PAPER-B OBILE : 9815527721, 4617721 IIT–JEE · 2010-12-26 · l.k.gupta (mathematic classes) . mobile: 9815527721, 4617721 pioneer education (the best way to success): s.c.o. 320,

L.K.Gupta (Mathematic Classes) wwwwww..ppiioonneeeerrmmaatthheemmaattiiccss..ccoomm.. MOBILE: 9815527721, 4617721

PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH 1

PAPER-B

IIT–JEE (2011)

(Integral calculus)

“TOWARDS“TOWARDS“TOWARDS“TOWARDS IITIITIITIIT---- JEE IS NOT A JOURNEY,JEE IS NOT A JOURNEY,JEE IS NOT A JOURNEY,JEE IS NOT A JOURNEY,

IT’S A BATTLE, ONLY THE TOUGHEST WILL SURVIVE”IT’S A BATTLE, ONLY THE TOUGHEST WILL SURVIVE”IT’S A BATTLE, ONLY THE TOUGHEST WILL SURVIVE”IT’S A BATTLE, ONLY THE TOUGHEST WILL SURVIVE” TIME: 60 MINS MAX. MARKS: 75

MARKING SCHEME

1. For each question in Section I : you will be awarded 5 marks if you have darkened

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

darkened. In all other cases, minus two (−−−−2) mark will be awarded.

2. For each question in Section II : 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 incorrect answers 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 bubbles are

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

4. For each question in Section IV : you will be awarded 2 marks for each row in

which you have darkened the bubble (s) corresponding to the correct answer. Thus, each

question in this section carries a maximum of 8 marks. There is no negative marks

awarded for incorrect answer (s) 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 6 multiple choice questions. Each question has 4 choices A), B), C) and D)

out of which ONLY ONE is correct.

1. Which of the following functions are differentiable in (-1,2) ?

(a) 2x

2

x

(log x) dx∫ (b) 2x

x

sin xdx

x∫ (c) x 2

2

0

1 t tdt

1 t t

− +

+ +∫ (d) none of these.

Sol.

2x

2dx

x

f(x) (log x)= ∫2 2f '(x) 2(log2x) (log x)= − f '(x)∵ does not exist for all x in (-1, 2)

let 2x

x

sin xg(x) , dx

x= ∫ ⇒

sin2x sin xg'(x) 2

2x x

= −

∴ g '(x) does not exist at x = 0 and so is not differentiable in (-1, 2) .

Lat h(x) = x 2

2

0

1 t tdt

1 t t

− +=

+ + ∫ Then,

2

2

1 x xh'(x) ,

1 x x

− +=

+ + which is defined for all x in (-1,2)

as 21 x x 0.+ + ≠

2. The slope of the tangent to the curve y = x

3

0

dx

1 x+∫ at the point where x= 1 is.

(a) 1/4 (b) 1/2 (c) 1 (d) none of these .

Sol.

3

x 1

dy 1 dy 1,

dx 1 x dx 2=

= =

+

3. Differential equation of the family of circles touching the line y = 2 at (0, 2) is

(a). 2 2 dyx (y 2) (y 2) 0

dx+ − + − = (b). 2 dx

x (y 2) 2 2x y 0dy

+ − − − =



(c). 2 2 dxx (y 2) y 2 (y 2) 0

dy

+ − + + − − =

(d). None of these

Sol.

Equation of circle will be 2 2x (y 2) λ(y 2) 0+ − + − =

Differentiating, we get 2x + 2 (y − 2) dx dy

λ 0dy dx

+ =

2 2 dxthe equation is x (y 2) (y 2) 2x 2y 4 0

dy

∴ + − − − + − =

4. The solution of the equation dy x(2 log x 1)

dx sin y y cos y

+=

+ is

(a). 2

2 xy sin y x log x c

2= + + (b) 2y cos y x (log x 1) c= + +

(c). 2

2 xy cos y x log x c

2= + + (d) 2y sin y x log x c= +

Sol.

(y cos y sin y) dy (2x log x x)dx+ = +

y sin y sin y dy sin y dx− +∫ ∫

2 2 1x log x x dx x dx c

x= − + +∫ ∫

2y sin y x log x c∴ = +

5. The slope of the tangent at (x, y) to a curve passing through a point (2, 1) is 2 2x y

2xy

+,

then the equation of the curve is

(a) 2 22(x y ) 3x− = (b) 2 22(x y ) 6y− =

(c) 2 2x(x y ) 6− = (d) 2 2x(x y ) 10+ =



Sol:

2 2dy x y

dx 2xy

+= ….. (1)

dy dvPut y vx v x

dx dx= ∴ = +

equation (1) transforms to∴

2 2 2 2dv x v x 1 vv x

dx 2x vx 2v

+ ++ = =

2 2dv 1 v 1 vx v

dx 2v 2v

+ −⇒ = − =

2

2vdv dx

1 v x⇒ =

−

2log x log(1 v ) log C⇒ + − =

2x(1 v ) C⇒ − =

2

2

yx 1 C

x

⇒ − =

2 2x y Cx⇒ − =

It passes through (2, 1)

34 1 2C C

2∴ − = ⇒ =

2 2 2 23x y x 2(x y ) 3x

2∴ − = ⇒ − =

6. The area bounded by 1 1y sec x,y cosec− −= = x and line x − 1 = 0 is

(a) π

log(3 2 2)2

+ − sq. units (b) π

log(3 2 2)2

− + sq. units

(c) e

π log 3− sq. units (d) None of these.

Sol.



integrating along x-axis, we get

2

1 1

1

A (cosec x sec x) dx− −= −∫

Integrating along y-axis, we get

π/4

0

A 2 (sec y 1)dy= −∫

π/4

02 [log sec y tan y y ]= + −

π π2 log 2 1 log (3 2 2) sq.units.

4 2

= + − = + −

SECTION – II (Integer Type)

This section contains 5 questions. The answer to each question is a single-digit integer,

ranging from 0 to 9. Correct digit below the question no. in the answer sheet is to be

bubbled.



7. If 1 tan x 1( tan x cot x )dx a tan c

b tan x

− −

+ = +

∫ , then the value of 4 5a b 30+ − must be

Sol. 0006

Let I ( tan x cot x ) dx= +∫

(sin x cos x)dx

(sin x cos x

+= ∫

Put sin x cos x t− =

21 sin 2x t⇒ − =

(cos x sin x)dx dt∴ + =

1 1

22

dt tThen, I 2 sin t 2 tan c

1 t1 t

2

− −

= = = + − −

∫

1 sin x cos x2 tan c

sin 2x

− −

= +

1 tan x 12 tan c

2tan x

− −

= +

We get a 2, b 2= =

4 5Then, a b 4 32 36+ = + =

⇒ 4 5a b 30+ − = 36 – 30 = 6

8. If 2 2tan x b tan xsin 4x. e dx acos x.e c= +∫ , then the value of 2ba 255− must be

Sol. 0001

2tan xLet I sin 4x . e dx= ∫

2tan x2 sin 2x cos 2x. e dx= ∫



22

tan x

2

1 tan x4 sin xcos x. . e dx

1 tan x

−=

+ ∫

22 6 2 tan x4 tan x sec x cos x(1 tan x).e dx= −∫

2Put tan x t=

22tan x sec x dx dt=

{ } tt

3 3

(t 1) 2 e dt(1 t)e dtThen, I 2 2

(1 t) (1 t)

+ −−= = −

+ +∫ ∫

t

2 3

1 22 e dt

(1 t) (1 t)

= − −

+ + ∫

t

2

2ec

(1 t)= − +

+

24 tan x2cos x.e c= − +

2b 8a 2, b 4, then a ( 2) 256∴ = − = = − =

9. If 2x (1 x )

0f(t) dt x

+

=∫ , then the value of 5.f(2) must be

Sol. 0001

Differentiating both sides w.r.t. x, then

2 2f (x (1 x)) (2x 3x ) 1+ × + =

At x 1=

1f (2)

5=

15f(2) 5 1

5∴ = × =

10. If I = 2

0x[2x] dx∫ , where [.] denotes the greatest integer function, then the value of



12I

17

Must be

Sol. 0003

2

0I x [2x] dx= ∫

1/2 1 3/2 2

0 1/2 1 3/20.dx xdx 2xdx 3x dx= + + +∫ ∫ ∫ ∫

1 22 2

2 3/2

1

1/2 3/2

x 3x0 [x ]

2 2

= + + +

1 1 9 3 91 1 4

2 4 4 2 4

= − + − + −

3 5 21

8 4 8= + +

34 17

8 4= =

17128I 128

4∴ = ×

32 17 544= × =

12 173

17 4∴ × =

11. If ( )

sin2α

2 2

0

π 2π dxα and I

2 3 4cos α x< < =

−∫ then the value of

I α

π

+

must be

Sol. 0001

sin 2α

1 1

0

x sin 2αI sin sin

2 cos α 2 cos α

− − = =

1sin (sin α)−=



2ππ α π α

3

= − < <

∵

⇒ I α π α α

1π π

+ − + = =

SECTION −−−− III (Paragraph Type)

This section contains 2 paragraphs. Based upon each of the paragraphs 3 multiple choice

questions have to be answered. Each of these question has four choices A), B), C) and D)

out of which Only One is correct.

Paragraph for questions 12 −−−− 14

Two curves 2/3 1/3

1C [f (y)] [ f (x)] 0 and≡ + =

2/3 2/3

2C [ f(y)] [ f(x)] 12≡ + = ,

Satisfying the relation: 2 2f (x y) f (x y) (x y) f (x y) 4xy (x y )− + − + − = − .

12. The area bounded by C1 and C2 is

(a). 2π 3− sq. units (b). 2π 3+ sq. units

(c). π 6+ sq. units (d). 2 3 π− sq. units

13. The area bounded by the curve C2 and x y 12+ = is

(a). 12π 2 12− sq. units (b). 6 12− sq. units

(c). 2 12 6− sq. units (d). None of these

14. The area bounded by C1 and x + y + 2 = 0 is

(a). 5/2 sq. units (b) 7/2 sq. units

(c). 9/2 sq. units (d). None of these



Sol.

12. (b) Given

2 2(x y)f (x y) (x y)f (x y) 4xy(x y )− + − + − = −

2 2 2 2(x y )[(x y) (x y) ]= − + − −

3 3(x y)(x y) (x y)(x y)= − + − + −

3 3 3f (x y) (x y) f (x) x ,f (y) y⇒ + = + ⇒ = =

Now equations of given curves are

2y x 0+ = …. (1)

2 2x y 12+ = … (2)

Solving equations (1) and (2), we get x 3, y 3= − = ± The area bounded by curves

3 0

2

32 3

A 2 12 x dx x dx−

−−

= − + −

∫ ∫

π/33

2 2

1

π/22 3

I 2 12 x dx 2 12 cos θdθ

−−

−−

= − =∫ ∫

π/3

π/2

12 (1 cos 2θ)dθ

−

−

= +

∫



π/3

π/2

sinθ π 3 π12 θ 12

2 3 4 2

−

−

= + = − − +

π 312 2π 3 3.

6 4

= − = −

0 3/2 03/23

2

3

2[( x) ] 4I 2 x dx [0 3 ]

3/ 2 3−

−

−= − = = − −

−∫

4 3 .=

A 2π 3 3 4 3 2π 3= − + = + sq. units.

13. (a). The required area is =area of circle –area of square

12 24 sq units= π −

14. (c). The Required area



( )( )2

2

1

22 3

1

y y 2 dy

y y2y

2 3

4 8 1 14 2

2 3 2 3

9sq units

2

−

−

= − − − −

= + −

= + − − − +

=

∫

Paragraph for questions 15 −−−− 17

A certain radioactive material is known to decay at a rate proportional to the amount

present. Initially there is 50 kg of the material present and after two hours it is observed

that the material has lost 10 percent of its original mass. Based on these data answer the

following questions.



15. The expression for the mass of the material remaining at any time t

(a). (1/2)(ln0.9) tN 50e−= (b). (1/4)(ln9) t50e −

(c). (ln0.9) tN 50e−= (d). None of these

16. The mass of the material after four hours

(a). 0.5ln 950− (b). 2ln 950e−

(c). 2ln 0.950e− (d). None of these

17. The time at which the material has decayed to one half of its initial mass.

(a). (ln 1 / 2)/ (1/ 2 ln 9) hr (b). (ln 2) /( 1 / 2 ln 0.9)hr−

(c). (ln 1 / 2) / ( 1 / 2 ln 0.9)hr− (d). None of these

Sol.

15. (a). Let N denote the amount of material present at time t. Then,

dNkN 0

dt− =

This differential equation is separable and linear, its solution is N = cekt …. (1)

At t = 0, we are given that N = 50. Therefore, from equation

(1), 50 = cek (0) , or c = 50. or c = 50.

Thus, N = 50 ekt ….. (2)

At t = 2, 10 percent of the original mass of 50 mg or 5 mg, has decayed.

Hence, at t = 2, N = 50 − 5 = 45.

Substituting this value into equation (2) and solving for k, we have

2k 1 4545 50e or k log

2 50= =

Substituting this value into (2), we obtain the amount of mass present at any time t as

N = 50 (1/2)(ln0.9)te− … (3)

where t is measured in hours.

16. (c). We require N at t = 4. Substituting t = 4 into (3) and then solving for N , we find

N = 50e− 2ln 0.9



17. (c). We require t when N = 50/2 = 25. Substituting N = 25 into equation (3) and solving

for t, we find 25 = 50 (1/2)(ln0.9)te− or t = (ln 1/2) / (−1/2 ln 0.9) hr.

SECTION −−−− IV (Matrix type)

This section contains 2 questions. Each question has four statements (A, B, C and D) given

in Column I and five statements (p, q, r, s and t) in Column II. Any given statement in

Column I can have correct matching with one or more statements(s) given in Column II.

For example, if for a given question, statement B matches with the statements given in q

and r, then for that particular question, against statement B, darken the bubbles

corresponding to q and r in the ORS.

18. If at every point x of an interval [ a, b] ,the inequalities g(x) ≤ f(x) ≤ h(x) are fulfilled ,

then ( ) ( ) ( )b b b

a a ag x dx f x dx h x dx ,a b≤ ≤ <∫ ∫ ∫

Match the entries from the following two columns:

Column –I Column –II

(a) If µ <

( )

71

0 8

x dxλ ,then

3 1 x<

+∫

(p) [λ +µ ] = 2, Where [.] denotes the greatest

integer function

(b) If µ <

( )

1

0 6

dxλ ,then

1 x<

+∫

(q) [λ +µ ] = 4, Where [.] denotes the greatest

integer function

(c) If µ <

( )

1

0 2 3

dxλ ,then

4 x x<

− −∫

(r) [λ -µ ] = 0, Where [.] denotes the greatest

integer function

(s) [λ -µ ] = 3, Where [.] denotes the greatest

integer function

(t) [λ +µ ] = 0, Where [.] denotes the greatest

integer function

Sol:



Since

( ) ( )

1 1 17 77 7

8 83 30 0 0

x x0 x 0 x 1 Then 0dx dx x dx

1 x 1 x< < ∀ < < < <

+ +∫ ∫ ∫

( )

1 7

830

x dx 1 1Hence,0 λ ,μ 0

8 81 x< < ∴ = =

+∫ [λ μ] 0,[λ μ] o(R,T)+ = − =

Since ( ) ( ) ( )2 6 21 x (1 x ) 1 x x 0,1− < + < + ∀ ∈

( ) ( ) ( )2 6 2

1 1 1x (0,1)

1 x 1 x 1 x⇒ > > ∀ ∈

− + +

( ) ( ) ( )

1 1 1

2 6 20 0 0

dx dx dx

1 x 1 x 1 x⇒ < <

+ + −∫ ∫ ∫ ( )

( )

1

2 1

60

1 1dxln{x 1 x } {sin x}

0 01 x

−⇒ + + < <+

∫

( )

1

60

dx π πln2 λ 1.57μ ln2 0.693[λ μ 2[λ μ] 0(p,r)

2 21 x⇒ < < ∴ = ≈ = ≈ + = − =

+∫ Since ,

2 2 3 24 x 4 x x 4 2x x (0,1)− > − − > − ∀ ∈

( ) ( ) ( )( ) ( ) ( )

2 2 3 2

2 2 3 2

1 1 14 x 4 x x 4 2x x (0,1) x (0,1)

4 x 4 x x 4 2x⇒ − > − − > − ∀ ∈ ⇒ < < ∀ ∈

− − − −

( ) ( ) ( )

1 1 1

2 2 3 20 0 0

dx dx dx

4 x 4 x x 4 2x∴ < <

− − − −∫ ∫ ∫

( ) ( )

1 1

1 1

42 3 2 30 0

1 1x dx 1 x π dx πsin sin

2 2 60 02 24 x x 4 x x

− − < < ⇒ < <

− − − −∫ ∫

4

π πμ π0.52 λ 4.43and [λ μ] 4,[λ μ] 3(q,s)

6 2= ≈ ∴ = ≈ + = − =

19. Match the statements in Column –I with the Column –II.



Column –I Column –II

(a) ( )

( )( )

4 2 21/3 1

22

t t 1 2t 1tan x dx A ln Btan c

3t 1

−− + −

= + + +

∫

Where t = tan1/3x

(p) A =

1

4

(b) 3sin x sin x 2cosx 1dx Acos x Bln c

cos2x 2cosx 1

+ += + + −

∫ (q)

A = 1

3

(c)

( )( )1 1

2 2

dx xA tan x Btan c

2x 1 x 4

− − = + +

+ + ∫

(r) A =

1

2

(s) B =

3

2

(t) B=

3

4 2

Ans. (a) Let ( )

21/3 3 2 2

6

3tI (tan x) dx put tan x t sec xdx 3t dt dx

1 t= = ∴ = ⇒ =

+∫

(b) let ( ) ( )

( )

3 2

2

sin x sin x dx 1 sin x sin xdxI

cos2x 2cos x 1

+ += =

−∫ ∫

2

2

(2 cos x)sin x dtdx put 2cosx t sin xdx

2( 2cos x) 1

−= = ∴ = −

−∫ then,

2

2

2 22

t dt2

2 1 (t 4)2I dt

(t 1) (t 1)2

− −

− = =− −∫ ∫

22 2

1 3 1 3 t 11 dt t In| | c

t 1 2 t 12 2

− = − = − +

− + ∫



( )2 4

1 3 2cos x 12cos x In| | c

2 2 2cos x 1

−= − +

+

4 4

1 3 2cosx 1 1 3cos x In| | c A (r),B (t)

2 22 2cosx 1 2

−= − + ∴ = =

+

3 2

6 2 3

t dt t .dtI 3 3

(1 t ) 1 (t )

∴ = =

+ + ∫ ∫ Put

2

3 2

3 zdz 3 zdzt z 2tdt dz than I

2 1 z 2 (1 z)(1 z z )= ⇒ = = =

+ + − +∫ ∫

2

2 2

1 3( 1 2z)

1 1 1 1 3 dz2 2In(1 z) dz In(1 z) In(1 z z )2 2 (1 z z ) 2 4 4 1 z z

− + += − + + = − + + − + +

− + − +∫ ∫2

22

1 1 3 dzIn(1 z) In(1 z z )

2 4 4 1 3z

2 2

= − + + − + +

− +

∫

2 4 2 21 1

2 2 2

1 1 z z 3 2 2z 1 1 t t 1 3 2t 1In . tan c In tan c

4 (1 z) 4 4 (t 1) 23 3 3

− − − + − − + − = + + = + + + +

1 3A (P):B (S)

3 2∴ = =

(c) Let

1 1

2 2 2 2

dx 1 1 1 1 1 xI dx tan x tan c

(x 1)(x 4) 3 x 1 x 4 3 2 2

− − = = − = − +

+ + + + ∫ ∫

1 11 1 x 1 1tan x tan c A (Q),B

3 6 2 3 6

− − = − + ∴ = = −

Documents

PAPER-B OBILE : 9815527721, 4617721 IIT–JEE · 2010-12-26 · l.k.gupta (mathematic classes) . mobile: 9815527721, 4617721 pioneer education (the best way to success): s.c.o. 320,