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1

PAPER –B

IIT–JEE (2012)

(Calculus)

“TOWARDS IIT– JEE IS NOT A JOURNEY, IT’S A BATTLE, ONLY THE TOUGHEST WILL SURVIVE”

TIME: 60 MINS MAX. MARKS: 80 MARKING SCHEME

In Section I (Total Marks: 24), for each question you will be awarded 3 marks if you darken ONLY

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

other cases, minus one (1) mark will be awarded.

In Section II (Total Marks: 16), for each question you will be awarded 4 marks if you darken ALL

the bubble(S) corresponding to the correct answer(s) ONLY and zero marks otherwise. There are

no negative marks in this section.

In Section III (Total Marks: 24), for each question you will be awarded 4 marks if you darken ONLY

the bubble corresponding to the correct answer and zero marks otherwise. There are no negative

marks in this section.

In Section IV (Total Marks: 12), for each question you will be awarded 2 marks for each row in

which you have darkened ALL the bubble(s) corresponding to the correct answer(s) ONLY and zero

marks otherwise. Thus, each question in this section carries a maximum of 6 marks. There are no

negative marks in this section.

NAME OF THE CANDIDATE CONTACT NUMBER

L.K. Gupta (Mathematics Classes)

FOR SOLUTIONS KINDLY VISIT

www.pioneermathematics.com (In latest Updates)




2

Section– I (Total Marks: 24) (Single Correct Answer Type)

This section contains 8 multiple choice questions. Each question has four choices (a), (b), (c), (d)

out of which ONLY ONE is correct.

1. If f(x) n

x x xLt ...... tonterms

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

, then range of f(x) is

(a) {0, 1} (b) {–1, 0} (c) {–1, 1} (d) [–1, 1] Sol: (a)

Sn = 1 1 1 1 1

11 x 1 x 1 2x 1 2x 1 3x

+ …… +

1 1

1 (n 1)x 1 nx

= 1 – 1

1 nx

But n

if x 0

lt nx if x 0

0 if x 0

f (x) = nl t

Sn =1 When x 0

0 When x 0

Range of f = {0, 1} 2. Let f be a continuous function on R such that

fn

1

4

= (sin en)2

2n

2

ne

n 1

then f (0) =

(a) 1 (b) 0 (c) 1 (d) 1

4

Sol: (a) As f continuous on R, so

nlim

f (1/4n) = f (0)

and nlim

2

2n n

2

n(sine )e

n 1

= nlim

2

n

2n

sine 1

1 1 / ne

= 1

Since |sin en| ≤ 1 and 2ne 0 as n ∞, thus f (0) = 1




3

3. If f(x) = cos 3[x] x2

π

, –1 < x < 2 and [x] is the greatest integer less than or equal to x,

then f’ 3

2

π

is equal to

(a) 0 (b) 1 (c) 1

2 (d)

1

2

Sol: (a) As we know

1 < 3

2

π < 2 If x = 3

2

π ⇒ [x] = 1

So, f (x) = cos 3x2

π

= sin x3

⇒ f ’(x) = cos x3 . 3x2

f ’ 3

2

π

= 3 2/3

2

π

. Cos 2

π = 0

⇒ f ’ 3

2

π

= 0

4. If the relation between subnormal SN and subtangent ST at any point S on the curve by2 =

(x+a)3 is p(SN) = q (ST)2, then p

q is equal to

(a) a

27b (b)

8a

27b (c)

8b

27b (d)

8b

27

Sol: (d) Here by2 = (x + a)3, differentiating both sides, we get

2by dy

dx= 3 (x + a)2.1 ⇒

dy

dx =

3

2

2(x a)

by

Length of subnormal ⇒ SN = ydy

dx =

3

2

2(x a)

b

… (1)

and length of subtangent ⇒ ST = y dy

dx =

2

2

2by

3(x a) ….(2)

2p (ST)

q (SN) (given)

⇒ p

q =

2

2 2 2

(2by ).2b

{3(x a) } .3(x a) {using (i)and (iii)}




4

= 3 2

6

8b {(x a) }

27 (x a)

{using, by2 = (x+a)3} =

8b

27

p

q =

8b

27

5. With the usual meaning for a, b, c and s if ∆ be the area of a triangle, then the error in ∆, (δ∆) resulting from a small error in the measurement of c (δ c) is given by

(a) δ∆ = c

4s

δ (b) δ∆ =

Δ. c

4s

δ (c) ∆ =

2

(abc)sc

4 s (d) none of these

Sol: (d)

Since ∆ = s(s a)(s b)(s c) .

= {s (s – a) (s – b)(s – c)}1/2 Taking logarithm of both sides, we get

in ∆ = 1

2{In s + In (s – a) + In (s – b) + In (s – c)

1 dΔ 1 1 ds 1 d(s a) 1 d(s b) 1 d(s c)

. . .Δ dc 2 s dc s a dc (s b) dc s c dc

…(1)

But s = 1

2 (a + b + c)

Now ds 1 d(s a) ds da 1

,dc 2 dc dc dc 2

– 0 =

1

2,

d(s b) ds db 1 10

dc dc dc 2 2

and d(s c) ds 1 1

1 1dc dc 2 2

Now from (1),

1

Δ.

Δ 1 1 1 1 1 1 1 1 1. . . .

c 2 s 2 (s a) 2 (s b) 2 (s c) 2

δ

δ

= 1 1 1 1 1

4 s (s a) (s b) (s c)

Hence δ∆ = Δ 1 1 1 1

4 S s a s b s c

δc

6. Let f(x) = In x and g(x) = x2. If c ∈ (4, 5) then c In25

16

4

5

equals to

(a) c In 5 – 8 (b) 2 (c2 In 4 – 8) (c) 2(c2 In 5 – 8) (d) c In 4 – 8 Sol: (b) Let ϕ(x) = x2 In (4) – 16 In x, which is continuous [4, 5] and differentiable on (4, 5), so by LMVT,




5

(4)

5 4

ф(5) ф = Φ ‘(c), c ∈ (4, 5)

Now, Φ (5) – Φ (4) = In 25

16

4

5

and Φ ‘(c) = 2

c (c2 In 4 –8 ) ⇒ c In

25

16

4

5

= 2 (c2 In 4 – 8)

7. When the determinant

2

2 2

2

cos2x sin x cos4x

sin x cos2x cos x

cos4x cos x cos2x

is expanded in power of sin x, then the

constant term in that expansion is

(a) 1 (b) 0 (c) –1 (d) 2

Sol: (c)

f(x)

=

2 2 2 2

2 2 2

2 2 2 2

1 2sin x sin x 1 8sin x(1 sin x)

sin x 1 2sin x 1 sin x

1 8sin x(1 sin x) 1 sin x 1 2sin x

⇒ The required constant term is

f(0) =

1 0 1 1 0 0

0 1 1 0 1 1

1 1 1 1 1 0

= 1(0 – 1) = – 1

8. The product of the matrices

A = 2

2

cos cos sin

cos sin sin

θ θ θ

θ θ θ

and

B = 2

2

cos cos sin

cos sin sin

ф ф ф

ф ф ф is a null matrix if – ϕ =

(a) (2n+1) 2

π (b) n π (c) 2n π (d) n

2

π




6

Sol: (a)

AB = 2

2

cos cos sin

cos sin sin

θ θ θ

θ θ θ

2

2

cos cos sin

cos sin sin

ф ф ф

ф ф ф

=2 2

2 2

cos cos cos cos sin sin

cos cos sin sin sin sin

θ θ θ

θ θ θ

ф ф ф

ф ф ф

2 2

2 2

cos cos sin sin sin cos

cos cos sin sin sin sin

θ θ θ

θ θ θ

ф ф ф

ф ф ф

cos cos (cos cos sin sin )

sin cos (cos cos sin sin )

θ θ θ

θ θ

ф ф ф

ф ф+ ф ф

cos sin (cos cos sin sin )

sin cos (cos cos sin sin )

θ θ θ

θ θ θ

ф ф ф

ф ф ф

= cos cos cos( ) cos sin cos( )

sin cos cos( ) sin sin cos( )

θ θ θ θ

θ θ θ θ

ф ф ф ф

ф ф ф ф

Clearly AB is the zero matrix if cos ( – Φ) = 0 i.e.

– Φ is an odd multiple of 2

π.

Section – II (Total Marks: 16)

(Multiple Correct Answer (s) Type)

This section contains 4 multiple choice questions. Each question has four choices (a), (b), (c), and

(d) out of which ONE or MORE may be correct.

9. Let f : D R be defined by f (x) = In (In (In (In x))) then (a) f (x) is into (b) f (x) is one – one (c) f (x) is onto (d) D = (ee , )

Sol: (b, c, d) For f (x) to be real x > 0, In x > 0, In (In x) > 0 and In (In x)) >0 ⇒ x > 0, x > 1, x > e and x > ee ⇒ D = (ee, ) Clearly Range of f (x) = R ⇒ f (x) is onto

Also, f’ (x) = e10 if x e

x In(x)In(Inx)

f (x) in one–one in its domain.




7

10. The function f (x) = 2

2

1x

x

, x ≠ 0 is ([x] represent the greatest integer ≤ x )

(a) continuous at x = 1 (b) discontinuous at x = – 1 (c) Discontinuous at infinitely many points (d) Continuous everywhere

Sol: (b, c) We have f (1) – f(–1) = 1

Let x >1 then 0 < 2

1

x < 1

⇒ 2

1

x

= 0 ⇒ f(x) = 0 ∀ x >1

Also, if x < – 1, then x2 > 1 ⇒ 0 < 2

1

x <1

⇒ f (x) = 0 ∀ x < – 1 Hence f (1) = 1, f (–1) = 1 and f (x) = 0 If |x| > 1 f (x) cannot be continuous at x = 1 and x = – 1

Again Let 1

2 < x2<1 ⇒ 1 <

2

1

x < 2

⇒ 2

1

x

= 1 ⇒ 1

2 < x2

2

1

x

< 1

2

2

1x

x

= 0

⇒ f (x) = 0 if x ϵ 1 1

1, ,12 2

Next, Let 1

3< x2 <

1

2 ⇒ 2 <

2

1

x < 3

⇒ 2

1

x

= 2 ⇒ 2

3 < x2

2

1

x

< 1

2

2

1x

x

= 0 ⇒ f (x) = 0 if

x ϵ 1 1 1 1

, ,2 3 3 2

At x = ± 1

2, x2 =

1

2.

2

1

x = 2 ⇒ f(x) = 1

Similarly at x = 1 1 1

, ,23 5

…….




8

f (x) is discontinuous at infinite number of points

given by x ϵ 1

, n Nn

.

11. If F (x) = f (x) g(x) and f’ (x) g’(x) = c, then

(a) F’ = c f g

f ' g'

(b)

" "F" f g 2c

F f g fg (c)

F''' f ''' g'''

F f g (d)

F''' f ''' g'''

F" f " g"

Sol: (a, b, c)

Given F(x) = f(x). g(x) …. . (1)

Differentiating both sides w.r.t., x we get

F’ (x) = f’ (x). g(x) + g’(x).f’(x)

⇒ F’(x) = f’ (x) g’(x) f(x) g(x)

f '(x) g'(x)

⇒ F’ = c f g

f ' g'

⇒ (a) is correct

Again differentiating both sides w.r.t., x we get

F” (x) = f”(x).g(x)+g”(x).f(x)+2f’(x).g’(x)

⇒ F”(x) = f”(x).g(x)+g”(x). f(x) + 2c …(2)

Dividing both sides by F(x) = f(x) . g(x)

{∵ f’(x).g’(x) = c}

then F"(x) f "(x) g"(x) 2c

F(x) f(x) g(x) f(x)g(x)

or F" f " g" 2c

f f g fg ⇒ (b) is correct.

Again given f’(x) g’(x)= c


f’(x)g”(x) + g’(x) f”(x) = 0

From (2), F” (x)= f”(x).g(x) + g”(x).f(x)+2c


F”(x) = f”(x). g’(x) + f’’’(x) .g(x)+g”(x).f’(x)+f(x).g’’’(x)+0

=f’’’(x).g(x)+g’’’(x).f(x)+(0) [from (3)]




9

Now dividing both sides by F(x) =f(x) g(x)

Then F'''(x) f '''(x) g"'(x)

F(x) f(x) (x)

or F"' f "' g"'

F f g

12. Let f (n) = n n 1 n 2n n 1 n 2

n n 1 n 2n n 1 n 2

n n 1 n 2

P P p

C c C

where the symbols have their usual meanings. The f(n) is

divisible by

(a) n2 + n + 1 (b) (n + 1) ! (c) n! (d) none of these

Sol: (a, c)

f(n) =

n n 1 n 2

n! (n 1)! (n 2)!

1 1 1

= 3 3 2

2 2 1

C C C

C C C

n 1 1

n! nn! (n 1)(n 1)! using

1 0 0

= (n+1)(n+1)!–nn! = n! = n![(n+1)2 – n] = n! (n2 + n + 1) Thus, f(n) is divisible by n! and n2+ n+ 1.




10

Section – III (Total Marks: 24) (Integer Answer Type)

This section contains 6 questions. The answer to each of the questions is a single–digit integer,

ranging from 0 to 9.The bubble corresponding to the correct answer is to be darkened in the

Answer sheet.

13. Let f x y f(x) f (y)

2 2

∀ x, y ∈ R

If f’ (0) exists and equals – 1 and f(0) = 1, then the value of f (–1) is equal to Sol: (2) Put y = 0 in the given relation, then

fx f(x) f(0)

2 2

2f x

2

= f (x) + 1 ⇒ f (2x) = 2f(x) –1 …(1)

Now,

f’(x) h 0 h 0

f(x h) f(x) 1 2(x h)lim lim f f(x)

h h 2

h 0lim h 0

1 f(2x) f(2h) 2f(x) f(2h) 1lim

h 2 2h

[from (1)]

h 0

f(2h) f(0)lim

2h

= f’(0) = –1

f(x) = –x + c. Put x = 0 ⇒ 1 = 0 + c ⇒ c = 0 f(x) = –x + 1 ⇒ f(–1) = 2

14. If y3 – y = 2x then y

12

2

2

1 d y dyx x

27 dx dx

is equal to

Sol: (9) Given y3 – y = 2x Differentiate both sides with respect to x, we get

(3y2 – 1) dy

dx = 2 ⇒

2

dy 2

dx (3y 1)

…(1)

Again differentiating both sides with respect to x, we get

2

2 2 2

dy2.6y

d y dxdx (3y 1)

Using (1) we get




11

2

2 2 3

d y 24y

dx (3y 1)

…(2)

Now, 2 1x

27

2

2

d y dyx

dx dx

= 2

2 3 2

1 24 2xx

27 (3y 1) (3y 1)

[From (1) and (2)]

= 2 2 2 2

2 3 2

y (y 1) 1 24y y(y 1)

4 27 (3y 1) (3y 1)

(∵ y3 – y = 2x)

= 2 2 2 2

2 3 2

{27y (y 1) 4} ( 24y) y(y 1)

108 (3y 1) (3y 1)

= 2 2 2 2

2 3 2

y 54y (y 1) 8 9(y 1)

9 (3y 1) (3y 1)

= 2

3

y 2(1 )( 2) 8 3( 2) y

9 9

α α α

α α

( = 3y2 –1)

15. If the acute angles between the curves y = |x2 – 1| and y = |x2– 3| at their points of

intersection be such that tan = m

7 then m2 –30 is equal to

Sol: (2) Given curves are y = |x2 – 1| …(1) y = |x2 – 3| …(2)

y = 2

2

x 1, x 1or 1

1 x , 1 x 1

…. (3)

and y = 2

2

x 3, x 3or x 3

3 x , 3 x 3

…(4)

Equating the two values of y from (1) and (2) we get |x2 – 1| = |x2 – 3|

or x2 – 1 = ± (x2 – 3) ⇒ x = ± 2

From (1), when x = ± 2 , y = 1

Let A ≡ ( 2 , 1) and B ≡ (– 2 , 1) Here A and B are the points of intersection of curves (1) and(2)

Angle of intersection between curves (1) and (2) at A ( 2 , 1):

From (3), l

at( 2 ,1)

dy(2x 0) 2 2 m

dx

(say)




12

From (4), 2

at( 2 ,1)

dy( 2x) 2 2 m

dx

(say)

Let be the acute angle between curves (1) and (2) at A, then

tan = 1 2

1 2

m m 2 2 2 2 4 2

1 m m 1 8 7

⇒ m = 4 2

16. If f (x) = cos–1 1

13 (2 cos x – 3 sin x) + sin–1

1

13 × (2cos x + 3 sin x) w.r.t. 21 x

then find 3 df(x)

7dx

Sol: (3)

f (x) = cos–1 1

13 (2 cos x – 3 sin x) + sin–1

1

13 (2cos x + 3 sin x)

= cos–1 11 313cos x tan

213

+ sin–1 11 2

13sin x tan313

= cos–1 1 1 13 2cos x tan sin sin x tan

2 3

⇒ f’ (x) =

1

2 1

3sin x tan

2

31 cos x tan

2

+

1 1 1

1 12 1

2 3 2cos x tan sin x tan cos x tan

3 2 3

3 22 sin x tan cos x tan1 sin x tan2 33

⇒ f’ (3/4) = 1 + 1 = 2

Now let g(x) = 21 x ⇒ g’(x) = 2

x

1 x

⇒ g’ (3/4) = 3/5 ⇒ f’ (3/4)/g ‘(3/4) = 10/3.




13

17. If there is an error of k% in measuring the edge of a cube, then the percent error in

estimating its volume nk then what is value of n?

(a) k (b) 3 (c) k

3 (d) None of these

Sol: (3)

V = x3 and the percent error in measuring x = dx

x × 100 = k

The percent error in measuring volume = dV

V × 100

Now, dV

x = 3x2

⇒ dV = 3x2 dx ⇒ dV

V =

2

3

3x dx dx3

x x

dV

V × 100 = 3

dx

x × 100 = 3k

18. Given a matrix A =

a b c

b c a

c a b

where a, b, c are real positive numbers, abc =1 and ATA = I,

then find the value of a3 + b3 + c3. Sol: ATA = I

⇒

a b c

b c a

c a b

a b c

b c a

c a b

=

1 0 0

0 1 0

0 0 1

⇒

2 2 2

2 2 2

2 2 2

a b c ab bc ca ab bc ca

ab bc ca a b c ab bc ca

ab bc ca ab bc ca a b c

=

1 0 0

0 1 0

0 0 1

⇒ a2 + b2 + c2 = 1 (1) and




14

ab + bc + ca = 0 (2) Now, (a3 + b3 + c3) = (a + b + c)(a2 + b2+ c2 – ab – bc – ca) + 3 abc = (a + b + c) + 3 (3) Now, (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) = 1 + 2 × 0 = 1 ⇒ a + b + c = 1 (Since a, b, c are real positive numbers) Now from Eq. (3), a3 + b3 + c3 = 1 + 3 = 4 Alternative solution: ATA = I ⇒ ATA | = |I| ⇒ |AI2 = 1 ⇒ (a3 + b3 + c3 – 3abc)2 = 1 ⇒ a3 + b3 + c3 – 3abc = 1 (since a, b, c are positive real number) ⇒ a3 + b3 + c3 ≥ 3abc (∵ A.M. ≥ G.M.) ⇒ a3 + b3 + c3 = 4

Section – IV (Total Marks: 12) (Matrix–Match Type)

This section contains 2 questions. Each question has three statements (a, b and c) given in Column I and five statements (p, q, r, s and t) in Column II. Any given statements 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 the particular question, against statement B, darken the bubbles corresponding to q and r in the ANSWER SHEET.

19. Observe the following columns :

Column – I Column – II

(a) 1

x 0

tan xlim

x

,

where [.] denotes the greatest integer function, is equal to

p. 1

(b) If the 3x 0

sin2x a sin xlim

x

be a

finite number then a can be equal to

q. 0

(c) If f (x) = g (x) 1/x 1/x

1/x 1/x

e e

e e

,

x ≠ 0, where g(x) = xnh (x),

h (x) being a continuous function, then n

r. 1




15

can be equal to (d) Let f (x) = –1 + |x – 1|, g (x) = 2 – |x + 1|, then go f (x) is continuous if x is equal to

s. 2

t.

3

2

Sol: A–q; B–s; C–r,s,t,;D–p,q,r,s,t)

(A) Comparing the graphs of y = x and y = tan–1x, we get that if x > 0, x > tan–1 x > 0 ⇒ 0 < 1tan x

x

< 1

Also, if x < 0 then x < tan–1 x < 0 ⇒ 0 1tan x

x

< 1.

So, 1

x 0

tan xlim

x

= 0

(B) 3 2x 0 x 0

sin2x asin x 2cos2x acosxlim lim

x 3x

In order that the limit exits, 2 – a = 0 ⇒ a = 2.

then limit =

x 0 x 0

4sin2x asin x 8cos2x acosxlim lim

6x 6

which is a finite number (= –1) if a = 2.

(c) ∵ 1/x 1/x 2/x

1x 1/x 2/xx 0 x 0

e e 1 elim lim

e e 1 e

= and

1/x 1/x 2/x

1/x 1/x 2/xx 0 x 0

e e e 1lim lim

e e e 1

= –1

Clearly 1/x 1/x

1/x 1/xx 0

e elimg(x)

e e

can exist if and only if

x 0lim

g(x) = 0 ⇒ n ≥ 1




16

(D) f (x) = x 2 if x 1

x if x 1

, g (x) = 1 x, if x 1

x 3, if x 1

g(f(x)) = 1 f(x) if f(x) 1

f(x) 3 if f(x) 1

1 (x 2) if x 2 1 and x 1

1 ( x) if x 1 and x 1

x 2 3 if x 2 1 and x 1

x 3 if x 1 and x 1

⇒ g (f(x)) = 3 x if x 1

1 x if x 1

Clearly g(f(x)) is continuous everywhere.

20. Observe the following columns :

Column – I Column – II (a) The normal line to y = be–x/a where it crosses y–axis, has slope equal to p.

2

2

a

b

(b) Subnormal length to x y = a2b2 at

any point (x, y) is p then 1

p|y3| is equal

to

q. a

b

(c) The length of sub–tangent at any

point (x, y) on the ellipse 2 2

2 2

x y

a b = 1 is

p then2

p| x|

y is equal to

r. a2 b2

(d) If m be slope of tangent at any point

(x, y) on the curve 2 2

2 2

x y

a b = 1 then

my

x

is equal to

s. 2

2

b

a




17

Sol: A – q; B – r; C – p; D – s

(A) Point of intersection (0, b), dy

dx = be–x/a

1;

a

m = (0,b)

dy b

dx a

⇒ Slope of normal = a

b

(B) d y

,x x

Subnormal = 2 2 3

2 2 2 2

dy . y y y | y |y y.

a bdx x x a b

y

(c) m = 2

2

dy xb

dx ya ;

Length of subtangent = 2 2

2 2

2

y y y a

dy xb | x | bdx ya

(D) 2 2

2 2

x y

a b = 1 ⇒

2 2

2x 2y dy

a b dx = 0

⇒ 2

2

dy b x

dx a y

iit–jee - pioneermathematics.com 2012... · 2011-10-05 · iit–jee (2012) (calculus) “towards...

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