OBILE +2 {JEE Main – 2012, 13} Pioneer Education … · 2012-08-26 · It can be easily see that...

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nL.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 URL: WWW.PIONEERMATHEMATICS.COM

+2 {JEE Main – 2012, 13} 26 Aug 2012

Name: __________________________Batch (Day) ________________________ Phone No.____________________

“IT IS NOT ENOUGH TO HAVE A GOOD MIND, THE MAIN THING IS TO USE IT WELL”

Marks: – 240 Time: – 2hr.

[Each right answer carries 4 marks and wrong – 1]

1. Consider a function f (n) defined for all n N. The function satisfies the following two

conditions

(i) f(1) + f(2) + f(3) + …..to 1.

(ii) f(n) = {(1 – p) p–1} {f(n + 1) + f(n + 2) + …. to }, where 0 < p < 1. Then f(2), is equal to

(a) p (1 – p) (b) 1 – p (c) 1 + p (d) none of these

Sol:

We have,

f(n) = {(1 – p) p–1} {f (n+ 1) + f(n + 2) + ………. to∞}

Put n = 1

f(1) = {(1 – p)p–1 } {f (2) + f(3) + …. to ∞}

= {(1 – p) p–1} {1 – f(p)}

= (1 – p) p–1– (1 – p)p–1 f(1).

f (1) {1 + (1 – p)p–1} = (1 – p)p–1

f(1) . p–1 = (1 – p) p–1

f(1) = 1 – p

Put n = 2

f(2) = {(1 – p) p–1} {f (3) + f(4) + …}

f(2) = {(1 – p) p–1} {1 – 1 + p – f(2)}

= ((1 – p)p–1) (p – f(2))

= (1 – p) – (1 – p) p–1 f(2)

f(2) = {1 + (1 – p)p–1 } = 1 – p

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f(2) = p(1 – p).

Hence (a) is the correct answer.

2. If

2

xlim ( x x 1 ax b) 0, then the values of a and b are given by :

(a) a = – 1, b = 1

2 (b) a = 1, b =

1

2 (c) a = 1, b =

1

2 (d) none of these

Sol:

We have, 2

xlim x x 1 ax b 0

2

ylim y y 1 ay b 0

[Putting x = –y; as x – , y ]

1/2

2y

1 1lim y 1 ay b 0

y y

2y

1 1 1lim y 1 ... ay b 0

2 y y

y

1 1lim y(1 a) b ... 0

2 2y

1 + a = 0 and 1

2 – b = 0

a = – 1 and b = 1

2.


3. Let f (a) = g(a) = k and their nth derivatives fn (a), gn(a) exist and are not equal for some n.

Further if

x a

f a g x f a g a f x g alim 4,

g x f x

then the value of k is :

(a) 4 (b) 2 (c) 1 (d) 0

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Sol: x a

f(a) g(x) f(a) g(a) f(x) g(a)lim 4

g(x) f(x)

Applying L’Hospital rule

x a

f(a)g'(x) g(a)f '(x)lim 4

g'(x) f '(x)

x a x a

kg'(x) kf '(x) k[g'(x) f '(x)]lim 4 lim 4

g'(x) f '(x) g'(x) f '(x)

k = 4.


4. If f(x) =

[x] xe 2

, x 0([.] denotes[x] | x| 1

1, x 0

the greatest integer function), then

(a) f(x) is continuous at x = 0 (b) x 0lim f x 1

(c)

x 0lim f x 1

(d) none of these

Sol:

f(x) =

[x] |x|e 2, x 0

[x] | x|

1 , x 0

[x] |x| 1

x 0 x 0

e 2 e 2lim f(x) lim

[x] | x| 1

[x] |x| x

x 0 x 0 x 0

e 2 e 2lim f(x) lim lim

[x] | x| x

Hence (d) is the correct answer.

5. If f(x) = 1

,1 x

then the points of discontinuity of the function f[f{f(x)}] are

(a) {0} (b) {0, 1} (c) {1, – 1} (d) none of these

Sol:

We have, f(x) =1

1 x.

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As at x =1, f (x) is not defined, x = 1 is a point of discontinuity of f(x).

If x ≠ 1, f [f(x)] = f1 1 x 1

.1 x 1 1/(1 x) x

x =0, 1 are points of discontinuity of f[f(x)].

If x ≠ 0, x ≠ 1,

f [f {f(x)}] = f x 1 1

x.(x 1)x 1

x

Hence (b) is the correct answer.

6. The function f(x) = 2tan 2x

sin2x is not defined at x π / 4. The value of f π / 4 so that f is

continuous at x π / 4, is

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

Sol:

f is continues at x = π/4, if x π/4lim

f(x) = f (π/4)

Now, L = x π/4lim

2tan 2x

sin 2x

log L = x π/4lim

tan2 2x log sin 2x

= x π/4lim

2

log sin2x

cot 2x

= x π/4lim

2

2cot 2x 1.

2cot 2x cosec 2x.2 2

L = e–1/2 f(π/4) = e–1/2 = 1 / e


7. If f(x) = cos (log x), then 1 x

f x .f y f f xy2 y

is equal to :

(a) 2 (b) 1 (c) 0 (d) none of these

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Sol: We have,

1 x

f x .f y f f xy2 y

= cos (log x) . cos (log y)

– 1 x

cos log cos{log(xy)}2 y

=cos (log x ). cos (log y) – 1

2{cos(log x – log y) + cos (log x + log y)}

= cos (log x) . cos (log y) – 1

2 . 2 cos (log x) cos (log y)

= cos (log x) . cos (log y) – cos (log x) . cos (log y)

= 0

8. If f(x) = 1

,1 x

then the points of discontinuity of the function f3n(x) is/are where

fn = fof ….of (n times), are

(a) x = 2 (b) x = {0, 1} (c) x = – 1 (d) continuous everywhere

Sol:

Clearly, x = 1 is a point of discontinuity of the function f(x) = 1

1 x.

If x ≠ 1, then (fof) (x) = f[f(x)] = f 1 x 1

1 x x

,

which is discontinuous at x = 0

if x ≠ 0 and x ≠ 1, then

(fofof) (x) = f[(fof) (x)] = fx 1

x.x

which is continuous everywhere.

So, the only points of discontinuity are x = 0 and x = 1.


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9. If f(x) = 2

1,

x 17x 66 then

2f

x 2

is discontinuous at x is equal to

(a) 7 25

2, ,3 11

(b) 8 24

2, ,3 11

(c) 7 24

2, ,3 11

(d) none of these

Sol:

Let u = 2

x 2

f(u) = 1

(u 6)(u 11)

f(u) is undefined when u is undefined.

u = 6, u = 11

x = 2, 2 2

6, 11x 2 x 2

x = 2, 7 24

x , x3 11

Hence (c) is the correct answer.

10. If f(x) =

x, x 1

1 x

x, x 1

1 x

, then f(x) is

(a) discontinuous and non–differentiable at x = –1, 1 and 0

(b) discontinuous and non–differentiable at x = – 1, where as continuous and differentiable

x = 0 and x = 1

(c) discontinuous and non–differentiable at x = – 1 and x = 1, whereas continuous and

differentiable at x = 0

(d) none of these

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Sol:

we have, f(x) =

x, x 1

1 x

x, 1 x 0

1 x

x, 0 x 1

1 x

x, x 1

1 x

It can be easily see that f(x) is discontinuous and so non–differentiable at x = –1 and x = 1,

whereas it is continuous and differentiable at x = 0.


11. The left hand derivative of f(x) = [x] sin πx at x = k, an integer and [x] = greatest

integer x, is

(a) k

1 k 1 π (b) k 1

1 . k 1 π

(c) k

1 .k π (d) k 1

1 .k π

Sol:

f’ (k – 0) = h 0

[k h]sin π(k h) [k]sin πklim

h

= k 1

h 0

( 1) (k 1)sin πh k 0lim

h

k 1

h 0

( 1) .(k 1)sin π hlim

h

= (–1)k . (k – 1) π


12. Range of f(x) = [1 + sin x] + x x x

2 sin 3 sin .... n sin x 0, π ,2 3 n

where

[.] denotes the greatest integer function, is

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(a) 2 n n 1n n 2

,2 2

(b) n n 1

2

(c) 2 2n n 1n n 2 n n 2

, ,2 2 2

(d) 2n n 1 n n 2

,2 2

Sol:

n(n 1) x xf(x) [sin x] sin .... sin

2 2 n

This range of (x) =

n(n 1) n(n 1), 1 asx [0, π].

2 2

Hence (d) is correct answer.

13. The period of the function f(x) = cos2πx πx

sin5 4

is :

(a) 5 (b) 8 (c) 12 (d) 40

Sol:

Since cos 2πx

5 is a periodic function with period

2π5

2π /5 and sin

π x

4 is a periodic function

with period 2π

π / 4 = 8.

f (x) is a periodic with period = L.C.M. (5, 8) = 40

14. If f(x) = 22x x x , 1 x 1, then f(x) is

(a) continuous but not differentiable in [–1, 1]

(b) continuous as well as differentiable in [–1, 1]

(c) differentiable but not continuous in [–1, 1]

(d) neither differentiable nor continuous in [–1, 1]

Sol:

we have f(x) = 2x + |x – x2|, –1 x 1

= 2x + |x||1–x|,–1x1

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=

2x ( x)(1 x) 1 x 0

2x x(1 x), 0 x 1

= 2

2

x x , 1 x 0

3x x , 0 x 1

Since the polynomial functions are continuous as well as differentiable everywhere, so the

only doubtful point is the point x = 0.

L f’(0) =

2

h 0 h 0

f(0 h) f(0) ( h h ) 0lim lim

h h

= h 0lim

(1 – h) = 1.

Rf’ (0) = 2

h 0 h 0

f(0 h) f(0) (3h h ) 0lim lim

h h

= h 0lim

(3 – h) = 3.

Since Lf’(0) ≠ Rf’ (0), therefore f(x) is not differentiable at x = 0 but f(x) is continuous at x

= 0 (as Lf’ (0) and Rf’ (0) are finite).

Hence f (x) is continuous but not differentiable in the interval [–1, 1].


15. If f(x) = 3

1

2

3x xcot

1 3x

and g(x) =

21

2

1 xcos ,

1 x

then

x a

f x f a 1lim , 0 a

g x g a 2

is :

(a) 2

3

2 1 a (b)

2

3

2 1 x (c)

3

2 (d)

3

2

Sol: (d)

16. Let f: R R, g : R R be two functions given by f(x) = 2x – 3, g(x) = x3 + 5. Then

(fog)–1 (x) is equal to :

(a) 1/3

x 7

2

(b) 1/2

x 7

2

(c) 1/3

7x

2

(d)

1/3x 2

7

Sol:

Let y = (fog) (x) = f [g (x)] = f (x3 + 5)

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= 2(x3 + 5) – 3 = 2x3 + 7.

x3 =y 7

2

x = 1/3 1/3

1y 7 y 7(fog) (y)

2 2

(fog)–1 (x) =

1/3x 7

2

17. If f(x) = 3

3

164x

x and a, b are the roots of

14x 3,

x then :

(a) f(a) = 12 (b) f(b) = 11 (c) f(a) = f(b) (d) none of these

Sol:

We have,

f(a) = 64a3 + 3

1

a = (4a)3 +

3

1

a

= 3

1 1 14a 3.4a. 4a

a a a

= (3)3 – 12.3 = 27 – 36 = –9.

1Since a,b are roots of 4x 3

x

14a 3

a

Similarly, f(b)=–9

f(a) = f(b) = – 9.

18. The inverse of the function y = x x

x x

10 101

10 10

is :

(a) 10

1 xy log

2 2 x

(b) y = log10

x

2 x (c) 10

1 xy log

2 1 x

(d) none of these

Sol:

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we have , y =

x x

x x

10 101

10 10

y – 1 = 2x

2x

10 1

10 1

2x 2x

2x 2x

y 1 1 10 1 10 1

y 1 1 10 1 10 1

[Using componendo and dividendo]

2x

2xy 2.10 y10

y 2 2 2 y

2x = log10 y

2 y

x = 110

1 ylog f (y)

2 2 y

Hence, the inverse of the given function is

y = 10

1 xlog .

2 2 x

19. If T1 is the period of the function 3 x [x]y e

T2 is the period of the function 3x [3x]y e

([.] denotes the greatest integer function), then :

(a) T1 = T2 (b) 21

TT

3

(c) 1 2T 3T (d) none of these

Sol: Let g (x) =e3(x) T1 = 1

and f(x) = e{3x} T2 = 1/3

T1 = 3T2

20. The values of constants a and b so as to make the function 2

1, x 1

xf x

ax b, x 1

,

continuous as well as differentiable for all x, are

(a) 1 3

a , b2 2

(b) 1 3

a , b2 2

(c) 1 3

a , b2 2

(d) none of these

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Sol:

Since f (x) is continuous for all x, therefore, it is continuous at x = 1 also.

f(1) = h 0lim

f(1–h) 1 = h 0lim

[a(1 – h)2 + b]

a + b = 1

Also, f(x) is differentiable at x = 1.

h 0 h 0

f(1 h) f(1) f(1 h) f(1)lim lim

h h

2

h 0 h 0

11

a(1 h) b 1 |1 h|lim lim

h h

2

h 0 h 0

(a b 1) (h 2h)a 1 1 hlim lim

h h(1 h)

2a = – 1 (Using a + b = 1)

a = –1

2.

Hence a + b = 1 b = 1 – a = 3

2.


21. The value of tan x

ex 0lim log sin x

is :

(a) 1 (b) – 1 (c) 0 (d) none of these

Sol:

tan xe e

x 0 x 0lim log (sin x) lim tan x.log sin x(0. form)

= e

x 0

log sin xlim

cot x form

= 2x 0

cot xlim

cosec x [Using L'Hospital's Rule]

= x 0lim

(–cos x . sin x) = 0

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22. The function: f(x) =max. {(1 – x), (1 + x), 2}, x , , is

(a) continuous at all points (b) differentiable at all points

(c) differentiable at all points except at x = 1 and x = – 1

(d) continuous at all points except at x = 1 and x = – 1, where it is discontinuous.

Sol:

From the graph, f(x) =

1 x, x 1

2, 1 x 1

1 x, x 1

which clearly shows f(x) is continuous at all points and differentiable everywhere except x =

–1 and x = 1.

Graph of f(x) is shown as;

Hence (a) and (c) are correct answer.

23. If f(x) = [tan2 x] (where [.] denotes the greatest integer function), then :

(a) x 0lim

f(x) does not exist (b) f(x) is continuous at x = 0

(c) f(x) is non–differentiable at x = 0 (d) f(0) = 1.

Sol: (b )

24. If 2x 0

4 sin2x Asin x B cos xlim

x

exists, then the value of ‘A’ and ‘B’ are

(a) – 2 and – 4 (b) – 4 and – 2 (c) – 3 and – 2 (d) none of these

Sol:

Since the given limit exists and denominator approaches zero as x 0.

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Numerator must approach zero as x 0 which is possible if 4 + B = 0 which is obtained by

substituting zero in numerator in place of x.

B = – 4

2x 0

4 sin2x A sinx 4 cos xlim

x

[0/0 form]

x 0

2cos2x Acosx 4sin xlim

2x

[By L.H. Rule]

Again Dr 0 while Nr = 2 + A

Hence for above limit to exist A + 2 = 0, A = –2

x 0

2cos2x 2cosx 4sinxlim

2x

[0/0 form]

x 0

4sin2x 2sinx 4cosxlim

2

= 4

2 = 2.

A = –2 and B = – 4.


25. If f(x) = [x] sin x

,[x 1]

where [.] denotes the greatest integer function, then the

points of discontinuity of f in the domain are :

(a) I (b) I – {0} (c) R – [– 1, 0) (d) none of these

Sol :

[x + 1] = 0, if 0≤ x + 1 < 1 i.e., –1 ≤ x < 0

Thus domain of f = R – [–1, 0).

We have, sin π

[x 1]

continuous at all points of R – [–1, 0) and [x] continuous on R–I, where

I denotes the set of integers. Thus the points where f can possibly be discontinuous are …, –3,

–2, 0, 1, 2, ….

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For 0 ≤ x< 1, [x] = 0 and sin π

[x 1]

is defined.

f (x) = 0 for 0 ≤ x < 1.

Also, f is not defined on {–1, 0), so the continuity of f and 0 mean continuity of f from right at

0. Since f is consitnuous from right at 0, so f is continuous at 0. Hence the set of points of

discontinuity of f is I – {0}.

26. If the function f is defined by y = f(x), where x = 2t – t y = 2t t t , t R, then :

(a) f(x) is differentiable as well as continuous at x = 0

(b) f(x) is continuous but not differentiable at x = 0

(c) f(x) is differentiable but not continuous at x = 0

(d) none of the above

Sol:

When t ≥ 0, |t| = t

x = 2t – t and y = t2 + t . t = 2t2

y = 2x2, when x ≥ 0 [As t ≥ 0 x ≥0]

Also, for t ≤ 0, |t| = – t

x = 2t + t = 3t and y = t2 + t(– t) = 0

y = 0 when x ≤ 0 [As t ≤ 0 x ≤ 0]

Hence the function is defined as:

y = f(x) = 22x , x 0

0, x 0

Now, Lf’ (0) = h 0 h 0

f(0 h) f(0) 0lim lim 0

h h

and Rf’ (0) = 2

h 0 h 0

f(0 h) f(0) 2hlim lim 0

h h

Since Lf’ (0) = R f’ (0), therefore, f(x) is differentiable and Hence continuous at x = 0.

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27. If f(x) = p 1

x cos , x 0, thenat x 0, f x is :x

0, x 0

(a) continuous, if p > 0 and differentiable, if p>1

(b) continuous , if p > 1 and differentiable, if p > 0

(c) continuous and differentiable, if p > 0 (d) none of the above

Sol: Continuity at x = 0 :

LHL = p

h 0 h 0

1lim f (0 h) lim( h) cos 0,if p 0

h

RHL = p

h 0 h 0

1lim f(0 h) lim h cos 0, if p 0

h

and f(0) = 0

f(x) is continuous at x = 0, if p > 0

Differentiable at x = 0 :

Lf’ (0) = h 0

f(0 h) f(0)lim

h

=

p

h 0

1( h) cos 0

hlimh

= p

h 0

1lim ( h) cos

h = 0 if p – 1 > 0, i.e., p > 1;

Rf’ (0) =

p

h 0 h 0

1h cos 0f(0 h) f(0) hlim lim

h h

= p 1

h 0

1lim h cos

h

= 0 if p > 1.

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28. If f(x) =

tan [x], [x] 0

[x]

0, [x] 0

where [x] denotes the greatest integer less than or equal to

x, then x 0lim f x

equals :

(a) 1 (b) – 1 (c) 0 (d) does not exist

L.H.L. = h 0 h 0

tan[ h]lim f(0 h) lim

[ h]

= h 0

tan( 1)lim tan1

( 1)

R.H.L. – h 0 h 0

tan[h]limf(0 h) lim

[h] = 0

Since L.H.L. ≠ R.H.L. h 0lim

f(x) does not exist.

29. If α and β be the roots of ax2 + bx + c = 0, then

1/ x α2

x αlim 1 ax bx c is :

(a) log a α β (b) a α βe

(c) a β αe

(d) none of these

Sol:

x αlim

(1 + ax2 + bx + c)1/(x–α)

= 2

x α

1lim [(1 ax bx c) 1]

(x α )e

[Using x alim g(x )[f(x ) 1]g( x)

x alim[f(x)] e

Provided

f(x) 1 and g (x) as x a]

=

2

x α x α

(ax bx c) a(x α)(x β)lim lim

(x α) (x α)e e

[ α , β are roots of ax2 + bx + c = 0]

= e(α–β).

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30.

21 cos x 1

4 2x 1

2x 1

x x x 1lim

x x 1

is equal to :

(a) 1 (b) 1/2

2

3

(c) 1/2

3

2

(d) 1/2e

Sol:

21 cos x 1

4 2x 1

2x 1

x x x 1lim

x x 1

=

2

2

x 12sin

24 2

x 1

2x 1

x x x 1lim

x x 1

2

1/2x 1

sin1 22

4 2 x 12 3

22x 1

x x x 1lim

x x 1

31. If f(x) =

1 1

x xxe x 0,

0, x 0

then f(x) is :

(a) continuous as well as differentiable for all x

(b) continuous for all x but not differentiable at x = 0

(c) differentiable for all x but not continuous at x = 0 (d) none of the above

Sol:

We have,

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f(x) =

1 1,

x x

1 1

x x

2/x

xe X 0

x, x 0xexe , x 0

x 00, x 0

x 00,

f (x) is differentiable as well as continuous everywhere except possibly at x = 0.

Now,

R f’ (0) = h 0

f(0 h) f(0)lim

h

2h

2/h

h 0 h 0

he 0lim lim e 0

h

h 0 h 0

f(0 h) f(0) h 0lim lim 1.

h h

Since Lf’ (0) ≠ R f’ (0), therefore f(x) is not differentiable at x = 0 but f(x) is continuous at x = 0

[as L f’ (0) and R f’ (0) are finite].

32. If ω 1 is a cube root of unity, then

100 200 2

101 202

2 100 200

1 2ω ω ω 1

A 1 1 ω 2ω ω

ω ω 2 ω 2ω

(a) A is singular (b) A = 0 (c) A is symmetric (d) none of these

Sol : (a), (b)

We have

2 2

2

2 2

1 2ω ω ω 1

A 1 1 ω 2ω ω

ω ω 2 ω 2ω

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2

2

2

ω ω 1

1 ω ω [ 1 ω ω 0]

ω ω ω

ω ω 1

A w 1 1 ω 0

ω ω ω

[taking ω common from C2] Thus, A = 0 and hence A is singular.

33. If A β

cosα sinα 0

α, β sinα cosα 0 ,

0 0 e

then 1

A α, β

is equal to

(a) A α, β (b) A α, β

(c) A α, β (d) A α, β

Sol : (b)

We have

1

A α, β

=

β β

β β

β

e cos α e sinα 01

e sinα e cos α 0e

0 0 1

A α, β

34. Let a, b, c R be such that a + b + c > 0 and abc = 2. Let

a b c

A b c a

c a b

If A2 = I, then value of a3 + b3 + c3 is

(a) 7 (b) 2 (c) 0 (d) – 1.

Sol : (a)

A2 = I

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2 2 2a b c = 1 and bc + ca + ab = 0

Also, 2A 1 A 1.

But A = 3abc – (a3 + b3 + c3)

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

= – (a + b + c)

As a + b + c > 0, we get A 1.

a3 + b3 + c3 = 7

35. If A =

1 2 1

0 1 1

3 1 1

, then A3 – 3A2 – A – 9I is equal to

(a) O (b) I (c) A (d) A2

Sol : (a)

Calculate directly.

36. If α, β, γ are the roots of x3 + px2 + q = 0, where q 0, then

1 / α 1 /β 1 / γ

1 /β 1 / γ 1 / α

1 / γ 1 / α 1 /β

equals

(a) – p/q (b) 1/q (c) p2/q (d) 0

Sol : (d)

We have βγ γα αβ 0.

We can write as

3 3 3

βγ γα αβ1

γα αβ βγα β γ

αβ βγ γα

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= 3 3 3

βγ γα αβ γα αβ1

γα αβ βγ αβ βγα β γ

αβ βγ γα βγ γα

[Using 1 1 2 3C C C C ]

3 3 3

0 γα αβ1

0 αβ βγα β γ

0 βγ γα

= 0 [all zero property]

37. If p a, q b, r c and the system of equations

px + ay + az = 0

bx + qy + bz = 0

cx + cy + rz = 0

has a non-trivial Solution, then the value of p q r

p a q b r c

is

(a) – 1 (b) 0 (c) 1 (d) 2

Sol : (d)

As the given system of equations has non-trivial Solution

p a a

b q b 0

c c r

Applying 2 2 1C C C and 3 3 2C C C , we get

p a p a p

b q b 0 0

c 0 r c

Expanding along C3, we get

b q b p a p

a p r c 0c 0 b q b

a p c q b r c [p q b b a p } 0

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p a q b c p r c q b b r c p a 0

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

c p b0

r c p a q b

p q r q b r c2

p q q b r c q b r c

38. For a fixed positive integer n, let

n 1 ! n 1 ! n 3 !/ n n 1

D n 1 ! n 3 ! n 5 !/ n 2 n 3

n 3 ! n 5 ! n 7 !/ n 4 n 5

then

D

n 1 ! n 1 ! n 3 ! is equal to

(a) – 8 (b) – 16 (c) – 32 (d) – 64

Sol : (d)

Taking (n – 1) ! Common from R1, (n + 1)! From R2 and (n + 3)! from R3, we get

1D n 1 ! n 1 ! n 3 ! D

where D1 =

1 n 1 n n 3 n 2

1 n 3 n 2 n 5 n 4

1 n 5 n 4 n 7 n 6

Applying 3 3 2R R R and 2 2 1R R R , we get

1

1 n 1 n n 3 n 2

D 0 0 4n 6 4n 14

0 4n 14 4n 22

4n 6 4n 14 4n 6 8

4n 14 4n 22 4n 14 8

2 2 1[C C C ]

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4n 6 864

8 0

Thus,

D64

n 1 ! n 1 ! n 3 !

39. If α, β and γ are real numbers, then

1 cos β α cos γ α

cos α β 1 cos γ β

cos α γ cos β γ 1

is equal to

(a) – 1 (b) cos α cos β cos γ

(c) cos α cos β cos γ (d) none of these

Sol : (d)

We can write as a product of two determinants as follows:

cos α sin α 0 cos α sin α 0

cos β sinβ 0 cos β sinβ 0 0

cos γ sin γ 0 cos γ sin γ 0

40. Let

sinθcos sinθ sin cos θ

cos θ cos cos θ sin sin θ

sin θ sin sinθ cos 0

then

(a) is independent of θ (b) is independent of

(c) is a constant (d) θ π/2

d0

dθ

Sol : (b), (d)

Applying 1 2 2C C cot C , we get

0 sin θ sin cosθ

0 cos θ sin sinθ

sinθ/ sin sin θcos 0

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2 2sin θ[ sin sin θ cos θ sin

sin

[expand along C1]

sin θ

which is independent of .

θ π/2

d dAlso, cos θ cos π /2 0

dθ dθ

41. The graph of the function y = f(x) has a unique tangent at the point (a, 0) through which

the graph passes. Then

e

x a

log 1 6f xlim

3f x

is

(a) 1 (b) 0 (c) 2 (d) none of these

Sol :

42. Consider the system of linear equations in x, y and z:

(sin 3θ) x – y + z = 0

(cos 2θ) x + 4y + 3z = 0

2x + 7y + 7z = 0

The value of θ for which the system of equations has a non-trivial Solution are

(a) nπ :n I

(b) mmπ 1 π/6:m I

(c) nnπ 1 π/3: n I

(d) none of these

Sol : (a), (b)

The given system of equations will have a non-trivial solution if

sin3θ 1 1

cos 2θ 4 3 0

2 7 7

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Applying 2 2 1 3 3 1R R 4R and R R 7R , we get

sin 3θ 1 1

cos 2θ 4 sin3θ 0 7 0

2 7 sin 3θ 0 14

Expanding along C2, we get

7 /2 cos 2θ 4 sin 3θ 2 7 sin θ ] 0

2 cos 2θ sin 3θ 2 0

sin θ 2 sinθ 1 2 sin θ 3 0

As 2 sin θ 3 can never be zero, we get

sin θ 0 or sin θ 1 / 2 sin π /6

m

θ n π or θ mπ 1 π / 6 n, m I

43. Let 2 3 2 3

3

tan[e ]x tan[ e ]xf x , x 0

sin x

The value of f(0) for which f(x) is continuous is

(a) 15 (b) 12 (c) – 12 (d) 14

Sol : (a)

Since 27 e 8, so [e2] = 7 and [- e2] = – 8

so 3 3

3x 0 x 0

tan 7 x tan 8 xf 0 lim f x lim

sin x

3 3 3 3

3 3 3 3x 0

tan7x x tan 8x xlim 7 8

7x sin x 8x sin x

= 7 + 8 = 15

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44. If [.] denotes the greatest integer function then

2n

[x] [2x] ..... [nx]lim

n is

(a) 0 (b) x (c)x

2 (d)

2x

2

Sol :

nx 1 [nx] nx. Putting n = 1, 2, 3, …, n and adding them,

x n n [nx] x n

2 2 2

n [nx] n1x . x.

n n n n

……(1)

Now, 2 2n n n

n n1 1 xlim x . x. lim lim

n n n n 2

As the two limits are equal, by (1)

2n

[nx] xlim .

n 2


45. If A = x

2x 2 x

tan π x 1lim lim 1

x 2 x

then

(a) A > 3 (b) A > 4 (c) A < 4 (d) A is a transcendental number

Sol : (a), (b), (d)

x 2 x 2

tan π x 2tan π xlim lim

x 2 x 2

y

π tan ylim π[y π x 2 ]

y

and 2 2

x

1x 1/x x lim

x

2 2x x

1 1lim 1 lim 1

x x

= e0 = 1

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x

2x 2 x

tan π x 1A lim lim 1 π 1 4

x 2 x

and is a transcendental number.

46. Let f(x) = 2n

2nn

x 1lim ,

x 1

then

(a) f(x) = 1 for x > 1 (b) f(x) = - 1 for x < 1

(c) f(x) is not defined for any value of x (d) f(x) = 1 for x = 1

Sol : (a), (b)

If x > 1, then 2n

nlim x .

Thus for 2n

2nn

1 1/ xx 1, f x lim 1.

1 1 / x

If x 1 , then 2n

nlim x 0,

therefore for x 1, f x 1.

If x 1 then

x2n = 1 for any n and therefore f(x) = 0

47. f(x) = [sin x] + [cos x], x [0, 2π], where [.] denotes the greatest integer function. Total

number of points where f(x) is non-differentiable, is equal to

(a) 2 (b) 3 (c) 5 (d) 4

Sol :

[sin x] is non-differentiable at π

x , π, 2π2

and [cos x] is non-differentiable at x = 0,

π 3π, , 2π.

2 2 Thus f(x) is definitely non-differentiable at x = 0,

π 3π, , 2π.

2 2 Thus f(x) is

definitely non-differentiable at x = π , 3π

, 0.2

Also π π

f 1, f 0 0, f 2π 1,2 2

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f 2π 0 1. Thus f(x) is also non-differentiable at π

x2

and 2 π .


48. If 1 1p qcos cos α,

a b

then 2 2

2

2 2

p qk cos α sin α,

a b where k is equal to

(a) 2pq

ab (b) –

2pq

ab (c)

pq

ab (d)

pq

ab

Sol :

2 21

2 2

p q p qcos . 1 1 α

a b a b

2 2

2 2

pq p q1 1 cos α

ab a b

2 2 2 2 2

2 2 2 2

pq p a p qcos α 1

ab a b a b

2 2 2 2 2 22

2 2 2 2 2 2

p q 2pq p q p qcos α cos α 1

a b ab a b a b

2 2

2 2

2 2

p 2pq qcos α 1 cos α sin α

a ab b

2pqk .

ab


49. Total number of positive integral values of ‘n’ so that the equations

2

21 1 nπ

cos x sin y4

and 2

21 1 π

sin y cos x16

are consistent, is equal to

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

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Sol :

We have 2

1 24n 12 sin y π

16

224n 1 π

0 π32 4

also, 1 24n 1 1 72 cos x π n

16 4 4

also, 1 2 24n 1 4n 12 cos x π 0 n π

16 32

1 8 1n

4 π 4

n 1.


50. Let α π /5 and cos α sin α

A ,sin α cos α

then B = A + A2 + A3 + A4 is

(a) singular (b) non-singular (c) skew-symmetric (d) B 1

Sol : (a), (b)

We have

2 3cos2α sin 2α cos3α sin3αA , A

sin2α cos2α sin3α cos 3α

and 4 cos 4α sin 4αA

sin4α cos 4α

We have cos α cos 2α cos 3α cos 4α

cos α cos2α cos π 2α cos π α [ 5α π]

cos α cos 2α cos 2α cos α 0

and sin α sin 2α sin3α sin4α

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sin α sin 2α sin π 2α sin π α

2[sin α sin2α]

3α α 3π π2 2 sin cos 4 sin cos

2 2 10 10

π π π4 sin cos

2 5 10

π π

4 cos cos a say5 10

Thus, B = 0 a

a 0

B is skew-symmetric.

Also, 2 2 2π πB a 16 cos cos 0

5 10

B is non-singular.

51. The complete Solution set of 1sin (sin 5) > x2 – 4x is

(a) x 2 9 2π (b) x 2 9 2π (c) x 9 2π (d) x 9 2π

Sol:

1 2sin sin 5 x 4x

1 2sin [sin 5 2π ] x 4x

2 2x 4x 5 2π x 4x 2π 5 0

2 9 2π x 2 9 2π .


Each question in section has four choices (a),(b),(c) and (d) out of which only one is correct.

Mark your choices as follows:

(a) STATEMENT -1 is True, STATEMENT – 2 is True; STATEMENT – 2 is correct

explanation for STATEMENT – 1

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(b) STATEMENT – 1 is True, STATEMENT – 2 is True; STATEMENT – 2is not a correct

explanation for STATEMENT -1

(c) STATEMENT – 1 is True, STATEMENT -2 is False

(d) STATEMENT -1 is False, STATEMENT – 2 is True

52. Statement-1 : The range of the function f(x) = sin2x + α sin x β where α > 2 will be

real numbers between 2α

β4

and β α 1.

Statement-2 : The function g(t) = t2 + α t + 1 where t [-1, 1] and α > 2 will attain

minimum and maximum values at – 1 and 1.

Sol: (d)

2 2f x sin x α /2 β α /4

If α 2 so α /2 > 1 but sin x 1

so max f(x) 2sin x α sin x β

1 α β, sin x 1

min f(x) = min 2sin x α sin x β

1 α β, sin x 1

range f(x) = [1 α β, 1 α β]

Clearly the statement 2 is true.

53. A function y of x is represented by equation y2 – 1 + log2(x – 1) = 0

Statement-1 : The domain of y is [1, 3]

Statement-2 : The domain of inverse of y is R.

Sol : (b)

The given equation makes sense if x > 1.

Moreover 2y 1 log x 1 makes sense if 21 log x 1 2 x 1 x 3.

Hence the domain of y is [1, 3]. Also 1 – y2 = log2 (x – 1)

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31 yx 1 2 .

Hence the inverse of the function defined by is 21 y1 2 whose domain is R.

54. Statement-1 If a and b are positive and [x] denotes the greatest integer x, then

x 0

x b blim .

a x a

Statement-2

x

xlim 0,

x where {x} denotes fractional part of x.

Sol : (a)

x 0 x 0

x b x b blim lim

a x x x x

x 0

b / xb blim

x a b / x

x

yb blim

a a y

b

a

Since, x 1

0 x 1 sox x

for x . 0.

Hence

x

xlim 0

x

55. Let f(x) =

1 x x 0

1 [x] sin x, 0 x π / 2

3 x π /2

Statement-1 f is continuous of R {1}.

Statement-2 The greatest integer function is discontinuous at every integer.

Sol : (b)

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x 0 x 0

f 0 1 lim f x lim 1 x 1

x 0 x 0lim f x lim 1 [x] sin x 1

So f is continuous at x = 0.

Since [x] is right continuous but not left continuous at x = 1 so also is f.

x π/2

f π / 2 3 lim f x ,

x π/2 x π/2lim f x lim 1 [x] sin x

= x π/2 x π/2

1 lim [x] lim

sin x

= 1 + 1 + 1= 3.

So f is continuous at x = π /2. Thus f is continuous on R ~ {1}. Also [x] is not continuous at

ever x I.

56. Suppose X = a b

c d

satisfies the equation X2 – 4X + I = O.

Statement-1 : if a + d 4, then there are just two such matrix X.

Statement -2: There are infinite number of matrices X, satisfying X2 – 4X + 3I = O.

Sol : (b)

2X 4X 3I O X I X 3I O

a 1 a 3 bc b a d 4 0 0

c a d 4 d 1 d 3 bc 0 0

If a + d 4, then b = 0, c = 0.

(a – 1) (a – 3) = 0 (d – 1) (d – 3) = 0.

a 1, , d 1, 3

As a + d 4, a = 1, d = 1 or a = 3, d = 3.

Thus, X = 1 0 3 0

or X0 1 0 3

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If a + d = 4, we get

a 1 a 3 bc 00.

0 1 a 3 a bc

There are infinite number of matrices satisfying X2 – 4X + 3I = O

57. Let A be a 2 × 2 matrix with real entries. Let I be the 2 × 2 identity matrix. Denote by

tr(A), the sum of diagonal entries of A. Assume that A2 = I.

Statement-1 : If A I and A - I, then det(A) = - 1.

Statement-2: If A I and A – I, then tr(A) 0.

Sol: (c)

Let A = a b

c d

.

Now, 2 2A I det A 1

2

det A 1 det A 1.

Also, 2 1A I A A

a b d b1

det ac d c a

If detA = 1, then

a = d, b = – b, c = – c a = d, b = c = 0.

In this case A = a 0

0 a

2A 1 a 1 a 1.

A = I or A = - I. A contradiction.

This, det(A) = – 1.

a b d b d b

c d c a c a

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a d tr A a d 0.

Statement-1: is true and stamen-2 is false.

58. Let

n n 3 n 6

n n 1 n 4 n 7

n 2 n 5 n 8

a a a

a a a

a a a

Statement-1 : If ka 0 k 1 and a1, a2, a3, … are in G.P. then n 0 n 1.

Statement-2 : If a1, a2, a3 … are in A.P. then n 0 1.

Sol : (b)

Let k 1ka ar k 1, then

n n n 3 n 6

2 2 3

1 1 1

a a a r r r 0

r r r

Next, if ak = b + (k – 1)d, then using 2 3 1C C C and 2C 2 1C C , we get

n

n n 1

n 2

a 3d 3d

a 3d 3d 0

a 3d 3d

[ C2 and C3 are identical]

59. Let f(x) =

20 cos x sin x

sin x 0 cos x

cos x sin x 0

.

Statement-1 : If sin 2x = 1, then f(x) = 2/3

Statement-2 : f(x) = 0 if sin x = cos x

Sol : (d)

Multiplication of two determinants leads us

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21 y y

f x y 1 y

y y 1

where y = sin x cos x

Using 1 1 2 2 2 3C C C , C C C , we get

2 2

1 0 y 1 0 y

f x 1 y 1 1 y 1 y 0 1 2y

0 1 1 0 1 1

2

1 y 1 2y

When sin 2x = 1, y = 1/2 and f(x) = 0

When sin x = cos x, 1 – 2y = 1 – 2 sin2x = cos 2x = 0

f(x) = 0.

60. Statement-1: Let f(x) be a function such that f(x – 1) + f(x + 1) = 2 f(x) for all x R

then f is periodic.

Statement-2: P There exists a function with period 8.

Sol : (b)

Replacing x – 1 by x, we get

f x f x 2 2 f x 1

Again replacing x by x + 2

f x 2 x 4 2 f x 3

Adding (1) and (2), we get

f x 2f x 2 f x 4 2 f x 1 f x 3

2 2 f x 2

2f x 2

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f x f x 4 0

f x 4 f x 8 0

Subtracting (4) and (3) we obtain

f(x) = f(x + 8)

Thus f is periodic.

We know that the period of the function kx – [kx] is 1/k.

A function with period 8 is 1 1

x x .8 8

OBILE +2 {JEE Main – 2012, 13} Pioneer Education … · 2012-08-26 · It can be easily see that...

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Transcript of OBILE +2 {JEE Main – 2012, 13} Pioneer Education … · 2012-08-26 · It can be easily see that...