PAPER-A Integral calculus Sol - … cal… · L.K.Gupta (Mathematic Classes) . MOBILE:...

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PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH 1111

PAPER–A

IIT–JEE (2011)

(Integral Calculus solutions)

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MARKING SCHEME

1. For each question in Section I, you will be awarded 3 marks if you have darkened

only the bubble corresponding to the correct answer and zero mark if no bubble is

darkened .In all other cases, minus one (–1) mark will be awarded.

2. For each question in Section II, you will be awarded 3 marks if you darken only the

bubble corresponding to the correct answer and zero mark if no bubble is darkened.

Partial marks will be awarded for partially correct answers. No negative marks will be

awarded in this section.

3. For each question in Section III, you will be awarded 3 marks if you darken only the

bubble corresponding to the correct answer and zero mark if no bubble is darkened. In

all these cases, minus one (–1) mark will be awarded.

4. For each question in Section IV, you will be awarded 3 marks if you darken the

bubble corresponding to the correct answer and zero mark if no bubble is darkened. No

negative marks will be awarded for in this section. NAME OF THE CANDIDATE PHONE NUMBER

L.K. Gupta (Mathematics Classes)

Pioneer Education (The Best Way To Success)

S.C.O. 320, Sector 40– D, Chandigarh Ph: – 9815527721, 0172 – 4617721.



Section I

This section contains 8 multiple choice questions. Each question has four choices A),

B), C) and D) out of which ONLY ONE is correct.

1. nlim

→∞

1/n 2/n (n 1)/n1 e e e......

n n n n

− + + + +

equals

(a) 0 (b) 1 (c) e–1 (d) none of these

Sol:

1/n 1/n 2 1/n n 1

n

1 e (e ) .......(e )lim

n

−

→∞

+ + +

= 1/n n

1/nn

1.[(e ) 1]lim

n(e 1)→∞

−

−

=1/nn

1(e 1)lim (e 1) 1 e 1

e 1

1 / n

→∞− = − × = −

−

2. 2 2

cc

sin t sin t

x yx 0y

1lim e dt e dt

x +→

−

∫ ∫ is equal to(where c is a constant)

(a) 2sin ye (b)

22 sin ysin ye (c) 0 (d) none of these

Sol:

2 2c c

sin t sin t

x 0y x y

1lim e dt e dt

x→+

−

∫ ∫ =

2 2x yc

sin sin t

x 0y c

1lim e dt e dt

x

+

→

+

∫ ∫ =

2x y

sin t

y

x 0

e dt

limx

+

→

∫

=

2sin (x y )

x 0

e 0lim

1

+

→

− =

2sin ye

3. Area enclosed by the curve y = f (x) defined parametrically as 2

2 2

1 t 2tx , y

1 t 1 t

−= =

+ + is

equal to



(a) π sq. units (b) π / 2 sq.units (c)3π

sq.units4

(d) 3π

sq.units2

Sol:

Clearly t can be any real number

Let t = 2

2

1 tan θtan θ x

1 tan θ

−⇒ =

+

x cos 2 θ, and⇒ =

2

2 tan θy sin 2 θ

1 tan θ= =

+

2 2x y 1⇒ + =

Thus, required area = π sq.units.

4. { } { }15

1

The value of sgn( x )dx, where . denotes the fractional part function, is−

∫

(a) 8 (b) 16 (c) 24 (d) 0

Sol:

15 16

1 0

sgn({x})dx sgn({x 1})dx−

= −∫ ∫ (by property)

=16 1 1 1

0 0 0 0

sgn({x})dx 16 sgn({x})dx 16 sgn(x)dx 16 1.dx 16= = = =∫ ∫ ∫ ∫

5. The slope of the tangent to the curve y=

2x

1 2

x

cos t dt−

∫ at 4

1x

2=

(a) 4 8 3

π2 4

−

(b)

4 8 1π

3 4

−

(c)

4 8 1π

3 4

−

(d) none of these

Sol:



2x

1 2

x

y cos t dt−= ∫∵

1 4 1 2dycos (x ).2x cos (x ).1

dx

− −∴ = −

⇒4

1 1

1 4x2

dy 1 2 1| cos cos

dx 2 2 2

− −

=

= −

= 3/4π π.2

3 4− =

4 8 1π

3 4

−

6. 12 9

5 3 3

(2x 5x )dx

(x x 1)

+

+ +∫ is equal to.

(a) 2

5 3 2

x 2xc

(x x 1)

++

+ + (b)

10

5 2

xc

2(x 3x 1)+

+ +

(c)In 5 3 7 4| x x 1| (2x 5x ) c+ + + + + (d) none of the above

Sol:

Let I = 12 9

5 3 3

2x 5xdx

(x x 1)

+

+ +∫ = 3 6

3

2 5

2 5

x xdx

1 11

x x

+

+ +

∫

Put 2 5

1 11 t

x x+ + = ∴

3 6

2 5dx dt

x x

− − =

Then I = 3 2

dt 1c

t 2t− = +∫ =

2

2 5

1c

1 12 1

x x

+

+ +

= 10

5 3 2

xc

2(x x 1)+

+ +



7. If { }21

f(x)cos xdx f(x) c2

= +∫ then f(x) is

(a) x + c (b) sin x + c (c) cos x + c (d) c

Sol:

'1f(x)cos x .2f(x)f (x)

2=

Then f '(x) cosx= ∴ f(x) sin x c= +

8. If

π/2

1

0

I cos(sin x)dx;= ∫π/2

2

0

I sin(cosx)dx= ∫ and

π/2

3

0

I cosxdx,= ∫ then.

(a) 1 2 3I I I> > (b) 2 3 1I I I> > (c) 3 1 2I I I> > (d) 1 3 2I I I> >

Sol:

x 0>∵ sin x x cos(sin x) cosx ....(1)∴ < ⇒ >

Also π

0 x2

< < 1 cosx 0∴ > > , sin(cosx)<cosx From Eqs (1)and (2) we get (2)

cos (sin x) > cos x >sin (cos x) Or

π/2 π/2 π/2

0 0 0

cos(sin x)dx cos x dx sin(cosx)dx> >∫ ∫ ∫

⇒ 1 3 2I I I> >

Section II.

This section contains 5 multiple choice questions. Each question has four choices A),

B), C) and D) out of which ONE OR MORE may be correct.

9. Which one of the following functions is/are homogeneous?

(a). 2 2

x yf (x,y)

x y

−=

+



(b) 1 2

13 3x

f (x,y) x y tany

−−=

(c) 2 2 x/yf (x,y) x (ln x y ln y) ye= + − +

(d) 2 2

22x y x 2yf (x,y) x ln ln(x y) y tan

x 3x y

+ += − + +

−

Sol:

(a). 1

2 2 2

λ(x y)f (λx,λy) λ f (x,y)

λ (x y )

−−= =

+

⇒homogeneous of degree (−1).

(b). 1/3 2/3 1 xf (λx,λy) (λx) (λ y) tan

y

− −=

1/3 1/3 2/3 1 xλ x y tan

y

− − −=

1

3λ f (x,y)−

=

⇒homogeneous

(c). ( )2 2 2 x/yf (λx,λy) λx ln λ (x y ) ln λy λ ye= + − +

2 2

x/yλ (x y )λx ln λye

λy

+ = +

( )2 2 x/yλ x ln x y ln y ye = + − +

λ f (x,y)=

⇒homogeneous

(d). 2 2 2 2

2 22λ x λ y x 2yf (λx,λy) λx ln λ x tan

λ x λ(x y) 3x y

+ += +

+ −



2 22 22x y x 2y

λx ln λ x tanx(x y) 3x y

+ += +

+ −

⇒non homogeneous

10. If f (x), g(x) be twice differential functions on [0, 2] satisfying f ‘’ (x) = g’’ (x), f ‘(1)

=2g’(1) = 4 and f (2) = 3g(2) = 9, then

(a). f(4) −g(4) = 10 (b). f (x) g(x) 2 2 x 0− < ⇒ − < <

(c). f (2) g(2) x 1= ⇒ = − (d). f (x) g(x) 2x− = has real root

Sol: (a), (b), (c)

We have f’’(x) = g’’(x). On integration, We get f’(x) = g ‘(x) + C …. (i)

Putting x = 1, we get

f’ (1) = g’ (1) + C 4 2 C C 2⇒ = + ⇒ =

f '(x) g'(x) 2∴ = +

Integrating w.r.t. x, we get f(x) = g(x) + 2x + c1 ….. (ii)

Putting x = 2, we get

f (2) = g (2) + 4 + c1 1 19 3 4 c c 2⇒ = + + ⇒ =

f(x) g(x) 2x 2. Putting x 4, we get f(4) g(4) 10∴ = + + = − =

f (x) g(x) 2 2x 2 2 x 1 1 2 x 0− < ⇒ + < ⇒ + < ⇒− < <

Also f(2) g(2) x 1= ⇒ = −

f (x) g(x) 2x− = has no solution.

11. Identify the statement (s) which is/are true.

(a). y/x yf (x,y) e tan

x= + is homogeneous of degree zero.

(b). 2

1y y yx ln dx sin dy 0

x x x

−+ = is homogeneous differential equation.

(c). 2f (x,y) x sin x cos y= + is not homogeneous.



(d). 2 2 2 3(x y )dx (xy y )dy 0+ − − = is a homogeneous differential equation.

Sol: (a), (b), (c)

12. The solution of 2 2

2 2

xdx ydy 1 x y

xdy ydx x y

+ − −=

− +is

(a). { }2 2 1x y sin tan (y / x) C−+ = + (b). { }2 2 1x y cos (tan y / x) C−+ = +

(c). 2 2 1x y (tan (sin y / x) C)−+ = + (d). ( )1 2 2y x tan c sin x y−= + +

Sol: (a), (b)

The D. E can be re-written as

( ) 2 22 2

x dx y dy x dy y dx

x y1 x y

+ −=

+− +

( )1 2 2

2 2

xdy ydxSince d tan ( y / x) ,and d x y

x y

− −= +

+

2(xdx y dy),= +

( )

( )

2 2

2 22 2 2 2

1d x y

xdy ydx2we havex yx y 1 x y

+−

∴ =++ − +

{ }1d tan (y / x)−=

2 2 2

2

t dtPut x y t in t he L.H.S and get

t 1 t+ =

−

{ }1d tan (y / x)−=

Integrating both sides, we get

1 1sin t tan (y / x) c− −= +

( )1 2 2 1i.e.,sin x y tan (y / x) c− −+ = +



13. Let f(x) = x t

2

1

3

1 t+∫ dt, where x > 0, then

(a). for0 α β,f(α) f(β)< < < (b). for0 α β,f(α) f(β)< < >

(c). 1f(x) π / 4 tan x, x 1−+ < ∀ ≥ (d). 1f(x) π / 4 tan x, x 1−+ > ∀ ≥

Sol.

x x

2 2 2

3 3 1f '(x) 0 x 0 f '(x) , x 1

1 x 1 x 1 x= > ∀ > ⇒ = > ∀ ≥

+ + +

x x

2

1 1

1f '(x) dx dx

1 x⇒ >

+∫ ∫

1 1 1f(x) tan x tan 1 f(x) π / 4 tan x− − −⇒ > − ⇒ + >

Section III [Linked comprehension type]

This section contains 2 paragraphs. Based upon the first paragraph 3 multiple choice

questions and based upon the second paragraph 2 multiple choice questions have to be

answered. Each of these questions has four choices A), B), C) and D) out of which ONLY

ONE is correct.

Paragraph for question 14 to 16

Let the definite integral be defined by the formula b

a

b af (x)dx (f (a) f(b)).

2

−= +∫ For

more accurate result for c (a,b)∈ , we can useb c b

a a cf(x)dx f (x)dx f (x)dx F(c)= + =∫ ∫ ∫ so

that forb

a

a b b ac , we get f (x)dx (f (a) f(b) 2f(c))

2 4

+ −= = + +∫

14. π/2

0∫ sin xdx is equal to

(a) π

(1 2)8

+ (b)π

(1 2)4

+ (c) π

8 2 (d)

π

4 2



15. If

x

a

3x a

x af (x)dx (f (x) f (a))

2lim 0, then f (x)

(x a)→

− − + =

−

∫is maximum degree

(a) 4 (b) 3 (c) 2 (d) 1

16. If f’’(x) 0 x (a,b)< ∀ ∈ and c is a point such that a < c < b, and (c, f (c)) is the point lying

on the curve for which F (c) is maximum, then f’ (c) is equal to

(a) f (b) f (a)

b a

−

− (b)

2(f (b) f (a))

b a

−

− (c)

2f(b) f(a)

2b a

−

− (d) 0

Sol.

14. (a) π/2

0

π0

π π2sin x dx sin 0 sin 2 sin

4 2 2

− = + +

∫

( )π1 2

8= +

15. (d).

x

a

3x a

x af (x) dx (f (x) f (a))

2lim 0

(x a)→

− − + =

−

∫

a h

a

3h 0

hf (x) dx (f (a h) f (a))

2lim 0h

+

→

− + +=

∫

2h 0

1 hf (a h) [f (a) f (a h)] ( f '(a h))

2 2lim3h→

+ − + + − +⇒ 0=

[ Using L’ Hospital’ s Rule]

2h 0

1 1 hf (a h) f (a) f '(a h)

2 2 2lim 03h→

+ − − +⇒ =

h 0

1 1 hf '(a h) f '(a h) f ''(a h)

2 2 2lim 06h→

+ − + − +⇒ =

[ Using L’ Hospital’ s Rule]



h 0

f ''(a h)lim 0

12→

− +⇒ =

f ''(a) 0, a R⇒ = ∀ ∈

⇒ f (x) must be of max. degree 1.

16. (b). f ''(x) 0, x (a,b),for c (a,b)< ∀ ∈ ∈

c a b cF(c) ( f (a) f (c)) ( f (b) f (c))

2 2

− −= + + +

b a c a b cf (c) f (a) f (b)

2 2 2

− − −= + +

b a 1 1F'(c) f '(c) f(a) f (b)

2 2 2

−⇒ = + −

1[(b a)f '(c) f(a) f (b)]

2= − + −

1F''(c) (b a)f ''(c) 0

2= − < [ f ''(x) 0, x (a,b) and b a]< ∀ ∈ >∵

∴ F (c) is max. at the point (c, f (c)) where

f (b) f (a)F'(c) 0 f '(c) 2

b a

−= ⇒ =

−

Paragraph for question 17 to 18

Integrals of the form 2R(x, ax bx c) dx+ +∫ are calculated with the aid of one of the

three Euler substitutions

1. 2ax bx c t x a if a 0;+ + = ± >

2. 2ax bx c tx c if c 0;+ + = ± >

3. 2ax bx c) (x α)t if a<0,c<0+ + = −

whereα is a real root of ax bx c a(x )(x )2 0+ + = = − α − β



17. Which of the following functions does not appear in the primitive of

2

1

1 x 2x 2+ + + if t is a function of x ?

(a) e

log t 1+ (b)e

log t 2+ (c)1

t 1+ (d) None of these

18.

( )3

2

xdx

7x 10 x− −∫ can be evaluated by substituting for x as

(a) 2

2

5 2tx

t 1

+=

+ (b)

2

2

5 tx

t 2

−=

+ (c)

2

2

2t 5x

3t 1

−=

− (d) None of these

Sol.

17. (d)

Here a = 1 > 0, therefore we make the substitution 2x 2x 2 t x+ + = − . Squaring both

sides of this equality and reducing the similar terms, we get

2 22

2

t 2 t 2t 22x 2tx t 2 x dx dt ;

2(1 t) 2 (1 t)

− + ++ = − ⇒ = =

+ +

2 22 t 2 t 4t 4

1 x 2x 2 1 t2(1 t) 2 (1 t)

− + ++ + + = + − =

+ +

Substituting into the integral, we get

2 2

2 2 2

2(1 t)(t 2t 2) (t 2t 2)l dt dt

(t 4t 4)2 (1 t) (1 t) (t 2)

+ + + + += =

+ + + + +∫ ∫

Now let us expand the obtained proper rational fraction into per tail fractions :

2

2 2

t 2t 2 A B D

(t 1) (t 2) t 1 t 2 (t 2)

+ += +

+ + + + +.

18. (a) In this case a < 0 and c < 0, therefore neither the first, nor the second Euler

substitution is applicable. But the quadratic 7x − 10 − x2 has real roots α 2,β 5= = ,

therefore we use the third Euler substitution :



27x 10 x (x 2)(5 x) (x 2)t− − = − − = −

25 x (x 2)t ;⇒ − = −

2

2

5 2tx

t 1

+⇒ =

+

Section IV (Integer type)

This section contains TEN questions. The answer to each question is a single–digit

integer, ranging from 0 to 9. The correct digit below the question number in the ORS is to

be bubbled.

19. 1

2

2x 2If sin dx

(4x 8x 13)

− + + +

∫1 22x 2

(x 1)tan λ ln(4x 8x 13) c,3

− + = + + + + +

then

the value of − 4 λ must be

Sol. 0003

1

2

2x 2I sin dx

(4x 8x 13)

− +

= + +

∫

1

2 2

2x 2I sin dx

(2x 2) 3

− +

= + +

∫

Put 2x 2 3 tanθ+ =

22dx 3 sec θ dθ∴ =

23 sec θ dθThen, I θ.

2= ∫

{ }3

θ. tan θ ln secθ c2

= − +

2

13 2x 2 2x 2 2x 2. tan ln 1 c

2 3 3 3

− + + +

= − + +



1 22x 2 3(x 1)tan ln(4x 8x 13) c

3 4

− + = + − + + +

3Here, λ

4= −

Then, 4λ 3− =

20. If π/2

8 4

0

14πsin xcos xdx

λ=∫ then the value of ( )λ 4090−

Sol. 0006

π/28 4

0

(7.5. 3. 1)(3. 1) πsin x cos x dx .

12.10. 8. 6. 4. 2 2=∫

(by Wallis’ formula)

7π 14π

2048 4096= =

λ 4096=

⇒ ( )λ 4090− = 4096 – 4090 = 6.

21. If 7 6 5 3 2

2

22

(2x 3x 10x 7x 12x x 1)dx

(x 2)−

+ − − − + +

+∫2

(5π λ)20

= − , then the value of

( )λ 60− must be

Sol. 0004

7 6 5 3 22

22

(2x 3x 10x 7x 12x x 1)Let I dx

(x 2)−

+ − − − + +=

+∫

7 5 32

22

(2x 10x 7x x)dx

(x 2)−

− − +=

+∫

6 22

22

(3x 12x 1)dx

(x 2)−

− ++

+∫



2 42

20

3x (x 4) 10 2

(x 2)

− += +

+∫ (by property)

22 2

20

12 3x (x 2) dx

x 2

= − +

+ ∫

25

3 1

0

3x 1 x2 2x tan

5 2 2

− = − +

12 2 1 π2 4 2 .

5 42

= − +

24 2 π8 2 2.

5 4= − +

2(5π 64)

20= −

λ 64∴ =

⇒ ( )λ 60− = 64 – 60 = 4.

22. If 1

49

0I x (1 x) dx,= −∫ then the value of

12545

I− = must be

Sol. 0005

149

0I x(1 x) dx= −∫∵

149

0(1 x)(1 (1 x)) dx= − − −∫ [By property]

1 149 49 50

0 0(1 x)x dx (x x )dx= − = −∫ ∫

150 51

0

x x

50 51

= −



1 1 1

50 51 2550= − =

12550

I∴ =

⇒ 1

2545I

− = 2550 – 2545 = 5

23. If area between the curves y = xex and y = xe−x and the line x = 1 is λ sq unit, then

the value of e( λ ) = 2 must be

Sol. 0002

Required area 1

x x

0λ (xe xe ) dx−= −∫

{ }1

x x x x

0x (e e ) (e e )− −= + − +

{ } { }1 1(e e ) (e e ) 0 (1 1)− −= + − − − − −

2

e=

⇒2

128eλ 128e 256e

= × =

⇒ e( λ ) = 2



24. If area enclosed between the curves y = ln(x + e) and x = ln1

y

and the axis of is x

is λ sq unit, then the value of 2λ6 1292−

Sol. 0004

1y ln(x e) and x ln

y

= + =

x x1e y e

y

−= ⇒ =

0x

1 e 0Required area ln (x e)dx e dx

∞−

−∴ = + +∫ ∫

ex

1 0ln x dx e dx

∞−= +∫ ∫ ( by property)

{ } { }e x

1 0x ln x x e

∞−= − −

{ } { }(e e) (0 1) 0 1= − − − − −

2 sq unit=

λ 2∴ =

2λ 4Then, 6 6 1296= =

⇒ 2λ6 1292− = 4.



25. The equation of a curve whose slope at any point is thrice its abscissa and which

passes through 2( 1, 3)is 2y λ(x 3)− − = − , then the value of λ must be

Sol. 0003

dy3x

dx=∵

dy 3x dx⇒ =

23xint ergrating, we get y c

2= +

Since, it passes through (− 1, − 3)

then 3

3 c2

− = +

9c

2∴ = −

223x 9

y 2y 3(x 3)2 2

∴ = − ⇒ = −

λ 3∴ =

26. If the solution of the differential equation 2 3dysec y 2x tan y x

dx+ = is

2 tan y = 22 xλ(x 1) ce ,−− + c is arbitrary constant, then the numerical value of λ must be

Sol. 0001

2 3dysec y 2x tan y x

dx+ =∵ …..(i)

let tan y v=

2 dy dvsec y

dx dx∴ =

Then from Eq. (i),

3dv2vx x

dx+ =



2x

e

IF 2xdx e∴ = =∫

Solution is∴

3v. (IF) x . (IF)dx c= +∫

2 2x 3 xtan y . e x .e dx c⇒ = +∫

2 dtPut x t xdx

2= ⇒ =

2x t1tan y . e te dt c

2∴ = +∫

t t1( te e ) c

2= − +

2x1tan y (t 1) ce

2

−⇒ = − +

22 x1or tan y (x 1) ce

2

−= − +

22 xor 2 tan y (x 1) 2ce−= − +

22 x(x 1) ce (Replacing 2c by c)−= − +

Hence, λ 1=

27. A particle moves in a straight line with a velocity given by dx

x 1dt

= + (x is the

distance travelled). If the time taken by a particle to traverse a distance of 99 m is

λ , then the value of ( )e

1020λ log 35− must be

Sol. 0005

dx dxx 1 dt

dt x 1= + ⇒ =

+

ln(x 1) t c⇒ + = +



Putting t 0, x 0, we get c 0= = =

t ln(x 1)⇒ = +

efor x 99, t ln 100 2 log 10= = =

10 e 1020λ log e 20 2 log 10 log e= × ×

40=

⇒ ( )e

1020λ log 35− = 40 – 35 = 5.

28. The differential equation whose solution represents the family

3x 5xy ae be= + is 2

2

d y dy8 λy 0

dx dx− + = , then the value of λ 8− must be

Sol. 0007

3x 5x 3x 5xy ae be or ae be y 0= + + − =∵ … (i)

3x 5xdy3ae 5be

dx∴ = +

3x 5x dyor 3ae 5be 0

dx+ − = …. (ii)

Again differentiating both sides w.r.t. x, then

23x 5x

2

d y9ae 25be 0

dx+ − = …. (iii)

From Eqns. (i), (ii) and (iii), we get

2

2

1 1 y

dy3 5 0

dx

d y9 25

dx

=

Expanding w.r.t. R 1, then



2 2

2 2

d y dy d y dy5 25 1 3 9 y (75 45) 0

dx dx dx dx

− − − + − =

⇒2

2

d y dy2 16 30y 0

dx dx− + =

∴ 2

2

d y dy8 15y 0

dx dx− + =

∴ λ 15=

PAPER-A Integral calculus Sol - … cal… · L.K.Gupta (Mathematic Classes) . MOBILE:...

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Transcript of PAPER-A Integral calculus Sol - … cal… · L.K.Gupta (Mathematic Classes) . MOBILE:...