JEE (MAIN + ADVANCED) Examination (2018) (Page # 35)

TM JEE ADVANCED 2018 [PAPER - 2]

Hindi _ [MATHEMATICS]

SECTION 1 (Maximum Marks: 24)

• This section contains SIX (06) questions.
• Each question has FOUR options for correct answer(s). ONE OR MORE THAN ONE of these four
option(s) is (are) correct option(s).
• For each question, choose the correct option(s) to answer the question.
• Answer to each question will be evaluated according to the following marking scheme:
Full Marks : .... If only (all) the correct option(s) is (are) chosen.
Partial Marks : .... If all the four options are correct but ONLY three options are chosen.
Partial Marks : .... If three or more options are correct but ONLY two options are chosen, both of
which are correct options.
Partial Marks : .... If two or more options are correct but ONLY one option is chosen and it is a
correct option.
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered).
Negative Marks : .... In all other cases.
• For Example: If first, third and fourth are the ONLY three correct options for a question with
secondoption being an incorrect option; selecting only all the three correct options will result in
+4 marks. Selecting only two of the three correct options (e.g. the first and fourth options),
without selecting any incorrect option (second option in this case), will result in +2 marks.
Selecting only one of the three correct options (either first or third or fourth option) ,without
selecting any incorrect option (second option in this case), will result in +1 marks. Selecting any
incorrect option(s) (second option in this case), with or without selection of any correct option(s)
will result in .2 marks.

1. For any positive integer n, define fn : (0, )  R as

n
1  1 
f (x) =  tan  1   x  j  x  j  1  for all x  (0, ).
n

j1 

   
(Here, the inverse trigonometric function tan–1x assumes values in   ,  
 2 2 
Then, which of the following statement (s) is (are) TRUE ?
5 10

(A)  tan2 fj 0  = 55
  (B)  1  fj' 0  sec2  fj 0 =0
j 1 j1

(C) For any fixed positive integer n, xlim tan  fn  x   = 1

 n
2
(D) For any fixed positive integer n, lim sec  fn  x   = 1
x 
1. fdl h/kukRed i w
. kkZ
a
d n dsfy , ] fn : (0, )  R l Hkhx  (0, ) dsfy , ,
n
 1 
fn(x) =  tan1  1   x  j  x  j  1  ds: i esi fj Hkkf"kr gS
A
j1

    
(; gk¡,i zfr y kse f=kdks.kfer h; Qy u tan–1x dk eku   2 , 2  esekur sgS
)
 
r c,fuEu esl sdkS ul k@dkS ul sdFku l R; gS
@gksa
xs\

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5 10

(A)  tan2 fj  0 = 55
  (B)  1  fj' 0  sec2  fj 0 = 10
j1 j1

(C) fdl hfuf' pr /kukRed i w d n dsfy , ] xlim

. kkZ
a tan  fn  x   = 1 gS
A
 n

(D) fdl hfuf' pr /kukRed i w d n dsfy , ] xlim

. kkZ
a sec2  fn  x   = 1 gS
A


Sol. D
n   x  j   x  j  1 
1
f (x) =  tan  1   x  j  x  j  1 
n
j1  

n
 1 1 
f (x) =   tan  x  j  tan  x  j  1 
n
j1
fn(x) = tan–1(x + n) – tan–1x
 tan(fn(x)) = tan[tan–1(x + n) – tan–1x]
 x  n  x n
tan(fn(x)) = 1  x x  n  tan(fn(x)) =
  2
1  x  nx
2
 n 
 sec2(fn(x)) = 1 + tan2(fn(x))  sec2(fn(x)) = 1   2 
 1  x  nx 
2
lim sec2  fn  x   = lim 1   n 
x  2  =1
x   1  x  nx 

2. Let T be the line passing the points P(–2, 7) and Q(2,–5). Let F1 be the set of all pairs of circles
(S1, S2) such that T is tangent to S1 at P and tangent to S2 at Q, and also such that S1 and S2 touch
each other at a point, say M. Let E1 be the set representing the locus of M as the pair (S1, S2)
varies in F1. Let the set of all straight line segments joining a pair of distinct points of E1 and
passing through the point R(1,1,) be F2. Let E2 be the set of the mid-points of the line segments
in the set F2. Then, which of the following statment(s) is (are) TRUE ?
4 7
(A) The point (–2,7) lies in E1 (B) The point  ,  does NOT lie in E2
5 5

1   3
(C) The point  ,1  lies in E2 (D) The point  0,  does NOT lie in E1
2   2
2. ekukT , d j s[kkgSt ksfcUnq v ksP(–2, 7) r FkkQ(2,–5) l sgksd j xq t j r hgS
AekukF1 mu l HkhoÙ̀k; q Xeks(S1, S2) dkl eq Pp;
bl i zd kj gSfd T, P i j S1 dhLi ' kZ j s[kk r Fkk Q i j S2 dhLi ' kZ
j s[kk gSA r Fkk bl i zd kj HkhS1 r FkkS2 , d & nw l j sdks, d
fcUnqM i j Li ' kZdj r sgS
At c ; q Xe (S1,S2), F1 esfopfj r (varies) dj r k gSr ksekukfd l eq Pp; E1, fcUnqM dsfcUnq i Fk
dksn' kkZ
r k gS
A ekuk fd E1 dsnksfHkUu&fHkUu fcUnq v ksds; q
Xe dkst ksM+ r hgSeku fd E2, l eq Pp; F2 j s[kk[ k.Mksds e/;
fcUnq
v ksdk l eqPp; gS A r c fuEu esl sdkS ul k@dkS ul sdFku l R; gS @gksaxs
4 7
(A) fcUnq(–2,7), E1 esfLFkr gS
A (B) fcUnq 5 , 5  , E2 esfLFkr ughgS
A
 

1   3
(C) fcUnq 2 ,1  , E2 esfLFkr gS
A (D) fcUnq 0, 2  ,E1 esfLFkr ughgS
A
   
Sol. D

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 JEE (MAIN + ADVANCED) Examination (2018) (Page # 37)

AP = AQ = AM
Locus of M is a circle having PQ as its diameter
Hence, E1 : (x – 2)(x + 2) + (y – 7)(y + 5) = 0 and x  ±2
Locus of B(midpoint)
is a circle having RC as its diameter

E2 : x(x – 1) + (y – 1)2 = 0

Now, after checking the options, we get (D)

b1 
 
b
3. Let S be the set of all column matrices  2  such that b1, b2, b3  R and the system of equastions
b 
 3
(in real variables)
–x + 2y + 5z = b1
2x – 4y + 3z = b2
x – 2y + 2z = b3
has at least one solutions. Then, which of the following system (s) (in real variables) has (have)
b1 
 
b
at least one solutions for each  2   S ?
b 
 3
(A) x + 2y + 3z = b1, 4y + 5z = b2 and x + 2y + 6z = b3
(B) x + y + 3z = b1, 5x + 2y + 6z = b2 and – 2x – y – 3z = b3
(C) –x + 2y – 5z = b1, 2x – 4y + 10z = b2 and x – 2y + 5z = b3
(D) x + 2y + 5z = b1, 2x + 3z = b2 and x + 4y – 5z = b3
b1 
 
3. ekukS l HkhLr EHk vkOgksb2 
;w dk l eq
Pp; bl i zd kj gSfd b1, b2, b3  R gSr Fkk l ehdj .k fudk; ks(okLr fod pj kses)
b 
 3
–x + 2y + 5z = b1
2x – 4y + 3z = b2
x – 2y + 2z = b3
b1 
 
dsde l sde , d gy gS
A r c] fuEu esl sdkS
u l sfudk; @fudk; ks(okLr fod pj kses) dsi zR; sd b2   S dsfy , de l s
b 
 3
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de , d gy gksxk@gksa
xs?
(A) x + 2y + 3z = b1, 4y + 5z = b2 r Fkkx + 2y + 6z = b3
(B) x + y + 3z = b1, 5x + 2y + 6z = b2 r Fkk– 2x – y – 3z = b3
(C) –x + 2y – 5z = b1, 2x – 4y + 10z = b2 r Fkkx – 2y + 5z = b3
(D) x + 2y + 5z = b1, 2x + 3z = b2 r Fkkx + 4y – 5z = b3

Sol. A,C,D
We find D = 0 & since no pair of planes are parallel, so there are infinite number of solutions.
Let P1 + P2 = P3
 P1 + 7P2 = 13P3
 b1 + 7b2 = 13b3
(A) D  0  unique solution for any b1, b2, b3
(B) D = 0 but P1 + 7P2  13P3
(C) D = 0 Also b2 = – 2b1, b3 = – b1
Satisfies b1 + 7b2 = 13b3 (Actually all three planes are co-incident)
(D) D  0
1
4. Consider two straight lines, each of which is tangent to both the circles x2 + y2 = and the
2
parabola y2 = 4x. Let these lines intersect at the point Q. consider the ellipse whose center is at
the origin O(0,0) and whose semi - major axis is OQ. If the length of the minor axis of this ellipse
is 2 , Then which of the following statment(s) is (are) TRUE ?
1
(A) For the ellipse, the eccentricity is and the length of the latus rectum is 1
2
1 1
(B) For the ellipse, the eccentricity is and the length of the latus rectum is
2 2
1 1
(C) The area of the region bouded by the ellipse between the lines x  and x = 1 is    2
2 4 2
1
(D) The area of the region bounded by the ellipse between the lines x = and x = 1 is
2

   2
16
1
4. nksl j y j s[kkvksi j fopkj dhft , ] ft l esl si zR; sd oÙ̀k x2 + y2 = r Fkk i j oy ; y2 = 4x nksuksdsl kFkLi ' khZgS A ekuk
2
; sj s[kk, ¡ fcUnqQ i j i zfr PNsn dj r h gS
A , d nh?kZ oÙ̀k i j fopkj dj r sgSft l dk dsUnzew
y fcUnqO(0,0) i j gSr Fkk ft l dh
v) Z&nh?kZv{kOQ gS A; fn bl nh?kZoÙ̀kdhy?kq &v{kdhya ckbZ 2 gSAr c fuEu esl sdkS
ul k@dkSul sdFku l R; gS
@gksa
xs\
1
(A) nh?kZ
oÙ̀k dsfy , ] mRdsUnzrk gSr Fkk ukfHky a
c dhy a
ckbZ1 gS
A
2
1 1
(B) nh?kZ
oÙ̀k dsfy , ] mRdsUnzrk gSr Fkk ukfHky a
c dhy a
ckbZ gS
A
2 2
1 1
(C) j s[kkvks x  r Fkkx = 1 dse/; nh?kZ
oÙ̀k } kj k i fj c) {ks=k dk {ks+
=kQy    2 gSA
2 4 2
1 
(D) j s[kkvks x = r Fkkx = 1 dse/; nh?kZ
oÙ̀k } kj k i fj c) {ks=k dk {ks=kQy    2  gS
A
2 16
Sol. A,C

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 JEE (MAIN + ADVANCED) Examination (2018) (Page # 39)

1
Let equation of common tangent is y = mx +
m

1
00
m 1
 =  m4 + m2 – 2 = 0  m = ±1
1  m2 2

Equation of common tangents are y = x + 1 and y = – x – 1

Point Q is (–1,0)
x2 y2
 Equation of ellipse is  =1
1 1/2
1 1 2b2
(A) e = 1 = and LR = =1
2 2 a

(C)

1 1
1 x 1
2  . 1  x2 dx 2

1  x2  sin1 x 
Area 2 =
 2 2 1 / 2
1/ 2

   1      2
= 2      = 2   =
 4  4 8   8 4  4 2

5. Let s,t, r be non - zero complex numbers and L be the set of solutions z = x + iy x, y  R,i   1 
of the equation sz + tz + r = 0, where z = x – iy. Then, which of the following statement (s) is
(are ) TRUE ?
(A) If L has exactly one element, then |s|  |t|
(B) If |s| = |t|, then L has infinitely many elements
(C) The number of elements in L  z : z  1  i  5 is at most 2
(D) If L has more than one element, then L has infinitely many elements

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5. ekuks,t, r v' kwU; l fEeJ l a [ ; k, gSr FkkL l ehdj .k sz + tz + r = 0 dsgy ks z = x + iy  x, y  R,i  1  dk

l eq
Pp; gS ] t gk¡ z = x – iy gS A r c] fuEu esl sdkSul k@dkS
ul sdFku l R; gS
@gksa
xs\
(A) ; fn L dk ds oy , d vo; o gS ] r c |s|  |t|
(B) ; fn |s| = |t| gS , r c L dsvU kUr vo; o gS
A
(C) L  z : z  1  i  5
esvo; oksdhl a [ ; k vf/kd l svf/kd 2 gS
A
(D) ; fn L ds, d l svf/kd vo; o gS
] r c L dsvuUr vo; o gS A

Sol. Given
sz  tz  r  0 ......(1)
on taking conjugate sz  tz  r  0 .....(2)
from (1) and (2) elliminating z


z s
2
 t
2
 = r t  rs
(A) = If |s|  |t| then z has unique value
(B) If |s| = |t| then r t  rs may or may not be zero so L may be empty set
(C) locus of z is noll set or singleton set or a line in all cases it will intersect given circle at most
two points.
(D) In this case locus of z is a line so L has infinite elements

f  x  sin t  f  t  sin x
6. Let f : (0,)  R be a twice differentiable function such that lim = sin2x for all
t x tx
x  (0, ).
  
If f   =  , then which of the following statement (s) is (are) TRUE ?
6 12

 
(A) f   =
4 4 2

x4
(B) f(x) <  x2 for all x  (0,)
6
(C) There exists   (0,) such that f'() = 0
 
(D) f "    f   = 0
2
  2

f  x  sin t  f  t  sin x
6. ekukf : (0,)  R , d nq
xq
ukvody uh; Qy u bl i zd kj gSfd l Hkhx  (0, ) dsfy , lim
t x tx
= sin x 2
gS
A
   
; fn f  6  =  gS
, r c fuEu esl sdkS
ul k@dkS
ul sdFku l R; gS
@gksa
xs\
  12

 
(A) f   =
4 4 2

x4
(B) l Hkhx  (0,) dsfy , f(x) <  x2
6
(C)   (0,) bl i zd kj fo| eku gSfd f'() = 0 gS
A
 
(D) f "    f   = 0
2 2

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 JEE (MAIN + ADVANCED) Examination (2018) (Page # 41)

Sol. B,C,D
f  x  sin t  f  t  sin x
lim   sin2 x
xx tx
by using L Hopital
f  x  cos t  f '  t  sin x
lim   sin2 x  f(x) cos x - f’(x) sin x = sin2x
xx 1

 f '(x) sin x  f(x) cos x 

   1
 sin2 x 

 f x  
 d  sin x   1
 

f x
 xc
sin x

   
Put x = & f   
6 6 12
 c=0  f(x) = –x sin x
    1
(A) f   
4 4 2
(B) f(x) = – x sin x
x3 x4 x4
as sin x > x – , x sin x  x2   f  x   x 2  x  0,  
6 6 6
(C) f’(x) = – sin x –x cos x
f’(x) =0  tan x = –x  there exist
  0,   for which f’(  ) = 0

(D) f’’ (x) = – 2 cos x + xsinx

     
f ''    , f      f ''    f   0
2 2 2 2 2 2

SECTION 2 (Maximum Marks: 24)

• This section contains EIGHT (08) questions. The answer to each question is a NUMERICAL
VALUE.
4 For each question, enter the correct numerical value (in decimal notation, truncated/
rounded.off to the second decimal place; e.g. 6.25, 7.00, .0.33, ..30, 30.27, .127.30) using
the mouse and the onscreen virtual numeric keypad in the place designated to enter the
answer.
• Answer to each question will be evaluated according to the following marking scheme:
Full Marks : ..3 If ONLY the correct numerical value is entered as answer.
Zero Marks : 0 In all other cases.

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
2
1 3
7. The value of the integral  1
dx
is
0
 x  1 2 6 4
1  x  

2
1 3
7. l ekdy u  1
dx
dk eku gS
A
0
 x  1 2
1  x  6 4

Sol. 2
1 1
2 1  3  2 1  3  dx
 1/ 4
dx
  1/ 4
0
 x  1 1  x 
2 6
 0
 1  x  1  x
2
2 4

 12 
 1 
=  0
dx  1  3  


1  x  2
1  x  
put x = sin 

 6 
 cos d 
=  0 cos  1  sin   
1 3  
 
 

 6 
 1  sin  d 
=  0 cos2  
1 3  
 
 

 6 
6

 
= 
0
sec 2
 d   
0
tan  sec d   1 3  
 
 

 
 6 6

tan  |  sec  |  1  3
= 0 0
 
 

1  2 
=   1
3  3 

=  3 1  3 1 
2
=  3 1 = 2

8. Let P be a matrix of order 3 × 3 such that all the entries in P are from the set {–1, 0 , 1}. Then,
the maximum possible value of the determinant of P is ........
8. ekukP dksfV 3 × 3 dk, d vkO
;w
g bl i zd kj gSfd P esfLFkr l Hkhi zfof"V; k¡l eq
Pp; {–1, 0 , 1} l sgS
A r c P dsl kj f.kd
dk vf/kdr e l a
Hka
o eku gksxk&
Sol. 4
Set A = {-1,0,1}

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for obtaining the determinant having maximum possible value

aii  min{1, 0,1}
aij  max{1, 0,1}

 1 1 1 
 
1 1 1 
Matrix p = 
 1 1 1

p  1 1  1   1  1  1 1  1

p 4
which is maximum possible value.

9. Let X be a set with exactly 5 elements and Y be a set with exactly 7 elements. If  is the number
of one - one functions from X to Y and  is the number of onto functions from Y to X, then the
1
value of
5!
    is ..........
9. ekukX , d l eq
Pp; gSft l dsBhd 5 vo; o gSr Fkk Y , d l eq
Pp; gSft l dsBhd 7 vo; o gS
A; fn , X l sY r c , dS
dh
1
Qy uksdhl a
[ ; k gSr Fkk  ,Y l sX r d dsvkPNknd Qy uksdhl a
[ ; k gS
, rc     dk eku gS
A
5!
Sol. 119
n(X) =5
n(Y) =7
  Number of one-one function = 7 C5  5!
  Number of onto funtion Y to X

a1 b1
a2 b2

a7 b7

1,1,1,1,3 1,1,1,2,2

7! 7!
 5! 3
 5!
3! 4! 2!  3!
=  7

C3  3.7 C3 5!  4  7C3  5!


 4 7 C3 7 C5  4  35  21  119
5!

10. Let f : R  R be a differentiable function with f(0) = 0. If y = f(x) satisfies the differentiable
equation
dy f  x  is
= (2 + 5y)(5y – 2), then the value xlim
dx 

10. ekuk f : R  R , d vody uh; Qy u gSt gk¡ f(0) = 0 gS

A ; fn y = f(x), vody l ehdj .k
dy f  x  dk eku gS
= (2 + 5y)(5y – 2) dksl a
rq , r c xlim
"V dj r k gS &
dx 

Sol. 0.4
dy
 25y2  4
dx

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dy
so,  dx
25y2  4

2
y
1 1 5 xc
 n
Integrating, 25 2 2
2 y
5 5

5y  2
 n  20  x  c 
5y  2
Now, c = 0 as f(0) = 0
5y  2
 e 
20x
Hence
5y  2

5f  x   2
 lim e
20x 
lim
x  5f  x   2 x 

Now, RHS =0  xlim


5f  x   2  0
2
 lim f(x) 
x  5

11. Let f : R  R be a differentiable function wich f(0) = 1 and satisfying the equation
f(x + y) = f(x) f'(y) + f'(x)f(y) for all x, y  R Then, the value of loge(f(4)) is ......
11. ekukf : R  R , d vody uh; Qy u gSt gk¡f(0) = 1 r Fkk l ehdj .k f(x + y) = f(x) f'(y) + f'(x)f(y)
dksl Hkh x, y  R dsfy , l a
rq
"V dj r k gS
] r c loge(f(4)) dk eku gS
A
Sol. 2
P(x,y) : f(x+y) = f(x) f’(y) + f’(x) f(y)  x,y  R
P(0,0) : f(0) = f(0) f’(0)+f’ (0) f(0)
 1= 2f’(0)
1
 f’(0) =
2
P(x,0) : f(x) = f(x). f’(0) + f’(x) . f(0)
1
 f(x) = f(x)  f '(x)
2
1
 f '(x)  f(x)
2
1
x
 f(x)  e2

 In (f(4))=2

12. Let P be a point int he first octant, whose image Q in the plane x + y = 3 ( that is, the line
segment PQ is perpendicualr to the plane x + y = 3 and the mid - point of PQ lies in the plane x
+ y = 3) lies on the z - axis. Let the distance of P from the x - axis be 5. If R is the imags of P in
the xy - plane, then the length of PR is ......
12. ekuk P i zFke v"Bka
' k (octant) es, d fcUnqgS, ft l dk i z
fr fca
c Q l er y x + y = 3 es( vFkkZ r ~] j s[kk[ k.M PQ l er y
x + y = 3 dsy a cor ~gSr Fkk PQ dk e/; fcUnql er y x + y = 3 esfLFkr gS ) z -v{k i j fLFkr gSA ekuk P dhx - v{k
l snw
j h5 gS A; fn R, xy l er y esP dk i zfr fca
c gS ] r c PR dhy a ckbZgS
A

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Sol. 8
Let

P  , ,  
Q 0, 0,   &
R  ,,   

  
Now, PQ || ˆi  ˆj  ˆi  ˆj || ˆi  ˆj 

Also, mid point of PQ lies on the plane
 
  3
2 2
    =6
 = 3
Now, distance of point P from X- axis is 2   2  5

 2  2  25

  2  16
as     3
as   4
Hence, PR = 2  =8

13. Consider the cube in the first octant with sides OP, OQ and OR of length 1, along the x-axis,
1 1 1
y - axis, and z- axis, respectively, where O(0,0,0) is the origin, Let S  , ,  be the centre of
2 2 2
the cube and T be the vertex of the cube opposite to the origin O such that S lies on the diagonal
           
OT. If p = SP , q = SQ , r = SR and t = ST , then the value 
alue of p  q  r  t is ..... 
13. i zFke vkW
DVsa
V (octant)es, d ?ku i j fopkj dhft , ft l dhHkq
t kvks OP, OQ r FkkOR dhy a
ckbZ1 gSr Fkkt ksØe' kx-v{k,
1 1 1
y- v{k, r Fkkz- v{k dsvuq
fn' k gS
, t gk¡O(0,0,0) ew A ekuk S  2 , 2 , 2  , ?ku dk dsUnzgSr Fkk T ew
y fcUnqgS y fcUnqO ds
 
       
l Eeq
[ k ?ku dk ' kh"kZbl i zd kj gSfd S,fod.kZ OT i j fLFkr gS
A ; fn p = SP , q = SQ , r = SR r Fkk t = ST
   
, r c p  q  r  t  dk eku gS&
gS
Sol. (0.5)

T
S

O y
Q

P
x

   1 1 1 1 ˆ ˆ ˆ
p  SP   ,  ,   
2 2 2 2
i  jk  
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   1 1 1 1 ˆ ˆ ˆ
q  SQ    , ,   
 2 2 2 2
i  j  k  
   1 1 1 1 ˆ ˆ ˆ
r  SR    ,  ,  
 2 2 2 2
i  j  k  
   1 1 1  1
t  ST   , ,  
2 2 2 2
ˆ
ˆi  ˆj  k  
ˆi ˆ
ˆj k ˆi ˆj kˆ
    1 1
 
p  q   r  t  4 1  1  1  4  1  1 1
1 1 1 1 1 1

1 ˆ
k 1
16

2ˆi  2ˆj   2ˆi  2ˆj  
2

2


2 2 2 2
14. Let X =  10
C1  + 2  10
C2  + 3  10
C3  + ..... 10  10
C10  .

1
where 10
Cr, r {1,2,.....10} denote binomial coefficient. Then, the value of X is ____
1430
2 2 2 2
14. ekukX =  10
C1  + 2  10
C2  + 3  10
C3  + .....+ 10  10
C10  gS
,

1
t gk¡ 10Cr, r {1,2,.....10} f} i } xq
. kka
d dksn' kkZ
r k gS
] rc X dk eku gS&
1430
Sol. (646)
n
n 2
x  r.( C ) ;n  10
r 0
r

n
x  n. (n Cr )n1Cr 1
r 0

n
x  n. n Cnr .n1Cr 1
r 1

x  10.19 C9
x 1 19
 . C9
1430 143
x  n.2n1 Cn1;n  10
= 646

SECTION 3 (Maximum Marks: 12)

• This section contains FOUR (04) questions.
• Each question has TWO (02) matching lists: LIST.I and LIST.II.
• FOUR options are given representing matching of elements from LIST.I and LIST.II. ONLY
ONE of these four options corresponds to a correct matching.
• For each question, choose the option corresponding to the correct matching.
• For each question, marks will be awarded according to the following marking scheme:
Full Marks : ..3 If ONLY the option corresponding to the correct matching is chosen.
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered).
Negative Marks : ..1 In all other cases.

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 x     x  
15. Let E1 = x  R : x  1 and  0 and E = x  E1 : sin1  loge    is real number  .
 x 1  2
   x  1  
  
(Here, the inverse trigonometric function sin–1x assumes values in  ,  )
 2 2
 x 
Let f : E1  R be the function defined by f(x) = loge  
 x  1

1   x 
and g : E2  R be the function defined by g(x) = sin  loge   .
  x  1
List - I List - II
 1   e 
P. The range of f is 1.  , 1  e    e  1 ,  
   
Q. The range of g contains 2. (0,1)
 1 1
R. The domain of f contains 3.  2 , 2 
 
S. The domain of g is 4.  , 0   0,  
 e 
5.  , e  1 
 
1 e 
6.  , 0    ,
 2 e  1 
The correct option is :
(A) P  4; Q  2; R  1; S  1
(B) P  3; Q  3; R  6; S  5
(C) P  4; Q  2; R  1; S  6
(D) P  4; Q  3; R  6; S  5
 x   1     x   
15. ekukE1 = x  R : x  1 r Fkk x  1  0  r FkkE2 = x  E1 : sin  loge  x  1   , d okLr fod l a[; k gS
.
      
  
(; gk¡, i zfr y kse f=kdks.kfer h; Qy u sin–1x dk eku  2 , 2 esekur sgS
)
 
  x
ekukf : E1  R, f(x) = loge  x  1  } kj k i fj Hkkf"kr Qy u gS
A
 
   x 
r Fkkg : E2  R, g(x) = sin1  loge   } kj k i fj H
kkf"kr Qy u gS
A
x  1
  
lw
ph- I lw
ph- II
 1   e 
P. f dk i fj l j gS
& 1.  , 1  e    e  1 ,  
   
Q. g dk i fj l j j [ kr k gS& 2. (0,1)
 1 1
R. f dk i zkUr j [ kr k gS
& 3.  2 , 2 
 
S. g dk i zkUr gS
& 4.  , 0   0,  
 e 
5.  , e  1 
 

1 e 
6.  , 0    , 
 2 e  1

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l ghfodYi gksxk&
(A) P  4; Q  2; R  1; S  1
(B) P  3; Q  3; R  6; S  5
(C) P  4; Q  2; R  1; S  6
(D) P  4; Q  3; R  6; S  5

Sol. (A)
x
E1 : 0
x 1
+

0 1

 E1 : x   , 0   1,  

 x 
E2: 1  n   1
 x  1

1  x 
 e
e  x  1 

x 1
Now  0
x 1 e


 e  1 x  1  0
e  x  1
+

-1/(e-1) 1

 1 
 x   ,   1,  
 1  e

x
also  e  0
x 1
 e  1 x  e  0
x 1
+

1 e/(e-1)

 e 
 x   ,1   , 
 e  1 

 1   e 
So E2 :  ,  , 
 1  e   e  1 
x
as Range of is R +- {1}
x 1
 Range of f is R - {0} or  , 0  0,  

        
Range of g is  ,  \{0} or   , 0    0, 
 2 2  2   2
Now P  4, Q  2, R  1, S  1
Hence A is correct

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16. In a high school, a commintee has to be formed from a group of 6 boys M1, M2, M3, M4, M5, M6 and
5 girls G1, G2, G3, G4, G5.
(i) Let 1 be the total number of ways in which of the commitee can be formed such that the
commitee has 5 members, having exactly 3 boys and 2 girls.
(ii) Let 2 be the total number of ways in which the commitee can be formed such that the
committee has at least 2 members, and having an equal number of bosy and girls.
(iii) Let 3 be the total number of ways in wihc the committee can be formed such that the
commitee has 5 members, at least 2 of them being girls.
(iv) Let 4 be the total number of ways in which the committee can be formed such that the
committee has 4 members, having at least 2 girls and such that both M1 and G1 are NOT in the
committee together.

List - I List - II
P. The value of 1 is 1. 136
Q. The value of 2 is 2. 189
R. The value of 3 is 3. 192
S. The value of 4 is 4. 200
5. 381
6. 461

The correct option is :

(A) P  4 ; Q  6; R  2; S  1
(B) P  1 ; Q  4; R  2; S  3
(C) P  4 ; Q  6; R  5; S  2
(D) P  4 ; Q  2; R  3; S  1
16. , d mPPkfo| ky ; es]6 y M+
d ksM1, M2, M3, M4, M5, M6 r Fkk5 y M+
fd; ksG1, G2, G3, G4, G5 ds, d l ew
g esl s, d l ehfr
cuk; ht kr h gS A
(i) ekuk1 dq y r j hdksdhl a [ ; k gSft l esl fefr bl i zd kj l scuk; ht k l dr hgSfd l fefr es5 l nL; gSft uesl sBhd
3 y M+d sr Fkk 2 y M+ fd; k¡ gS
A
(ii) ekuk2 dq y r j hdksdhl a [ ; k gSft l esl fefr bl i zd kj l scuk; ht k l dr hgSfd l fefr esU; w ur e 2 l nL; gksr Fkk
y M+d sr Fkk y M+
fd; ksdhl a [ ; k cj kcj gksA
(iii) ekuk3 dq y r j hdksdhl a [ ; k gSft l esl fefr bl i zd kj l scuk; ht k l dr hgSfd l fefr es5 l nL; gS sft l esde
l sde 2 y M+ fd; k¡ gksA
(iv) ekuk4 dq y r j hdksdh l a [ ; k gSft l esl fefr bl i zd kj l scuk; h t k l dr h gSfd l fefr ess4 l nL; gSs] ft uesl s
U; w
ur e 2 y M+ fd; k¡ gksr Fkk M1 r FkkG1 nksuksl fefr esl kFk&l kFk u gksA

lw
ph- I lw
ph- II
P. 1 dk eku gS
A 1. 136
Q. 2 dk eku gS
A 2. 189
R. 3 dk eku gS
A 3. 192
S. 4 dk eku gS
A 4. 200
5. 381
6. 461

l ghfodYi gS:
(A) P  4 ; Q  6; R  2; S  1
(B) P  1 ; Q  4; R  2; S  3
(C) P  4 ; Q  6; R  5; S  2
(D) P  4 ; Q  2; R  3; S  1

Sol. (C)
 6  5 
(1) 1       200
3   2
So P  4
(2)

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 6   5   6   5   6   5   6  5   6   5 
2                         
1   1   2   2   3   3   4   4   5   5 

11
   1
5 
= 46!
So Q  6
 5   6  5   6  5   6   5   6 
(3) 3                    
 2   3   3   2   4  1   5   0 

 11   5   6   5   6 
=          
 5   0   5  1   4 
= 381
So R  5
 5   6   4  5   5   6   4  1  5 
(4) 2                       
 2   2  1  1   3  1   2   1   4 
= 189
= So S  2

x2 y2
17. Let H :   1 , where a > b > 0, be a hyperbola in xy - plane whose conjugate axis LM
2
a b2
subtends an angle of 60° at one of its vertices N. Let the area of the trignale LMN be 4 3 .
List - I List - II
P. The length of the conjugate axis of H is 1. 8
4
Q. The eccentricity of H is 2.
3

2
R. The distance between the foci of H is 3.
3
S. The length of the latus rectum of H is 4. 4

The correct option is :

(A) P  4; Q  2 ; R  1; S  3
(B) P  4; Q  3 ; R  1; S  2
(C) P  4; Q  1 ; R  3; S  2
(D) P  3; Q  4 ; R  2; S  1

x2 y2
17. ekukH :   1 gS
,t gk¡a > b > 0 gS
, xy -l er y es, d vfr i j oy ; gSft l dhl a
;q
Xehv{kLM bl ds' kh"kksZN
a2 b2
esl s, d i j 60° dk , d dks.k cukr h gS
A ekuk f=kHkq
t LMN dk {ks=kQy 4 3 gS
A
lw
ph- I lw
ph- II
P. H dhl a
;q
Xehv{k dhy a
ckbZgS
& 1. 8
4
Q. H dhmRdsUnzrk gS
& 2.
3

2
R. H dhukfHk; ksdse/; nw
j hgS
A 3.
3
S. H dsukfHky a
c dh y a
ckbZgS
& 4. 4

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 JEE (MAIN + ADVANCED) Examination (2018) (Page # 51)

l ghfodYi gS
&:
(A) P  4; Q  2 ; R  1; S  3
(B) P  4; Q  3 ; R  1; S  2
(C) P  4; Q  1 ; R  3; S  2
(D) P  3; Q  4 ; R  2; S  1
Sol. (B)
L
300

30 0
O a 300 N

b
tan 300 =
a ab 3
1
Now area of LMN  .2b.b 3  4 3  3b2
2
b2 2
 b=2 & a = 2 3  e  1

a2 3
P. Length of conjugate axis = 2b = 4
So P  4
2
Q. Eccentricity e 
3
So Q  3
R. Distance between foci = 2 ae
 2 

2 2 3  
 3
8
So R  1
2
2b2 2 2  4
S. Length of latus rectum =  
a 2 3 3
So S  2

  
     1, e 2  2 
18. Let f1 : R  R, f2 :  ,   R; f3 : 
 2 2   R and f4 : R  R be functions defined by
 
  x2 
(i) f1(x) = sin  1  e 
 
 sin x
 if x  0
1
(ii) f2(x) =  tan x , where the inverse trigonometric function tan–1x assumes values in
 1 if x  0


  
 2 ,2,
 
(iii) f3(x) = [sin(loge(x +2))], where for t  R, [t] denotes the greates integer less than or equal
to t,

 2 1
 x sin   if x  0
(iv) f4(x) =  x
 0 if x  0


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List - I List - II
P. The function f1 is 1. NOT continuous at x = 0
Q. The function f2 is 2. Continuous at x = 0 and NOT differentiable at
x=0
R. The function f3 is 3. differentiable at x = 0 and its derivative is NOT
continuous at x = 0
S. The function f4 is 4. differentiable at x = 0 and its derivative is
continuous at x = 0

The correct option is :

(A) P  2, Q  3; R  1; S  4
(A) P  4, Q  1; R  2; S  3
(A) P  4, Q  2; R  1; S  3
(A) P  2, Q  1; R  4; S  3
  
     1, e 2  2 
18. ,
ekukf1 : R  R, f2 :  2 2   R; f3 :  r Fkkf4 : R  R fuEu } kj k i fj Hkkf"kr Qy u gS
&
   R
 

  x2 
(i) f1(x) = sin  1  e 
 

 sin x
 1
; fn x  0    
(ii) f2(x) =  tan x , t gk¡ i zfr y kse f=kdks.kfer h; Qy u tan–1x dk eku   2 , 2  esekur sgS
,
 1 ; fn x  0  

(iii) f3(x) = [sin(loge(x +2))] gS
,t gk¡ t  R dsfy , , [t], t l sde ; k cj kcj egÙke i w
. kkZ
a
d dksn' kkZ
r k gS
A
 2 1
x sin   ; fn x  0
(iv) f4(x) =  x
 0 ; fn x  0

lw
ph- I lw
ph- II
P. Qy u f1 gS& 1. x = 0 i j l r r ~ughgS &
Q. Qy u f2 gS
& 2. x = 0 i j l r r ~gSy sfdu x = 0 i j vody uh; ughgS
A
R. Qy u f3 gS
& 3. x = 0 i j vody uh; gSr Fkk bl dk vody t
x = 0 ij l r r ~ughgS &
S. Qy u f4 gS
& 4. x = 0 i j v od y uh; gS r Fkk bl d k v od y t
x = 0 ij l r r ~gS
&

l ghfodYi gS
&
(A) P  2, Q  3; R  1; S  4
(A) P  4, Q  1; R  2; S  3
(A) P  4, Q  2; R  1; S  3
(A) P  2, Q  1; R  4; S  3

Sol. (D)
2
(i) f(x)  sin 1  e x

1
f1' (x)  cos 1  e x .
2

2 1e  x2
0  e  x2
.  2x  
at x=0 f1' (x) does not exist
So. P  2

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 sin x
 ,x  0
f2 (x)   tan1 x
(ii)
0 x=0


sinx x
lim 1
x0 x tan1 x
 f2  x  does not continuous at x = 0
so Q  1
(iii) f3  x   sin n  x  2    0
1< x + 2 < e / 2

 0< n x  2  
2

 0<sin( n(x  2))  1

 f3  x   0
So R  4
 2 1
x sin , x0
(iv)  f4  x    x
 0 , x=0

So S  3

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