x 1  x  0

tan x
(B) f (0) = 0and Lim 1  not continuous at x = 0
x 0 x
 1, x  I
(C) f x     not continuous at x = 0
 0, x  I
x  sin x 0  x 1
(D) f x     continuous & differentiable atx = 0
 x  sin x 1  x  0

17. Which of the following statements are True?

| 2x  1 |  | 2x  1 |
(A) Lim does not exist.
x 0 x
2x if x  1

(B) The function f (x) =  is derivable at x = 1.
 2
x  3x  2 if x  1
2
(C) The equation e x = x has a unique root in (0, 1)
 (D) The function f (x) = x5 + 10x3 + 20x – 18 is strictly increasing  x  R.
Ans: C, D
2 1
 if x 
 x 2


 4x 1 1
Sol. (A) f (x) =    x , x0
 x 2 2

 2 1
 x
x 2
Henceexists and equal – 4
(B) f is discontinuous at x = 1non derivable
2
(C) Use IVT for f (x) = e x – xin (0, 1)
(D) f ' (x) = 5x4 + 30x2 + 20 > 0x  R

18. Let f be a twice differentiable function on [0, 2] such that f (0) = 0, f (1) = 2, f (2) = 4, then
(A) f     2 for some    0,1 (B) f      2 for some   1, 2 
(C) f      0 for some    0, 2  (D) none of these
Ans: A, B, C
Sol. Let g(x) = f(x) – 2x so that
g(0) = f(0) = 0
g(1) = f(1) – 2 = 0g(0) = g(1) = g(2) = 0 ;Alsog ' (x) = f ' (x) – 2
and g(2) = f(2) – 4 = 0
Rolle's for g(x) in [0, 1]  there exists some  (0, 1) such that g'() = 0 f '() = 2
Similarly Rolle's for g(x) in [1, 2]  there exists some  (1, 2) such that g'() = 0
 f '() = 2
again using Rolle's for g'(x) in [, ] where g '(x) =f '(x) – 2
some  (, ) such that g "(x)=0 f "()=0
where  (0, 2).

19. f(x) is continuous and twice differentiable function such that f(-1) = 5, f(0) = 2 and f(1) = 3.
Then
(A) there exist at least two roots of equation f   x   4 x  1 in  1,1
(B) there exist at least one c  1,1 such that f(c) = 4
(C) Equation f  x   2 x 2  x  2 has at least 3 roots in  1,1
(D) none of these
Ans: A, B, C
Sol. Let g(x) = f(x) – [2x2 – x + 2]
g(–1) = g(0) = g(1) = 0
Applying rolle’s theorem on g(x) between (–1, 0) and (0, 1).
We get g(c1) = g(c2) = 0 where c1(–1, 0) and c2(0, 1).
Now applying rolle’s theorem on g(x) between (c1, c2), we get g(c) = 0 where c(c1, c2)
 f(c) = 4 where c(–1, 1).

20. Let f(x) be twice differentiable function such that f"(x) in [0, 2]. Then
(A) f (0)  f (2)  2f (c) , for at-least one c, c  (0, 2)
 (B) f (0)  f (2)  2f (1)
(C) f (0)  f (2)  2f (1)
2
(D) 2 f (0)  f (2)  3 f  
3
Ans: C, D

21. Let f be real-valued function on R defined as f(x) =x4 (1 – x)2 , then which of the following
statement(s) is(are) correct ?
(A) f '(c) = 0 for some c  (0, 1). (B) f "(x) vanishes exactly twice in R.
 2
(C) f (x) is an even function. (D) Monotonic increasing in  0,   1,   .
 3
Ans: A, D
Sol. We havef(x) = x4 (1 – x)2,x  [0. 1]
f(x) being polynomial, is continuous and also differentiable.
Now, f '(x) = 4x3 + 6x5 – 10x4
and f(0) = 0 = f(1) Rolle's Theorem is applicable.
so, there exist at least one c  (0, 1) such that f '(c) = 0
4c3 + 6c5 – 10c4 = 0
2
2c3 (2 + 3c2 – 5c)c = 0, 1,
3
2
As c =  (0, 1)
3
y

x
O 2/3 1
Also, f "(c) = 0
12c + 30c – 40c = 02c2 [6 + 15c2 – 20c] = 0
2 4 3

 c = 0 or15c2 – 20c + 6 = 0, will have two distinct roots.

Clearly,f "(c) = 0 has three distinct solutions option (B)is false.
Now, verify alternatives.

22. Which of the following is/are correct?

(A) Between any two roots of ex cos x = 1, there exists at least one root of tan x = 1.
(B) Between any two roots of ex sin x = 1, there exists at least one root of tan x = – 1.
(C) Between any two roots of ex cos x = 1, there exists at least one root of ex sin x = 1
(D) Between any two roots of ex sin x = 1, there exists at least one root of ex cos x = 1.
Ans: A, B, C
Sol. (A) Letf (x) = ex cos x – 1
 f ' (x) = ex(cos x – sin x) = 0
 tan x = 1, between two roots of f (x). (Rolle's theorem)
(B) Letg (x) = ex sin x – 1,
g ' (x) = ex(sin x + cos x) = 0
 tan x = – 1,between two roots of g (x).
(C) Leth (x) = e–x – cos x,
h ' (x) = – e–x + sin x = 0
 e–x = sin x, between two roots of h (x).
23. Which of the following statement is/are CORRECT?
(A) All zeroes of the cubic f (x) = 12x3 + 24x2 + 11x – 1are confined in [–1, 1].
1 1 
(B) LMVT is applicable for the function f(x) =x – in the interval  , 3
x 2 
(C) Let f(x) = x + 2|x + 1| + 2|x – 1|. If f(x) = k has exactly one real solution, then k = –1
(D) Let f (x) be a real differentiable function such that f (x) + f ' (x)  1 x R and f (0) = 0,
1
then the greatest value off (1) is 1  .
e
Ans: B, D
Sol. (A) f(x) = (x + 1) (12x2 + 12x – 1)
 12  192
x = –1,
24
 12  192
Clearly < –1 (A) is false.
24
1 1 
(B) Clearly f(x) =x – is continuous as well as differentiable in the interval  , 3 Hence
x 2 
LMVT is applicable .
(C) Let f (x) = x + | 2(x + 1) | + 2 | x – 1 |
 x  2( x  1)  2( x  1), x  1   3x , x  1
 
Then f (x) = x  2( x  1)  2( x  1),  1  x  1 f (x) = x  4 ,  1  x  1
x  2( x  1)  2( x  1), x 1  5x , x 1
 
Graph of y = f(x) is as shown.
y
= 5x
f ( x)

f(x)

(1,5)
= –3

(–1,3) +4
x

x
)=
f (x
x
O

Clearly y = k can intersect

y = f(x) at exactly one point only if k = 3 Ans.]
(D) Letg(x) = ex f(x) ; then g'(x) = ex (f(x) + f '(x))  ex.
Integrating with respect to x from 0 to 1,we get
1 1
(e  1)
g(1) – g(0) =  g ' ( x )dx   e x dx  e  1 . But g(1) – g(0) = e · f(1), so we get f (1) 
0 0
e
Alternatively:
We havef(x) + f '(x)  1  x  R
ex( f(x) + f '(x))  exx R ( Multiplying both the sides by ex)

 
1 1
1
  e x (f ( x )  f ' ( x )) dx   e x dx  e x f ( x ) 0  e  1  e f(1) – f(0)  e – 1
0 0
(e  1)
 f (1)  (As f(0) = 0)
e
1
Hence the greatest value off (1) = 1  .
e
Numerical Answer Type Questions

24. Number of roots of the equation x2 . e2–|x| = 1is

Ans: 4
Sol. draw graphs of y = x2 . e2–| x |andy = 1
y
(0, 4)

y=1
x
–2 0 2
x 2 ·e 2 x if x  0
y = 

x 2 ·e 2 x if x  0

25. Suppose that f (0) = – 3 and f ' (x)  5 for all values of x. Then the largest value which f (2) can
attain is
Ans: 7
Sol. Using LMVT in [0, 2]
f ( 2)  f ( 0)
= f ' (c)where c  (0, 2)
20
f ( 2)  3
5
2
f (2)  7

26. The graphs y = 2x3 – 4x + 2 and y = x3 + 2x – 1 intersect in exactly 3 distinct points. The slope
of the line passing through two of these points
Ans: 8
Sol. Let (x1, y1) and (x2, y2) are two of these points given y = x3 + 2x – 1 and
y=2x3 – 4x + 2
 y1 = 2x13 – 4x1 + 2 ....(1)
y
R

and 2y1 = 2x13 + 4x1 – 2 ....(2)

(2) – (1)
y1 = 8x1 – 4 ....(3)
|||ly y2 = 8x2 – 4 ....(4)
—————
y2 – y1 = 8(x2 – x1)
y 2  y1
=8
x 2  x1
27. If f(x) is a twice differentiable function such that f(a) = 0. f(b) = 2. f(c) = –1, f(d) = 2,
f(e) = 0, where a < b < c < d < e, find the minimum number of zeroes of
g  x    f   x    f  x  f   x  in the interval [a, e].
2

Ans: 6
Sol. Graph of f(x)
d
g(x) = ( f  x f  x )
dx a b c d
Let h(x) = f  x  f   x 
from the graph f(x) is zero at at least four places and f   x  is –1
zero at at least three places
hence f  x  f   x  is zero in at least 7 places.
hence h  x  is zero in at least 6 places.
h  x  = g(x) = 0 has minimum 6 solutions.

3 x0
2
28. If the function f (x) =  x  3x  a 0  x  1 satisfy the hypothesis of the mean value theorem
mx  b 1 x  2
for the interval [0, 2] then find the value of a + m+ b.
Ans: 8
3 x0
2
Sol. f (x) =  x  3x  a 0  x  1
mx  b 1 x  2
function should be continuous in [0, 2] and for continuous at x = 0
f (0) = Lim f ( x )
x  0
3=a a = 3
for continuous at x = 1
f (1) = Lim f ( x )
x 1
m+b=–1+3+3
m+b=5 ......(1)
f (x) should be derivable in (0, 2)
 for derivability at x = 1
f '(1–) = f '(1+)
–2 + 3 = m
 m = 1 and m+b=5

29. Let a > 0 and f be continuous in [–a, a]. Suppose that f ' (x) exists and f ' (x)  1 for all
x  (–a, a). If f (a) = a and f (– a) = – a, find value of f (0) .
Ans: 0
Sol. Consider y = f (x) in [–a, 0]
Since f continuous and has a derivative hence using LMVT in [–a, 0],  some c  (–a, 0)
f ( 0)  f (  a )
s.t. f ' (c) = 1
a
or f (0) + a a  f (0)  0 ....(1)
|||ly applying LMVT in [0, a]
f ( a )  f ( 0)
f ' (a) =  1 or0 – f (0)  a
a
  f (0)  0 ....(2);from (1) and (2)f (0) = 0
ba
30. Find the greatest possible integral value of , where 0 < a < b < 3 .
tan b  tan 1 a
1

Ans: 3
Sol. Letf (x) = tan–1 x,x Î [a, b]
tan 1 b  tan 1 a 1
UsingLMVT, we get = ,where0 < a < c < b < 3
ba 1  c2
 ba 
So, 1 <  1 1 
<4
 tan b  tan a 
1 1
As,  1
4 1  c2
 The greatest possible integral value is 3.

Passage Type Questions

Paragraph for question nos.31 & 32

Let f be a function such that it is differentiable wherever it is continuous.

If f '() = f '() = 0 and f "() · f "() < 0 where , Rare such that f () = 4,
f () = 0 and <. Then

31. If f is continuous in [, ] and f "() – f "() < 0, then minimum number of solutions of
equation f '(x) = 0 in [ – 1,  + 1] is
(A) 2 (B) 3 (C) 4 (D) 6
Ans: A

32. If f is continuous in [, ] and f "() – f "() > 0. Letn1 and n2be minimum number of solutions
of f '(x) = 0 and f (x) = 0 in [ – 1,  + 1] respectively, then the value of n1 + n2 is
(A) 4 (B) 5 (C) 6 (D) 7
Ans: C
Sol. (33-34)
(i) f "() – f "() < 0f "() < f "()f "() < 0andf "() > 0
Hence, maxima at x = and minima at x = 

 
(ii) f "() > f "()f "() > 0, f "() < 0
x = is point of local minima and x = is a point of local maxima

' 
 '

minimum number of roots of f '(x) = 0 in [ – 1,  + 1] is 4

and minimum number of roots of f(x) = 0 in [ – 1,  + 1] is 2.
 Paragraph for question nos. 33 & 34

Let f (x) be a cubic polynomial with leading coefficient unity such that f (a) = b and
f ' (a) = f '' (a) = 0. Suppose g (x) = f (x) –f (a) + (a – x) f ' (x) + 3 (x – a)2 for which conclusion
of Rolle's theorem in [a, b] holds at x = 2, where 2  (a, b).

33. The value of f " (2) is equal to

(A) 2 (B) 3 (C) 4 (D) 6
Ans: D

b
34. The value of definite integral  f (x) dx is equal to
a
123 213 321 481
(A) (B) (C) (D)
64 64 64 64
Ans: C
Sol. (35 and 36)
We have, f (x) =(x – a)3 + b .........(i)
Since, Rolle's theorem is applicable to g (x) at x = 2 ,so
g (a)=g (b) and g' (2) = 0
 0 =f (b) – f (a) +(a – b) f ' (b) + 3 (b – a)2
and (a – 2) f " (2) +6 (2 – a) =0
f ( b)  f (a )
 f '(b) = + 3 (b – a) .........(ii)
ba
and f " (2)= 6 .........(iii)(As a  2, think)
(i) As, f " (2) = 6 [using (iii) in (i)]
6 (2 – a) = 6  a = 1.
(ii) Using (i) in (ii), we get
3 5
(b – a)=  b= .
2 2
5
2
 5 321
   ( x  1)3   dx = ]
1  2  64

Matching Type Questions

35. 35. Match the column:

Column I Column II
(A) Number of positive roots of the equation(x – 1) (x – 2) (x – 3) + (x – 1) (P) 1
(x – 2) (x – 4) + (x – 2) (x – 3) (x – 4) + (x – 1) (x – 3) (x – 4) = 0, will be

(B)  x2  (Q) 2
If the function g(x) = 2f   + f(6–x2)xRis increasing in the interval
2
(a,), where f (x) > 0 xR then value of ‘a’ is equal to
(C) If f(x) = ex  x [0, 1] & f(1) – f(0) = f (c) where c (0, 1) then (R) 3
n (ec + 1) is equal to
(D) If langranges mean value theorem is applicable for the function (S) 4
 mx  c, x  0
f(x) =  x in [– 2, 2] then m + 3c is equal to
 e , x0
Ans: AR; BQ; C P; D S
Sol. (A) L.H.S. of given equation is f (x) where f(x) = (x – 1) (x – 2) (x – 3) (x – 4)
 f(x) is continuous & derivable and f(1) = f(2) = 0
 f (x) = 0 has at least one root in (1, 2)
Similarly f (x) = 0 has at least one root in (2, 3) and(3, 4)
 f (x) is a cubic function hence it has exactly one root in (1, 2), (2, 3), (3, 4)
x  2 
(B) g(x) = 2f   + f(6 – x2)
 2 
  x2  
g (x) = 2x f     f (6 – x 2 )
  2  
x2
> 6 – x2 x < – 2 or x > 2
2
x2
for > 6 – x2
2
 x2 
f    – f (6 – x2) > 0
 2 
{ f (x) > 0  f (x) is increasing}
– + – +
–2 0 2
 g (x) > 0 x  (2, )  a = 2
ec  1
(C) Applying LMVT in [0, 1] = ec for some c  (0, 1)ec + 1 = e n (ec + 1) = 1
1 0
mx  c, x  0
(D) f(x) =  x
 e , x0
for application of LMVT function must be continuous and derivable in [– 2, 2]
f(0+) = f(0–)  c = 1 and f (0+) = f (0–)  m = 1  m + 3c = 4

36. Match the column:

Column I Column II
(A) The equation x log x = 3 – x has at least one root in (P) (0, 1)
(B) If 27a + 9b + 3c + d = 0, then the equation (Q) (1, 3)
4ax3 + 3bx2 + 2cx + d = 0has at least one root in
(C) 1 (R) (0, 3)
If c = 3 and f(x) = x + , then interval of x in which LMVT is
x
applicable for f(x), is

(D) 1 (S) (–1, 1)

If c = and f(x) = 2x – x2, then interval of x in which LMVT is
2
applicable for f(x), is
(T) (3, 1)
Ans: AQ ; BR; CQ ;D  P
3
Sol. (A) f (x) = log x  + 1
x
 f(x) = (x – 3) log x + c
 f(1) = f(3)
(B) f (x) = 4ax3 + 3bx2 + 2cx + d
f(x) = ax4 + bx3 + cx2 + dx + e
 f(0) = f(3)27a + 9b + 3c + d = 0
f (b)  f (a) 2 ab  1 2
(C)
ba
 
f 3  
3 ab

3
f (b)  f (a) 1
(D)  f    a + b = 1
ba  2
AOD-5:
Single Correct Answer Type Questions

1. Side of an equilateral triangle expands at the rate of 2 cm/s. The rate of increase of its area
when each side is 10 cm, is
(A) 10 2 cm2/sec (B) 10 3 cm2/sec (C) 10 cm2/sec (D) 5 cm2/sec
Ans: B
da
Sol.  2cm / s
dt
3 2 dA 3 da 3
A a    2a   2 10.2  10 3cm 2 / s
4 dt 4 dt 4

2. A ladder 5 m long is leaching against a wall. The bottom of the ladder is pulled along the
ground away from the wall, at the rate of 2m/sec. How fast its height on the wall decreasing
when the foot of the ladder is 4 m away from the wall -
4 8 10 6
(A) m/sec (B) m/sec (C) m/sec (D) m/sec
3 3 3 3
Ans: B
Sol. at time t

y 5m

x
x2 + y2 = 52
dx dy
2x + 2y =0
dt dt
dx dy
Given , x, y, we have to find .
dt dt

3. On dropping a stone in stationary water circular ripples are observed. Rate of flow of ripples is
6 cm/sec. when radius of the circle is 10cm, then fluid rate of increase in its area is :
(A) 120 cm2/sec (B)  cm2/sec (C) 120cm2/sec (D) None
Ans: C
dr
Sol. = 6cm/sec
dt
A = r2
dA dr
= 2r . = 2 × 10 × 6= 120 cm2/sec
dt dt

4. The value of (127)1/3 to four decimal places is -

(A) 5.0267 (B) 5.4267 (C) 5.5267 (D) 5.001
Ans: A
Sol. Let y = x1/3, x = 125 and x + x = 127. Then,
dy 1
= 2/3 and x = 2
dx 3x
When, x = 125, we have
dy 1
y = 5 and =
dx 75
 dy 1 2
y = x y = ×2=
dx 75 75
2 8 1
(127)1/3 = y +y  5 + 5+ ×
75 3 100
(2.6667)
(127)1/3 = 5 + = 5.02667 = 5.0267
100
Hence (A) is the correct answer.

5. A balloon is pumped at the rate of a cm3/minute. The rate of increase of its surface area when
the radius is b cm, is
2a 2 a 2a
(A) 4
cm2/min (B) cm2/min (C) cm2/min (D) none of these
b 2b b
Ans: C
3 
d  r 3 
dV 4   4r 2 dr  a (given)
Sol.  
dt dt dt
dr a dS d(4r 2 ) dr a 2a
  ,   8r   8r  
dt 4r dt 2
dt dt 4r 2
r
dS  2a
   .
dt  r  b b

6. Two racers start a race at the same moment and finish in a tie. Which of the following must be
true?
(A) The racers speeds at the end of the race must have been exactly the same
(B) The racers had to have the same speed at some moment but not necessarily at exactly the
same time
(C) At some point during the race the two racers were not tied
(D) The racers must have had the same speed at the exactly the same time at some point in the
race
Ans: D
Sol. Let trajectories are f(t) and g(t)
since f(t) – g(t) = 0 has two roots 0 & T
so f '(t) – g '(t) = 0 must have at least one root in (0, T) 0 T

7. 7. A nursery sells plants after 6 year of growth. Two seedlings A and B are planted each of
dh A dh B
height 5 inches whose growth rates are = 0.5 t + 2 and = t + 1 where heights hA and
dt dt
hB are in cms and t is the time in years. Then -
(A) the height of the plants are equal at t = 3 (in years)
(B) the height of the plants are equal at t = 4 (in years)
(C) when the plants are sold, their heights are 26 cms and 30 cms for A & B respectively
(D) when the plants are sold, their heights are 26 cms and 29 cms respectively
Ans: B
t2
Sol. hA = + 2t + 5  hA = 5 at t = 0
4
t2
hB = +t+5
2

8. A truck is to be driven 300 km on a highway at a constant speed of x kmph. Speed rules of the
highway required that 30  x  60. The fuel costs Rs. 10 per litre and is consumed at the rate of
 x2
2+ litres per hour. The wages of the driver are Rs. 200 per hour. The most economical
600
speed to drive the truck, in kmph is
(A) 30 (B) 60 (C) 30 3.3 (D) 20 3.3
Ans: B
300
Sol. Time taken by the truck = hours
x
 x 2  300
 petrol consumed =  2   litre
 600  x
 expenses on traveling
300  x 2  3000 60000 6000 66000
E = 200 × 2  =  + 5x = + 5x
x  600  x x x x
dE 66000
 =– + 5 < 0 for all x  [30, 60]
dx x2
 most economical speed is 60 kmph

9. The radius of a right circular cylinder increases at the rate of 0.1 cm/min, and the height
decreases at the rate of 0.2 cm/min. The rate of change of the volume of the cylinder, in
cm3/min, when the radius is 2 cm and the height is 3 cm is
8 3 2
(A) – 2 (B) – (C) – (D)
5 5 5
Ans: D
Sol. Given,V = r2h
Differentiating both sides
dV  dh dr   dh dr 
=   r2  2r h  = r  r  2h 
dt  dt dt   dt dt 
dr 1 dh 2
= and =–
dt 10 dt 10
dr   2  1   r
= r  r     2h    = ( r  h)
dt   10   10   5
Thus, when r = 2 and h = 3,
dV (2) 2
= (2  3) =
dt 5 5

10. The altitude of a cone is 20 cm and its semi-vertical angle is 30°. If the semi-vertical angle is
increasing at the rate of 2° per second, then the radius of the base is increasing at the rate of -
160
(A) 30 cm/sec (B) cm/sec (C) 10 cm/sec (D) 160 cm/sec
3
Ans: B
Sol. Let  be the semi-vertical angle and r be the radius of the cone at time t. Then,
r = 20 tan 
dr d
 = 20 sec2
dt dt
 V



20 cm
A r r B
O
dr  d 
 = 20 sec2 30° × 2    30 and  2
dt  dt 
dr 4 160
 = 20 × × 2 cm/sec = cm/sec
dt 3 3
Hence (B) is the correct answer.

11. A rectangle with one side lying along the x-axis is to be inscribed in the closed region of the xy
plane bounded by the lines y = 0, y = 3x, and y = 30 – 2x. The largest area of such a rectangle
is
135 135
(A) (B) 45 (C) (D) 90
8 2
Ans: C
Sol. A = (x2 – x1)y
y = 3x1andy = 30 – 2x2
y

y=3x
y=30–2x

y y
x1 x2 x
O
 30  y y 
A (y) =   y
 2 3
6A(y) = (90 – 3y – 2y)y = 90y – 5y2
6A' (y) = 90 – 10y = 0 y = 9 ; A''(y) = – 10 < 0
21
x1 = 3 ; x2 =
2
 21  15 ·9 135
Amax=   3  9 = =  (C)
 2  2 2

12. Coffee is draining from a conical filter, height and diameter both 15 cms into a cylinderical
coffee pot diameter 15 cm. The rate at which coffee drains from the filter into the pot is 100 cu
cm /min.The rate in cms/min at which the level in the pot is rising at the instant when the
coffee in the pot is 10 cm, is
9 25 5 16
(A) (B) (C) (D)
16 9 3 9
Ans: D
Sol. For cylindrical pot
V = r2h
 15cm

15

dV  dh dr  dr
=  r 2  h·2r  (r = constant, = 0)
dt  dt dt  dt
dh
hence,100 = r2
dt
225 dh 15
100 =  · · (r = cm)
4 dt 2
dh 400 16
= = cm/min
dt 225 9

13. The lower corner of a leaf in a book is folded over so as to just reach the inner edge of the
page. The fraction of width folded over if the area of the folded part is minimum is :
(A) 5/8 (B) 2/3 (C) 3/4 (D) 4/5
Ans: B
x
Sol. 2A = xy sin ; cos  =
y
 x2 
4A2 = x2 y2 sin2 = x2y2[1 – cos2] = x2y2 1  2 
 y 
2 2 2 2
4A = x [y – x ] ....(1)
1 x
now – cos 2 =
x
1 x
1 – 2 cos2 =
x
2
x 1 x 2 2x 3
1–2· 2 =  y =
y x 2x  1
 2x 3  x4
from (1)4A2 = x2   x 2  ;4A2 = now proceed ]
 2x  1  2x  1

14. A particle move from right to left along the parabola y =  x in such a way that x-coordinate
(measured in meters) decreases at the rate of 8 m/sec. At the moment when x = – 4 the rate at
which the angle of inclination  of the line joining the particle to the origin is changing, is
(A) – 0.4 (B) – 0.2 (C) – 0.5 (D) – 1
Ans: A
y
Sol. y =  x ; tan =
x
x 1
tan = =
x x
d 1 1 dx 1
sec2 = + = ·(8) (where x = – 4)
dt 2 ( x)3 2 dt 16
 (x, y)

d 1
sec2 =– ....(1)
dt 2
2 1
Also where x = – 4, tan = =–
4 2
d 1
 (1 + tan2) =–
dt 2
5 d 1 d  2
=–  = = – 0.4.
4 dt 2 dt 5

15. The bottom of the legs of a three legged table are the vertices of an isoceles triangle with sides
5, 5 and 6. The legs are to be braced at the bottom by three wires in the shape of a Y. The
minimum length of the wire needed for this purpose, is
(A) 4 + 3 3 (B) 10 (C) 3 + 4 3 (D) 1 + 6 2
Ans: A
Sol. L = 4 – x + 2 9  x 2 where 0  x  4
dL 2x
 1 0
dx 9  x2
A

5 4–x 5
2 Ö9
x +x 2
Ö9+
x
B C
3 M 3
6
4x2 = 9 + x2
x= 3
now  (0) = L (4) = 10
 
 3 = 4 – 3 + 2 · 3 = 4 + 3 Ans.

16. An inverted cone has a depth of 10cm & a base of radius 5cm. Water is poured into it at the
rate of 1.5 cm3/min. The rate at which level of water in the cone is rising, when the depth of
water is 4cm
3 3 3 3
(A) cm/min (B) cm/min (C) cm/min (D) cm/min
8 4 16 
Ans: A
dh  1
Sol.  ? Now Volume = r2h
dt  ( h 4cm) 3
5 10 1 h2
= Volume =  h
r h 3 4
 5

r
10cm
1.5cm3 /min h

1 3
i.e. h = 2r = h
12
dh dr dV 1 2 dh
=2 = h
dt dt dt 4 dt
1 dh
 1.5 =  × 42
4 dt
dh 1.5
 =
dt 4

17. If the radius of a sphere is measured as 9 cm with an error of 0.03 cm, the approximate error in
calculating its volume is
(A) 9cm3 (B) 9.72 cm 3 (C) 10.72cm 3 (D) none of these
Ans: B
Sol. r = 9 cm; dr = 0.03
4
V = r 3
3
4 3
dV = 3r 2dr = 4r2dr = 4(81)(0.03) = (324) = 9.72 cm3
3 100

N
18. Let N be any four digit number say x1 x2 x3 x4. Then maximum value of is
x1  x 2  x 3  x 4
equal to
1111
(A) 1000 (B) (C) 800 (D) none of these
4
Ans: A
N 1000x1  100x 2  10x 3  x 4  900x 2  990x 3  999x 4 
Sol.  = 1000 –
x1  x 2  x 3  x 4 x1  x 2  x 3  x 4  x1  x 2  x 3  x 4 
N
 maximum value of = 1000 .
x1  x 2  x 3  x 4
Hence (A) is correct.

19. Two villages A and B are on the same side of a straight river. A pump set is to be installed by
the river side at a point P. Then if the villages are situated at a distance c then
(A) minimum value of PA + PB is c 2  2ab
(B) minimum value of PA + PB is c 2  4ab
(C) minimum value of PA + PB is c 2  ab
(D) none of these
Ans: B
Sol. Take river as x axis, line joining origin and village A as x axis
 B(b, k)
P c
A'
(–a, 0) A(a, 0)

k2 = c2 – (b – a)2
image of A in the river, this is A' and BA' must be minimum value of PA + PB
BA' = (b  a) 2  k 2
= (b  a) 2  c 2  (b  a) 2 = c 2  4ab

20. If the radius of circles C1 : x2+ y2 = 1 and C2 : (x – 6) 2 + (y – 8) 2 = 4 are increasing at the rate
of 0.3 cm/sec and 0.4 cm/sec then which of following is false
(A) C1 and C2 touch each other internally at time t = 1.5 min
(B) C1 and C2 touch each other externally at time t = 10 sec
(C) C1 and C2 will be orthogonal at time t (20, 90)
(D) C1 and C2 will be orthogonal at time t (10, 20)
Ans: C
dr1
Sol. = 0.3  r1= 0.3t + C1 at t = 0, r = 1 C1 = 1 r1 = 0.3t + 1
dt
dr2
= 0.4  r2= 0.4t + C2 at t = 0, r = 2 C2 = 2r2 = 0.4t + 2
dt
Now, two circles C1 and C2 will touch internally if C1 C2 = |r1 – r2|
10 = | 0.1 t + 1 |  0.1t + 1 = 10  t = 90 sec = 1.5 min
two circles will touch externally if C1C2 = r1 + r2
10 = 0.7t + 3  7 = 0.7t  t = 10sec
2g1g2 + 2f1f2 = C1 + C2
(0.4t + 2)2 – 100 + (0.3t + 1)2 = 0
(4t + 20)2 – 10000 + (3t + 10)2 = 0
25t2 + 220t – 9500= 0
f(t) = 5t2 + 44t – 1900 = 0
f(10) = negative, f(20) = positive
t (10, 20)

Numerical Answer Type Questions

21. The figure shows a right triangle with its hypotenuse OB

y
B

h
A(t,t2 )

x
O
along the y-axis and its vertex A on the parabola y = x2. Let h represents the length of the
hypotenuse which depends on the x-coordinate of the point A. The value of Lim ( h ) equals
x0
Ans: 1
1
Sol. Let A = (t, t2); mOA = t;mAB = –
t
 y
B
1
1 y=x
2
t
equation of AB, y – t2 = – (x – t) h
t t
A(t,t2 )

x
O
put x = 0
h = t2 + 1 (as x  0 then t  0)
now Lim ( h ) = Lim (1  t 2 ) = 1 Ans.
t 0 t 0

22. The ends A and B of a rod of length 5 are sliding along the curve y = 2x2. Let xA and xB be
the x-coordinate of the ends. At the moment when A is at (0, 0) and B is at (1, 2) the derivative
dx
9. B has the value equal to -
dx A
Ans: 1
Sol. Given, y = 2x2
Now, (AB)2 = (XB – XA) + (2xB2 – 2xA2 ) = 5
or (xB – xA)2 + 4 (xB2 – XA2)2 = 5
dx
On differentiating wrt XA and denoting B =D
dx A
2 2
2(xB – xA) (D – 1) + 8(xB – xA ) (2xBD – 2xA) = 0
On putting xA = 0; xB = 1, then
 2(1 – 0) (D – 1) + 8(1 – 0) (2D – 0) = 0
 2D – 2 + 16D = 0
1
D= .
9

23. The pressure p and the volume v of a gas are connected by the relation pv1.4  k, where k is a
constant. If there be an increase of 0.7 percent in the pressure, then decrease percentage in
volume is  find value of 2 .
Ans: 1
Sol.  pv1.4  k  log p  1.4 log v  log k
1 1 100 100 1.4
Hence p  1.4  v  0 or p  v  0
p v p v
100 p
By hypothesis,  0.7
p
100 v
Hence 0.7   1.4  0
v
100 v
or  0.5
v
Negative sign shows that there is decrease of 0.5 percent in the volume.

24. A square tank of capacity 250 cubic m. has to be dug out. The cost of land is Rs. 50 per sq.m.
The cost of digging increases with the depth and for the whole tank is 400 (depth)2 rupees.
Find the side of the tank for the least total cost.
Ans: 10
Sol. x 2 h  250 , where x the side of square base and h is the depth of tank.
Area of base = x 2 , cost of land = 50x 2
Cost of digging = 400h2
 250
E  400h 2  50x 2  400h 2  50.
h
Now solving for minimum E, we have
h  2.5 meters and x = 10 meters

25. If in a triangle ABC, the side 'c' and the angle 'C' remain constant, while the remaining
da db
elements are changed slightly, then the value of  is.........
cos A cos B
Ans: 0
Sol. a = 2R sin A
da = 2R cos A dA
da db
db = 2R cos B dB     2R (dA  dB)
 cos A 2 cos B
c
now = 2R
sin C
A+B=–C
da db
dA + dB = 0as C is constant, hence  =0
cos A cos B

26. An open Can of oil is accidently dropped into a lake; assume the oil spreads over the surface as
a circular disc of uniform thickness whose radius increases steadily at the rate of 10 cm/sec. At
the moment when the radius is 1 meter, the thickness of the oil slick is decreasing at the rate of
4 mm/sec, find how fast is it decreasing in mm/sec when the radius is 2 meters.
Ans: 2
dr
Sol. Given = 10 cm/sec
dt
when r = 100 cm
dh
= – 0.4 cm/sec
dt
dh
= ? where r = 200 cm
dt
we have V = r2h where V is constant
dh dh 2V dr
h= ; =– 3 ....(1)
dt dt r dt
2V 1
– 0.4 = – · 10
 106
V
 = 2 × 104 cu. cm

again using (1)
4
dh 4  10 1
=– ·10 = – = – 0.05 cm/sec
dt 8  10 6 20

Passage Type Questions

Paragraph for questionnos. 27 to 28

A lamp post of length 10 meter placed at the end A of a ladder AB of length 13 meter, which is
leaning against a vertical wall as shown in figure and its base slides away from the wall. Atthe
instant base B is 12m from the vertical wall, the base B is moving at the rate of 5m/sec. Aman
 (M) of height 1.5 meter standing at a distance 15m from the vertical wall, then answer
thefollowing question.

A
13

 
W BM

27. Rate at which  decreases, when the base B is 12 m from the vertical wall, is-
1
(A) 1 rad/sec (B) 2 rad/sec (C) 5 rad/sec (D) rad/sec
2
Ans: A
x
Sol. Let WB = x at time t, then = cos 
13
dx d 5 d d
 = 13 (– sin ) thus when WB = 12, then5 = – 13. . i.e. = –1 rad/sec.
dt dt 13 dt dt

28. Rate at which length of shadow of the man increases, when the base B is 12m from the vertical
wall, is
40 15
(A) 15 m/sec (B) m/sec (C) m/sec (D) 5 m/sec
27 2
Ans: B
Sol. WM = 15m
Height of the lamp post at time t = 169  x 2 + 10 m
Let  be the length of shadow at time t, then
169  x 2 10 15   15
= = +1
1.5  
169  x 2 + 10 = 22.5 + 1.5

1 2x dx 22.5 d
 =
2 169  x 2 dt 2 dt
d 12 2
=
dt 22.5
by the given condition,
15 15
we have + 1= = 10
 15
2
d 12  15  120 40
= ×   = =
dt 22.5 9 81 27

Paragraph for questionnos. 29 to 30

If Q is a quantity that varies with time, then the derivative dQ/dt gives the rate of change of that
quantity with respect to time. A conical paper cup 20 cm across the top and 15 cm deep is full of
water. The cup spring a leak at the bottom and loses water at 5 cu cm per minute.

29. How fast is the water level dropping at the instant when the water is exactly 7.5 cm deep ?
 (A) 1/ cm/min (B) 1/5 cm/min (C) 1/2 cm/min (D) 2/3 cm/min
Ans: B
R = 10

r
15
h
Sol.
2
1 1 2  4
V = r2h =   h  h  h 3
3 3 3  27
dv 4 2 dh dv 4 dh
 h    (7.5)2
dt 9 dt dt 9 dt

30. The amount of water when the height of water is 3 cm is (in cm3)
(A) 4 (B) 3 (C) 27 (D) 2
Ans: A
Sol. v(h = 3)  4
Q.1 Function f (x) , g (x) are defined on [–1, 3] and f ''(x) > 0, g ''(x) > 0 for all x  [–1, 3],
then which of the following is always true?
(A) f (x) – g(x) is concave upwards on (–1, 3)
(B) f (x) g(x) is concave upwards on (–1, 3)
(C) f (x) g(x) does not have a critical point on (–1, 3)
(D) f (x) + g(x) is concave upwards on (–1, 3)
Sol. D
(A) take f (x) = x2, g (x) = 2x2
f '' (x) – g '' (x) need not be greater than 0.
(B) try f (x) = (x – 1)2, g (x) = e–x
(C) try f (x) = x2, g (x) = x2.
ax 3
Q.2 The set of value(s) of 'a' for which the function f (x) = + (a + 2) x2 + (a  1) x + 2
3
possess a negative point of inflection .
(A) (,  2)  (0, ) (B) { 4/5 }
(C) ( 2, 0) (D) empty set
Sol. A
b
For negative point of inflection  < 0 ;
2a
y
f '(x)
x

–2 0
2
d y
alternatively = 0, get 'x' and put x < 0
dx 2
f  (x) = a x2 + 2 (a + 2) x + (a  1)
a 2 a 2
f  (x) = 2 ax + 2 (a + 2) = 0  x =  <0= >0
a a

Q.3 Let h be a twice continuously differentiable positive function on an open interval J. Let
g(x) = ln for each x  J
Suppose h' ( x) 2 > h''(x) h(x) for each x  J. Then
(A) g is increasing on J (B) g is decreasing on J
(C) g is concave up on J (D) g is concave down on J
[Sol. D
Given g(x) = ln h ( x ) 
h' (x )
g ' (x) =
h(x)
h ( x )h ' ' ( x )  (h ' ( x )) 2
g''(x) = <0 (given)
h 2 (x )
  g''(x) <0  g (x) is concave down
Q.4 For all x  (0, 1)
(A) ex < 1 + x (B) loge(1 + x) < x (C) sin x > x (D) loge x > x

Sol. B
(a) (A) x  (0, 1) (B) ln (1 + x) < x
ex < 1 + x f(x) = ln(1 + x) – x
1
f(x) = ex – x – 1 f (x) = –1
1 x
x
f (x) ex – 1 which is > 0  x  (0, 1) = < 0 in x  (0, 1)
1 x
so f(x) in (0, 1) i.e.f (x) is  is (0, 1)
i.e. f(x) > f(0) f(x) < f(0)f(x) < 0 is (B)
f(x) > 0
(C) f (x) = sin x – x (D) ln x > x
f ' (x) = cos x – 1 < 0 in   (0, 1) f (x) = ln x – x
1 1 x
f (x) is decreasing in x  (0, 1) f'(x) =  1 = >0 in (0,1)
x x
f (x) < f (0) f (x) is increasing in (0, 1)
f (x) < 0 i.e. sin x – x < 0; sin x < x f (x) < f (1) ]
 
Q.5 If ,   ,   and  < , then which one of the following is true?
2 
 
(A) e cos  – cos  (B) e cos  – cos  
 
 
(C) e cos  – cos < (D) e cos  – cos <
 
Sol. B
e cos x
Consider f(x) =
x
cos x
e (1  x sin x )
Now, f '(x) =
x2
As f '(x) < 0  f() > f()
cos 
e e cos  
So,   e cos  – cos  > . Ans.]
  
Q.6 The positive real number a such that xa  ax, for all positive real numbers x, is
1 1
(A) e (B) e (C) e  (D)
2 e

[Sol. A
a ln x  x ln a
 ln x ln a
  x > 0
x a
y (e, 1/e)

1 e x
O

 We have f ( x )  f (a ) is true  x  0 
ln x so, a is the point where f ( x ) atttains 
f(x) =
x its maximum value. 
 
ln a 1
   
a e
  a = e. Ans.
Q.7 The range of real constant 't' such that (1 – t) sin + t tan  >  always holds   
 
 0,  is
 2
1  1 
(A) [1, ) (B)  ,   (C*)  ,   (D) None
2  3 
Sol. C
 
(1 – t) sin + t tan  >      0, 
 2
  sin   
t>     0, 
tan   sin   2
  sin   
 t > maximum value of     0, 
tan   sin   2
 
Now, in  0,  tan  >  and the same quantity is subtracting from Nr and Dr .
 2
Hence, maximum value occur at  0+.
1
Hence, t  .
3
 
Aliter: Let f() = (1 – t) sin  + t tan  –  in  0,  [12th, 19-09-2010, Gems-3]
 2
f '() = (1 – t) cos  + t sec2 – 1
(1  t ) cos 3   cos 2   t
f '() =
cos 2

f '() =
  =

t 1  cos3   cos 2  1  cos  1  cos3   t  cos 2  

cos 2  cos

2
  cos 2   cos   1
 ve in 0,  2 
  
 
 1   1 
f '() =  t  2

 =t  2 
 sec   sec   1    sec   1   3 
 2  4 
 

2
 1 3 12 1 1
  sec       3  2

 2 4 4  1 3 3
min  sec    
 2 4 max
1
 If t   then f ' () > 0
3
 
 f   0,   f () > f (0) but f(0) = 0
 2
 
 f () > 0   0,  ]
 2
Q.8 Let f and g be two continuous function such that gof is increasing and g is
 
decreasing. If f g ( x 2  2 kx  3) > f g(k  3)   x  R, then number of possible integral
values of k is
(A) 3 (B) 4 (C) 5 (D) 6
[Sol. B
 
f g ( x 2  2 kx  3) > f g(k  3)   x  R
 g (x2 – 2kx + 3) < g(k – 3)  x  R
 x2 – 2kx + 3 > k – 3  x  R
 x2 – 2kx + 6 – k > 0  x  R
D < 0  4k2 – 4 (6 – k) < 0
 k2 + k – 6 < 0
 (k + 3) (k – 2) < 0
 k  (– 3, 2 )
 Number of integral values of k are 4 i.e. – 2, – 1, 0, 1.
3
sin
sin 1 2 and  = sin 2 , then which one of the following inequalities is
Q.9 Let  = ,=
sin 2 5 sin 3
sin
2
correct?
(A)  <  <  (B)  <  <  (C)  <  <  (D)  <  < 
[Sol. A
sin x sin ( x  1) cos x  sin x cos ( x  1) sin 1
f(x) =  f '(x) = =
sin ( x  1) 2
sin ( x  1) 2
sin ( x  1)
sin 1
Then f '(x) = > 0 i.e. f is increasing
sin 2 ( x  1)
 3
sin
sin 1 2  sin 2 i.e.  <  < .]
Now, the answer that we need to find is 
sin 2 sin 5 sin 3
2
Q.10 If the equation 4x + 5x + k = 0 (k  R) has a negative real root then
3

(A) k = 0 (B) – < k < 0 (C) 0 < k <  (D) – < k < 
[Sol. C
Let f (x) = 4x3 + 5x + k
f '(x) = 12x2 + 5 > 0  x  R
 f (x) is strictly increasing on R.
y
(0, f(0) = k)
x
x = x0

So,for f (x) = 0 to have a negative real root,f(0) > 0  k >0.]

Q.11 If the equation x4 + 8x3 + 18x2 + 8x + a = 0 has four distinct real roots, then the range of
a is
(A) (0, 9)
(B) (–9, 0)
(C) (–8, 1)
(D) (–1, 8)

[Sol. C
(x2 + 4x + 1)2 + a – 1 = 0
y

x
2 3 – 2 O 2 3
Graph of f(x)

 (x2 + 4x + 1)2 = 1 – a

f(x) = (x2 + 4x + 1)2

for four distinct solutions

1 – a  (0, 9)
 –a  (–1, 8)

a  (–8, 1). Ans.]

Q.12 Number of integral values of p for which the cubic 2x3 – 3x2 + p = 0 has 3 real roots
(not necessarily distinct), is
(A) 1 (B*) 2 (C) 3 (D) 4
Sol. B
Draw graph of y = p and y = 3x2 – 2x3

Hence, for three roots p  [0, 1]. Ans.

y

y=p

x
O
Aliter: Let f(x) =2x3 – 3x2+p
f '(x) = 6x (x – 1)
0
 f '(x) = 0 1
Now, f (0) · f(1)  0  p (p – 1)  0  p  [0, 1]. Ans.]
Q.13 Let f (x) = x3 – 3x2 + 2x. If the equation f (x) = k has exactly one positive and one
negative solution then the value of k equals
2 3 2 2 1
(A) – (B) – (C) (D)
9 9 3 3 3 3
[Sol. A
f (x) = x(x2 – 3x + 2)
f (x) = x(x – 2)(x – 1)
y

1+(1/Ö 3)
x
O 1–(1/Ö 3) 1 2
y=k
graph of y = f (x) is as shown
now f (x) = k to have
exactly one positive and
negative solution
 1 
we have, k = f 1   (think !) are the roots of f ' (x) = 0
 3
 1  1  1   1  1   2
 k = 1    1   =   1  = Ans. ]
 3   3   3   3  3  3 3
    
x x 2 x 1
Q.14 Consider the cubic equation x3 + px2 + qx + r = 0, where p, q, r are real numbers.
Which of the following statement is correct?
(A) If p2 – 2q < 0, then the equation has one real and two imaginary root.
(B) If p2 – 2q  0, then the equation has all real roots.
(C) If p2 – 2q > 0, then the equation has all real and distinct roots.
 (D) If 4p3 – 27q2 > 0, then the equation has real and distinct roots.
[Sol. A
Let f (x) = x3 + px2 + qx + r
 f '(x) = 3x2 + 2px + q
Disc. = 4p2 – 12 q = 4(p2 – 3q) = 4(p2 – 2q – q)
 If p2 < 2q  p2 < 3q
So, the equation f(x)=0 has one real and two imaginary roots.]
Q.15 If the equation x3 – 3x + 1 = 0 has three real roots x1, x2, x3, where x1 < x2 < x3, then the
value of is equal to
[Note: {x} denotes the fractional part of x.]
3 5
(A) (B) 1 (C) 2 (D)
2 2
[Sol. f(x) = x – 3x + 1.
3

f '(x) = 3(x2 – 1) = (x + 1) (x – 1)
f(x) is increasing in (–, –1) (1, ) and decreasing in (–1, 1)
Now, f(–2) = –1 f(1) = –1
f(–1) = 3 f(2) = 3
f(0) = 1
f(–2) · f(–1) < 0  Hence one root lies in (–2, –1)
  [x1] = –2
Now f(0) · f(1) < 0  Hence one root lies in (0, 1)
  [x2] = 0
Also f(1) · f(2) < 0  Hence one root lies in (1, 2)
 [x3] = 1
x1
x2
x3
So, x3 – 3x + 1 = 0
 x1 + x 2 + x 3 = 0
 {x1} + {x2} + {x3} = x1 – [x1] + x2 – [x2] + x3 – [x3]
= (x1 + x2 + x3) – ([x1] + [x2] + [x3])
= 0 – (–2 + 0 + 1) = 1. ]
Q.16 If the point (1, 3) serves as the point of inflection of the curve y = ax3 + bx2 then the
value of a + b is......

[Sol. 3
f  (x) = 3ax2 + 2bx and f  (x) = 6ax + 2b
 f (1) = 0  3a + b = 0 ...(1) also (1, 3) lies on the curve  a + b = 3....(2)
3 9
solving (1) and (2) we get a = – and b = . ]
2 2
Q.17 If f(x) = 1 – 3 e x – ex3 then find least integral solution satisfying the inequality
 1  
  e10  
    
e
  
1 / x
f  f  sgn 1  sgn | e x  1 | 
  
 x 2 8
 
 
 
.
[Note : sgn(y) denotes the signum function of y.]

[Sol. 3
f (x) = 1 – 3 e x – ex3
f ' (x) =  3 e – 3 e x2 < 0  f (x) is decreasing
 1   1  1 1
 ex   e 10  ex e10
 f  < f 1  >
 x  2   8  x2 8
   
 x  (2, 10)
 least integral solution = 3 .
a
Q.18 If the least positive value of p for which |6x + 25p| (4 + x2)  50 x  x  R is   , where
b
a, b  N, then find the least value of (b – 5a).

[Sol. 5
50x
| 6 x  25p | 
4  x2
3 p x
x 
25 2 4  x2
x (4  x 2 )1  x ·2 x 4  x2
f(x) =  f (x) = =
4  x2 (4  x 2 ) 2 (4  x 2 ) 2
k 0 4  h2 3
mAP = = 
25p 2 2
(4  h ) 25
h
6
 4  h2 3
2

( 4  h ) 25
y
, k)
  25p 
 , 0 P (h
 6 

–2 A O 2 x

100 – 25h2 = 3(16 + 8h2 + h4)

3h4 + 49h2 – 52 = 0
(3h2 + 52) (h2 – 1) = 0  h=±1
 h = 1 ; k = 1/5
1
k 3 5 3
Now =  
5p 25 25p 25
h 1
6 6
25p 4 a
 5=3+ p= 
2 25 b
b – 5a = 25 – 20 = 5. ]
Q.19 Let f : R  R be a function defined by f(x) = x3 – 3x2 – 32x + 33. Find the number of integral
value of x satisfying the relation f 3(x) – 3f 2(x) – 32f(x) + 33 < f(x3 – 4x2 – 3x + 19).

[Sol. 2
f '(x) = 3(x2 – 2x – 3) = 3 (x + 1) (x – 3)
Case-I: f '(x) > 0 i.e. f(x) x > 3 or x < –1
f(f(x)) < f(x3 – 4x2 – 3x + 19)
f(x) < x3 – 4x2 – 3x + 19
x2 – 6x + 8 < 0 x (2, 4)
x (3, 4)
Case-II: f '(x) < 0 i.e. f(x) x (–1, 3) 
 f(x) > x3 – 4x2 – 3x + 19 x < 2 or x > 4
x (–1, 2)
So, x (–1, 2) (3, 4)
Hence number of integral values of x is 2 i.e. x = 0, 1.]
Q.20 Let f (x) = x3 – 3x2 + [a]. If a [, ) be the complete range of values of 'a' for which f
(x) = 0 has three real and distinct solutions, then (– ) is equal to......
[Note: [k] denotes greatest integer function less than or equal to k.]
[Sol. 3f ' (x) = 3x2 – 6x
= 3x (x – 2)
f (0) = [a] and f (2) = [a] – 4
If f (0) · f (2) < 0  0 < [a] < 4
 1a<4 ]
Q.21 Let f be a polynomial function defined over R by f(x) = 4x3 – 6x2 + 1, then number of
distinct real solutions of f f ( x )  = 0 equals.......
(A) 9 (B*) 7 (C) 5 (D) 3
[Sol. 7
f '(x) = 12x (x – 1)
+ – +
f(0) = 1, f(1) = –1 0 1
f(–1) = negative, f(2) = positive
1  (–1, 0), 2  (0, 1)
y
1
–1 1
x
1 O 2 3 2
–1
3  (1, 2)
f(f(x)) = 0  f(x) = 1, 2, 3
f(x) = 1 has 3 distinct real solution
f(x) = 2 has 3 distinct real solution
f(x) = 3 has only real solution
 = 0 has 7 real solutions. ]
k
Q.22 The number of integral values of k for which the equation ex = has exactly two
x 3
solutions is

[Sol. 7
ex(x – 3) = k
Let f(x) = ex(x – 3)
f '(x) = ex · 1 + (x – 3) ex = ex (x – 2)
From the graph it is clear that, for exactly two roots k  (–e2, 0)
 number of integral values of k is, 7.
y

O x
2 3

y=k
2
–e
Q.23 For the function f(x) = x4 (12 ln x  7)
(A*) the point (1,  7) is the point of inflection (B*) x = e1/3 is the point of minima
(C*) the graph is concave downwards in (0, 1) (D*) the graph is concave upwards in
(1, )
[Sol. ABCD
dy d2y
= 16 x3 (3 ln x  1) & 2
= x2 (9 ln x) ]
dx dx
Q.24 Let f : R+  R is a strictly decreasing function for all x R+ such that
f (k2 – 2k) > f (3k – 4), then k can be
5 10 11 21
(A) (B) (C) (D)
2 3 4 5
[Sol. ABC
Given k2 – 2k > 0  k(k – 2) > 0
 k  (–, 0)  (2, ) ……(i)
+ +
0 – 2
4 4 
 3k – 4 > 0  k>  k  ,  ……(ii)
3 3 
 f is a decreasing function
 k2 – 2k < 3k – 4  k2 – 5k + 4 < 0
 1<k<4 ……(iii)
Common value of k from (i), (ii) and (iii) is
2<k<4
Correct value of k  (2, 4). So option A, B, C contains (2, 4). Ans.]

1  3x  1 
Q.25 Consider f : (–, 0)   ,    R, defined by f (x) = ln  e   then
 3e  2  3x 
(A) f (x) has no point of inflection. (B) f (x) is surjective but not injective.
1
(C) f (x) is bijective function. (D) f (x) = has two distinct solutions.
2e
[Sol. ABD
3x  1 
f (x) = ln e  
2  3x 
1 
Df = (–, 0)   ,  
 3e 
Lim f ( x )  (–)
 
x  31e


Lim f ( x ) = 0–
x 0 
 y

3x  1
f ( x)  ln  e  
2  3x 

O x
1 
 , 0
 3e 
  1
 0, 
 2e 

Lim f ( x )  –
x (  )
Lim f ( x )  
x 
3x 1  1  3  1 
f '(x) = ·   2   ·ln  e  
2  1   3x  2  3x 
e  
 3x 
3 3  1 
f '(x) =  ln  e  
2(3ex  1) 2  3x 
f '(x) > 0  x  Df
x (e x  e  x )  2
Q.26 Consider f(x) = , x  0 Identify which of the following statement(s)
2x
is/are correct-
(A) f(x) has no critical points
(B) f(x) has exactly two critical points
(C) f(x) has exactly one point of inflection
(D) f(x) is concave downwards in (–, 0) and concave upwards in (0, )
[Sol. BD
x (e x  e  x )  2 e x  e  x 1
f (x) = = +
2x 2 x
x x
e e 1
f '(x) =  2
2 x
e  ex
x
1
f '(x) = 0  = 2
2 x
 y

x
x1 x2

 x = x1, x 2
ex  e x 2
f "(x) =  3
2 x
 f "(x) > 0  x  (0, )
and f "(x) < 0  x  (– , 0)
Q.27 If f(x) = – x3 + 6x2 – 6x – 4 and g(x) = k(x– 2), (k > 0) intersect each other at three
distinct points then possible integral value(s) of k is/are
(A) 3 (B) 4 (C) 5 (D) 6
[Sol. ABC
f (x) = – x3 + 6x2 – 6x – 4
f '(x) = – 3x2 + 12x – 6
f "(x) = – 6x + 12
f "(x) = 0  x = 2
y = 6(x – 2)
f (x)
y = 5(x – 2)
y = 4(x – 2)

Tangent to f(x) at x = 2
f '(2) = 6, f(2) = 0
y = 6(x – 2)
From the graph it is clear that
possible integral values of k are 3, 4, 5
Q.28 Let f '(x) > 0 and f "(x) > 0 for all x1 < x2, then

 x1  x 2  f ( x1 )  f ( x 2 )  x1  2x 2  f ( x1 )  2f ( x 2 )
(A) f   > (B) f   >
 2  2  3  3
  x1  3x 2  f ( x1)  3f ( x 2 )
(C) f   > (D) None of these
 4  4

12. [A,B,C]

f '(x) > 0 it means curve is increasing f "(x) >0 it means curve is concave up.

[x2, f(x2)]

f ( x1)  f ( x 2 )
[x1, f(x1)]  x  x2 
f 1 
2
 2 

x1 x1  x 2 x2
2

 x1  x 2  f ( x1 )  f ( x 2 )
f  >
 2  2

y coordinate on the curve > y coordinate of points divides the at point line segment
x1  x 2
joining (x1, f(x1) and (x2, f(x2))
2

same we can proof option (B,C,D).

Q.29 Which of the following inequalities always hold good in (0, 1)

x2
(A) x > tan–1x (B) cos x < 1 –
2
2
x
(C) 1 + x ln  x  1  x 2  > 1 x 2 (D) x – < ln(1 + x)
  2
[Sol. ACD
(A) f (x) = x – tan–1x
1 x2
f ' (x) = 1 –  >0  f is increasing in (0, 1)
1 x2 1 x2
f (x) > f (0) but f (0) = 0
f (x) > 0  x > tan–1x in (0, 1)
2
x
(B) f (x) = cos x – 1 +
2
f ' (x) = – sin x + x = x – sin x > 0 in (0, 1)  (B) is not correct
 (C) f (x) = 1 + x ln  x  1  x 2  – 1 x 2
 
 1 2x 
1 · 
 2 1 x2  x
f ' (x) = x
  ln x  1  x 2  
2   
 x  1 x  1 x2
 
x x
=  ln x  1  x 2   > 0  x  R
1 x2   1 x2
 (C) is true
x2
(D) f (x) = x –  ln (1  x )
2
1 (1  x 2 )  1 x2
f ' (x) = (1 – x) – = =– < 0 for x > –1  (D) is correct
1 x 1 x 1 x
hence f (x) is decreasing in (0, 1)
 f (x) < f (0)
x2
f (x) < 0  x– < ln(1 + x) ]
2
Q.30 Let f (x) = x3 + ax2 + bx + c where a, b, c  R, then which of the following
statement(s) is(are) correct?
(A) If the equation f (x) = 0 has exactly one real root then f (x) must be strictly
increasing on R.
(B) If f (x) has a negative point of local minimum then both roots of equation f '(x) = 0
must be negative and distinct.
(C) If f (x1) · f (x2) < 0, x1 < x2 , then the equation f(x) = 0 has atleast one real root in
(x1, x2).
(D) f (x) possesses exactly one point of inflexion.
[Sol. BCD
(A)
y

x
(0, 0)

 False
(B)
 y

x
(0, 0)

 True
(C)
y
x = x1

x = x2
O × x

  True
a
(D) f '' (x) = 2(3x + a) = 0 x = True
3
and f " (x) changes sign also. ]

Paragraph for question nos. 31 to 32

If a continuous function f defined on the real line R, assumes positive and negative values in R
then the equation f (x) = 0 has a root in R. For example, if it is known that a continuous function
f on R is positive at some point and its minimum value is negative then the equation f (x) = 0 has
a root in R.

Consider f (x) = kex – x for all real x where k is a real constant.

Q.31 The line y = x meets y = kex for k  0 at

(A) no point (B) one point

(C) two points (D) more than two points

Q.32 For k > 0, the set of all values of k for which kex – x = 0 has two distinct roots is

(A) 0, 1 e (B) 1 e , 1 (C) 1 e ,   (D) (0, 1)

[Sol.

31. [B] Solving y = x are y = kex

x = kex  k = xe–x k  0  (B)

32. [A] kex = x  k = xe–x

y

1/e
xe–x
1 e ,  x
O

y = k if k < 0
1

 1
k > 0 for two distinct roots k   0,  ]
 e

Paragraph for question nos. 33 to 34

f: R  R, f (x) is a differentiable function such that all its successive derivatives exist. f(x) can
be zero at discrete points only and f(x) f(x)  0  x  R.

Now answer the following questions based on the above comprehension:

33. If  and  are two consecutive roots of f (x) = 0, then

(A) f() = 0   (, ) (B) f() = 0   (, )

(C) f() = 0   (, ) (D) f() = 0   (, )

Q.34 If f(x)  0, then maximum number of real roots of f(x) = 0 is/are

(A) no real root (B) one

(C) two (D) three

Sol. 33. B 34. B
Q.35 Column-I Column-II
(A) Let f (x) = x + ax + ax + 1 has local extrema at x = , 
3 2

where  < . (P) 9/2

If f () + f () = 2 then the value of 'a' equals
(B) Value of p + q for which f (x) = x3 + px2 + qx + r,
where p, q, r  R (Q) 11/2
is monotonically decreasing in largest possible interval (– 5/3, – 1), is
(C) If the equation e 2 x  k x has exactly 2 distinct solutions (R) 7
then 'k' can be equal to (S) 9

[Ans. (A) P ; (B) S; (C) P, Q, R, S]


Sol. (A) f ' (x) = 3x2 + 2ax + a = 0 ;  +=–
2a
and  =
a
3 3
given f () + f () = 2
(3 + a2 + a + 1) + (3 + a2 + a + 1) = 2
(3 + 3) + a(2 + 2) + a( + ) = 0
( + )3 – 3( + ) + a[( + )2 – 2] + a( + ) = 0
8a 3  2a   4a 2 2a  2a 2
– – a  + a    – =0
27  3   9 3 9
4a 3 2a 2
 0 (a  0, think ! )
27 3
4a 2 9
  a= Ans.
27 3 2
(B) f (x) = x3 + px2 + qx + r
 5
f ' (x) = 3x2 + 2px + q  3  x   (x + 1) = 3x2 + 8x + 5
 3
2p = 8  p = 4; q = 5
p + q = 9 Ans.
1
2x 2x 2 x e2x  e2x 2x
(C)
e
k; let y =
e
; x > 0;
dy
 2 x ; dy  (4 x  1) e
x x dx x dx 2x x

D I
2Ö e Minima
O x=1/4 0 1/4
1
at x = ; y = 2 e 2 × 2.75 = 5.50
4
for exactly one solution k = 2 e
for exactly 2 solution k > 2 e  Q, R, S
 for no solution k < 2 e  5.5 ]
Q.36 Column I contains function and Column II contains behaviours of function in their
domain. Entry of column I are to be matched with one or more than one entries of column
II.
Column-I Column-II
5
(A) f (x) = x  5 (P) monotonic
ln x
(B) g (x) = (Q) non monotonic
x
(C) h (x) = x sgn x, where sgn x denotes signum
function of x. (R)possesses extremum point
1
tan x , if x  0
(D) k (x) = 2 , if x  0 (S) possesses critical point
cot 1 x , if x  0
(T) possesses inflection point.

[Ans. (A) P, S, T; (B) Q, R, S, T; (C) Q, R, S ; (D) Q, R, S]

[Sol.
(A) We have f(x) = 5 x + 5
Domain of f (x) = R
1
Also, f '(x) = 4 5 > 0  x  R0.
5x
y
(0, 5)

x
O

Graph of f(x) = 5 x  5  f(x) is strictly increasing on R.

Clearly, f '(0) does not exist, although x = 0 is in the domain of function.

So, x = 0 is critical point of f(x).
5
Also, f "(x) changes sign about x = 0  x = 0 is point of inflection of f(x) = x + 5.
lnx
(B) Given, g(x) =
x
Domain of g(x) = R+
1  lnx
g'(x) = .
x2
 y
 1
 e, 
 e

x
O x =e

lnx
Graph of g(x) =
x
Now verify alternatives.
 x ; x0

(C) We have h(x) = x sgn x 0 ; x  0 = | x |  x  R.
x ; x0

y

y =– x y=x
x
O
Graph of h(x) = x sgn x
tan 1 x ; x0

(D) We have k(x) = 2 ; x0 .
 1
cot x ; x0
y

 y = cot–1x
(0, 2)
y =  /2
 y = tan–1 x
x
O

Graph of k(x) ]