rt-1 maths paper-1 & 2

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XII (F)RT-1_JEE ADVANCE_PAPER-I PAGE # 1

PART-A[SINGLE CORRECT CHOICE TYPE]

Q.1 to Q.4 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct. [3 Marks]

Q.1 Let f (x) =4

7x

2

3 and g(x) be the inverse function of f (x) then the value of (f1 og1)(17) is

equal to

(A)2

613 (B) 242 (C*) 17 (D)

2

613

[Sol. (f 1 og1)(x) = f1 )x(g 1 = f1 )x(f = x ; )17(ogf 11 = 17 Ans. ]

Q.2 Let A, B where 0 A, B 180 and sin A + sin B = 2

3

, cos A + cos B = 2

1

, then the value of

(A + B) is equal to

(A) 30 (B) 60 (C) 90 (D*) 120

[Sol. sin A + sin B =2

3, cos A + cos B =

2

1

BcosAcos

BsinAsin

=

2

BAcos

2

BAcos2

2

BAcos

2

BAsin2

=

2

BAtan

2

BAtan = 3 A + B = 120. Ans.]

Q.3 Let f : X Y be a function such that f(x) = 2x + x4 , then the set of X and Y for whichf(x) is both injective as well as surjective, is

(A) [2, 4] and 2,2 (B*) [3, 4] and 2,2(C) [2, 4] and [1, 2] (D) [2, 3] and [1, 2]

[Sol. Clearly domain is x [2, 4]Now, f '(x) = 0 x = 3 and f(3) = 2

Range 2,2

x

y

(2, 2) (4, 2)

(3, 2)

O(0, 0) x = 2 x = 4

Graph of f(x) = x 2 + 4 x in [2, 4]

But, domain has to be restricted in either [2, 3] or [3, 4] for function to be invertible. Ans.]

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Q.4 Let the quadratic equation jx2 + (j 3) x + 1 = 0 has two equal roots for two values of j. If these two

values of j are A and M, with A < M, then the value of (tan1A + tan1M) equals

(A*)

4

5tan 1 (B)

4

5tan

1(C)

4

5tan 1 (D)

4

5tan 1

[Hint: jx2 + (j 3) x + 1 = 0

D = 0

j2 10j + 9 = 0 j = 1, 9 A = 1 and M = 9

Now, tan1A + tan1M = tan11 + tan19 = +

91

91tan 1 =

4

5tan

1. Ans.]


Q.5 Let a sequence of number is as follows

1

3 5

7 9 1113 15 17 19

21 23 25 27 29

...............................................................

.............................................................................

If tn

is the first term of nth row then ntLim nn

is equal to

(A)2

1(B*)

2

1(C) 1 (D) 1

[Sol. Sn

= 1 + 3 + 7 + 13 + 21 + + tn

Sn = 1 + 3 + 7 + 13 + + tn1 + tn

0 = 1 + 2 + 4 + 6 + 8 + + (tn t

n1) t

n

tn

= 1 + 2 + 4 + 6 + 8 + + (tn t

n1)

= 1 +2

1n [2 2 + (n 2) 2]

= 1 + (n 1)n = n2 n + 1

n1nnLim

2

n=

n1nn

n1nnLim

2

22

n

=

2

1 Ans. ]

Q.6 If p (0, ), then the complete set of values of p for which sin p cos3p > sin3p cos p always holdgood, is

(A*)

4

3,

24,0 (B)

8,0

(C) (0, ) (D) none

[Sol.ph-1

sin p cos3p > sin3p cos p sin p cos p(cos2p sin2p) > 0 2

1sin 2p cos 2p > 0

sin 4p > 0 0 < 4p < or 2 < 4p < 3

4

3,

24

,0 Ans.]

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Q.7 If a sin1 x b cos1 x = c, then the value of a sin1 x + b cos1 x (whenever exists) is equal to

(A) 0 (B)ba

)ab(cab

(C)

2

(D*)

ba

)ba(cab

[Sol.70/itf

We have b sin1 x + b cos1 x =2

b...... (1)

and a sin1

x

b cos1

x = c ...... (2) (given)

On adding (1) and (2), we get (a + b) sin1 x =2

b+ c

sin1 x =ba

c2

b

. Similarly cos1 x =ba

c2

a

Hence (a sin1 x + b cos1 x) =ba

)ba(cab

]

[PARAGRAPH TYPE]

Q.8 to Q.10 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.

Paragraph for question nos. 8 to 10

Consider, f(x) = tan1

x

1x1 2

and g(x) = cosec1

x

x1 2, x 0

Q.8 The number of solution(s) of the equation

x2 = | f(x) g(x) | is

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

Q.9 If f(x) + g(x) =8

then x equals

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

1

Q.10 The value of

x

)x(g)x(f6Lim

0xequals

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

(A) 9 (B*) 8 (C) 6 (D) 5

Sol. Putting x = tan x R {0}

2,

2 {0}

f(x) = tan1

tan

1tan1 2

= tan1

2tan =

2

=

2

1tan1 x

g(x) = cosec1

tan

tan1 2

= cosec1(cosec ) = = tan1 x

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(i) f(x) g(x) = 2

1tan1 x

x2 = | f(x) g(x) |

/4

y

x

/4

clearly number of solutions are 2.

(ii) f(x) + g(x) = 8

2

3tan1x =

8

tan1x =

12

x = 32

(iii)

x

xtan9Lim

x

xtan2

36

Lim1

0x

1

0x= 8

1x

xtan 1

[MULTIPLE CORRECT CHOICE TYPE]

Q.11 to Q.14 has four choices (A), (B), (C), (D) out of which ONE OR MORE may be correct.

Q.11 Which of the following function (s) is/are Transcendental?

(A*) f (x) = 5 sin x (B*) f (x) =2 3

2 12

sin x

x x

(C) f (x) = x x2 2 1 (D*) f (x) = (x2 + 3).2x

[Hint:504/func

functions which are not algebraic are known as transcidental function]

Q.12 Which of the following statements is(are) correct ?[Note: [x] and {x} denote greatest integer less than or equal to x and fractional part of x respectively.]

(A*) f(x) = [ln x] + }xn{l , x > 1 is continuous at x = e.

(B) Discontinuity of f(x) at x = a non existence of limit at x = a.(C*) f : [1, 1] [1, 1], f(x)= x2 sgn(x) is a bijective function, where sgn x denotes signum function

of x.

(D*) If f is continuous on [1, 1] , f(1) = 4 and f(1) = 3, then there exists a number r such that 1r

and f(r) = .Sol.

(A) f(e) = 1 ; f(e+) = 1 ; f(e

) = 1

(B) Here, a = 13, l =3

1; so a +

l

1= 13 +

3

1

1= 13 3 = 10.

(C) f(x) = x2 sgn x =

0x,x

0x,0

0x,x

2

2

.

y

x(0,0)

(1,1)

(1,1)

x= 1 x=1

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(D) As, f is continuous on [1, 1] and f (1) = 4, f (1) = 3

there exists a number r such that | r | < 1 and f (r) = .]

Q.13 Which of the following function(s) have the same domain and range ?

(A) f(x) = 2x1 (B*) g(x) = x

1(C*) h(x) = x (D) l (x) = x4

[Sol.521/func (A) [1, 1] and [0, 1](B) (, 0) (0, ) & (, 0) (0, )(C) [0, ) and [0, )(D) (, 4] and [0, ) ]

Q.14 The maximum value of the function defined by f (x) = min (ex , 2 + e2x, 8) is then integral value

of x satisfying the inequality

12x][x

][xx2

< 0, is

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

(A*) 1 (B) 3 (C*) 5 (D*) 6[Sol. f (x) = min (ex , 2 + e2 x , 8)

From the graph it is clear that maximum

value of f (x) is, = e2

[] = [e2] = 7

12x7x

)7x(x2

< 0

)4x)(3x(

)7x(x

< 0

O

e2

x = 2x

y

0 3 4 7 ]

PART-B[MATRIX TYPE]

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

Q.1 Column-I Column-II

(A) If

3

2x

x22

x

tan1

)xsin1(2

xtan1

Lim

is equal tok

1, (P) 9

then the value of k where k N is

(B) If the range of the function f(x) = 3 22

x16

sin

is [a, b], (Q) 12

then the value of 2 (a2 + b2) is

(C) If the sum of all possible values of x (0, 2) satisfying the equation (R) 16

2cos x cosec x 4 cos x cosec x = 2 is equal to4

k(k N),

then the value of k is (S) 32

[Ans. (A) S; (B) P ; (C) Q]

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[Sol.

(A) 3

2x x2

)xsin1(2

x

4tan

Lim

x = h2 2

h42

x

=

30h h8

xcos12

htan

Lim

=

20h h162

h

xcos12

htan

Lim

=

32

1 k = 32 Ans.

(B) Domain is16

x2 0

Hence, x

4,

4

Since, f(x) is even

In

4,0 range of f(x) is

2

3,0

a = 0, b =2

3

Hence 2(a2

+ b2

) = 2

2

9

0 = 9 Ans.

(C) 2 cos x(cosec x 2) = (cosec x 2)

(2 cos x 1) (cosec x 2) = 0

cos x =2

1or sin x =

2

1

sum = 3 =4

12

4

k k = 12 Ans.]

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PART-C[INTEGER TYPE]

Q.1 to Q.3 are "Integer Type" questions. (The answer to each of the questions are upto 4 digits)

Q.1 Graph of a function y = f (x) is shown in the adjacent figure.

Four limits l1, l

2, l

3and l

4are given as :

l1

= ]2)x(f[Lim0x

l2

= )0(f]x[fsinsinLim 10x

4

3

2

22

f (x)

0x

then find the value of (l1

+ l2

).

[Note : [k] denotes greatest integer function less than or equal to k.] [Ans. 0002]

[Sol. l1

= 2)x(fLim0x

= 1 { f (0+) < 2}

l2

= )0(f]x[fsinsinLim 10x

= )0(f)0(fsinsin 1 = 3)3(sinsin 1 = [ 3 + 3] = 3 l

1+ l

2= 1 + 3 = 2. Ans. ]

Q.2 A cubic equation x3 + ax2 + bx + c = 0 with real coefficients has 1 and 1 i as two roots . Then find the

value of a + b. [Ans: 0001]

[Hint: 3rd root is 1 + i.]

Q.3 Find the smallest positive integral value of n for which

(n 2) x2 + 8x + n + 4 > sin1 (sin 12) + cos1 (cos 12) x R. [Ans: 0005][Sol: We have

sin1 (sin 12) + cos1( cos 12) = (4 12) + (4 12) = 0 (n 2)x2 + 8x + n + 4 > 0 x R (n 2) > 0 n 3 and (8)2 4 (n 2) (n + 4) < 0 or n2 + 2n 24 > 0 n > 4 n 5So, n

smallest= 5. Ans.]

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XII (F) RT-1_JEE ADVANCE_PAPER-II PAGE # 1

PART-A[SINGLE CORRECT CHOICE TYPE]


Q.1 If x6tanx5tanx4tan

x3tanx2tanxtanLim

333

333

0x

=q

p(p, q N), then least value of (p + q) is

(A) 45 (B) 47 (C*) 49 (D) 51

[Sol.

3

33

3

33

3

33

3

33

3

33

3

33

)x6(

x6tanx216

)x5(

x5tanx125

)x4(

x4tanx64

)x3(

x3tanx27

)x2(

x2tanx8

x

xtanx

=21612564

2781

=

405

36=

459

49=

45

4Ans.]

Q.2 Which one of the following formulas describes all solutions to the equation

log 2 + log (sin ) + log (cos ) = 0, where is in radians?

(A*) = 2k +4

(B) = k +

4

(C) = k +

2

(D) = 2k

4

[Note: Where k I ][Sol. We have log (sin 2) = log (2sin cos ) = 0

So, we need 2 = 2n +2

.

However, we also need to be in the first quadrant as both sin and cos must be positive.

So, = 2k +4

, k I Ans.]

Q.3 Let f (x) =

2

1

x1

x2sin and g (x) =

1x

1xcos

2

21

, then the value of )100(g)10(f is equal to

(A) 2 )100(tan)10(tan 11 (B) 0

(C*) 2 )10(tan)100(tan 11 (D) 2 )100(tan)10(tan 11

[Sol. f (x) =

2

1

x1

x2sin = 2 tan1x, for x 1

and g (x) =

1x

1xcos

2

21

=

2

21

x1

x1cos = 2 tan1x, for x 0

Now f (10) g (100) = )100(tan2)10(tan2 11 2 )10(tan)100(tan 11 Ans.]

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Q.4

6

7coscos

1is equal to

(A)6

7(B*)

6

5(C)

3

(D)

6

[Sol.

6

7

coscos

1

=

2

3

cos

1

=

2

3

cos

1

=

6

= 6

. Ans.]


Q.5 Let g : R

2

,6

is defined by g(x) = sin1

2

2

x1

cx.

Then the possible values of k for which g is surjective function, is

(A)

2

1(B)

2

1,1 (C*)

2

1(D)

1,2

1

Sol.84/itf/SC

We have2

1 2

2

x1

cx

< 1

2

1 1

1x

)1c(2

< 1 Rx c + 1 > 0

So c >1 and1x

1c2

2

1

x2 + 1 2c + 2So x2(2c + 1) 0 Rx 4(2c + 1) 0

c

2

1.

Hence c =2

1Ans.]

Q.6 Let f(x) =x

1x2 . The function )x(ff will be defined for

(A) | x | > 1 (B) | x | 1(C) | x | < 1, x 0 (D*) no real values of x

[Sol. f 2(x) = 2

x

1x =

2

x

11

Now, )x(ff =)x(f

1)x(f2 =

x

1x

x

1

2

2

, which cannot be defined for any real value of x.]

Q.7 Let y = cos x (cos x cos 3 x) . Then y is(A) 0 only when x 0 (B) 0 for all real x(C*) 0 for all real x (D) 0 only when x 0

[Hint29/ph-1

y = cos x (2 sin 2x sin x) = sin22x

y 0 x RAliter: 2y = 2cos2x

2cos x cos 3x = (1 + cos 2x)

(cos 4x + cos 2x) = 1

cos 4x = 2sin22x

y = sin22x 0 x R (C) ]

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[PARAGRAPH TYPE]

Q.8 to Q.10 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.

Paragraph for question nos. 8 to 10

Let f(x) =x

3xcos5xcos22

Q.8 The value of )x(fLim0x

is

(A) 2 (B*) 0 (C) 5 (D)5

Q.9 Difference between the maximum and minimum values of x f(x) is equal to

(A) 2 (B) 3 (C) 5 (D*) 10

Q.10 The value of

3

f is equal to

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

2

[Sol. f(x) =x

axcos)2a()x2cos1(

(i)x

)x(fLim

0x=

2

2

0x x

)xcos1(axcos2xcos2Lim

=20x20x x

)xcos1(xcos2Lim

x

)xcos1(aLim

=

2

12

2

a=

2

a1

But2a

1 =

21 2

a =23 a = 3 Ans.

(ii) Now, x f(x) = 2cos2x5cos x + 3 = 2

2

3xcos

2

5xcos2 = 2

16

25

2

3

4

5xcos

2

Maximum when cos x =1 2

16

1

16

81= 10

Minimum when cos x = 1. Minimum value = 0 (Maximum value Minimum value) = 10. Ans.

(iii)

3

f =

3

32

5

4

1

=

3

1

= 3

3

= 3 Ans.]

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PART-B[MATRIX TYPE]

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

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

Q.1 Column-I Column-II

(A) Let f (x) = |x + 4| + |x + 1| + |x| + |x

5| + |x

9|, then largest integral (P) 10value of k for which equation f(x) = k has no solution is

(B) The value of

1n 41

n41

1sin

40(Q) 18

(R) 25

(C) Let {an} be a geometric sequence with a

7= 50 and a

11= 3 5250 .

If the value of a3

is 3 k2 (k N), then the value of k, is (S) 40

(D) Let k 1

and k2

are two values of k for which the equation

4x2

4(5x + 1) + k2

= 0 has one root equals to two more (T) 50than the other, then the value of 2

221 kk , is

[Ans. (A) Q; (B) P (C) R ; (D) T]

[Sol.

(A)

14 0 5 9

36

x

20

19

(B)

1n 41

n41

1sin =

1n2

1

n2

1tan =

1n

2

1

1n41

2tan

)1n2(tan1n2tan 11n

1

442

(C) a3

=11

27

a

a= 3

2

5250

50= 3 5

10= 3 252 k = 25 Ans.]

(D).12/qe/SC

4x24(5x + 1) + k2 = 0

4x220x + (k24) = 0

two roots are , + 2

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2 + =4

20= 5 + 1 =

2

5 = 1

2

5 =

2

3

( + ) =4

4k2

2

3

22

3= 4

4k2

23

2

7= 4

4k2

21 = k24

k2 = 25 k = 5

2221 kk = 50 Ans.]]PART-C

[INTEGER TYPE]

Q.1 to Q.7 are "Integer Type" questions. (The answer to each of the questions are upto 4 digits)

Q.1 Consider the following equation (x22x + m) (x22x + n) = 0 (where m and n are real numbers).

Let the roots , , , ( < < < ) of the equation form an arithmetic sequence with the first term

equal to4

1. If the value of | mn | can be expressed as

q

pwhere p and q are in their lowest form,

then find the value of (p + q). [Ans. 3]

[Sol. The four roots are 0.25, k + 0.25, 2k + 0.25, 3k + 0.25.

Since, the coefficient on x is 2 in both factors, one of the two quadratic expressions factors as

(x0.25) (x(3k + 0.25)) and the other as (x(k + 0.25)) (x(2k + 0.25)).

In addition, 3k + 0.5 = 2. So, k =2

1.

If we let m = (k + 0.25) (2k + 0.25) and n = 0.25 (3k + 0.25).

We have | mn | =16

7

16

15 =

2

1 p + q = 3. Ans.]

Q.2 Let f be a function defined for all x > 1 such that 4 f(x1 + 1) + 8 f(x + 1) = log12

x,

then find the value of 4 )17(f)13(f)10(f . [Ans. 3]

[Sol. 4

1x

1f + 8f(x + 1) = log

12x .......(1)

x x1

8

1x

1f + 4f(x + 1) =

log12x .......(2)

2 (1) 8

1x

1f + 16f(x + 1) = 2log

12x .......(3)

(3)(2) ___________________________

12 f(x + 1) = 3log12

x

4f(x + 1) = log12

x

put x = 9 4f(10) = log12

9

x = 12 4f(13) = log12

12

x = 16 4f(17) = log12

16

____________

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4[f(10) + f(13) + f(17) = log12

(9 12 16) = log12

123 = 3

f(10) + f(13) + f(17) =4

3 4 )17(f)13(f)10(f = 3. Ans.]

Q.3 Let p and q (p < q) be two prime numbers such that their difference is odd and

8tanlog qp9 = 83 . Find the value of p2 + q2. [Ans. 53][Sol. Difference of two prime numbers is odd one of them must be 2.

p = 2

8tanlog

qp

9 = 223

8

cotlogqp

9 = 223

9logqp

8cot

=

2

12

9log qp12 = 212 9log

qp = 2 9 = p + q q = 7

Now, p2 + q2 = 4 + 49 = 53. Ans.]

Q.4 Let Pn

=

n

2n

2

)1n(n

21 (n N). If n

nPLim

=

b

a(where a and b are coprime), then find the value

of (a + b). [Ans. 0010]

[Sol. Pn

=

n

2n

2

)1n(n

21 =

n

2n

2

)1n(n

)1n()2n(

Pn

=

n

2n

2n

2n

2

n

1n

1n

2n

Pn

=

2

2

2

2

2

2

2

2

2

2

2

2

n

)1n(.......

3

2

2

1

)1n(

)2n(.......

4

5

3

4= 2

2

n

1

9

)2n(

nn

PLim

=9

1

b

a (a + b) = 10. Ans.]

Q.5 The smallest integral value of x for which the function f(x) = log (log(x1)) exists is [Ans: 12]

[Sol. log(x1) > 0 x1 > 10 x > 11Smallest integral value = 12 ]

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Q.6 Let

r

1r

q

1q

p

1p < 0 where p, q, r R{0} and =

r

|r|

q

|q|

p

|p| .

If the equation x2 + (m2)xn = 0 is satisfied by distinct values of '' then find the value of (m + n).[Ans. 7]

[Sol.

r

1

rq

1

qp

1

p < 0

p, q, r all three are negative or exactly one of then is negative =3 or 1

Now, x2 + (m2)xn = 03

1

(m2) =2 m = 4n =3 n = 3

m + n = 7 Ans. ]

Q.7 Find the value of tan (1 + sec 2) (1 + sec 22) (1 + sec 23) when =32 . [Ans. 1]

[Sol.

2cos

2cos1

cos

sin= tan 2

tan 2 (1 + sec 4)

4cos

4cos1

2cos

2sin= tan 4

tan 4 (1 + sec 8)

8cos8cos1

4cos4sin = tan 8

If =32

4tan

= 1. Ans.]

rt-1 maths paper-1 & 2

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