OBILE IIT–JEE - PioneerMathematics.Com 2012 calculus...PAPER –A IIT–JEE (2012) (Calculus)...

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PAPER –A

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: 21), 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 the section.

In Section III (Total Marks : 15), 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 IV (Total Marks: 28), 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

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: 21)

(Single correct answer choice)

This section contains 7 multiple choice questions. Each question has 4 choices.(a), (b), (c)

and (d), out of which ONLY ONE is correct.

1. If [x] denotes the integral part of x, then the domain of f(x) = cos–1 (x + [x])is

(a) (0, 1) (b) [0,1) (c) [0, 1] (d) [–1, 1]

Sol: (b)

For f (x) to be defined, –1 ≤ x + [x] ≤ 1 …(i)

When x ∈ I, Let x = k, then from (i), –1 ≤ 2k ≤ 1

⇒ – 1

2 ≤ k ≤

1

2 ⇒ k = 0 ⇒ x = 0.

When x ∉ I, let x = k + f, where k is the integral part of x and 0 < f < 1.

From (i), –1 ≤ 2k + f ≤ 1

⇒ 1 f 1 f

k2 2

. ⇒ k = 0 ⇒ 0 < x < 1

1 f 1 f 11 and

2 2 2

So, all possible values of x are given by 0 ≤ x < 1.

Domain of f = [0, 1)

2. The value of x 0

100x 99sinxlim

sinx x

where [.] represent greatest integral function, is

(a) 199 (b) 198 (c) 0 (d) none of these

Sol: (b)

From the graph we note that

0 < sin x < x and x < sin x <0

⇒ 0 < sin x

x < 1 and

x

sinx >1



3

Thus x 0

sinxlim

x

= 1– and x 0

xlim

sinx

= 1+

x 0

xlim 100

sinx

= 100 and x 0

sinxlim 99

x

= 98

Desired limit = 100 + 98

3. Suppose the function f satisfies the equation f(x + y) = f(x) f(y) for all x and y and f(x) = 1

+ xg (x) where x 0

Lim

g(x) = log a If fn (x) = Kf (x), then K =

(a) log a (b) n log a (c) (log a )n (d) n(log a)n

Sol: (c)

f’ (x) = h 0

f(x h) f(x)Lim

h

= h 0 h 0

f(x){f(h) 1} f(x) (1 hg(h) 1)Lim Lim

h h

= h 0

Lim

f (x) g (h) = log a f (x)

Hence f ” (x) = (log a) (f’ (x)) = (log a)2 f (x)

Thus f n (x) = (log a)n f (x) so K = (log a)n

4. The tangent at any point on the curve x = a cos3 , y = a sin3 meets the axes in P and Q.

The locus of the mid point of PQ is

(a) x3/2 + y3/2 = a3/2 (b) x2/3 + y2/3 = a2/3 (c) 4(x + y) =a (d) 4(x2 + y2) =a2



4

Sol: (d)

We have, 2

2

dydy 3asin cosd

dxdx 3acos ( sin )d

θ θθθ θ

θ

= – tan

Equation of tangent at ‘’ is y – a sin3 = – tan (x – acos3 )

⇒ x y

acos a sinθ θ = 1

P is (a cos , 0) and Q (0, a sin ). If mid point of PQ is (h, k), then

h = acos asin

, k2 2

θ θ Eliminating , we get h2 + k2 =

2a

4

focus is x2 + y2 = 2a

4

5. x and y are the sides of two squares such that y = x – x2. The rate of change of the area

of the second square with respect to the area of the first square is

(a) x2 – x + 1 (b) 2x2 + 2x – 1 (c) 2x2 – 3x + 1 (d) x2 + x – 1

Sol: (c)

Given x and y are sides of two squares thus the area of two squares are x2 and y2

We have to obtain 2

2

dy2y

d(y ) y dydx .d(x ) 2x x dx

… (i)

where the given curve is, y = x – x2

⇒ dy

dx = 1 – 2x … (ii)

Thus 2

2

d(y ) y

d(x ) x (1 – 2x) [from (i) and (ii)]

or 2 2

2

d(y ) (x x )(1 2x)

d(x ) x

= 2x2 – 3x + 1

So, the rate of change of the area of second square with reject to first square is (2x2 – 3x + 1).



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6. If f’and g’ exist for all x ∈ [a, b] and if g’ (x) ≠ 0 ∀ x ∈(a, b) then for some c ∈ (a, b),

f(c) f(a)

g(b) g(c)

=

(a) f’(c)g’(c) (b) f '(c)

g'(c) (c) f’(c) –

1

g'(c) (d)

f(a)f '(c)

g(b)g'(c)

Sol: (b)

Let h(x) = {f(x)} – f(a)} {g(b) – g(x)}

Then h(x) is continuous and differentiable and h(a) = h(b) = 0

Therefore h (x)satisfies the conditions of Rolle’s MVT.

Therefore for some c ∈ (a, b), f’(c) = 0

⇒ f’ (c) {g (b) – g (c)} – g’ (c) {f (c) – f(a)} = 0

⇒ f(c) f(a)

g(b) g(c)

=

f '(c)

g'(c)

7. Let 2

x 2 x

x x 6

x x 6

= Ax4 + Bx3 + Cx2 + Dx + E. Then the Value of 5A + 4B + 3C + 2D + E is

equal to

(a) zero (b) –16 (c) 16 (d) –17

Sol: (d)

Let given determinant is equal to ∆ (x).

Then 5A + 4B + 3C + 2D + E = ∆ (1) + ∆’(1)

Now, ∆(1) = 0 as R2 and R3 are identical

∆’ (x) = 2 2

1 0 1 x 2 x x 2 x

x x 6 2x 1 0 x x 6

x x 6 x x 6 1 1 6

∆’ (1) =

1 2 1

2 1 0

1 1 6

= 1 (6 – 0) – 2 (12 – 0) + 1 (2 – 1) = – 17



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Section – II (Total Marks: 16)

(Multiple Correct Answers 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.

8. Let n be a positive integer with f(n) =1! + 2 ! +3 ! + … + n! and P(x) and Q(x) be

polynomial in x such that f (n+2) = P (n) f(n+1)+Q(n)f(n) for all ≥ 1, then

(a) P(x) = x + 3 (b) Q(x) = – x – 2 (c) P(x) = – x – 2 (d) Q(x) = x + 3

Sol: (a, b)

we have f (n + 2) – f (n +1) = (n + 2) !

= (n + 2) (n + 1)! = (n + 2) [f(n + 1) – f(n)]

⇒ f (n + 2) = (n + 3) f (n + 1) – (n + 2) f (n)

P(x) = x + 3 and Q (x) = – x – 2

9. If f(x) = [x] + x [x] , where [.] denotes the greatest integer function then

(a) f is continuous for all x ∈ R

(b) f if differentiable for all x ∈ R

(c) f is not continuous at integral points

(d) f is differentiable for all x ∈ R – I, (I is the set of integers)

Sol: (a, d)

f(x) = [x] + x [x]

Let n < x <n + 1, where n ∈ I then [x] = n and hence f (x) = n + x n .

Clearly f (x) is continuous and differentiable ∀ x ∈ (n, n + 1)

Now f(n) = n

x nlim

f (x) =

x nlim

(n – 1 + x n 1 ) = n

x nlim

f (x) =

x nlim

(n + x n ) = n

f (x) is continuous at x = n ∈ I

⇒ f (x) is continuous for all x ∈ R



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Again f’ (n–0) = x nlim

n 1 x n 1 n

x n

= x nlim

(x n 1) 1 1

(x n)(x n 1 1) 2

f ’ (n + 0) = x nlim

n x n n

x n

= x nlim

1

x n= +

f (x) is not differentiable at x = n ∈ 1

10. If f(x) =

cos(x ) cos(x ) cos(x )

sin(x ) sin(x ) sin(x )

sin( ) sin( ) sin( )

α β γ

α β γ

β γ γ α α β

then

(Given ≠ ≠ )

(a) 2f() = f() + f()

(b) 2f() + f () = 3f()

(c) f’ () = 0

(d) f () = 0

Sol: (a, b, c)

f ’ (x) =




α β γ

α β γ

β γ γ α α β



8

+




α β γ

α β γ

β γ γ α α β

= 0 + 0 ⇒ f (x) is a constant function.

Thus f () = f () = f ()

11. If 1 tan 1 tan a b

tan 1 tan 1 b a

θ θ

θ θ

then

(a) a = cos 2 (b) a = 1 (c) b = sin 2 (d) b = 1

Sol: (a, c)

11 tan 1 tan

tan 1 tan 1

θ θ

θ θ

= 2

1 tan 1 tan1

1 tantan 1 tan 1

θ θ

θθ θ

= 2

2 2

1 tan 2tan1

1 tan 2tan 1 tan

θ θ

θ θ θ

= cos2 sin2

sin2 cos2

θ θ

θ θ

a = cos 2, b = sin 2

Section – III (Total Marks: 15)

(Paragraph Type)

This section contains two paragraphs. Based upon one of the paragraph, 2 multiple choice questions and

based on the other paragraph 3 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.

Passage for questions 12 to 13.

Read the paragraph carefully and answer the following questions:

A determinant is called cyclic if it follows the arrangement symmetrically with a, b, c for



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example 2 2 2

1 1 1

a b c

a b c

is a cyclic determinant

Now, if we increase the degree of any row in this determinant symmetrically its value will be

multiplied by expression which is also cyclic and increases the degree of the value of

determinant.

On the basis of above information, answer the following questions:

12. The value of 2 2 2

1 1 1

a b c

bc ca ab

is equal to

(a) (a – b) (b – c) (c – a) (b) (a – b) (b – c) (c – a) (a + b + c )

(c) (a – b) (b – c) (c – a) (ab + bc + ca) (d) (a – b) (b – c) (c – a) abc

Sol: (b)

∆1 = 2 2 2

1 1 1

a b c

bc ca ab

= 3 3 3

a b c1

a b cabc

abc abc abc

= 3 3 3

a b c

a b c

1 1 1

3 3 3

1 1 1

a b c

a b c

∵ ∆1 is cyclic and obtained by increasing the degree of third row of ∆ by unity (1), of the

value of ∆ will be multiplied by a linear cyclic expression, i.e. (a + b + c), so, ∆1 = ∆ (a + b

+ c)

13. The value of 2 2 2

a b c

a b c

bc ca ab

is equal to

(a) (a – b) (b – c) (c – a) (a + b + c)2

(b) (a – b) (b – c) (c – a) (a + b + c)

(c) (a – b) (b – c) (c – a) (ab + bc + ca)

(d) (a – b) (b – c) (c – a) (a2 + b2 + c2 + ab + bc + ca)



10

Sol: (c)

∆2 = 2 2 2

a b c

a b c

bc ca ab

= 2 2 2

3 3 3

1 1 1

a b c

a b c

Now ∆2 is obtained by increasing the degree of second row and third row by unity each, so

the value of ∆ will be multiplied by a quadratic cyclic expression, say

{k1 (a2 + b2 + c2) + k2 (bc + ca + ab)}

So, ∆2 = ∆ [k1 (a2 + b2+ c2) + k2 (bc + ca + ab) ]

The values of k1 and k2 can be obtained by substituting suitable unequal values of k1 and k2.

For example.

Put a = 0, b = 1, c = –1 then we get 2k1 – k2 = – 1

Put a = 0, b = 1, c = 2 then we get 5k1 + 2k2 = 2

Thus k1 = 0 and k2 = 1

Thus ∆2 = ∆ (bc + ca + ab)

Passage for questions from 14 to 16.

Let f(x) be a real valued function satisfying the functional equation f(x) + f(1 – x) = k for all x

∈ Q, where k is a contant quantity. Let m be a positive integer. Put x = r

m 1 in the given

equation, we get f r m 1 r

f km 1 m 1

m m

r 1 r 1

r m 1 rf f

m 1 m 1

= mk

m m

r 1 t 1

r tf f

m 1 m 1

= mk (putting m+ 1– r = t)

m m

r 1 r 1

r r mk2 f mk f

m 1 m 1 2



11

14. If f (x) = x

x

4

4 2, where x ∈ Q, then

f1 2 2008

f ... f2009 2009 2009

equals to

(a) 1004 (b) 2008 (c) 2009 (d) none of these

Sol: (a)

If f (x) = x

x

4

4 2, where x ∈ Q

then f (x) + f (1 – x) = x 1 x

x 1 x

4 4

4 2 4 2

= x x

x x x x

4 4 4 4

4 2 4 2.4 4 2 2 4

= 1 = k

Now, f

1 2 2008f ...... f

2009 2009 2009

= 2008

r 1

r 2008 1f

2009 1 2

= 1004

15. If f (x) = x 3

1 x x

3

3 3

for all ∈ Q, then the value of the sum

f1 2 54

f ... f55 55 55

is

(a) 1 (b) 27 (c) 54 (d) 55

Sol: (a)

f (x) + f (1 – x) = x 3 x 2

1 x x x 1 x

3 3

3 3 3 3

= 3–3 x 1 x

3

1 x x

3 33

3 3



12

Thus, f 1 2 54

f ....... f55 55 55

= 354

r 1

r 54.3

54 1 2

= 1

16. If f(x) = x

x

a

a a (a > 0), then

2n 1

r 1

r2f

2n

is equal to

(a) 1 (b) 2n (c) 2n – 1 (d) (2n 1)a

2

Sol: (c)

f(x) = x

x

a

a a

f (x) + f (1 – x) = x 1 x

x 1 x

a a

a a a a

= x

x

a

a a +

x

a

a a a

= x

x

a

a a +

x

a

a a = 1 = k

So, 2n 1 2n 1

r 1 r 1

r r2f 2f

2n 2n 1 1

= (2n – 1). 1 = 2n – 1

Section IV (Total Marks: 28)

(Integer Answer Type)

This section contains 7 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.

17. Let f: [0, 1] [0, 1] defined by f(x) = 1 x

1 x

, for 0 ≤ x ≤ 1 and leg g : [0, 1] [0, 1]

defined by g (x) = 4x (1 – x), 0 ≤ x ≤ 1. If range of fog (x) is [, ], then + =



13

Sol: (1)

fog (x) = f(g(x)) = f (4x(1–x))

⇒ 1 4x (1 x)

1 4x (1 x)

when 0 ≤ 4x (1 – x) ≤ 1 and 0 ≤ x ≤ 1

But 4x – 4x2 ≥ 0 ⇒ 0 ≤ x ≤ 1

4x – 4x2 ≤ ⇒ (2x – 1)2 ≥ 0 ⇒ x ∈ R

Hence fog (x) = 2

2

1 4x 4x

1 4x 4x

, 0 ≤ x ≤ 1

Let y = 2

2

4x 4x 1

(4x 4x) 1

, 0 ≤ x ≤ 1

Put 4x2 – 4x = t t ∈ [–1, 0]

y = 2

1 t dy 1 t 1 t,

1 t dt (1 t)

> 0

Range of fog (x) = [0, 1]

⇒ + = 1

18. Let (tan ) x + (sin ) y and ( cosec ) x + cos y =1 be two variable straight

lines, being the parameter. Let P be the point of intersection of the lines. If the coordinates

of P in the limiting position when 0 be (h, k) then h – k is equal to

Sol: 3

Here two straight line, (tan ) x + (sin )y = and ( cosec ) x + (cos ) y = 1 have their

point of intersection as,

x = cos sin

sin

α α α

α α

y =

x tan

sin

α α

α

when 0, we obtain the point P.

i.e. 0 0

cos sin 0limx lim form

sin 0α α

α α α

α α

= 0

sin cos coslim

cos 1α

α α α α

α

(applying L– Hospital’s rule)



14

= 20 0 2

2sin cossin 2 2

lim lim2sin / 2 2sin

2

α α

α αα

α α

αα

= 0 0

22lim lim 2

tan /2 tan2

α α

αα

αα

Again, 0

limα

y = 0

limα

0

x tan xlim

sin sin cosα

α α α

α α α

0 0

xlim lim

sin cosα α

α

α α = 1 – 2 = –1 [ ∵

0limα

x = 2]

⇒ 0

limα

y = – 1

Hence, in limiting position P(2, –1) ⇒ h – k = 2 + 1 = 3

19. If the tangent at (a, b) to the curve x3 + y3 = c3 meets the curve again at (a1, b1), then

1 1a b

a b + 1 is equal to

Sol:

Given curve is x3 + y3 = c3

2 2

(a ,b)2 2

dy x dy a

dx y dx b …. (1)

and tangent at (a, b) cuts the curve again at (a1, b1)

slope of tangent = 1

1

b b

a a

= –

2

2

a

b{From (1)} ….(2)

Also a3 + b3 = c3 … (3)

and 3 31 1a b = c3 … (4)

Subtracting (4) from (3), we get (a3 – 31a ) + (b3 – 3

1b ) = 0

⇒ (a – a1) (a2 + aa1 + 21a ) + (b – b1) (b2 + bb1 + 2

1b ) = 0



15

⇒ 2 2

1 1 12 2

1 1 1

b b a aa a

a a b bb b

… (5)

From (2) and (5), 2 2 2

1 12 2 2

1 1

a a aa a

b b bb b

⇒ a2b2 + a2bb1 +a2 21b = a2b2 + ab2 a1 + 2

1a b2

⇒ ab (ab1 – ba1) + a2 21b – 2

1a b2 = 0

⇒ ab (ab1 – ba1) + (ab1 + a1b) (ab1 –a1b) = 0

⇒ (ab1 – a1b) (ab + ab1 + a1b) = 0

If ab1 – a1b = 0 then 3 33

3 331 1 1 1

a b a b c

a b ca b

= 1

(Law of proportion)

a = a1 and b = b1 which is impossible.

ab1 – a1b ≠ 0

Hence, ab + ab1 + a1b = 0 or 1 1a b

a b = –1

20. Let f x x 1 .

2x + 1

12x

2x ........

Compute the value of f(50). f’ (50)–50 = ?

Sol: 0

1 1

f x x x1 x f (x)x x

12x

2x ........

⇒ f(x) – x = 1

x f(x)

⇒ f2 (x) – x2 = 1

differentiating w.r.t.x



16

⇒ 2f (x). f ’ (x) – 2x = 0

⇒ f (x) . f ’ (x) = x

⇒ f (50). F ‘ (50) = 50

21. A man is moving away from a tower 41.6 m high at a rate of 2 m/s. If the eye level of

the man is 1.6 m above the ground, then the rate at which the angle of elevation of the top of

the tower changes, when he is at a distance of 30m from the foot of the tower, is –

n

125radian/s. what is the value of n.

Sol: (3)

Let CD be the position of man at any time t. Let BD = x, then EC = x. Let ∠ ACE =

Given, AB = 41.6m, CD = 1.6 m and dx

dt = 2m/s.

AE = AB– EB = AB – CD = 41.6 – 1.6 = 40m

we have to find d

dt

θ when x = 30 m

From ∆ AEC, tan = AE 40

EC x …(1)

Differentiating w.r.t. to t, sec2 2

d 40 dx

dt x dt

θ

⇒ sec2 d

dt

θ =

2

40

x

× 2

⇒ d

dt

θ =

2

80

x

cos2 =

2

80

x

2

2 2

x

x 40

2 2

xcos

x 40θ

⇒ 2 2

d 80

dt x 40

θ

(2)

when x = 30 m, d

dt

θ = –

2 2

80

30 40 = –

4

125 radian/s.



17

22. If a ≠ p, b ≠ q, c ≠ r and

p b c

a q c

a b r

= 0, then find the value of p q r

p a q b r c

.

Sol:

Given,

p b c

a q c

a b r

= 0

Applying R1 R1 – R2 and R2 R2 – R3 reduce the determine into

p a b q 0

0 q b c r

a b r

= 0

⇒ (p – a) (q – b) r + a (b – q) (c – r) –b (p – a) (c – r) = 0

Dividing throughout by (p – a) (q – b) (r – c), we get

r a b

r c p a q b

= 0

⇒ r a b

1 1r c p a q b

= 2

23. Let A = o

o o

α

and (A + 1)50 – 50 A =a b

c d

. Then the value of a + b + c + d is

Sol: (a)

We have, A2 = O, Ak = O, ∀ k ≥ 2

Thus,

(A + I)50 = I + 50A

⇒ (A + I)50 = I + 50A

⇒ a = 1, b = 0, c = 0, d = 1

OBILE IIT–JEE - PioneerMathematics.Com 2012 calculus...PAPER –A IIT–JEE (2012) (Calculus)...

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