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### Transcript of Teko Classes IIT JEE/AIEEE MATHS by SUHAAG SIR ... âˆ« = is (code-V2T5PAQ22) (a) e (b) e (c) e2...

• Teko Classes IIT JEE/AIEEE MATHS by SUHAAG SIR Bhopal, Ph. (0755)32 00 000 www.tekoclasses.com Question. & Solution. Int. Cal. Page: 1 of 26

THE “BOND” || Phy. by Chitranjan|| ||Chem. by Pavan Gubrele|| ||Maths by Suhaag Kariya||

TOPIC = INTEGRAL CALCULUS

Single Correct Type

Que. 1. Let 3 3 3 2 4

f (x) sin x sin x sin x 3 3

π π    = + + + +   

    then the primitive of f(x) w.r.t. x is

(a) 3sin 3x

C 4

− + (b) 3cos3x

C 4

− + (c) sin 3x

C 4

+ (d) cos3x

C 4

+

where C is an arbitrary constant. (code-V2T3PAQ5)

Que. 2. If the dependent variable y is changed to ‘z’ by the substitution y = tan z then the differential equa-

tion

22

2 2

d y 2(1 y) dy 1

dx 1 y dx

+   = +  

+   is changed to

2

2

d z

dx =

2

2 dzcos z k , dx

  +  

  then the value of k equals

(a) –1 (b) 0 (c)1 (d) 2 (code-V2T3PAQ10)

Que. 3. ( )xx n ex dx∫ � is equal to (code-V2T5PAQ1)

(a) xx C+ (b) x. n x� (c) ( )

x c n x C+� (d) None

Que. 4. The value of the definite integral 2008

x

2008

f '(x) f '( x) dx

(2008) 1 −

+ −

+∫ equals (code-V2T5PAQ3)

(a) f (2008) f ( 2008)+ − (b) f (2008) f ( 2008)− −

(c) 0 (d) f ( 2008) f (2008)− −

Que. 5. e

1

1 n x dx

xx n x

  +  

  ∫

� equals (code-V2T5PAQ4)

(a) e (b) 2e (c) 2 e (d) 2e

Que. 6. If x

4

0

g(x) cos t dt,= ∫ then g(x )+ π equals (code-V2T5PAQ5)

(a) g(x) g( )+ π (b) g(x) g( )− π (c) g(x)g( )π (d) [ ]g(x) g( )π

Que. 7. Let f be a positive function. Let ( ) ( ) k k

1 2

1 k 1 k

I x f x(1 x) dx; I f x(1 x) dx, − −

= − = −∫ ∫ where 2k 1 0.− >

Then 2

1

I

I is (code-V2T5PAQ7)

(a) k (b) 1/2 (c) 1 (d) 2

Que. 8. 2 2

dx

x 16 x− ∫ has the value equal to (code-V2T5PAQ8)

(a) 1 x

C arc sec 4 4

  −  

  (b)

1 x arc sec C

4 4

  + 

 

(c) 2

16 x C

16x

− − (d)

2 16 x

C 16x

− +

• Teko Classes IIT JEE/AIEEE MATHS by SUHAAG SIR Bhopal, Ph. (0755)32 00 000 www.tekoclasses.com Question. & Solution. Int. Cal. Page: 2 of 26

THE “BOND” || Phy. by Chitranjan|| ||Chem. by Pavan Gubrele|| ||Maths by Suhaag Kariya||

Que. 9. If ( )2 2x 2x 2x .e dx e ax bx c d,− −= + + +∫ then (code-V2T5PAQ10)

(a) 1 1 1

a ,b ,c 2 2 4

= − = − = − (b) 1 1 1

a ,b ,c 2 2 4

= − = − = −

(c) 1 1

a ,b 1,c 2 2

= − = − = − (d) 1

a 1,b 1,c 2

= = = −

Que. 10. If

1

-1

0

π 1 tan x dx= - ln 2

4 2∫ then the value of the definite integral ( ) 1

1 2

0

tan 1 x x dx −

− +∫ equals

(a) n 2� (b) n 2 4

π + � (c) n 2

4

π − � (d) 2 n 2� (code-V2T5PAQ11)

Que. 11.

2

n

nn 0

n lim x dx

2→∞ ∫ equals (code-V2T5PAQ12)

(a) 1 (b) 2 (c) 1

2 (d) 4

Que. 12. If f is continuous function and x 2

0 t

F(x) (2t 3). f (u)du dt  

= +    ∫ ∫ then F"(2) is equal to

(a) 7f (2)− (b) 7f '(2) (c) 3f '(2) (d) 7f (2) (code-V2T5PAQ15)

Que. 13. ( )( )sin x / 2

x 1 x cos x n x sin x dx

π

π

+ +∫ � is equal to (code-V2T5PAQ16)

(a)

2

2

π (b)

2

π (c)

24

4

π − π (d) 1

2

π −

Que. 14. Let f :[0, ) R∞ → be a continuous strictly increasing function, such that (code-V2T5PAQ17)

x

3 2

0

f (x) t.f (t)dt= ∫ for every x 0.≥ The value of f (6) is

(a) 1 (b) 6 (c) 12 (d) 36

Que. 15. If the value of definite integral

/ 4

x

/ 6

1 cot x dx,

e sin x

π

π

+ ∫ is equal to / 6 / 4ae be−π −π+ then (a + b) equals

(a) 2 2− (b) 2 2+ (c) 2 2 2− (d) 2 3 2− (code-V2T5PAQ18)

Que. 16. Let 3 0

n x J dx

1 x

= +∫ �

and 2 0

x n x K dx

1 x

= +∫ �

then (code-V2T5PAQ20)

(a) J K 0+ = (b) J K 0− = (c) J K 0+ < (d) none

• Teko Classes IIT JEE/AIEEE MATHS by SUHAAG SIR Bhopal, Ph. (0755)32 00 000 www.tekoclasses.com Question. & Solution. Int. Cal. Page: 3 of 26

THE “BOND” || Phy. by Chitranjan|| ||Chem. by Pavan Gubrele|| ||Maths by Suhaag Kariya||

Que. 17. The value of x > 1 satisfying the equation x

1

1 t n t dt ,

4 =∫ � is (code-V2T5PAQ22)

(a) e (b) e (c) 2e (d) e 1−

Que. 18. If

x

1

F(x) f (t)dt= ∫ where 2

t 4

1

1 u f(t) du

u

+ = ∫ then the value of F"(2) equals (code-V2T5PAQ23)

(a) 7

4 17 (b)

15

17 (c) 257 (d)

15 17

68

Que. 19. Let f be a continuous function on [a, b]. If ( )( ) x b

a x

F(x) f (t)dt f (t)dt 2x a b  

= − − +    ∫ ∫ then there exist

some ( )c a,b∈ such that (code-V2T8PAQ6)

(a)

c b

a c

f (t)dt f (t)dt=∫ ∫ (b) ( ) c b

a c

f (t)dt f (t)dt f (c) a b 2c− = + −∫ ∫

(c) ( )( ) c b

a c

f (t)dt f (t)dt f (c) 2c a b− = − +∫ ∫ (d) ( )( ) c b

a c

f (t)dt f (t)dt f (c) 2c a b+ = − +∫ ∫

Que. 20. The value of the definite integral ( ) ( ) / 2

x 2 2

0

x x I e cos sin x cos sin sin x sin dx,

2 2

π  

= +   

∫ is

(a) ( ) / 21 e cos1 sin1 1

2

π + −  (b) ( ) / 2e

cos1 sin1 2

π

+ (codeV2T10PAQ2)

(c) ( )/ 2 1

e cos1 1 2

π − (d) [ ]

/ 2e cos1 sin1 1

2

π

+ −

(a) ( ) / 21 e cos1 sin1 1

2

π + −  (b) ( ) / 2e

cos1 sin1 2

π

+

(c) ( )/ 2 1

e cos1 1 2

π − (d) [ ]

/ 2e cos1 sin1 1

2

π

+ −

Que. 21. A tank with a capacity of 1000 liters originally contains 100 gms of salt dissolved in 400 liters of

water. Beginning at time t 0= and ending at time t 100= minutes, water containing 1 gm of salt per

liters enters the tank at the of 4 liters per minute, and the well mixed solution is drained from the tank

at a rate of 2 liter/minute. The differential equation for the amount of salt y in the tank at time t is

(a) dy y

4 dt 400 2t

= − +

(b) dy y

4 dt 200 t

= − +

(code-V2T11PAQ2)

(c) ( )

dy y 4

dt 200 t 2 = −

+ (d) dy y

4 dt 500 t

= − +

• Teko Classes IIT JEE/AIEEE MATHS by SUHAAG SIR Bhopal, Ph. (0755)32 00 000 www.tekoclasses.com Question. & Solution. Int. Cal. Page: 4 of 26

THE “BOND” || Phy. by Chitranjan|| ||Chem. by Pavan Gubrele|| ||Maths by Suhaag Kariya||

Que. 22. Let y y(t)= be solution to the differential equation 2y ' 2ty t ,+ = then t

y lim

t→∞ is (code-V2T11PAQ4)

(a) zero (b) 1

2 (c) 1 (d) Non existent.

Que. 23. The area of the region bounded below by 1y sin x,−= above by 1y cos x−= and on the left by y-axis,

is

(a) 2 1− (b) 2 2− (c) 2 1+ (d) 2 (code-V2T11PAQ7)

Que. 24. 2

0

x dx

4 x−∫ is equal to (code-V2T13PAQ20)

(a) 2

π (b) 1

2

π − (c) 1π − (d) 2π −

Que. 25. If 2

4 2

2

x . 4 x dx −

−∫ has the value equal to kπ then the value of k equals (code-V2T14PAQ20)

(a) 0 (b) 2 (c) 8 (d) 4

Que. 26. Let q

2008 r

dx 1 x n C

x x p 1 x

  = + 

+ +  ∫ � where p,q.r N∈ and need not be distinct, then the value of

( )p q r+ + equals (code-V2T14PAQ25)

(a) 6024 (b) 6022 (c) 6021 (d) 6020

Que. 27. ( ) ( ) / 2

x

0

sin x n(sin x) x cot x dx

π

+∫ � is (code-V2T17PAQ7)

(a) – 1 (b) 1 (c) 0 (d) Indeterminant

Que. 28. Let ( ) 2

y n 1 cos x= +� then the value of 2

2 y / 2

d y 2

dx e + equals (code-V2T17PAQ8)

(a) 0 (b) 2

1 cos x+ (c)

4

(1 cos x)+ (d) 2

4

(1 cos x)

+

Que. 29. The value of definite integral 1

x 2

1

dx

(1 e )(1 x ) −

+ +∫ is (code-V2T18PAQ1)

(a) / 2π (b) / 4π (c) / 8π (d) /16π

Que. 30. The expression 2

3

2

d y y

dx on the ellipse 2 23x 4y 12+ = is equal to (code-V2T19PAQ5)