MATHEMATICS

The following questions given below consist The Reason (R) is true but it is not a correct
of an "Assertion" (A) and "Reason" (R) explanation for Assertion (A).
Type questions. Use the following Key to
choose the appropriate answer.
Q.2 Let f : [1, 13]  R be an integrable function
(A) If both (A) and (R) are true, and (R) is the
correct explanation of (A). with f "(x) > 0   R.
(B) If both (A) and (R) are true but (R) is not the Assertion :
correct explanation of (A).
(C) If (A) is true but (R) is false. 3 13 9
(D) If (A) is false but (R) is true.
1
f ( x ) dx  
11
f ( x ) dx   f ( x ) dx
5
Q.1 Assertion : If n is a positive integer, then
Reason : If a < b < c and f "(x) > 0, then
n
sin x 2 1 1 1 f(a – b + c)  f(a) + f(c) – f(b)

0
x
dx  1    .....  
 2 3 n Sol. [A]

  Let b = a + (1 –)c
Reason : Over the interval  0,  ,  f(b)f(a) + (1 –) f(c) …
 2
(i)
sin x 2
 .
x  1–  
Sol. [B]
n
sin t
LHS = 
0
t
dt a b c

 2 3
sin t sin t sin t
= 
0
t
dt  

t
dt  
2
t
dt  ...
Now, a – b + c = (a + c) – (a + (1 – )c)
=  (1 – ) a + c
f(a – b + c)  (1 – ) f(a) + f(c) …(ii)
  
sin u sin u sin u

0
u
du  
0
u
du  
0
u  2
du  ... On adding equations (i) and (ii), we get
f(a – b + c) + f(b)  f(a) + f(c)
 So Reason is true.
sin u
...+ 
0
u  ( n  1) 
du Now, let c = a + 10 and b = a + 4
 f(a + 6) + f(a + 4)  f(a) + f(a + 10)
In I2 put t –  = u ; In I3 put t – 2 = u and so on.
On integrating both sides in a  [1, 3]
Now, 3 3

 f (a  6) da +  f (a  4) da 
 
sin u sin u 
Ir =  u  (r  1) du     (r  1) du
0 0
1 1

(u < ) 3

2 1  f (a ) da +
= . 1
 r
3
2 1 1 1
 I1 + I2 + .... + In  1    ....  
 2 3 n  f (a  10) da
1

 Assertion is true.

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 9 7 Reason : The integrand in I becomes rational
  f ( x ) dx +  f ( x ) dx 
7 5

3 13 x
 f ( x ) dx +  f ( x ) dx by the substitution t = 3
1 x3
.
1 11
9 3
1 2 1 / 3
  f ( x ) dx   f ( x ) dx + Sol.[D] I = 
dx
= 
dt
5 1
0
3
1 x 3
0
1 t3
13 2 1 / 3
1 2t  1 1 1 t
 f ( x ) dx
1
= tan  ln
3 3 2 3
1 t3 0
11

Hence (A) is the correct answer. 1 1 3

I= tan 1  ln(21 / 3  1)
3 1 1 6 2
3

5 / 4
(sin x  cos x )
10
Q.6 Assertion : 
 3 / 4
 
 x 
dx = 0

 | cos x | dx  20  4
Q.3 Assertion : e 1
0
Reason : If f(x) is an odd function, then
b
Reason :  f ( x )dx  0, then f(x) 
a
a

0,x(a, b).
 a
f ( x ) dx = 0  x  [–a, a]

x
5 / 4
Sol.[C] Assertion : 10  | cos x | dx Sol.[B] I = 
(sin x  cos x )
  dx
0
 3 / 4  x 
 4
e 1
 / 2  
= 10 
 
cos xdx 
 
– cos dx  = 10.2 =
5 / 4
 
2 cos x  
 0 / 2   4

20
=

 3 / 4
 
 x 
dx
 4
e 1
3 / 4

Reason :  cos xdx = sin x |30 / 4 = put x –
4
= t  dx = dt
0
1  
2 cos t 2 cos t
2  I= 

t
e 1
dt = 

e t  1
dt
  3 
but cos x < 0,  ×   , 
2 4  b
(a f ( x ) dx =
b

 Reason is false.
 a
f (a  b  x ) dx )


Q.5 Assertion :
1
 2I = 2  cos t dt I=0
dx 1 3 1 
I= 
0
3
1 x 3
=
3
tan–1 3
1  32
–
2
n(
So Assertion is true and Reason is also true but
3
2 – 1) not the correct explanation of Assertion.

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 2 /2
1 x
ln( t ) [ln ( t )]2
Q.7 Assertion:  cot 2 x dx = 4  cot 2 x dx f(x) + f   =
x 
1 t
=
2
0 0

nT T
1 2
Reason :  f ( x ) dx = n  f ( x ) dx, where f(e) + f   = (ln(e))  1
0 0
e 2 2
n is a positive integer and T is a period of f(x) 10 

 | cos x | dx  20 .
2  / 2
Q.9 Assertion :
Sol.[B] 
0
cot 2 x dx = 2 0
cot 2 x dx = 4  0
cot 2 x
0
b
dx
so Assertion is true and Reason is also true it
Reason :  f ( x ) dx  0, then f(x)  0,
a

 x  (a, b).
does not explaing Assertion.


x
ln( t ) 
Sol.[C] Assertion : 10 | cos x | dx
Q.8 Assertion : For x > 0 let f(x) = 1 1 t
dt 0

 / 2  

1 

= 10  cos x dx   cos x dx  = 10.2
 0 / 2



then f(e) + f   = ½.
e = 20
3 / 4 3 / 4
1 1
Reason : f(x) + f   =x–e+
x 2
for all x > Reason : 
0
cos xdx  sin x
0
=

0. 1
2
Sol.[C] Assertion is true
  3 
but cos x 0  x   , 
x ln( t ) 2 4 
f(x) = 1 1 t
dt
 Reason is false.

1 1/ x ln( t ) 1
f  =
x 1 1 t
dt t
z
Questions Add (24–6-09)
1
dt = – dz
z2

  1
1 log    1   
f   = 1x  z 
  2  dz 
x 1
1
z  z  Q.15 Assertion : 
0
x sinx cos2x dx =
2  sin x
0
cos2x dx
1 x (log z) 1
f  =
x 1 (z  1)
×   dz z  t
z Reason :
b

 x f(x) dx =
ab
b

 f ( x) dx
2
a a

1 x log( t )  1  Sol.[C] Assertion is true by using

f  =
x 1   dt
( t  1)  t 
b b

property  a

f ( x )dx  f (a  b  x ) dx
a
1 x log ( t ) ln( t ) and hence Reason is incorrect by same property.

e
f(x) + f   = × dx
x
  1 ( t  1) t

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 x dt  A is true
Q.15 Assertion(A) :  2
t t2 –1
=
12
then x R given A
 (A) holds.
=2
1 dx
Reason(R) :  0
1– x 2
= /2
Q.17 Assertion (A) : If {.} represents fractional part
x dt
Sol.[B]  
= sec 1 t  x 5.5
t2 – 1
 {x} dx = 21/8.
2 2
t function, then
= sec–1x – sec–1 2 0

1 dx Reason (R) : If [.] and {.} represent greatest

also  0
1– x 2 
= sin 1 x  1
0
= /2 integer and fractional part functions

t
2
 {x} dx = [ t ]  {t} .
2
1  1 respectively, then
Q.11 Assertion:  x
cosec101  x   dx = 0
 x 0 2 2
1/ 2
t [t] t
 1
Reason: f(x) =
1
coasec101  x   is even
x
Sol.[A]  {x}dx =  {x}dx +  {x}dx
x  0 0 [t]
function 1 {t }
[t]
Sol.[B] Assertion is true
2
= [t] 
0
xdx +
 xdx
0
=
2
+
1  1
I=
1/ 2
 x
cos ec101  x   dx
 x
{t}2
1 2
let =t
x  Reason is true.
–1/x2dx = dt 5. 5
2 21
2 1  1  {x}dx = 5  (.5) =
I= 1/ 2

t
cosec101  t   dt
 t 0 2 2 8

 Assertion (A) is true and is explained by

I = – I, 2I = 0, I = 0
reason (R)
Reason true
f(x) = 1/x cosec101 (x – 1/x) Q.36 Statement-1 : Let f(x) be an even function
f(–x) = –1/x cosec101 (–x + 1/x) x

f(–x) = 1/x cosec101 (x – 1/x) = f(x)

which is periodic, then g(x) =  f ( t ) dt is
a
function is even function also periodic.
a
Statement-2: If (x) is a differentiable and
sin x  x 4 periodic function, then (x) is also periodic.
Q.13 Assertion (A):
a
 7 | x |
dx is the same

a
[D]
 x4
as 2 
0
7 | x |
dx. Q.37 Statement-1 : If {.} represents fractional part
function, then
sin x 5 .5
Reason (R): Since is an odd function 21
7 | x |  {x} dx = 8
a a 0
sin x  x 4  x4
so that
a
 7 | x |
dx = 2  0
7 | x |
dx. Statement-2 : If [.] and {.} represent greatest
integer and fractional part functions
4
x
Sol.[A] R is true [ is even function]
7 | x |
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 t 2a a
[t] {t}2
respectively, then  {x} dx =
0 2
+
2

0

f ( x )dx  2 f ( x ) dx if f(2a –
0
[A] x) = f(x)
Reason Point ‘c’ may be a point even outside
10  the interval
Q.38 Statement-1 :  | cos x | dx = 20 /2
2 x (1  sin 2 x )
b
0 Q.13 Assertion (A) : 
 / 2
1  cos 2 x
dx = 0

Statement-2 :  f ( x ) dx 0, / 2
a
then f(x)  0,  x  (a, b)
Reason (R) : 
 / 2
sin2x cos2x (sinx + cosx)dx =

[C]
4
2 15
 tan
2
Q.39 Statement-1 : x dx = 4 Sol. [C] S1 is true because integrand is odd function. also
0 / 2

 sin
2
/2 x cos2x (sin x + cos x) dx =
 tan x dx
2
 / 2
0 / 2 / 2

 sin 2 x cos3x dx = 2  sin

2
nT T x cos3x dx
Statement-2 :  f ( x ) dx = n  f ( x ) dx ,  / 2 0
0 0 1

e
x
where n is an integer and T is a period of f(x). Q.12 Consider I1 = cos 3 x dx and
[B] 0
1

e
5 x 2
1  cos 2 x I2 = cos 3 x dx
Q.13 Assertion(A) : 
0
2
dx = 15 0

Assertion : I1 < I2
n 

Reason(R) : 
0

f ( x )dx  n f ( x )dx
0
Reason : If a continuous function defined in [a,
b] has its range as [m, M], then
(nI) b

Sol.
where f(x + T) = f(x)
[D]
m(b – a)   f ( x )dx  M (b – a).
a
5 
Sol.[B] x > x x  (0, 1)
2

Assertion   | cos x | dx  5 | cos x | dx =

0 0
2
– x < – x2  e–x < e – x  e–x cos2x <
2
10 e – x cos 2 x
Reasonis property. 1 1

 e
–x2
 e – x cos 3 x dx < cos 3 x dx
 0 0
Q.15 Assertion(A) :  f (tan 2 x ) dx =  I1 < I2  Assertion is true
0 Reason is obviously true.
/2

 f (tan
2
2 x ) dx Q.59 Statement 1 : Let f:R  R, f(x) = x + sin x. If
0 f –1 (x) is the inverse function of f(x),
b c 
2
Reason(R) : 
a
f ( x ) dx = 
a
f ( x ) dx + then  f –1
( x )dx =  – 1.
2
0
b
Statement 2 : Graph of y = f –1 (x) is image of

c
f ( x ) dx where ‘c’ is necessarily a point graph of y = f(x) in the line x – y = 0.

within (a, b)
Sol.[C] Assertion using property
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 y=x b

f(x) Statement-II :  f ( x ) dx =
a

Sol.[D] f–1(x) b

  
 f (a  b  x ) dx
a

  
–1
f (x) f ( x )dx x dx 1
+ =2 2(1  x )  1
 tan
1
0

0

0 Sol. [A] I = dx
1  (1  x )  (1  x ) 2
 ( x  sin x )dx +  f
–1 0
( x )dx
 = 2 1
0 0 1  2x
 tan
  1
I= dx
f = – 
1 x  x2
–1
( x )dx ( x  sin x )dx 0
 2
0 0
I = –I
 2  2
  2  2I = 0  I = 0
=  –  2
2  =
 2 –2
3 Q.117 Statement I : Value of
Q.119 Statement-I:  tan (sin x ) dx  0 –5 –1

3 
–4
sin (x2 – 3) dx +  sin (x
–2
2
+ 12x + 33) dx is
Statement-II: If f(x) is odd function then
zero
a

 f ( x ) dx  0
a

a
Statement II :  f ( x ) dx
–a
= 0 if f(x) is odd

Sol. [A] function

Sol.[B] Do yourself
2
Q. 35 Statement I :  0
sin 3 x dx  0

Statement II : sin3 x is an odd function Q.118 Statement I : f(x) is symmetrical about x = 2,

2 a 2 a
2

Sol.[B]  sin
3
dx = 0 as sin3 (2 – x) = –sin3x then  f ( x ) dx = 2   f ( x ) dx
2–a 2
0
Statement II : If f(x) is symmetrical about
1 x = b then f(b – ) = f (b + )  R
Q. 36 Statement I : If  0
e sin x dx  , then
Sol.[A] Do yourself
200
 0
e sin x dx  200 2

 cos
99
Q.119 Statement I : Value of x dx is 0
Statement II :
0
na a
 0
f ( x ) dx  n 0
f ( x ) dx , n  I and 2a a

f(a + x) = f (x) Statement II : 

0
f ( x ) dx = 2  f ( x ) dx
0
200

e
sin x
Sol.[D] Period of esinx is 2 dx  200 if f (2a – x) = f(x)
0
Sol.[A] Do yourself
Q.82 Statement-I : The value of
1
2x  1
 tan
1
dx = 0
0
1 x  x2 .

Q.

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