maths paper-1 & 2(xyz)

7/30/2019 Maths Paper-1 & 2(Xyz)

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XIII RT-2 [Paper-I] Page # 1

MATHEMATICS

PART-A

[SINGLE CORRECT CHOICE TYPE]

Q.1 to Q.20 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct. [20 × 3= 60]

Q.1 Which one of the following limit does not have value unity ?

(A)xsin

)x(tansinLim

0x(B)

xcos

)x(cossinLim

2x

(C)x

x1x1Lim

0x

(D*)2x

1

0x x

xsinLim

[Sol. (A)xsin

)x(tansinLim

0x

0

0=

xx

xsin

xtanxtan

)xsin(tan

Lim0x

=

x

xtanLim

0x = 1.

(B)xcos

)x(cossinLim

2x

0

0; Put x = h

2

, =

hsin

)hsin(sinLim

0h = 1.

(C)x

x1x1Lim0x

00 (Rationalise) = )x1x1(x

)x1()x1(Lim0x

=

2

2= 1.

(D)2x

1

0x x

xsinLim

(1) = eL, where L = 20x x

11

x

xsinLim

= 30x x

xxsinLim

= 3

53

0x x

x.......!5

x

!3

xx

Lim

=

6

1. Ans.]

Q.2 If the solutions of the equation x2 + px + q = 0 are the cubes of the solutions of the equation

x2 + mx + n = 0, then

(A) m3 – 3mn + p = 0 (B) m3 + 3mn + p = 0 (C*) m3 – 3mn – p = 0 (D) m3 + 3mn – p = 0

[Sol. Given, x2 + mx + n = 0

; x2 + px + q = 0

3

3

3 + 3 = – p ; 3 3 = q

and + = – m ; = n

Now, ( + )3 – 3 ( + ) = – p – m3 – 3n ( – m) = – p – m3 + 3mn = – p

m3 – 3mn – p = 0. Ans.]

Q.3 In an arithmetic progression, if Sn

= n(5 + 3n) and tn

= 32, then the value of n is

[Note : Sn

and tn

denote the sum of first n terms and nth term of arithmetic progression respectively.]

(A) 4 (B*) 5 (C) 6 (D) 7

[Sol. We have, tn

= Sn – S

n – 1= n (3n + 5) – (n – 1) (3n + 2) = 6n + 2 = 32 (Given) n = 5. Ans.]

Q.4 If , are the roots of equation x2 + x – 2 = 0, then the value of

22

4444 )1()1(

is equal to

(A) 2 (B) – 2 (C) 4 (D*) – 4




MATHEMATICS

[Sol. We have, x2 + x – 2 = 0

+ = – 1; = – 2

22

4444)1()1(

=

2)( 2

4545

=

2

)()( 4

=

2

)4

=

4

16

= – 4. Ans.

Aliter:

22

4444 )1()1(=

4

)(16

[Using 2 + = 2 + = 2 and + 1 =2

and + 1 = 2

]

=4

16= – 4. Ans.]

Q.5 If f (x) =

)1x(x)1x()2x)(1x(x)1x(x3

x)1x()1x(xx2

1xx1

then f (100) =

(A*) 0 (B) 100 (C) 1 (D) – 100

[Sol. Taking common factors x from C2, (x + 1) from C

3and (x – 1) from R

3, we have

f (x) = x (x2 – 1)322

233

CCC

CCC

x2xx3

x1xx2

111

= x (x2 – 1)

2)1x(2x3

1)1x(x2

001

= 0

f (100) = 0 ]

Q.6 Number of values of x satisfying the equation cos )1xcos(arc3 = 0 is equal to

(A) 0 (B) 1 (C) 2 (D*) 3

[Sol. Let = arc cos (x – 1)

Now, cos 3= 4 cos3 – 3 cos So, 4y3 – 3y = 0, where y = x – 1

y = ±2

3, 0

x = 1 ±23 , 1

Hence three values of x . Ans.

Aliter :

cos (3 cos – 1 (x – 1)) = 0

3 cos – 1 (x – 1) = (2n + 1)2

, n I

cos – 1 (x – 1) = (2n + 1)6

, n I




MATHEMATICS

cos – 1 (x – 1) =6

5,

2,

6

x – 1 =

2

3, 0,

2

3; x = 1 +

2

3, 1,

2

31 . Ans.]

Q.7 Let f(x) =

)3(tanx

2

1cosx

2

3sinsgn 2

.

If f(x) is identically zero for every x R, then the number of values of in [ – 2, 2], is

[Note: sgn k denotes the signum function of k.]

(A) 0 (B) 1 (C*) 2 (D) 3

[Sol. The equation 3tanx2

1cosx

2

3sin

2

= 0 must be an identity in x.

sin =2

3and cos =

2

1and tan = 3 =

3

or

3

5in [ – 2, 2]. Ans. ]

Q.8 The smallest integral value of p for which the inequality (p – 3)x2 – 2px + 3(p – 2) > 0

is satisfied for all real values of x, is

(A) 8 (B*) 7 (C) 6 (D) 5

[Sol. We have, (p – 3)x2 – 2px + 3(p – 2) > 0 is true for all x R, so

p – 3 > 0 ........(1)

and Disc. < 0 ........(2)

must be satisfied simultaneously.

p > 3 .......(1)

and D < 0 4p2 – 4(p – 3) (3p – 6) < 0f(x) = (p – 3)x – 2px + 3(p – 2)

2

v

x

2p2 – 15p + 18 > 0

2p2 – 12p – 3p + 18 > 0

2p (p – 6) – 3(p – 6) > 0

(2p – 3) (p – 6) > 0

p > 6 or p < 2

3

.......(2)

(1) (2) p (6, )

Hence, the smallest integral value of p equals 7. Ans.]

Q.9 If x = x0

is solution of the equation 0)x3()x2(3log2log 55 , then the value of

00

x

1x is equal to

(A*)6

37(B)

2log3log

3log2log

55

55

(C) log23 (D) 2

[Sol. We have 3log2log 55 )x3()x2(

Taking logarithm to the base 5 on both sides, we get

(log52) · (log

52 + log

5x) = (log

53) · (log

53 + log

5x)

– (log53 – log

52) · log

5x = (log

53 – log

52) · (log

53 + log

52)

x

1log5 = log

56 x =

6

1 x

0(Given)

Hence,

00

x

1x =

6

37. Ans. ]




MATHEMATICS

Q.10 Number of solutions of the equation xcos)x(sinsin 1 , for x

2

,2

is equal to

(A) 0 (B*) 2 (C) 4 (D) 6

[Sol. As, )x(sinsin 1= x , for x

2

,2

From above graph, the equation

y=cosx

0,2

0,2

Y

X

y= – x y=x

O

)x(sinsin 1 = cos x has two solutions, in 2,

2. Ans.]

Q.11 If e

1

x

a2Lim

a2

xtana

ax

, then a is equal to

(A) – (B) (C*)2

(D)

2

[Sol.e1)form1(

xa2Lim a2

xtana

ax

(Given)

Let

a2

xtan·a

ax x

a2Lim = eL,

where L =

a2

xtan·a

x

a1Lim

ax

Put x = a + h

=

)ha(a2tan·hLim0h =

a2

h

2tanhLim0h =

a2

htan

hLim0h =

a2

a2

xtan·a

ax x

a2Lim =

a2

e e – 1 (Given)

Hence, a =2

. Ans.]

Q.12 The true solution set of inequality log2(sin ) > log

2(cos ) is equal to

(A) In 4

5n2,4

n2

(B*)

In 2n2,

4n2

(C) In 4

n2,n2

(D)

In 4

7n2,

4n2

[Sol. Given, log2(sin ) > log

2(cos )

sin > 0, cos > 0 and sin > cos Hence, In 2

n2,4

n2

. Ans.]




MATHEMATICS

Q.13 The value of

n

3k 2

2

n k

1k Lim is equal to

(A*)3

2(B)

3

1(C)

2

1(D) 2

[Sol.

n

3k

2n

k

11Lim =

n

3k

n

3k

n k

1k ·

k

1k Lim =

3

1n

n

2Limn

=3

2. Ans.]

Q.14 Which one of the following function contains only one integer in its range?

[Note: sgn k denotes the signum function of k.]

(A) f(x) =

2

21

x1

x1cos

2

1(B) g(x) =

x

1xsgn

(C) h(x) = sin2x + 2sin x + 2 (D*) k(x) = cos – 1(x2 – 2x + 2)

[Sol.

(A) f(x) =

2

1cos – 1

2

2

x1

x1

Df = R

As, 0 cos – 1

2

2

x1

x1<

Rf =

2

,0 .

(B) g(x) = sgn

x

1x

Dg

= ( – , 0) (0, )

Rg

= { – 1, 1}

(C) h(x) = sin2x + 2 sin x + 2

Dh

= R

Also, h(x) = (sin x + 1)2 + 1

Rh

= [1, 5].

(D) k(x) = cos – 1 (x2 – 2x + 2) = cos – 1 1)1x( 2 ; Dk

= {1} ; Rk

= {0}. Ans.]

Q.15 Which one of the following function defined below is continuous at origin?

(A) f(x) =

0x,0

0x,x

1sin

(B*) g(x) =

3,2x,1

3,2x,6x5x

)6x5xcos(2

2

(C) h(x) =

0x,1

0x,x

1tanx 1

(D) k(x) =

0x,1

0x,1x

)1xsin(




MATHEMATICS

[Sol.

(A)x

1sinLim)x(f Lim

0x0x = does not exist,

becausex

1sin oscillates from – 1 to 1 in the neighbourhood of x = 0. So f(x) is discontinuous at x = 0.

(B) )x(gLim0x

=6

1cos 6 = g(0) g(x) is continuous at x = 0.

(C)

x

1tan·xLim)x(hLim

1

0x0x= 0 ×

2

or2

= 0

But, h(0) = 1

So, h(x) is discontinuous at x = 0.

(D) )x(k Lim0x

= sin 1

But k(0) = 1

So, k(x) is discontinuous at x = 0. Ans.]

Q.16 If 4x

–

2x + 1

+ 2 + | a | = cos y where x, y, a R, then the value of (a + x + y) can be

(A)2

(B) (C*) 2 (D) 3

[Sol. We have (2x – 1)2 + | a | + 1 = cos y

As, L.H.S. 1 and R.H.S. 1

x = 0; a = 0 and cos y = 1 i.e. y = 2n n I. ]

Q.17 If

x

xsinLim

1

0x+

x

x2sin212

+

x

x3sin312

+ …… +

x

nxsinn12

= 100,

then the value of n, is

[Note : [k] denotes the greatest integer less than or equal to k.]

(A) 2 (B) 3 (C*) 4 (D) 5

[Sol. In vicinity of x = 0 , |sin – 1 x| > |x|

x

xsin1

> 1, in vicinity of x = 0.y = x

y = sin x – 1

(0,0)x

y

l = 13 + 23 + 33 + .... + n3

=

2

2

)1n(n

= 100 n = 4. Ans.]

Q.18 If cos (22) =)(cos)(cos

31

31 , then

(A) tan 1, tan

2, tan

3are in A.P. (B*) tan

1, tan

2, tan

3are in G.P.

(C) tan 1, tan

2, tan

3are in H.P. (D) tan

1, tan

2, tan

3are NOT in A.P./G.P./H.P.

[Sol. cos (22) =

)(cos

)(cos

31

31

1

)2(cos 2 =)(cos

)(cos

31

31

(Using dividendo and componendo)

1)2(cos

1)2(cos

2

2

=31

31

coscos2

sinsin2

tan22

= tan 1

tan 3 tan

1, tan

2, tan

3are in G.P..Ans.]




MATHEMATICS

Q.19 If the function f(x) = 2p [2x + 5] + q[3x – 7] is continuous at x = 1, then

[Note: [k] denotes the greatest integer less than or equal to k and p, q R.]

(A*) 2p + q = 0 (B) p + 2q = 0 (C) p + q = 0 (D) 10p – 7q = 0

[Sol. We have, f(x) = 10p + 2p [2x] – 7q + q[3x]

Now, f(1+) = 10p + 4p – 7q + 3q = (14p – 4q) = f(1)

Also,

f(1 – ) = 10p + 2p – 7q + 2q = (12 p – 5q)

As, f(x) is given continuous at x = 1, so

14p – 4q = 12p – 5q 2p + q = 0. Ans.]

Q.20 The maximum value of expression

x4

sinx4

cos22 is equal to

(A) 1 (B*) 2 (C)2

3(D)

2

[Sol. Let Expression (E) =

x4

sinx4

cos22

=

x42

cosx4

cos 22

=

x

4cos·2 2 =

x2

2cos1 = 1 – sin 2x

Hence, maximum value is 2. Ans.]

Q.21 to Q.30 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct. [10 × 4= 40]

Q.21 Let f (x) =)1x(·e1

)1xsin(·e)xcos(Lim

nx

nx

n

, then which one of the following is correct?

(A) The value of f(1) is equal to zero.

(B) The value of f(0+) is equal to – sin 1.

(C) The value of f(0 –

) is equal to – 1.

(D*) The value of )0(f )0(f is equal to 1sin1 .

[Sol. We have, f(1) = cos = – 1

Also, f (x) =

),0(x);xcos(

)0,(x;)1x(

)1xsin(

So, f (0+) = 1 and f(0 – ) = – sin 1.

)0(f )0(f = 1 – ( – sin 1) = (1 + sin 1). Ans.]

Q.22 The sequence a1, a2, a3, .... satisfies a1 = 19, a9 = 99, and for all n 3, an is the arithmetic mean of the first (n – 1) terms. Then a

2is equal to

(A*) 179 (B) 99 (C) 79 (D) 59

[Sol.33/seq/SC

n 3, a3

=2

aa21

....(1) [11th (PQRS) 15-10-2006]

a4

=3

a)aa(321

=

3

aa233

a

4= a

3

a5

=4

)aaaa(4321

=

4

aa344

= a

4 ;a

3= a

4= a

5= ......... = a

9= 99




MATHEMATICS

put in equation (1)

99 =2

a192

a

2= 179. Ans.]

Q.23 If the function f(x) = )bx(1ax

e

is non differentiable at exactly one point then the value of ab is

equals to

(A*) 1 (B) 2 (C) 3 (D) 4

[Sol: (ax – 1) (x – b) should be perfect square

D = 0 for ax2 – (ab + 1) x + b (ab + 1)2 – 4ab = 0 or (ab – 1)2 = 0 ab = 1 ]

Q.24 Let f : R R be defined by f(x) =

QRxif ,x

Qxif },x{.

If )x(f Limx

exist , then the true set of values of is

[Note : {k} denotes the fractional part of k and Q be the set of all rational numbers.]

(A) ( – 1, 1) (B) ( – 1, 0] (C*) (0, 1) (D) [0, 1)

[Sol. For limit to exist,

x – [x] = x, at x = + h or – h

x

Lim [ + h] = 0

Note that f(x) is discontinuous at x = 0,

because f(0 ) = 0, but f(0 )+ –

= 1 (via. x Q)

= 0 (via. x Q)

(0, 1). Ans.]

Q.25 The value of

1r 2

1

4

1r4

2tan is equal to

(A) tan – 11 (B) tan – 12 (C) tan – 13 (D*) tan – 14

[Sol.

1r 2

1

4

1r4

2tan =

1r

1

2

1r

2

1r4

2tan =

1r

1

22

1r

22

1r

1

2

1

tan

=

1r

1

2

2r

·2

2r

1

2

2

1r

2

2

1r

tan =

n

1r

11

n 2

1r

tan2

2

1r

tanLim




MATHEMATICS

=

4

1

2

ntan

4

1

2

ntan..........

4

5tan

4

7tan

4

3tan

4

5tan

4

1tan

4

3tanLim

11

111111

n

=

4

1tan

4

1

2

ntanLim 11

n=

4

1tan

2

1

=4

1cot 1

= tan – 14. Ans.]

Q.26 Let f(x) =

3n3x3;e2

3x3;x103x

2

l.

Which one of the following statement is incorrect?

(A) f(x) is continuous for all x ( – 3, 3 + ln 3].

(B*) Number of solutions of the equation f(x) = 1 is two.

(C) Range of f(x) is 10,1 .

(D) f(x) is not injective.

[Hint.

x= – 3

– 1

1

O(0, 0)

)10,0(

3 + n 3l

( – 3,1) (3,1)

x=3

x = 3 + n 2l

Y

X

(3+ n 3, – 1)l

]

Q.27 Let and be two real numbers such that = n and = 1 (where ).Then the value of )n(Lim

n

is equal to

(A) 0 (B)2

1(C)

3

2(D*) 1

[Sol. A quadratic whose roots are and is given by

x2 – nx + 1 = 0

x =2

4nn 2

As, so,=2

4nn2

2

)4nn(nLim)n(Lim

2

nn

)4nn(2

))4n(n(nLim

2

22

n

=)4nn(

n2Lim

2n =2

2= 1. Ans.]




MATHEMATICS

Q.28 Number of integral ordered pair(s) (x, y) satisfying the equation

tan – 1(x + 2011) + tan – 1

2012y

1= tan – 12, is equal to

(A) 1 (B) 2 (C*) 3 (D) 4

[Sol. Put X = x + 2011 and Y = y + 2012, so we get

tan

– 1

YX1

Y

1X

= tan

– 1

2 XY + 1 = 2 (Y –

X) Y = X2

X21

Y =X2

5

– 2 (X, Y) = (1, 3) (3, – 7), (7, – 3) and ( – 3, – 1)

(x, y) (x = – 2010, y = 2009), (x = – 2008,. y = – 2019), (x = – 2004, y = – 2015).

But (x = – 2014, y = – 2013) rejected.

[For Y to be an integer, 2 – x = ± 1 or ± 5 x = 1, 3, 7, – 3]. Ans.]

Q.29 If 0xLim 2

2x

1

x

ex4

tan

is equal to2

eq

p

, (p, q N), then the minimum value of (p + q) is

(A) 5 (B) 6 (C*) 7 (D) 9

[Sol.0x

Lim 2

2xtan1

xtan1n

x

1

x

ee

l

= e2

0xLim 2

2xtan1

xtan1·n

x

1

x

1e

l

= e2

0xLim 3x

x2xtan1

xtan1n

l

= e2

0xLim 3

x

x2)xtan1(n)xtan1(n ll

= e2

0xLim 3

3232

x

x2......3

xtan

2

xtanxtan......

3

xtan

2

xtanxtan

[Using series expansions]

= e2

0xLim 3

3

x

........xtan3

2)xx(tan2

= e2

3

e4

3

2

3

2 2

. Ans.]




MATHEMATICS

Q.30 The graph of functions f and g are shown below.

y=f(x)

O

– 1

1

2

– 2

– 1 – 2 21 x

y

y=g(x)

O

– 1

1

2

– 2

– 1 – 2 21 x

y

4/3

[Note : [k] denotes the greatest integer less than or equal to k.]

Consider the following statements

I. )x(g)x(f Lim1x

exist and is equal to 2.

II. )x(g)x(f Lim2x


III. )x(gf Lim0x


IV. )x(f gLim2x

exist and is equal to – 1.

Which of the statements I, II, III and IV given above are correct?

(A*) I, II and III (B) I, II, III and IV (C) I, II and IV (D) II, III and IV

[Sol.

I : )x(g)x(f Lim1x

= 2 True.

II : )x(g)x(f Lim2x

= 1 True.

III : )x(gf Lim0x = 1 True.

IV : )x(f gLim2x

= 0 False. ]



XIII RT-2 [Paper-II] Page # 1

MATHEMATICS

PART-A

[PARAGRAPH TYPE]

Q.1 to Q.5 has four choices (A), (B), (C), (D) out of which ONE OR MORE may be correct. [5 × 4 = 20]

Paragraph for question no. 1 to 3

Consider f, g and h be three real-valued continuous functions on R (the set of all real numbers)

defined by

f(x) =

x1,4qx

1x2

1,p

21x,3xx2

, g(x) =

0x,1a

0x,x

xcosba2 and h(x) =

1x,1x

1x,1x3

2

.

Q.1 Which of the following statement(s) is(are)correct?

(A*) The value of (p + q) equals2

3. (B) The value of (p + q) equals

2

5.

(C) The value of (a + b) equals 1. (D*) The value of (a + b) equals 0.


(A*) Number of real roots of equation f(x) = 0 is one.

(B*) Number of real roots of equation h(x) = 0 is zero.

(C*) The value of g() equals 2

4

.

(D) The range of function h(x) is R.


(A*) There exists some x0

> 1 such that h(x) > f(x) is true for all (x0,).

(B*) Both f(x) and h(x) are not injective.

(C) Number of real roots of equation g(x) = 0 in [0, 4] are 3.(D*) Range of f(x) is R.

[Sol.

2

1f =

2

1f =

2

1

4

1 + 3 =

4

1221 =

4

11and

2

1f = p

As, f is continuous at x =2

1, so p =

4

11.

Now, f(1 – ) = f(1) = p and f(1+) = q + 4

As, f is also continuous at x = 1, so p = q + 4 q =4

11 – 4 =

4

1611 q =

4

5

So, f(x) =

x1,4x4

5

1x2

1,

4

112

1x,3xx2




MATHEMATICS

2

1x

xx = 1

0,

5

16

y = 11/4

–

+

y = x – x + 32

4x4

5y

y

O

Graph of f(x)

As, g(x) is continuous at x = 0, so

20x x

xcosbaLim = a – 1 ......(1)

As, above limit exist, so a + b = 0 b = – a

a

20x x

xcos1

Lim = a –

1

2

a= a – 1 a = 2a – 2 a = 2

b = – 2

So, g(x) =

0x,1

0x,x

)xcos1(22

and h(x) =

1x,1x

1x,1x3

2

(0,1)y=x +1

3

(1,2)y=x +1

2

( – 1,0)

(1,0)

Graph of h(x)

[Note : g(x) = 0 cos x = 1

x = 2n , n I. But g(0) = 1 ]

Now, verify alternatives. ]

Paragraph for question no. 4 & 5

The graph of y = px2 + qx + r, x R is plotted in adjacent diagram. Given AM = 2 and CM = 1.

Where C is the vertex.

Y

XBA

C

(0, – 8)

O M

Q.4 Which of the following statement(s) is (are) correct?

(A*) The value of (4p – r) is equal to 7.

(B) The value of (4p – r) is equal to 5

(C) The sum of roots of equation px2 + qx + r = 0 is equal to 10.

(D*) The sum of roots of equation px2 + qx + r = 0 is equal to 12.




MATHEMATICS

Q.5 Which of the following statement (s) is (are) incorrect?

(A*) The value of )rqxpx(Lim 2

8x

is not equal to zero.

(B*) The inequality px2 + qx + r < 0 is true for all x (6,).

(C*) Harmonic mean of roots of the equation px2 + qx + r = 0 is3

32.

(D*) The value of q is equal to – 3.

[Sol. We have, y = px2 + qx + r, x R

Let OA = , OB = Also, y (x = 0) = – 8 r = – 8

Now, | | = 4 ()2 = 16

()2 – 4 = 16

p

r4

p

q2

= 16 q2 – 4pr = 16p2 .....(1)

As, ymax.

=p4

)pr4q( 2 = 1 (given)

q2 – 4pr = – 4p .....(2)

From (1) and (2), we get16p2 = – 4p

p =4

1(As, p 0)

Y

XA 1

C

(0, – 8)

O(6,0)(4,0) (8,0)

8x3x4

1y

2

Now, on putting the value of p =4

1and r = – 8 in equation (1),

we get, q2 + ( – 8) = 1 q = ± 3

But q = – 3 (reject)

q = 3

Hence y = 8x3x4

12

Now verify alternatives ]

[REASONING TYPE]

Q.6 to Q.9 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct. [4× 3= 12]

Q.6 Statement-1:

1n

1....

3n2n

4·3

2n2n

3·2

1n2n

2·1Lim

232323nis equal to

3

1.

Statement-2: Let f , g and h be three functions such that f (x) g (x) < h (x) for all x in some interval

containing the point x =c, and if )x(f Limcx = )x(hLimcx = L then )x(gLimcx = L.

(A*) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.

(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.

(C) Statement-1 is true, statement-2 is false. (D) Statement-1 is false, statement-2 is true.

[Sol. 1·2 + 2·3 + 3·4 + ............ + n(n + 1)

=

n

1n

)1n(n =

n

1n

2 )nn( =6

)1n2)(1n(n +

2

)1n(n =

3

)2n)(1n(n




MATHEMATICS

nn2n

)1n(·n....

nn2n

3·2

nn2n

2·1232323

<

nn2n

)1n(·n....

2n2n

3·2

1n2n

2·1232323

<1n2n

)1n(·n....

1n2n

3·2

1n2n

2·1232323

1n2n

3

)2n)(1n(n

nn2n

)1n(·n....

2n2n

3·2

1n2n

2·1

nn2n

3

)2n)(1n(n

2323232323

As,1n2n

3

)2n)(1n(n

Lim3

1

nn2n

3

)2n)(1n(n

Lim23n23n

So,

nn2n

)1n(·n....

3n2n

4·3

2n2n

3·2

1n2n

2·1Lim

23232323n=

3

1.

Hence, both S1

and S2

are true and S2

is explaining S1

also. Ans.]

Q.7 Statement-1: In triangle ABC, if cot A· cot C =2

1and cot B · cot C =

18

1, then cot C is equal to 4.

Statement-2: In triangle ABC, cot A · cot B + cot B · cot C + cot C · cot A equals 1.(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.


(C) Statement-1 is true, statement-2 is false.

(D*) Statement-1 is false, statement-2 is true.

[Sol. In ABC, we know that

cot A · cot B + cot B · cot C + cot C · cot A = 1

cot A · cot B +18

1+

2

1= 1 cot A · cot B =

9

4

Ccot18

1

Ccot2

1=

9

4 cot C =

4

1.

Hence, S1

is false but S2

is true.]

Q.8 Statement 1: The equation cos (x) + 2x – 3 = 0 has atleast one real root in (0, 2).

Statement 2: If f(x) is continuous function in [a, b] and f(a) · f(b) < 0, then there exist some c (a, b)

such that f(c) = 0.

(A*) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.


(C) Statement-1 is true, statement-2 is false.

(D) Statement-1 is false, statement-2 is true.

[Sol. Let f(x) = cos (x) + 2x – 3, x [0, 2]

f(x) = 1 + 1 –

3 = –

1, f(2) = 1 + 4 –

3 = 2.

As, f(x) is continuous function in [0, 2] and f(0) · f(2) < 0.

So, using intermediate value theorem, the equation f(x) = 0 has atleast one real root in (0, 2).

Also, S2

is obviously true and exaplaining S1

also. Ans.]




MATHEMATICS

Q.9 Statement-1: If 9x2x

9x2x2

2

9a2a

9a2a2

2

is true for all xR, then the sum of possible integral

values of a is equal to 2.

Statement-2: Let f be a real-valued function defined on R. If f(x) c is true for all xR,

(c is some finite real number) then c must be less than or equal to the

minimum value of f(x).

(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.


(C) Statement-1 is true, statement-2 is false. (D*) Statement-1 is false, statement-2 is true.

[Sol. Let y =9x2x

9x2x2

2

(y – 1) x2 + 2 (y + 1) x + 9 (y – 1) = 0

As, x is real, so put D 0 2

1 y 2

9a2a

9a2a

2

2

2

1

2a

2

–

4a + 18 a

2

+ 2a + 9 a

2

–

6a + 9 0

(a – 3)2 0

a = 3.

Hence, S1

is false, but S2

is true. Ans.]

PART-B

[MATRIX TYPE] [3 + 3 + 3 = 9]

Q.1 has three statements (A, B, C) given in Column-I and four statements (P, Q, R, S) given in Column-II.

Any given statement in Column-Ican have correct matching with one or more statement(s) given in Column-II.

Q.1 COLUMN-I COLUMN-II

(A) If x

2baxLim

3

0x

=

12

5, then the value of (b – a) is equal to (P) 0

(B) The smallest integer in the range of function (Q) 2

f(x) =1x2x

5x6x2

2

is equal to (R) 3

(C) If x [0, 4], y [0, 4], then the number of ordered pairs (x, y) (S) 4

of real numbers satisfying the equation sin – 1

(sin x) + cos – 1

(cos y) = 2

3

, is

[(A) R ; (B) P;(C) S]

[Sol. (A) We have,x

2baxLim

3

0x

=

12

5(exists)

3 b – 2 = 0 b = 8 .......(1)




MATHEMATICS

Now, we getx

2)8ax(Lim

3

1

0x

=

12

5

0

0

4)8ax(2)8ax(x

)2()8ax(Lim

3

1

3

2

3

0x=

12

5

12

a=

12

5 a = 5 .......(2)

Hence, (b – a) = 8 – 5 = 3. Ans.

(B) Let y =1x2x

5x6x2

2

yx2 + 2xy – x2 + 6x – 5 = 0

(y – 1)x2 + 2(y + 3)x + (y – 5) = 0

As x is real, so put D 0

++

Y

(0, 5)

O (1, 0)

x = 2

(5, 0)X

Graph of f(x) =x – 6x + 5

x + 2x +1

2

2

y = 1

x = – 1

2,–

13 4(y + 3)2 4(y –

1) (y –

5)

y 3

1[Also, y = 1 is possible when x =

2

1.]

So, minimum value of f(x) equals3

1, which occurs at x = 2.

The smallest integer in the range of function is 0.

(C) As, sin – 1 (sin x) 2

,

and cos – 1 (cos y) So, sin x = 1, cos y = – 1

Possible ordered pairs are

,

2;

3,

2;

,

2

5;

3,

2

5]

PART-C

[INTEGER TYPE]

Q.1 to Q.4 are "Integer Type" questions. (The answer to each of the questions are upto 4 digits) [4 × 5 = 20]

Q.1 Let f(x) =

x4,qx3

4x2,px

2x,5

and g(x) =

x1,1x

1x,2x3.

If )x(gf Lim1x

= 5, then find the value of (2p + q). [Ans. 0015]

[Sol. As, )x(gf Lim1x

= 5

15 – q = 5 q = 10 . ....(1)

and )x(gf Lim1x

= 5 2p = 5 p =2

5.....(2)

Hence, (2p + q) = 5 + 10 = 15. Ans.]




MATHEMATICS

Q.2 Let a1, a

2, a

3,.........., a

nare in arithmetic progression where n 10. If the sum of its first five even terms

is equal to 15 and the sum of the first three terms is equal to ( – 3) then find the seventh term

of arithmetic progression. [Ans. 0004]

[Sol. We have, first term = a1

and common difference = d

Now, a2

+ a4

+ a6

+ a8

+ a10

= 15

2

5(a

2+ a

10) = 15

a2

+ a10

= 6

(a1 + d) + (a1 + 9d) = 6 a

1+ 5d = 3 ......(1)

Also, a1

+ a2

+ a3

= – 3

a1

+ (a1

+ d) + (a1

+ 2d) = – 3

3a1

+ 3d = – 3

a1

+ d = – 1 .......(2)

From (1) and (2), we get

d = 1, a1

= – 2

Hence, a7

= a1

+ 6d = – 2 + 6(1) = 4. Ans.]

Q.3 If both roots of equation x2 – 2cx + c2 + c – 5 = 0 are less than 5, then find the largest integral

value of c.

[Ans. c < 4,c = 3 largest]

[Sol. We have, f(x) = x2 – 2cx + (c2 + c – 5)

x=5x

f(x)

or

x=5x

f(x)

Now, conditions for both roots of equation f(x) = 0 to be less than 5 are

(i) D 0 4c2 4 (c2 + c – 5) c 5 ......(1)

(ii) 5a2

b

5

2

c2 c < 5 .....(2) and

(iii) f(5) > 0 c2 – 9c + 20 > 0

(c – 4) (c – 5) > 0

c < 4 or c > 5 .....(3)

must be satisfied simultaneously.

(1) (2) (3)

c ( – , 4)

Hence, the largest integral value of c is 3. Ans.]

Q.4 Let g : RR be defined as g(x) = sgn (x2 –

5x + 6), then find the number of solutions of equation

sinx = )x(singcos 11 .[Note: sgn (k) denotes the signum function of k.] [Ans. 0001]

[Sol. As, g(sin – 1x) = )3x(sin·)2x(sinsgn11

= 1

We get, sin x = cos – 1(1) sin x = 0

x = n, n I

But domain of equation is [ – 1, 1].

The possible solution is x = 0.

Hence, the number of solutions of given equation are one. Ans.]




MATHEMATICS

Q.5 to Q.8 are "Integer Type" questions. (The answer to each of the questions are upto 4 digits) [4 × 6 = 24]

Q.5 Let a differentiable function f satisfying the functional equation

f (x + y) = f (x) f (y) V x, y.

Suppose f (5) = 2 and f '(0) = 3. find f '(5). [Ans: 0006]

[Sol: f ' (5) =h

)05()h5(Lim

0h

f f

f ' (5) =h

)0()5()h()5(Lim0h

f f f f

f ' (5) =

h

)0()h(Lim)5(

0h

f f f

= 2 f ' (0) = 2 × 3 = 6 ]

Q.6 Let f, g : RR be defined by f(x) = 3x – 1 + 1x2 and g(x) =5

1 5x25x3 , then at

how many points y = 10132xfog..,,.........xfog,xfog,xfog.min is not differentiable.

[Ans: 0003]

[Sol. f(x) =

2

1xx5

21x2x

, g(x) =

2

5x

5

x25x2x

– 1

1O

f(x) and g(x) are inverse of each other.

y = min {fog(x), (fog(x))2, (fog(x))3 ,........., (fog(x))2011}

y = min {x, x2, x3,........, x101} not differentiable at x = 0, ± 1.

Q.7 Let g (x) =

x4,4x

4x2,)2x(2x0,x

0x,2x

2

2

.

If the equation g (x) = k has four real and distinct roots, then find the sum of all possible integral values

of k. [Ans. 0001]

[Sol.

(4, 4)

y = x –

4(2, 2)

y = x

)0,2(

(0, – 2)

(2, 0) (4, 0)(0, 0)

y = x – 22

Graph of y = g(x), x

Y

X



MATHEMATICS

From above graph of g(x), the equation g(x) = k has four real and distinct roots if k [0, 2).

Hence, the sum of all possible integral values of k = 0 + 1 = 1. Ans.]

Q.8 Let f(x) = ]1x[ba2xsgn·)1aba(4 2 , xR.

If )x(f Lim2x

exists for some real value(s) of a, then find the smallest positive integral value of b.

[Note: [k] and sgn (k) denote the largest integer less than or equal to k and signum function of k

respectively.] [Ans. 0003]

[Sol. As, )x(f Lim)x(f Lim2x2x

4 + (a2 – ab + 1) + 2(a – b) = 4 – (a2 – ab + 1) + 3(a – b) 2a2 – a (2b + 1) + (2 + b) = 0Since a R, so put D 0.

(2b + 1)2 – 8 (2 + b) 0 4b2 – 4b – 15 0 b

,

2

5

2

3,

Hence, the smallest positive integral value of b is 3. Ans.]

maths paper-1 & 2(xyz)

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