MATHEMATICS DPP

TARGET : JEE (Advanced) 2015

Course : VIJETA & VIJAY (ADP & ADR)
TEST INF ORM ATION

Date : 01-05-2015
DPP DAILY PRACTICE PROBLEMS
NO.
08

TEST INFOR MATION

DATE : 03.05.2015 OPEN TEST (OT-02) ADVANCED
Syllabus : Full Syllabus

REVISION DPP OF
DEFINITE INTEGRATION & ITS APPLICATION AND INDEFINITE INTEGRATION

Total Marks : 149 Max. Time : 105.5 min.

Single choice Objective (–1 negative marking) Q. 1 to 14 (3 marks 2.5 min.) [42, 35]
Multiple choice objective (–1 negative marking) Q. 15 to 31 (4 marks, 3 min.) [64, 48]
Comprehension (–1 negative marking) Q.32 to 33 & Q.34 to Q.36 (3 marks 2.5 min.) [15, 12.5]
Single digit type (no negative marking) Q. 37 to 39 (4 marks 2.5 min.) [12, 7.5]
Double digit type (no negative marking) Q. 40 (4 marks 2.5 min.) [16, 2.5]

505  1007
1. If A = 
0
| cos x | dx and B = 
505 
| sin x | dx, then A + B is equal to

(A) 2013 (B) 2014

(C) 2015 (D) 2016

100
2. The least integer greater than   x  dx
0
is (where {.} is fractional part function)

(A) 50 (B) 51
(C) 52 (D) 53


ex 2 – x2 
3.  1– x  1– x 2
dx =

1– x 1 x
(A) ex . +c (B) ex +c
1 x 1– x

ex ex
(C) +c (D) +c
1– x 1 x

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4. Let f(x) = x
2
 
cos2 x 2x  6 tan x – 2x tan2 x dx and f(x) passes through the point (, 0), then the

3
number of solutions of the equation f(x) = x in [0, 2] is
(A) 1 (B) 2 (C) 3 (D) 4

2
5. Let f(x) is a continuous function symmetric about the lines x = 1 and x = 2. If  f(x) dx = 3 and
0

50

 f(x) dx  I , then  I  is equal to (where [.] is G.I.F.)

 
0

(A) 5 (B) 8 (C) 7 (D) 6

x 
6
6.  x 4  x2 2x 4  3x 2  6 dx is equal to

3/2 3/2

(A)
3x 6
 2x 4  6x 2  C (B)
2x 6
 3x 4  6x 2  C
18 24
3/2

(C)
 2x 6
 3x 4  6x 2  C (D) None of these
18

x2
7. For each positive integer n > 1, let Sn represents the area of the region bounded by  y 2  1 and
n2

y2
x2   1 , then lim Sn is equal to
n2 n

(A) 4 (B) 1 (C) 2 (D) 3

8x 43  13x38
8.  4
dx =
x13
 x5  1 
x39 x39
(A) 3
+c (B) 3
+c

3 x13  x 5  1   x13  x5  1 
x39 x52
(C) +c (D) +c

5 x13  x 5  1 
5

3 x13  x5  1 

x x
9. Let
 e (f(x) – f '(x))dx = (x), then  e f(x)dx is equal to

(A) (x) + ex f(x) + c (B) (x) – ex f(x) + c

1 1
(C) {(x) + ex f(x)} + c (D) ((x) + ex f(x)) + c
2 2

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10. Suppose f(x) is a real valued differentiable function defined on [1, ) with f(1) = 1. Further let f(x) satisfy
1
f(x) = , then the range of values of f(x) is
x2  f 2 (x)

(A) [1, ) (B) [1, 1 + /4)

(C) [1, /4) (D) [1 – /4, 1]

x
t2
xe
0
dt
11. The value of Lim is equal to
x0 1  x  ex

(A) 1 (B) 2 (C) –1 (D) –2

2
f(2t)
12. Let f(x) be a differentiable function such that f(0) = 0 and  f '(2t) e
0
dt  5 , then the value of f (4)

equals
(A) 2 n3 (B) n10 (C) n11 (D) 3 n2

 x 
13. The area enclosed by the curve y  4 – x2 , y  2 sin   and the x-axis is divided by y-axis in
2 2
the ratio

2 – 8 2 – 4 –3 2 2
(A) (B) (C) (D)
2  8 2  4 4 2   2 – 8

14. For any t  R and f being a continuous function.

1 cos2 t
Let I1 =  
xf x  2 – x  dx 
2
sin t

1 cos2 t
I2 =  f  x  2 – x   dx , then
sin2 t

(A) I1 = I2 (B) I1 = 2I2 (C) 2I1 = I2 (D) I1 + I2 = 0

x 2
x2
15. If  g  t  dt   t 2 g  t  dt , then equation g(x) =  has

2
2 x

1 1
(A) 2 solution if |  | < (B) 2 solution if |  | < &  0
2 2
1 1
(C) 1 solution if  = – (D) No solution if |  | >
2 2

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 [x]
16. If g(x) = {x} , where {.} and [.] represents fractional part and greatest integer function respectively and
k 1
f(k) =  g(x)dx (k  N), then
k

(A) f(1), f(2), f(3), . . . . . . are in H.P.


r 1
(B)   –1
r 1
f(r) = 1 – n 2


r 2
(C)   –1
r 1
f(r) = n  
e
n
 1 n(n  1)
(D)  f  r  =
r 0
2

17. If f(x) is a differentiable function such that f(x + y) = f(x) f(y)  x, y  R, f(0) 0 and
f(x)
g(x)= 2
, then
1 f x   
2015 2015
(A) 
–2014
g(x)dx  
0
g(x)dx

2015 2014 2015

(B) 
–2014
g(x)dx – 
0
g(x)dx  0
g(x)dx

2015
(C) 
–2014
g(x)dx  0

2014 2014
(D) 
–2014
2g(–x) – g(x) dx  2 
0
g(x)dx

2 7
 –1 1 –1  sin x
18. Let I = 
–1
 cot x  cot x  dx and J =
   | sin x | dx , then which of the following is/are correct ?
–2

(A) 2I + J = 6 (B) 2I – J = 3

2 2 2 I 5
(C) 4I + J = 26 (D) 
J 2

 1
 sin  n   x
 2
19. If In = 
0 sin
x
dx , where n  w, then

2
10 10
(A) In+2 = In (B) 
m1
Im  10 (C) I
m1
2m–1  10 (D) In+1 = In

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 1

x 4 1  x10065  1
20. If  2015
dx 
p
, then
0 1  x 5

(A) Number of ways in which p can be expressed as a product of two relatively prime factors is 8.
(B) Number of ways in which p can be expressed as a product of two relatively prime factors is 4.
(C) Number of ways in which p can be expressed as a product of two factors is 8.
(D) Number of ways in which p can be expressed as a product of two factors is 4.

3n–1 3n
r r
21. If Tn = r
r  2n
2
n 2
and Sn = 
r  2n1 r
2
 n2
 n  {1, 2, 3, . . . . }, then

(A) Tn > n 2 (B) Sn < n 2

(C) Tn < n 2 (D) Sn > n 2

1 1
1  x8 1  x9
22. If I1 =  1 x
0
4
dx and I2 =  1 x
0
3
dx , then

(A) I 2 < I1 < /4 (B) /4 < I2 < I1

(C) 1 < I1 < I2 (D) I2 < I1 < 1

23. Consider a continuous function 'f' where x4 – 4x2  f(x)  2x2 – x3 such that the area bounded by y = f(x),
g(x) = x4 – 4x2, the y-axis and the line x = t (0  t  2) is twice of the area bounded by y = f(x),
y = 2x2 – x3, y-axis and the line x = t (0  t  2) then
(A) f(2) = 0 (B) f(1) = 1/3
(C) f(1) = –2/3 (D) f(x) has two points of extrema

4
24. The value of the definite integral 2

x(3  x)(4  x)(6  x)(10  x)  sin x dx equals

(A) cos 2 + cos 4 (B) cos 2 – cos 4

(C) 2cos1 cos3 (D) 2sin1 sin 3

a
(sin1 e x  sec 1 e x )dx
25. The value of the definite integral

 (cot 1
ea  tan1 e x )(e x  e x )
(a  R) is

(A) Independent of a (B) dependent on a

  2 
(C)  n2 (D) –  n  tan1 ea 
2 2  

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 (k 1) 4
| sin2x |dx dx
26. Let I = 
k
| sin x |  | cos x |
, (k  N) and J =  sin x  cos x
0
, then which of the following hold(s)

good ?
/2
sin2x dx
(A) I = 2 
0
sin x  cos x
(B) I = 4 – 4J

(C) I = 4 – 2J (D) I = 2 – 2J

x8  4
27. If f(x) = x 4
– 2x 2  2
dx and f(0) = 0, then

(A) f(x) is an odd function (B) f(x) has range R

(C) f(x) = 0 has at least one real root (D) f(x) is a monotonic function

/2

n 1  x sin2   d, x  0, then
28. If f(x) = 
0
2
sin 


(A) f(x) =   x 1–1  (B) f'(3) =
4

(C) f(x) cannot be determined (D) f'(0) =
2

x
29. If f : R  R be a continuous function such that f(x) =  2tf(t)dt , then which of the following does not
1

hold(s) good?
2
(A) f() = e (B) f(1) = e
(C) f(0) = 1 (D) f(2) = 2

  3r 2
n 3 b
30. If lim
n
     2   f(x)dx , then
 n 
r 1  
n 
 0

(A) b = 1 (B) f(x) = 9x2 + 6

  3r 2
n 3 n   3r 2 3
(C) lim
n
     2 = 9
 n
r 1  
n
(D) lim
n
     2  = 15
 n
r 1  
n
 
1
31. 
A real valued function f(x) : R+  R+ satisfies f(tx)dt = nf(x) . If lim f(x) = g(x), g(1) = 2 and area
0
n

bounded by y = g(x) with x-axis from x = 3 to x = 7 is S, then

 8 8 8
(A) S   2,  (B) S   , 
 3 7 3

40
(C) S < (D) S > n4
21

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Comprehension # 1 (For Q. No. 32 to 33)

10
cos 4x cos5x cos6x cos7x
Consider the integral I = 
0
1  e2 sin 2x
dx

/2
32. If I = k .  cos 4x cos5x cos 6x cos7x dx , then 'k' is equal to
0

(A) 5 (B) 10 (C) 1 (D) 20

/4
33. If I = .  cos2x cos 4x cos 6x dx , then '' is equal to
0

(A) 5 (B) 20 (C) 10 (D) 5/2

Comprehension # 2 ( For Q. No. 34 to 36)

For i = 0, 1, 2, . . . . , n, let Si denotes the area of region bounded by the curve y = e–2x sin x with x-axis
from x = i to x = (i + 1).

34. The value of S0 is

1  e2 1– e –2
(A) (B)
5 5

1  e–2 1  e– 
(C) (D)
5 5

S2014
35. The ratio is equal to
S2015
–2 2 –
(A) e (B) e (C) 2e (D) e


36. The value of S
i 0
i is equal to

(A)

e  1  e  (B)

e2  e2   1 
5e 
– 1 
5 e 2

–1

e2  1 e2   1
(C) (D)

5 e2  – 1  e2  – 1

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  x  f(x)
37. Let f(x) be differentiable function satisfying the condition f     x, y  R – {0} and f(x)  0,
 y  f(y)

f(1) = 2. If the area enclosed by y  f(x) and x2 + y2  2 is A, then find [2A], where [.] represents G.I.F.


2
dx
38. The value of the definite integral  (sin x  cos x  2

sin x cos x ) sin x cos x
equals

1
39. A continuous real function 'f' satisfies f(2x) = 3 f(x) x  R. If  f(x)dx  1 , then compute the value of
0

2
definite integral  f(x)dx
1

1
1004 1004
x 1– x  dx
2010 0
40. If 2 1
=  then find the highest prime factor of .
1004
 x 1– x 
1004 2010
dx
0

ANSWER KEY
DPP # 7
REVISION DPP OF
VECTORS AND T HREE DIMENSIONAL GEOMETRY
1. (C) 2. (C) 3. (B) 4. (B) 5. (A) 6. (A) 7. (A)
8. (C) 9. (A) 10. (C) 11. (B) 12. (C) 13. (C) 14. (B)
15. (C) 16. (B) 17. (A) 18. (A,D) 19. (B,D) 20. (B,D)
21. (B,C,D) 22. (B,C) 23. (B,C) 24. (A,B,C) 25. (A,B,C,D)26. (A,B,D)
27. (A,C,D) 28. (A,C,D) 29. (A,B) 30. (A,C,D) 31. (A,C,D) 32. (A,B,D)
33. (A,C,D) 34. (C,D) 35. (B,D) 36. (A,B,D) 37. (A,D) 38. (D) 39. (C)
40. (B)

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 Solution of DPP # 8
TARGET : JEE (ADVANCED) 2015
Course: VIJETA & VIJAY ( ADP & ADR)

MATHEMATICS
505   1007  
1. A= 
0
| cos x | dx = 505 cos x dx = 1010 
0
B= 
505 
| sin x | = (1007 – 505)  | sin x | dx = 1004
0

100 100 100

 x3 / 2 
2. I=
 0
x –  x  dx = 
   3/2


0
–  
0
x  dx

4 9 16 25 36 49 64 81 100
2000  
= –  1.dx  2.dx  3.dx  4.dx  5.dx  6.dx  7.dx  8.dx  9.dx 
        
3 1 
 4 9 16 25 36 49 64 81 
2000 2000 155
= – [3 + 10 + 21 + 36 + 55 +78 + 105 + 136 + 171] = – 615 =
3 3 3

   
2 – x2 1 1– x 2 1 1 x
ex .  ex   dx =
 ex   dx
3. I=
 dx = 



1– x   1– x  2  1– x 1– x 2
  1– x  1– x 2 
  1  x  1  x
2 1 x


1 x
= ex. +c
1– x
3
4. f(x) =
 2x .cos2 x  6x 2 sin x cos x – 2x 3 sin2 xdx
3 2 3
=  2x .cos 2xdx   3
x sin
  = x
2 xdx
 .2cos 2x dx  sin 2x. x 3 – x3 .2cos 2x dx 
 
 f(x) = x3 sin2x +c  f(x) = x3 sin2x

5. f(x) = f(2 – x) & f(x) = f(4 – x)  f(x) = f(x + 2)  f(x) is periodic with period 2
50 2
Now I=
 f(x)
0
dx  25 f(x) dx = 75 
0

 x  2x  x 
6
6. =  x4  x2 4
 3 x 2  6 dx = 5
 x 3  x 2 x 6  3 x 4  6 x 2 dx

dt t3 / 2
Put 2x6 + 3x4 + 6x2 = t  =  t. = +c
12 18
7. When n  y2  1 & x2  1  –1y1 & –1x1
y
x = –1
y=1

y = –1
x=1
lim Sn = 4
n 

8x 43  13x38 8x –9  13x –14 –dt 1

8. I=  dx =  1 x dx Put 1 + x–8 + x–13 = t I =   +c
4 4
t4 3t 3
 x 13
 x 1 5
  –8
x –13

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 x
9.
e  f(x) – f '(x)  dx = (x) ...(i)
x
and
e  f(x)  f '(x) dx = ex f(x) ...(ii)
equation (i) & (ii)
2 e x f(x)dx = (x) + ex f(x)

10. f(x) > 0  f(x)   f(x)  1  x  1
x x
1 1
 f(x) 
1 x 2
x1  
1
f '(x)dx   1 x
1
2
dx

 
 f(x) – f(1)  tan x – tan 1 –1 –1
 f(x)  1 – + tan–1x  1 +
4 4
x x x
2 2
t2
x e t dt
 x e t dt
 e dt 2
0 0 0 0 0 ex
11. Let  = Lim   = Lim x
= 2Lim  0  = 2Lim = –2
x 0 (e  x  1)  0  x 0  ex  x  1  x 0 x   x 0 1
x2  
 x2
 
2
f(2t)
12. We have  f '(2t) e dt  5 Put ef(2t) = y  2f' (2t) ef(2t) dt = dy
0
ef ( 4 ) ef ( 4 )
1
Now  e y dy  5   e y dy  10  ef(4) – ef(0) = 10  ef(4) = 10 + e0 = 11
2 f (0 ) f ( 0)
e e
Hence f (4) = n 11
y

x
13. 1 22
2

Area to the left of y-axis = 

2
  x  
Area to the right of y-axis =   4 – x 2 – 2 sin    dx
0  2 2 
2
 x 4 – x2 4 2
x 4 x 
=   sin –1    cos  = 1 + /2 – 4/
 2 2 2  2 2 0
 0

1 cos2 t
14. I1 =  
xf x  2 – x  dx 
sin2 t
1 cos2 t
=  1  cos
2
2
 
t  sin2 t – x f 1  cos2 t  sin2 t – x 2 – 1  cos2 t  sin2 t – x    [P-5]
sin t
1 cos2 t 1 cos2 t
I1
= 2  f  2 – x  x dx –  
xf  2 – x  x dx  I1 = 2I2 – I1  2 I1 = 2I2 
I2
=1
sin2 t sin2 t

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 x
15. Differentiating both sides g(x) = x – x2 g(x)  g(x) =
1  x2
Now, graph of y = g(x) is
y

1/2
–1
x' x
1
–1/2

y'

k 1
k 1   x – k k 1 
k 1
16. f(k) =  x – k dx     f(k) =
 k 1  k 1
k  k

r 1
Now   –1
r 1
f(r) = f(1) – f(2) + f(3) + . . . . . . .

1 1 1 1
= –  – + . . . . . . = 1 – n 2
2 3 4 5

17. f(x + y) = f(x) f(y)  f(x) = ex

x
e 1
 g(x) = 2x
  g(x) is an even function
1 e e  e–x
x

0 2 6 7
3  5 sin x sin x
18. I=
–1
 2
dx 
2
dx  
2
0
J= 
–2 
| sin x |
dx   | sin x | dx
6
=0+=

 3  1
sin  n   x – sin  n   x

 2   2  dx
19. In+1 – In =
0
 sin
x
2
 2cos  n  1 x
x
sin
2
 In+1 –In
0 sin
 x
dx

2
 In+1 – In = 0

20. Put x5 = t
1
1 1  t2013
I=
5  1  t 
0
2015
dt

1 1
1 1 1 t –2
=
5  1  t  2015
dt 
5  2015
dt
0 0  t –1  1
1 1
= 
5 2014
 p = 5 × 2014 = 2 × 5 × 19 × 53

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 x
21. Consider f(x) =
1  x2
y
2/5

3/10

x
0 2 2+h 3–h 3
2+2h

3
x
Area bounded by f(x) with x-axis
x 2
 n 2
2
1
Clearly, h[f(2)+ f(2 + h) + . . . . + f(3 – h)] > n 2 > h[f(2 + h) + f(2 + 2h) + . . . . + f(3)]

22. For all x  (0, 1)

1 1
1 1  x9 1  x8 1

1  x2

1  x3

1  x4
1 
 1 x
0
2
dx < I2 < I1 < 1
0
dx  /4 < I2 < I1 < 1

2 3
y = 2x – x y = f(x)

23. –2
O 4 2
y = x – 4x

t t


0
  
 f(x) – (x 4 – 4x 2 ) dx = 2  2x 2 – x 3 – f(x) dx
 
0

on differentiating with respect to t.
1 4
 f(t) – (t4 – 4t2) = 2(2t2 – t3 – f(t))  f(t) = (t – 2t3)
3
4
24. We have  = 
2
x(3  x)(4  x)(6  x)(10  x)  sin x dx  ....(1)

4
Now = 
2
(6  x)  3  (6  x)  4  (6  x)  6  (6  x) 10  (6  x)   sin(6  x) dx 
b b 4
Applying

a

f(x) dx  f(a  b – x) dx
a
=
 (6  x)(x  3)(10  x)x(4  x)  sin(6  x)  dx ....(2)
2
 On adding (1) and (2), we get
4
4
2I =   sin x  sin(6  x) dx =   cos x  cos(6  x)
2
2
= – cos 4 + cos 2 + cos 2 – cos 4

= 2(cos 2 – cos 4) Hence I = cos 2 – cos 4 Ans.

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 a a
 sin1 e x  cos 1 e x   ex   1  ex 
25. We have I = 


cot 1 ea  tan 1 e x
  2x  dx =
 e  1 2


(cot e  tan 1 e x )
1 a
 2x  dx
 (e  1) 
ex
Put tan–1ex = t  2x
dx = dt
e 1
tan1 ea 1 a
 dt  tan e      2 
I=
2  (t  cot e ) 1 a
=
2
 n(t  cot 1 ea )
0
=  n
2   2  
 2  

  n cot 1 ea  = –  n  tan1 e a 

0

(k 1) 
| sin 2x |dx
26. We have I = 
k
| sin x |  | cos x |
; put x = k + t  dx = dt

 /2 /2
| sin 2x | dx sin2x dx (sin x  cos x)2  1
 I= 
0
| sin x |  | cos x |
= 2 
0
sin x  cos x
= 2 
0
sin x  cos x
dx

2 2 4
dx dx
= 2   sin x  cos x  dx  2 
0 0
sin x  cos x
=4–4  sin x  cos x
0
= 4 – 4J

2
x 8  4  4x 4 – 4x 4 x 4
2  – 4x 4 (x 4  2x 2  2)(x 4 – 2x 2  2)
27. f(x) =  4
x – 2x  2 2
dx =  4
x – 2x  2 2
dx =  (x 4 – 2x 2  2)
dx

5 3
x 2x
 f(x)= + + 2x
5 3

/ 2

n 1  x sin2   d ; / 2
1
28. f(x) =

0
sin2 
x0  f(x) =
 1  x sin
0
2

d

/2 2
sec  dq
 f(x) =
 1  1  x  tan
0
2

put tan = t


dt 1 
 f(x) =  1 2
 f(x) =
1 x
 tan  –1
1 x  t 
0  1 x t  0

 1
 f(x) = .  f(x) =  . 1 x  c put x = 0
2 1 x
 + c = f(0)  c = –  f(x) =   1 x – 1
x
f '(x)
nf(x) = x2 + nc
29. f(x) =
 2t
0
f(t)dt  f(x) = 2xf(x) 
f(x)
= 2x 

2
 f(x) = c.e x put x = 1 c.e = f(1) = 0  c=0  f(x) = 0 
1
  3r 2
n 3
 9x 
2
30. lim
n 
    2 
 n
r 1  
n
 2 .3dx
 0
x
31. tx = y   f(y) dy = xn f(x) 
0
f(x) = n[f(x) + xf'(x)]  f(x)(1 – n) = nx f(x)

f 'x  1 n  1  1– n 
  .  n(f(x)) =   nx + nc
f x  n  x  n 
1n
n c 2
 f(x) = c x as n  f(x) = cx–1 =  g(x) =
x x

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(32 to 33)
10 
cos 4x cos5x cos 6x cos7x
I=

0
1  e 2 sin 2x
dx

10 
cos 4x cos5x cos 6x cos7x
I= 
0
1  e –2 sin 2x
dx (from p-5)

10 
2I =  cos 4x cos5x cos 6x cos7x dx
0



2I = 10 cos 4x cos5x cos 6x cos7x dx
0
(from p-7)

/ 2
2I = 20
0
 cos 4x cos5x cos 6x cos7x dx (from p-6)

/ 2
I = 10  cos 4x cos5x cos 6x cos7x dx
0
 k = 10

Further,
/2
I=5  cos 4x.cos6x. cos12x  cos 2x 
0
 / 2 / 2 
I = 5  cos 4x cos 6x cos12x dx 
 cos 2x cos 4x cos 6x dx  
 
 0 0 
  / 4 
I = 5  0  2 cos 2x cos 4x cos 6x dx 
 (from p-6)
 
 0 
/4
I = 10  cos 2x cos 4x cos 6x dx
0
  = 10

–2x
y=e
y

(34 to 36) 3

 x
0 2

–2x
y = –e
i1 i1
–2x
 e –2x 
Now Si =
 e sin x dx  Si =   –2 sin x – cos x  
i  5 i

1 –2 i 1  e –2i

 Si =
5
e cos  i  1  – e–2i cosi  Si =
5

1  e –2 
1  e –2

1 e –2 
S 2014 5 e 2  1
(i) S0 =
5
(ii)
S 2015
= e2 (ii) i0
Si 
1– e –2


5 e 2 – 1 
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  x  f(x)  x   –x  –f(x)
37. f   Differentiable both side w.r.t. y f '  . 2   2 .f '(y)
 y  f(y)  
 y   y  f (y)
Put y = 1  f(x).x = 2f(x)  f(x) = x2
y
2
y= x

x
–1 1
2 2
x +y =2

1
1  2
A= 2 
0
2 – x2 – x2  dx  
3 2
 2A =
3
+   [2A] = 3

2 2
dx sec 2 x dx
38. Let I = 
 4 cos x(tan x  1  2 tan x ) tan x cos2 x
I=  (1 
4
tan x )2 tan x
 
2t dt  1   1
Put tan x = t2  sec2x dx = 2t dt I=  (t  1) ·t 2
I=–2   = – 2 0  2  = 1
1
 t  1 1  
1
39. We have f(2x) = 3 f(x) ....(1) and  f(x)dx  1
0
....(2)

1
1
From (1) and (2),
3 
f(2x) dx  1
0
2 2 1 2
1
Put 2x = t,
6 
f(t)dt  1
0
 
0
f(t) dt  6  
0

f(t) dt  f(t) dt  6
1
2 1
Hence  f(t) dt  6   f(t) dt = 6 – 1 = 5
1 0

1
1004
 x 1– x 
1004 2010
40. Consider I2 = dx Put x1005 = t  1005x1004dx = dt
0
1 1
1 1004 1 1004
1– 1– t 2 
 1– t 
2
So I2 =
1005
0
dt …(i) Also I2 =

1005  
0


dt …(ii)

1 1
1 1004 1 1004
 I2 =
1005   t  2 – t 
0
dt =
1005
t1004  2 – t 

0
dx Put t = 2y  dt = 2dy

1/ 2 1/ 2
1 1004 1004 1
So I2 =
1005   2y   2 – 2y  2dy =
1005
2.21004.21004 y
1004
1– y 1004 dy
0 0
1/ 2
1 1004
I2 = 22009 y
1004
1– y  dy …(iii)
1005
0
1 1/ 2
1004 1004
Now I1 =  x1004 1– x  dx = 2 x
1004
1– x  dx …(iv)
0 0
 From (iii) and (iv) we get
1 I I
I2 = 22010 1  22010 1 = 4020
1005 4 I2

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