Functions — Page 8 | JEE Mathematics

Q71

Let

f: \mathbb{R}-\{2,6\} \rightarrow \mathbb{R}

be real valued function defined as

f(x)=\dfrac{x^2+2 x+1}{x^2-8 x+12}

. Then range of

f

is

A

\left(-\infty,-\dfrac{21}{4}\right] \cup[1, \infty)

B

\left(-\infty,-\dfrac{21}{4}\right) \cup(0, \infty)

C

\left(-\infty,-\dfrac{21}{4}\right] \cup[0, \infty)

D

\left(-\infty,-\dfrac{21}{4}\right] \cup\left[\dfrac{21}{4}, \infty\right)

Correct Answer

Option C

Solution

$y=\dfrac{x^{2}+2 x+1}{x^{2}-8 x+12}$ $\Rightarrow(y-1) x^{2}-(8 y+2) x+12 y-1=0$ Let $y \neq 1$ , then $D \geq 0$

4(4 y+1)^{2}-4(y-1)(12 y-1) \geq 0

$\Rightarrow 16 y^{2}+1+8 y-\left(12 y^{2}-13 y+1\right) \geq 0$ $\Rightarrow 4 y^{2}+21 y \geq 0$ $\Rightarrow y \in\left(-\infty,-\dfrac{21}{4}\right) \cup[0, \infty)-\{1\}$

\text{ for } y=1 \text{, }

\begin{aligned} & -8 x+12=2 x+1 \\\\ & x=\frac{11}{10} \quad \therefore I \in R \end{aligned}

$\therefore$ Range $=\left(-\infty,-\dfrac{21}{4}\right] \cup[0, \infty)$

Q72

Let

f(x) = \left| \begin{array}{lll}{1 + {{\sin }^2}x} & {{{\cos }^2}x} & {\sin 2x} \\ {{{\sin }^2}x} & {1 + {{\cos }^2}x} & {\sin 2x} \\ {{{\sin }^2}x} & {{{\cos }^2}x} & {1 + \sin 2x} \end{array} \right|,\,x \in \left[ {{\pi \over 6},{\pi \over 3}} \right]

. If

\alpha

and

\beta

respectively are the maximum and the minimum values of

f

, then

A

{\alpha ^2} - {\beta ^2} = 4\sqrt 3

B

{\beta ^2} - 2\sqrt \alpha = {{19} \over 4}

C

{\beta ^2} + 2\sqrt \alpha = {{19} \over 4}

D

{\alpha ^2} + {\beta ^2} = {9 \over 2}

Correct Answer

Option B

Solution

f(x) = \left| \begin{array}{lll}{1 + {{\sin }^2}x} & {{{\cos }^2}x} & {\sin 2x} \\ {{{\sin }^2}x} & {1 + {{\cos }^2}x} & {\sin 2x} \\ {{{\sin }^2}x} & {{{\cos }^2}x} & {1 + \sin 2x} \end{array} \right|

$C_{1} \rightarrow C_{1}+C_{2}+C_{3}$

\begin{aligned} & = (2+\sin 2 x)\left|\begin{array}{ccc}1 & \cos ^{2} x & \sin 2 x \\1 & 1+\cos ^{2} x & \sin 2 x \\1 & \cos ^{2} x & 1+\sin 2 x\end{array}\right| \\\\ & R_{2} \rightarrow R_{2} \rightarrow R_{1} ; R_{3} \rightarrow R_{3} \rightarrow R_{1} \\\\ & = (2+\sin 2 x)\left|\begin{array}{ccc}1 & \cos ^{2} x & \sin 2 x \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right| \\\\ & f(x)=2+\sin 2 x ; x \in\left[\frac{\pi}{6}, \frac{\pi}{3}\right] \\\\ & f(x)_{\max }=2+1=3 \text{ for } x=\frac{\pi}{4} \\\\ & f(x)_{\min }=2+\frac{\sqrt{3}}{2} \text{ for } x=\frac{\pi}{6}, \frac{\pi}{3} \\\\ & \beta^{2}-2 \sqrt{\alpha}=4+\frac{3}{4}+2 \sqrt{3}-2 \sqrt{3} \\\\ & =\frac{19}{4} \end{aligned}

Q73

If

f(x) = {{{2^{2x}}} \over {{2^{2x}} + 2}},x \in \mathbb{R}

, then

f\left( {{1 \over {2023}}} \right) + f\left( {{2 \over {2023}}} \right)\, + \,...\, + \,f\left( {{{2022} \over {2023}}} \right)

is equal to

A 2011

B 2010

C 1010

D 1011

Correct Answer

Option D

Solution

\begin{aligned} & f(x)=\frac{4^x}{4^x+2} \\\\ & f(x)+f(1-x)=\frac{4^x}{4^x+2}+\frac{4^{1-x}}{4^{1-x}+2} \\\\ & =\frac{4^x}{4^x+2}+\frac{4}{4+2\left(4^x\right)} \\\\ & =\frac{4^x}{4^{\mathrm{x}}+2}+\frac{2}{2+4^{\mathrm{x}}} \\\\ & =1 \\\\ & \Rightarrow \mathrm{f}(\mathrm{x})+\mathrm{f}(1-\mathrm{x})=1 \\\\ & \text{ Now } \mathrm{f}\left(\frac{1}{2023}\right)+\mathrm{f}\left(\frac{2}{2023}\right)+\mathrm{f}\left(\frac{3}{2023}\right)+\ldots \ldots .+ \\\\ & \ldots \ldots \ldots . .+\mathrm{f}\left(1-\frac{3}{2023}\right)+\mathrm{f}\left(1-\frac{2}{2023}\right)+\mathrm{f}\left(1-\frac{1}{2023}\right) \end{aligned}

Now sum of terms equidistant from beginning and end is 1

\begin{aligned} & \text{ Sum }=1+1+1+\ldots \ldots \ldots+1 \text{ (1011 times }) \\\\ & =1011 \end{aligned}

Q74

If the domain of the function

f(x)=\dfrac{[x]}{1+x^{2}}

, where

[x]

is greatest integer

\leq x

, is

[2,6)

, then its range is

A

\left(\dfrac{5}{37}, \dfrac{2}{5}\right]-\left\{\dfrac{9}{29}, \dfrac{27}{109}, \dfrac{18}{89}, \dfrac{9}{53}\right\}

B

\left(\dfrac{5}{37}, \dfrac{2}{5}\right]

C

\left(\dfrac{5}{26}, \dfrac{2}{5}\right]

D

\left(\dfrac{5}{26}, \dfrac{2}{5}\right]-\left\{\dfrac{9}{29}, \dfrac{27}{109}, \dfrac{18}{89}, \dfrac{9}{53}\right\}

Correct Answer

Option B

Solution

$f(x)=\dfrac{k}{1+x^{2}}$ is a decreasing function where $k>0$

\begin{gathered} \therefore \quad x \in[2,3) \Rightarrow f(x)=\frac{2}{1+x^{2}} \in\left(\frac{2}{10}, \frac{2}{5}\right]=R_{1} \\\\ x \in[3,4) \Rightarrow f(x)=\frac{3}{1+x^{2}} \in\left(\frac{3}{17}, \frac{3}{10}\right]=R_{2} \\\\ x \in[4,5) \Rightarrow f(x)=\frac{4}{1+x^{2}} \in\left(\frac{4}{26}, \frac{4}{17}\right]=R_{3} \\\\ x \in[5,6) \Rightarrow f(x)=\frac{5}{1+x^{2}} \in\left(\frac{5}{37}, \frac{5}{26}\right]=R_{4} \\\\ \text{ Range }=R_{1} \cup R_{2} \cup R_{3} \cup R_{4} \\\\ =\left(\frac{5}{37}, \frac{2}{5}\right] \end{gathered}

Q75

The range of the function

f(x)=\sqrt{3-x}+\sqrt{2+x}

is :

A

[2 \sqrt{2}, \sqrt{11}]

B

[\sqrt{5}, \sqrt{13}]

C

[\sqrt{2}, \sqrt{7}]

D

[\sqrt{5}, \sqrt{10}]

Correct Answer

Option D

Solution

f(x) = \sqrt {3 - x} + \sqrt {x + 2}

y' = {{ - 1} \over {2\sqrt 3 - x}} + {1 \over {2\sqrt {x + 2} }} = 0

\Rightarrow \sqrt x + 2 = \sqrt 3 - x

\Rightarrow x = {1 \over 2}

y\left( {{1 \over 2}} \right) = \sqrt {{5 \over 2}} + \sqrt {{5 \over 2}} = \sqrt {10}

{y_{\min }}

at

x = - 2

or

x = 3

, i.e.,

\sqrt 5

$\therefore$

f(x) \in [\sqrt 5 ,\sqrt {10} ]

Q76

Consider a function

f:\mathbb{N}\to\mathbb{R}

, satisfying

f(1)+2f(2)+3f(3)+....+xf(x)=x(x+1)f(x);x\ge2

with

f(1)=1

. Then

\dfrac{1}{f(2022)}+\dfrac{1}{f(2028)}

is equal to

A 8000

B 8400

C 8100

D 8200

Correct Answer

Option C

Solution

f(1) + 2f(2) + 3f(3)\, + \,...\, + \,nf(n) = n(n + 1) + (n)

..... (i)

n \to n + 1

f(1) + 2f(2)\, + \,...\, + \,(n + 1)f(n + 1) = (n + 1)(n + 2)f(n + 1)

...... (ii) (i) and (ii) gives

3f(3) - 2f(2) = 0

4f(4) - 3f(3) = 0

$\vdots$

(n + 1)f(n + 1) - nf(n) = 0

\Rightarrow f(n + 1) = {{2f(2)} \over {n + 1}}

f(n) = {1 \over {2n}}

{1 \over {f(2022)}} + {1 \over {f(2028)}} = 8100

Q77

The domain of

f(x) = {{{{\log }_{(x + 1)}}(x - 2)} \over {{e^{2{{\log }_e}x}} - (2x + 3)}},x \in \mathbb{R}

is

A

( - 1,\infty ) - \{ 3\}

B

\mathbb{R} - \{ - 1,3)

C

(2,\infty ) - \{ 3\}

D

\mathbb{R} - \{ 3\}

Correct Answer

Option C

Solution

$x-2>0 \Rightarrow x>2$ $\mathrm{x}+1>0 \Rightarrow \mathrm{x}>-1$ $x+1 \neq 1 \Rightarrow x \neq 0$ and $x>0$ Denominator $\mathrm{x}^{2}-2 \mathrm{x}-3 \neq 0$ $(x-3)(x+1) \neq 0$ $\mathrm{x} \neq-1,3$ So Ans $(2, \infty)-\{3\}$

Q78

Let

f:R \to R

be a function such that

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

. Then

A

f(x)

is many-one in

( - \infty , - 1)

B

f(x)

is one-one in

( - \infty ,\infty )

C

f(x)

is one-one in

[1,\infty )

but not in

( - \infty ,\infty )

D

f(x)

is many-one in

(1,\infty )

Correct Answer

Option C

Solution

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

\Rightarrow f'(x) = {{({x^2} + 1)(2x + 2) - ({x^2} + 2x + 1)(2x)} \over {{{({x^2} + 1)}^2}}}

\Rightarrow f'(x) = {{2 - 2{x^2}} \over {{{({x^2} + 1)}^2}}}

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

$\therefore$

f'(x) = 0

at x = 1 and x = -1. So, x = 1 and x = -1 are point of maxima/minima. Option A :

f(x)

is many-one in

( - \infty , - 1)

. As x = -1 is a point of maxima/minima. So it is a boundary point in the range

( - \infty , - 1)

. So,

f(x)

is one-one in

( - \infty , - 1)

. $\therefore$ Option A is incorrect. Option B :

f(x)

is one-one in

( - \infty ,\infty )

. As, x = 1 and x = -1 are point of maxima/minima which is present inside the range

( - \infty ,\infty )

. So,

f(x)

is many-one function in

( - \infty ,\infty )

. $\therefore$ Option B is incorrect. Option C :

f(x)

is one-one in

[1,\infty )

but not in

( - \infty ,\infty )

. As x = 1 is a point of maxima/minima. So it is a boundary point in the range

[1,\infty )

. So,

f(x)

is one-one in

[1,\infty )

. As, x = 1 and x = -1 are point of maxima/minima which is present inside the range

( - \infty ,\infty )

. So,

f(x)

is many-one function in

( - \infty ,\infty )

. $\therefore$ Option C is correct. Option D :

f(x)

is many-one in

(1,\infty )

. As x = 1 is a point of maxima/minima. So it is a boundary point in the range

[1,\infty )

. So,

f(x)

is one-one in

[1,\infty )

.

$\therefore$ Option D is incorrect.

Note : Methods to Check One-One Function : (i) If a function is one-one, any line parallel to $x$ -axis cuts the graph of the function maximum at one point. (ii) Any continuous function which is entirely increasing or decreasing (no maxima/minima is present) in whole domain will always be one-one function.

Q79

The number of functions

f:\{ 1,2,3,4\} \to \{ a \in Z|a| \le 8\}

satisfying

f(n) + {1 \over n}f(n + 1) = 1,\forall n \in \{ 1,2,3\}

is

A 2

B 3

C 1

D 4

Correct Answer

Option A

Solution

$\because f:\{1,2,3,4\} \rightarrow\{a \in \mathbb{Z}:|9| \leq 8\}$ and $f(n)+\dfrac{1}{n} f(n+1)=1$ $\Rightarrow n f(n)+f(n+1)=n \quad \ldots$ (i) $\therefore f(1)+f(2)=1 \Rightarrow f(2)=1-f(1)$ But $f(1) \in[-8,8]$ Hence, $f(2) \in[-8,8] \Rightarrow f(1) \in[-7,8] \quad\ldots(\mathrm{A})$ and $2 f(2)+f(3)=2 \Rightarrow f(3)=2 f(1)$ $\therefore 2 f(1) \in[-8,8] \Rightarrow f(1) \in[-4,4] \quad\ldots(\mathrm{B})$ and $3 f(3)+f(4)=3 \Rightarrow f(4)=3-6 f(1)$ $\therefore f(1) \in\left[-\dfrac{5}{6}, \dfrac{11}{6}\right]\quad...(C)$ From (A), (B) and (C) : $f(1)=0$ or 1 $\therefore$ Only two functions are possible.

Q80

Let

f:\mathbb{R}\to\mathbb{R}

be a function defined by

f(x) = {\log _{\sqrt m }}\{ \sqrt 2 (\sin x - \cos x) + m - 2\}

, for some

m

, such that the range of

f

is [0, 2]. Then the value of

m

is _________

A 4

B 3

C 5

D 2

Correct Answer

Option C

Solution

We know that $\sin x-\cos x \in[-\sqrt{2}, \sqrt{2}]$

\begin{aligned} & \log _{\sqrt{M}}(\sqrt{2}(\sin x-\cos ) +M-2) \\\\ &\quad\quad\in {\left[\log _{\sqrt{M}}(M-4), \log _{\sqrt{M}} M\right] } \end{aligned}

$\Rightarrow \log _{\sqrt{M}}(M-4)=0 \Rightarrow M=5$