Functions — Page 5 | JEE Mathematics

Q41

The inverse function of f(x) =

{{{8^{2x}} - {8^{ - 2x}}} \over {{8^{2x}} + {8^{ - 2x}}}}

, x

\in

(-1, 1), is :

A

{1 \over 4}{\log _e}\left( {{{1 - x} \over {1 + x}}} \right)

B

{1 \over 4}\left( {{{\log }_8}e} \right){\log _e}\left( {{{1 - x} \over {1 + x}}} \right)

C

{1 \over 4}\left( {{{\log }_8}e} \right){\log _e}\left( {{{1 + x} \over {1 - x}}} \right)

D

{1 \over 4}{\log _e}\left( {{{1 + x} \over {1 - x}}} \right)

Correct Answer

Option C

Solution

f(x) =

{{{8^{2x}} - {8^{ - 2x}}} \over {{8^{2x}} + {8^{ - 2x}}}}

= y $\therefore$

{{y + 1} \over {y - 1}} = {{{{2.8}^{2x}}} \over { - {{2.8}^{ - 2x}}}}

$\Rightarrow$

{{1 + y} \over {1 - y}}

= 84x $\Rightarrow$

{\log _e}\left( {{{1 + y} \over {1 - y}}} \right)

= 4x

{\log _e}8

$\Rightarrow$ x =

{1 \over {4{{\log }_e}8}}{\log _e}\left( {{{1 + y} \over {1 - y}}} \right)

$\therefore$ f-1(x) =

{1 \over 4}\left( {{{\log }_8}e} \right){\log _e}\left( {{{1 + x} \over {1 - x}}} \right)

Q42

If g(x) = x2 + x - 1 and (goƒ) (x) = 4x2 - 10x + 5, then ƒ

\left( {{5 \over 4}} \right)

is equal to:

A

{1 \over 2}

B

{3 \over 2}

C -

{1 \over 2}

D -

{3 \over 2}

Correct Answer

Option C

Solution

Given, (goƒ) (x) = 4x2 - 10x + 5 $\Rightarrow$ g(f(x)) = 4x2 - 10x + 5 $\therefore$ g(f(

{5 \over 4}

)) =

4 \times {{25} \over {16}} - {{50} \over 4} + 5

=

- {5 \over 4}

...(1) Also given, g(x) = x2 + x - 1 $\therefore$ g(f(x)) = f2(x) + f(x) –1 $\Rightarrow$ g(f(

{5 \over 4}

)) = f2(

{5 \over 4}

) + f(

{5 \over 4}

) –1 ....(2) from (1) & (2) f2(

{5 \over 4}

) + f(

{5 \over 4}

) –1 =

- {5 \over 4}

$\Rightarrow$ f2(

{5 \over 4}

) + f(

{5 \over 4}

) +

{1 \over 4}

= 0 $\Rightarrow$ (f(

{5 \over 4}

) +

{1 \over 2}

)2 = 0 $\Rightarrow$ f(

{5 \over 4}

) = -

{1 \over 2}

Q43

Let f : R → R be defined as f (x) = 2x – 1 and g : R - {1} → R be defined as g(x) =

{{x - {1 \over 2}} \over {x - 1}}

. Then the composition function f(g(x)) is :

A one-one but not onto

B onto but not one-one

C both one-one and onto

D neither one-one nor onto

Correct Answer

Option A

Solution

Given, f(x) = 2x $-$ 1; f : R $\to$ R

g(x) = {{x - 1/2} \over {x - 1}};g:R - \{ 1) \to R

f[g(x)] = 2g(x) - 1

= 2 \times \left( {{{x - {1 \over 2}} \over {x - 1}}} \right) - 1 = 2 \times \left( {{{2x - 1} \over {2(x - 1)}}} \right) - 1

= {{2x - 1} \over {x - 1}} - 1 = {{2x - 1 - x + 1} \over {x - 1}} = {x \over {x - 1}}

$\therefore$

f[g(x)] = 1 + {1 \over {x - 1}}

Now, draw the graph of

1 + {1 \over {x - 1'}}

$\because$ Any horizontal line does not cut the graph at more than one points, so it is one-one and here, co-domain and range are not equal, so it is into.

Hence, the required function is one-one into.

Q44

Let f, g : N

\to

N such that f(n + 1) = f(n) + f(1)

\forall

n

\in

N and g be any arbitrary function. Which of the following statements is NOT true?

A If g is onto, then fog is one-one

B f is one-one

C If f is onto, then f(n) = n

\forall

n

\in

N

D If fog is one-one, then g is one-one

Correct Answer

Option A

Solution

f(n + 1) = f(n) + 1

f(2) = 2f(1)

f(3) = 3f(1)

f(4) = 4f(1)

.....

f(n) = nf(1)

f(x)

is one-one

Q45

A function f(x) is given by

f(x) = {{{5^x}} \over {{5^x} + 5}}

, then the sum of the series

f\left( {{1 \over {20}}} \right) + f\left( {{2 \over {20}}} \right) + f\left( {{3 \over {20}}} \right) + ....... + f\left( {{{39} \over {20}}} \right)

is equal to :

A

{{{39} \over 2}}

B

{{{19} \over 2}}

C

{{{49} \over 2}}

D

{{{29} \over 2}}

Correct Answer

Option A

Solution

f(x) = {{{5^x}} \over {{5^x} + 5}}

..... (i)

f(2 - x) = {{{5^{2 - x}}} \over {{5^{2 - x}} + 5}}

f(2 - x) = {5 \over {{5^x} + 5}}

.... (ii) Adding equation (i) and (ii)

f(x) + f(2 - x) = 1

f\left( {{1 \over {20}}} \right) + f\left( {{{39} \over {20}}} \right) = 1

f\left( {{2 \over {20}}} \right) + f\left( {{{38} \over {20}}} \right) = 1

\begin{aligned}& : \\ & :\end{aligned}

f\left( {{{19} \over {20}}} \right) + f\left( {{{21} \over {20}}} \right) = 1

and

f\left( {{{20} \over {20}}} \right) = f(1) = {1 \over 2}

$\therefore$ Sum =

19 + {1 \over 2} = {{39} \over 2}

Q46

Let x denote the total number of one-one functions from a set A with 3 elements to a set B with 5 elements and y denote the total number of one-one functions form the set A to the set A

\times

B. Then :

A 2y = 273x

B y = 91x

C 2y = 91x

D y = 273x

Correct Answer

Option C

Solution

Number of elements in A = 3 Number of elements in B = 5 Number of elements in A $\times$ B = 15 Number of one-one function x = 5 $\times$ 4 $\times$ 3 x = 60 Number of one-one function y = 15 $\times$ 14 $\times$ 13 y = 15 $\times$ 4 $\times$

{{14} \over 4}

$\times$ 13 y = 60 $\times$

{7 \over 2}

$\times$ 13 2y = (13)(7x) 2y = 91x

Q47

Let

A = \{ 1,2,3,....,10\}

and

f:A \to A

be defined as

f(k) = \left\{ \begin{array}{ll}{k + 1} & {if\,k\,is\,odd} \\ k & {if\,k\,is\,even} \end{array} \right.

Then the number of possible functions

g:A \to A

such that

gof = f

is :

A 55

B 105

C 5!

D 10C5

Correct Answer

Option B

Solution

f(1) = 2 f(2) = 2 f(3) = 4 f(4) = 4 f(5) = 6 f(6) = 6 f(7) = 8 f(8) = 8 f(9) = 10 f(10) = 10 $\therefore$ f(1) = f(2) = 2 f(3) = f(4) = 4 f(5) = f(6) = 6 f(7) = f(8) = 8 f(9) = f(10) = 10 Given, g(f(x)) = f(x) when x = 1, g(f(1)) = f(1) $\Rightarrow$ g(2) = 2 when, x = 2, g(f(2)) = f(2) $\Rightarrow$ g(2) = 2 $\therefore$ x = 1, 2, g(2) = 2 Similarly, at x = 3, 4, g(4) = 4 at x = 5, 6, g(6) = 6 at x = 7, 8, g(8) = 8 at x = 9, 10, g(10) = 10 Here, you can see for even terms mapping is fixed.

But far odd terms 1, 3, 5, 7, 9 we can map to any one of the 10 elements.

$\therefore$ For 1, number of functions = 10 For 3, number of functions = 10

\begin{aligned}& . \\ & . \\ & . \\ &\end{aligned}

for 9, number of functions = 10 $\therefore$ Total number of functions = 10 $\times$ 10 $\times$ 10 $\times$ 10 $\times$ 10 = 105

Q48

The range of a

\in

R for which the function f(x) = (4a

-

3)(x + loge 5) + 2(a

-

7) cot

\left( {{x \over 2}} \right)

sin2

\left( {{x \over 2}} \right)

, x

\ne

2n

\pi

, n

\in

N has critical points, is :

A [1,

\infty

)

B (

-

3, 1)

C

\left[ { - {4 \over 3},2} \right]

D (

-

\infty

,

-

1]

Correct Answer

Option C

Solution

f(x) = (4a - 3)(x + \ln 5) + 2(a - 7)\left( {{{\cos {x \over 2}} \over {\sin {x \over 2}}}.{{\sin }^2}{x \over 2}} \right)

f(x) = (4a - 3)(x + \ln 5) + (a - 7)\sin x

f'(x) = (4a - 3) + (a - 7)\cos x = 0

\cos x = {{ - (4a - 3)} \over {a - 7}}

- 1 \le - {{4a - 3} \over {a - 7}} \le 1

- 1 \le {{4a - 3} \over {a - 7}} \le 1

{{4a - 3} \over {a - 7}} - 1 \le 0

and

{{4a - 3} \over {a - 7}} + 1 \ge 0

\Rightarrow {{ - 4} \over 3} \le a \le 2

Q49

The inverse of

y = {5^{\log x}}

is :

A

x = {5^{\log y}}

B

x = {y^{{1 \over {\log 5}}}}

C

x = {5^{{1 \over {\log y}}}}

D

x = {y^{\log 5}}

Correct Answer

Option B

Solution

y = {5^{\log x}}

\Rightarrow \log y = \log x.log5

\Rightarrow \log x = {{\log y} \over {\log 5}} = {\log _5}y

\Rightarrow x = {e^{{{\log }_5}y}}

\Rightarrow x = {y^{{{\log }_5}e}}

\Rightarrow x = {y^{{1 \over {\log 5}}}}

Q50

If the functions are defined as

f(x) = \sqrt x

and

g(x) = \sqrt {1 - x}

, then what is the common domain of the following functions : f + g, f

-

g, f/g, g/f, g

-

f where

(f \pm g)(x) = f(x) \pm g(x),(f/g)x = {{f(x)} \over {g(x)}}

A

0 \le x \le 1

B

0 \le x < 1

C

0 < x < 1

D

0 < x \le 1

Correct Answer

Option C

Solution

f + g = \sqrt x + \sqrt {1 - x}

\Rightarrow x \ge 0

&

1 - x \ge 0 \Rightarrow x \in [0,1]

f - g = \sqrt x - \sqrt {1 - x}

\Rightarrow x \ge 0

&

1 - x \ge 0 \Rightarrow x \in [0,1]

f/g = {{\sqrt x } \over {\sqrt {1 - x} }}

\Rightarrow x \ge 0

&

1 - x > 0 \Rightarrow x \in [0,1)

g/f = {{\sqrt {1 - x} } \over {\sqrt x }}

\Rightarrow 1 - x \ge 0

&

x > 0 \Rightarrow x \in (0,1]

g - f = \sqrt {1 - x} - \sqrt x

\Rightarrow 1 - x \ge 0

&

x \ge 0 \Rightarrow x \in [0,1]

$\Rightarrow$

x \in (0,1)