Functions — Page 6 | JEE Mathematics

Q51

Let f : R

-

{3}

\to

R

-

{1} be defined by f(x) =

{{x - 2} \over {x - 3}}

. Let g : R

\to

R be given as g(x) = 2x

-

3. Then, the sum of all the values of x for which f

-

1(x) + g

-

1(x) =

{{13} \over 2}

is equal to :

A 3

B 5

C 2

D 7

Correct Answer

Option B

Solution

Finding inverse of f(x)

y = {{x - 2} \over {x - 3}} \Rightarrow xy - 3y = x - 2 \Rightarrow x(y - 1) = 3y - 2

$\therefore$

{f^{ - 1}}(x) = {{3x - 2} \over {x - 1}}

Similarly for

{g^{ - 1}}(x)

y = 2x - 3 \Rightarrow x = {{y + 3} \over 2} \Rightarrow {g^{ - 1}}(x) = {{x + 3} \over 2}

$\therefore$

{{3x - 2} \over {x - 1}} + {{x + 3} \over 2} = {{13} \over 2}

\Rightarrow 6x - 4 + {x^2} + 2x - 3 = 13x - 13

\Rightarrow {x^2} - 5x + 6 = 0

\Rightarrow (x - 2)(x - 3) = 0

$\Rightarrow$ x = 2 or 3

Q52

Let [ x ] denote the greatest integer

\le

x, where x

\in

R. If the domain of the real valued function

f(x) = \sqrt {{{\left| {[x]} \right| - 2} \over {\left| {[x]} \right| - 3}}}

is (

-

\infty

, a)

]\cup

[b, c)

\cup

[4,

\infty

), a < b < c, then the value of a + b + c is :

A 8

B 1

C

-

2

D

-

3

Correct Answer

Option C

Solution

For domain,

{{{\left| {[x]} \right| - 2} \over {\left| {[x]} \right| - 3}}}

$\ge$ 0 Case I : When

{\left| {[x]} \right| - 2}

$\ge$ 0 and

{\left| {[x]} \right| - 3}

> 0 $\therefore$ x

\in

( $-$ $\infty$ , $-$ 3) $\cup$ [4, $\infty$ ) ...... (1) Case II : When

{\left| {[x]} \right| - 2}

$\le$ 0 and

{\left| {[x]} \right| - 3}

< 0 $\therefore$ x

\in

[ $-$ 2, 3) .....

(2) So, from (1) and (2) we get Domain of function = ( $-$ $\infty$ , $-$ 3) $\cup$ [ $-$ 2, 3) $\cup$ [4, $\infty$ ) $\therefore$ (a + b + c) = $-$ 3 + ( $-$ 2) + 3 = $-$ 2 (a < b < c) $\Rightarrow$ Option (3) is correct.

Q53

Let

f:R - \left\{ {{\alpha \over 6}} \right\} \to R

be defined by

f(x) = {{5x + 3} \over {6x - \alpha }}

. Then the value of

\alpha

for which (fof)(x) = x, for all

x \in R - \left\{ {{\alpha \over 6}} \right\}

, is :

A No such

\alpha

exists

B 5

C 8

D 6

Correct Answer

Option B

Solution

f(x) = {{5x + 3} \over {6x - \alpha }} = y

..... (i)

5x + 3 = 6xy - \alpha y

x(6y - 5) = \alpha y + 3

x = {{\alpha y + 3} \over {6y - 5}}

{f^{ - 1}}(x) = {{\alpha x + 3} \over {6x - 5}}

...... (ii) fo

f(x) = x

f(x) = {f^{ - 1}}(x)

From eqn (i) & (ii) Clearly

(\alpha = 5)

Q54

Let g : N

\to

N be defined as g(3n + 1) = 3n + 2, g(3n + 2) = 3n + 3, g(3n + 3) = 3n + 1, for all n

\ge

0. Then which of the following statements is true?

A There exists an onto function f : N

\to

N such that fog = f

B There exists a one-one function f : N

\to

N such that fog = f

C gogog = g

D There exists a function : f : N

\to

N such that gof = f

Correct Answer

Option A

Solution

g : N $\to$ N g(3n + 1) = 3n + 2, g(3n + 2) = 3n + 3, g(3n + 3) = 3n + 1

g(x) = \left[ \begin{array}{ll}{x + 1} & {x = 3k + 1} \\ {x + 1} & {x = 3k + 2} \\ {x - 2} & {x = 3k + 3} \end{array} \right.

g\left( {g(x)} \right) = \left[ \begin{array}{ll}{x + 2} & {x = 3k + 1} \\ {x - 1} & {x = 3k + 2} \\ {x - 1} & {x = 3k + 3} \end{array} \right.

g\left( {g\left( {g\left( x \right)} \right)} \right) = \left[ \begin{array}{ll}x & {x = 3k + 1} \\ x & {x = 3k + 2} \\ x & {x = 3k + 3} \end{array} \right.

If f : N $\to$ N, if is a one-one function such that f(g(x)) = f(x) $\Rightarrow$ g(x) = x, which is not the case If f : N $\to$ N f is an onto function such that f(g(x)) = f(x), one possibility is

f(x) = \left[ \begin{array}{ll}x & {x = 3n + 1} \\ x & {x = 3n + 2} \\ x & {x = 3n + 3} \end{array} \right.

n

\in

N0 Here f(x) is onto, also f(g(x)) = f(x) $\forall$ x

\in

N

Q55

Let f : R

\to

R be defined as

f(x + y) + f(x - y) = 2f(x)f(y),f\left( {{1 \over 2}} \right) = - 1

. Then, the value of

\sum\limits_{k = 1}^{20} {{1 \over {\sin (k)\sin (k + f(k))}}}

is equal to :

A cosec2(21) cos(20) cos(2)

B sec2(1) sec(21) cos(20)

C cosec2(1) cosec(21) sin(20)

D sec2(21) sin(20) sin(2)

Correct Answer

Option C

Solution

f(x) = cos $\lambda$ x $\because$

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

= $-$ 1 So, $-$ 1 = cos

{\lambda \over 2}

$\Rightarrow$ $\lambda$ = 2 $\pi$ Thus f(x) = cos2 $\pi$ x Now k is natural number Thus f(k) = 1

\sum\limits_{k = 1}^{20} {{1 \over {\sin k\sin (k + 1)}} = {1 \over {\sin 1}}\sum\limits_{k = 1}^{20} {\left[ {{{\sin \left( {(k + 1) - k} \right)} \over {\sin k.\sin (k + 1)}}} \right]} }

= {1 \over {\sin 1}}\sum\limits_{k = 1}^{20} {(\cot k - \cot (k + 1)}

)

= {{\cot 1 - \cot 21} \over {\sin 1}} = \cos e{c^2}1\cos ec(21).\sin 20

Q56

Consider function f : A

\to

B and g : B

\to

C (A, B, C

\subseteq

R) such that (gof)

-

1 exists, then :

A f and g both are one-one

B f and g both are onto

C f is one-one and g is onto

D f is onto and g is one-one

Correct Answer

Option C

Solution

$\therefore$ (gof) $-$ 1 exist $\Rightarrow$ gof is bijective $\Rightarrow$ 'f' must be one-one and 'g' must be ONTO.

Q57

Let f : N

\to

N be a function such that f(m + n) = f(m) + f(n) for every m, n

\in

N. If f(6) = 18, then f(2) . f(3) is equal to :

A 6

B 54

C 18

D 36

Correct Answer

Option B

Solution

f(m + n) = f(m) + f(n) Put m = 1, n = 1 f(2) = 2f(1) Put m = 2, n = 1 f(3) = f(2) + f(1) = 3f(1) Put m = 3, n = 3 f(6) = 2f(3) $\Rightarrow$ f(3) = 9 $\Rightarrow$ f(1) = 3, f(2) = 6 f(2) . f(3) = 6 $\times$ 9 = 54

Q58

The absolute minimum value, of the function

f(x)=\left|x^{2}-x+1\right|+\left[x^{2}-x+1\right]

, where

[t]

denotes the greatest integer function, in the interval

[-1,2]

, is :

A

\dfrac{3}{4}

B

\dfrac{3}{2}

C

\dfrac{1}{4}

D

\dfrac{5}{4}

Correct Answer

Option A

Solution

$\mathrm{f}(\mathrm{x})=\left|\mathrm{x}^{2}-\mathrm{x}+1\right|+\left[\mathrm{x}^{2}-\mathrm{x}+1\right] ; \mathrm{x} \in[-1,2]$ Let $g(x)=x^{2}-x+1$

=\left(x-\frac{1}{2}\right)^{2}+\frac{3}{4}

\because\left|\mathrm{x}^{2}-\mathrm{x}+1\right| \text{ and }\left[\mathrm{x}^{2}-\mathrm{x}+1\right]

Both have minimum value at $\mathrm{x}=1 / 2$ $\Rightarrow$ Minimum $\mathrm{f}(\mathrm{x})=\dfrac{3}{4}+0$ $=\dfrac{3}{4}$

Q59

The range of the function,

f(x) = {\log _{\sqrt 5 }}\left( {3 + \cos \left( {{{3\pi } \over 4} + x} \right) + \cos \left( {{\pi \over 4} + x} \right) + \cos \left( {{\pi \over 4} - x} \right) - \cos \left( {{{3\pi } \over 4} - x} \right)} \right)

is :

A

\left( {0,\sqrt 5 } \right)

B [

-

2, 2]

C

\left[ {{1 \over {\sqrt 5 }},\sqrt 5 } \right]

D [0, 2]

Correct Answer

Option D

Solution

f(x) = {\log _{\sqrt 5 }}\left( {3 + \cos \left( {{{3\pi } \over 4} + x} \right) + \cos \left( {{\pi \over 4} + x} \right) + \cos \left( {{\pi \over 4} - x} \right) - \cos \left( {{{3\pi } \over 4} - x} \right)} \right)

f(x) = {\log _{\sqrt 5 }}\left[ {3 + 2\cos \left( {{\pi \over 4}} \right)\cos (x) - 2\sin \left( {{{3\pi } \over 4}} \right)\sin (x)} \right]

f(x) = {\log _{\sqrt 5 }}\left[ {3 + \sqrt 2 (\cos x - \sin x)} \right]

Since

- \sqrt 2 \le \cos x - \sin x \le \sqrt 2

\Rightarrow {\log _{\sqrt 5 }}\left[ {3 + \sqrt 2 \left( { - \sqrt 2 } \right) \le f(x) \le {{\log }_{\sqrt 5 }}\left[ {3 + \sqrt 2 \left( {\sqrt 2 } \right)} \right]} \right]

\Rightarrow {\log _{\sqrt 5 }}(1) \le f(x) \le {\log _{\sqrt 5 }}(5)

So, Range of f(x) is [0, 2] Option (d)

Q60

Let a function f : N

\to

N be defined by

f(n) = \left[ \begin{array}{ll}{2n,} & {n = 2,4,6,8,......} \\ {n - 1,} & {n = 3,7,11,15,......} \\ {{{n + 1} \over 2},} & {n = 1,5,9,13,......} \end{array} \right.

then, f is

A one-one but not onto

B onto but not one-one

C neither one-one nor onto

D one-one and onto

Correct Answer

Option D

Solution

When n = 1, 5, 9, 13 then

{{n + 1} \over 2}

will give all odd numbers.

When n = 3, 7, 11, 15 ..... n $-$ 1 will be even but not divisible by 4 When n = 2, 4, 6, 8 .....

Then 2n will give all multiples of 4 So range will be N.

And no two values of n give same y, so function is one-one and onto.