Functions

Let $x$ and $y$ be any two elements in the domain (Z), such that

$ \begin{equation*} f(x+y)=f(x)+f(y) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{equation*} $

Differentiating above expression w.r.t. $y$, keeping $x$ constant, we get

$ f^{\prime}(x+y)=f^{\prime}(y) $

Let $y=0 \Rightarrow f^{\prime}(x+0)=f^{\prime}(0)$

and Assume $f^{\prime}(0)=K$

$ \therefore \quad f^{\prime}(x)=K $

Integrating on both sides, we get

$ \begin{aligned} & f(x)=K x+C \\ & \quad\{C \rightarrow \text { integration constant }\}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{aligned} $

Now putting $x=y=0$ in Eq. (i), we get $f(0+0)=f(0)+f(0) \Rightarrow f(0)=2 f(0)$ or $f(0)=0$

Let $x=0$, from Eq. (ii)

$ \begin{aligned} f(0) & =K \times 0+C \text { or } 0=0+C \text { or } C=0 \\ \therefore f(x) & =k x \end{aligned} $

Case (i) If $k>0$, then $f(x)$ is strictly increasing.

Case (ii) If $k<0$, then $f(x)$ is strictly decreasing

So, function is injective

Also $f(x)$ where every element in the codomain is a valid output of the function i.e., range is equal to codomain.

So, function is surjective also. Therefore, there are two bijective functions.

Given that, $A_n=\{(n+1) k \mid k \in \mathbf{N}\}$

If $n=1, A_1=\{2 k \mid k \in \mathbf{N}\}$

for, $k=1,2,3 \ldots A_1=\{2,4,6,8, \ldots\}$

If $n=2, A_2=\{3 k \mid k \in \mathbf{N}\}$

for $k=1,2,3, \ldots \quad A_2=\{3,6,9, \ldots\}$

If $n=3, A_3=\{4 k \mid k \in \mathbf{N}\}$

for $k=1,2,3, \ldots A_3=\{4,8,12, \ldots\}$

Now, $X=\bigcup_{n \in \mathbf{N}} A_n$

or $X=A_1 \cup A_2 \cup A_3 \ldots$

or $\quad X=\{2,3,4,5,6, \ldots\}$

Here, $f: X \rightarrow N$ and $f(x)=x \forall x \in \mathbf{X}$

$\because f(x)$ is linear function i.e., it is one-one function but there is no value of $x$ in domain where $f(x)=1$

$\therefore f(x)$ is not onto function.

We have,

$ f(n)= \begin{cases}2 n, & \text { if } n>0 \\ 1, & \text { if } n=0 \\ -2 n-1, & \text { if } n<0\end{cases} $

When, $n>0$

$ f(n)=2,4,6,8 \ldots \ldots $

When, $n<0$

$ f(n)=1,3,5,7 \ldots . . $

∴ Range of $f(n)$ is $\{1,2,3,4, \ldots \ldots\}$.

$\therefore \quad$ Codomain $=$ Range

Hence, $f(n)$ is onto.

Here, $f(0)-f(-1)=1$

$\therefore f$ is not one one.

We have,

$ \begin{aligned} \frac{x^5-5}{x^5+x^2} & =f(x)+\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1} \\ \frac{x^5-5}{x^3+x^2} & =x^2-x+1+\frac{-x^2-5}{x^3+x^2} \\ \therefore \frac{-x^2-5}{x^3+x^2} & =\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1} \\ -x^2-5 & =A x(x+1)+B(x+1)+C\left(x^2\right) \\ -x^2-5 & =(A+C) x^2+(A+B) x+B \\ A+C & =-1, A+B=0, B=-5 \end{aligned} $

Solving we get, $A=5, B=-5, C=-6$

$ \begin{array}{ll} \therefore & f(x)=x^2-x+1 \\ \therefore & f(K)=K^2-K+1 \end{array} $

Given, $f(K)+A+B+C=1$

$ \begin{aligned} & & K^2-K+1+5-5-6 & =1 \\ \Rightarrow & & K^2-K-6 & =0 \\ \Rightarrow & & (K+2)(K-3) & =0 \\ \Rightarrow & & K & =3,-2 \end{aligned} $

Largest value of $K=3$

$ \begin{aligned} &\\ &\begin{aligned} & \text { We have, } f(x)=x-\frac{1}{x} \mathrm{gl} \\ & \therefore(f(x))^3=x^3-\frac{1}{x^3}-3(x)\left(\frac{1}{x}\right)\left(x-\frac{1}{x}\right) \\ & =f\left(x^3\right)-3\left(x-\frac{1}{x}\right) \\ & \Rightarrow \quad(f(x))^3=f\left(x^3\right)-3 f(x) \\ & \Rightarrow \quad 3 f(x)=f\left(x^3\right)-[f(x)]^3 \end{aligned} \end{aligned} $

We have,

$ f(x)=[x] \text { and } g(x)=3\left[\frac{x}{3}\right] $

Given, $f(x)=g(x)$

$ \therefore \quad[x]=3\left[\frac{x}{3}\right] $

Here, $[x]$ and $\left[\frac{x}{3}\right]$ is an integers

Let $\quad\left[\frac{x}{3}\right]=k$

$ \begin{array}{lc} \because {[x]=3 k} \\ \because x \in[3 k, 3 k+1) \\ \{x \in \mathbf{R} / 3 k \leq x<3 k+1, k \in \mathbf{Z}\} \end{array} $

$ \begin{aligned} & \text { Given, } f(x+y)=f(x) \cdot f(y) \\ & \therefore \,\,\,\,\,\,\,\,\,\,\,\,\quad f(x)=a^x \\ & \,\,\,\,\,\,\,\,\,\,\,\,f(1)=a^1=2 \Rightarrow a=2 \end{aligned} $

$ \begin{aligned} &\therefore \quad f(x)=2^x\\ &\text { Area enclosed by the lines }\\ &\begin{aligned} & 2|x|+5|y| \leq 4 \\ & 4\left(\frac{1}{2} \times 2 \times \frac{4}{5}\right)=\frac{16}{5} \end{aligned} \end{aligned} $

$ \begin{aligned} & =\frac{16}{1+(2)^2} \\ =\frac{(2)^4}{1+(2)^2}= & \frac{f(4)}{1+f(2)} \end{aligned} $

We have,

$ f(x)= \begin{cases}\frac{60-5 x}{3}, & 0 \leq x \leq 6 \\ 10, & 6 \leq x \leq 7 \\ 31-3 x, & 7 \leq x \leq 10\end{cases} $

Now, $f(x)=10 \forall x \in[6,7]$

So, $f(x)$ is not one-one.

Again,

$ \begin{array}{ll} & 0 \leq x \leq 6 \\ \Rightarrow & -30 \leq-5 x \leq 0 \\ \Rightarrow & 60-30 \leq 60-5 x \leq 60 \\ \Rightarrow & \frac{30}{3} \leq \frac{60-5 x}{3} \leq \frac{60}{3} \\ \Rightarrow & 10 \leq \frac{60-5 x}{3} \leq 20 \text { and } 7 \leq x \leq 10 \\ \Rightarrow & -30 \leq-3 x \leq-21 \\ \Rightarrow & 1 \leq 31-3 x \leq 10 \end{array} $

∴ Range of $f(x)=[1,20]$

It is given that co-domain of $f(x)=[1,20]$

∴ Range of $f(x)=$ co-domain of $f(x)$

So, $f(x)$ is onto.

We have,

$ f(x)=\sqrt{\log _{10}\left(\frac{5 x-x^2}{4}\right)} $

For $f(x)$ to be defined

$ \begin{aligned} & \log _{10}\left(\frac{5 x-x^2}{4}\right) \geq 0 \text { and } \frac{5 x-x^2}{4}>0 \\ \Rightarrow & \quad \frac{5 x-x^2}{4} \geq 1 \text { and } \frac{5 x-x^2}{4}>0 \\ \Rightarrow & \quad \frac{5 x-x^2}{4} \geq 1 \\ \Rightarrow & \quad 5 x-x^2 \geq 4 \\ \Rightarrow & x^2-5 x+4 \leq 0 \\ \Rightarrow & (x-4)(x-1) \leq 0 \end{aligned} $

$ \begin{aligned} &\therefore \quad x \in[1,4]\\ &\text { So, domain of } f(x) \text { is }[1,4] \text {. } \end{aligned} $

(a) $f(x) = \sin \left[ {{\pi \over 6}\sin \left( {{\pi \over 2}\sin x} \right)} \right],\,x \in R$

$ = \sin \left( {{\pi \over 6}\sin \theta } \right),\,\theta \in \left[ { - {\pi \over 2},{\pi \over 2}} \right]$,

where, $\theta = {\pi \over 2}\sin x$

$ = \sin \alpha ,\alpha \in \left[ { - {\pi \over 6},{\pi \over 6}} \right]$,

where, $\alpha = {\pi \over 6}\sin \theta $

$\therefore$ $f(x) \in \left[ { - {1 \over 2},{1 \over 2}} \right]$

Hence, range of $f(x) \in \left[ { - {1 \over 2},{1 \over 2}} \right]$

So, option (a) is correct.

(b) $f\{ g(x)\} = f(t),t \in \left[ { - {\pi \over 2},{\pi \over 2}} \right]$

$ \Rightarrow f(t) \in \left[ { - {1 \over 2},{1 \over 2}} \right]$

$\therefore$ Option (b) is correct.

(c) $\mathop {\lim }\limits_{x \to 0} {{f(x)} \over {g(x)}}$

$ = \mathop {\lim }\limits_{x \to 0} {{\sin \left[ {{\pi \over 6}\sin \left( {{\pi \over 2}\sin x} \right)} \right]} \over {{\pi \over 2}(\sin x)}}$

$ = \mathop {\lim }\limits_{x \to 0} {{\sin \left[ {{\pi \over 6}\sin \left( {{\pi \over 2}\sin x} \right)} \right]} \over {{\pi \over 6}\sin \left( {{\pi \over 2}\sin x} \right)}}\,.\,{{{\pi \over 6}\sin \left( {{\pi \over 2}\sin x} \right)} \over {\left( {{\pi \over 2}\sin x} \right)}}$

$ = 1 \times {\pi \over 6} \times 1 = {\pi \over 6}$

$\therefore$ Option (c) is correct.

(d) $g\{ f(x)\} = 1$

$ \Rightarrow {\pi \over 2}\sin \{ f(x)\} = 1$

$ \Rightarrow \sin \{ f(x)\} = {2 \over \pi }$ ..... (i)

But, $f(x) \in \left[ { - {1 \over 2},{1 \over 2}} \right] \subset \left[ { - {\pi \over 6},{\pi \over 6}} \right]$

$\therefore$ $\sin \{ f(x)\} \in \left[ { - {1 \over 2},{1 \over 2}} \right]$ ..... (ii)

$ \Rightarrow \sin \{ f(x)\} \ne {2 \over \pi }$, [from Eqs. (i) and (ii)]

i.e. No solution.

$\therefore$ Option (d) is not correct.

$\text { Given } f_2(x)=x^2 \text { for } x \in[0, \infty) \text { and co-domain } \mathrm{R} \text {. }$

It is clear that $f_2(x)$ is continuous, differential and one-one but not onto because range of $f_2(x)$ is $[0, \infty)$ and given co-domain is R.

Hence $S$ match with 4.

Given, $f_3(x)=\left\{\begin{array}{cc}\sin x, & \text { if } x<0 \\ x, & \text { if } x \geq 0\end{array}\right.$. Domain of $f_2(x)$ is $R$ and co-domain $R$.

Draw the graph of $f_3(x)$

Range of $f_3(x) \in(-\infty, 1] \neq$ co-domain of $f_3(x)$

So, $f_3(x)$ is not onto.

$\because \sin x$ is not one-one function for $x \geq 0$

$\therefore f_3(x)$ is not one-one function.

We know $y=\sin x$ and $y=x$ both are continuous in the given domain.

At $\quad x=0$

$\begin{aligned} \lim _{x \rightarrow 0^{+}} f_3(x) & =\lim _{x \rightarrow 0^{+}} \sin x=0 \\ \lim _{x \rightarrow 0^{-}} f_3(x) & =\lim _{x \rightarrow 0^{-}} x=0 \\ f_3(0) & =\sin 0=0 \\ \Rightarrow \lim _{x \rightarrow 0^{+}} f_3(x) & =\lim _{x \rightarrow 0^{-}} f_3(x)=f_3(0) \end{aligned}$

$\therefore f_3(x)$ is continuous for all $x \in \mathrm{R}$.

We know that $y=\sin x$ and $y=x$ both are differentiable in the given domain R.H.D at $x=0$ is

$\lim _\limits{h \rightarrow 0} \frac{f_3(0+h)-f_3(0)}{h}=\lim _\limits{h \rightarrow 0} \frac{\sin h-0}{h}=1$

L.H.D at $x=0$ is

$\lim _\limits{h \rightarrow 0} \frac{f_3(0-h)-f_3(0)}{-h}=\lim _\limits{h \rightarrow 0} \frac{-h-0}{-h}=1$

$\Rightarrow$ R.H.D $=$ L.H.D at $x=0$

$\therefore f_3(x)$ is differentiable for all $x \in R$.

Hence, Q match with 3.

Now, $f_2{ }^{\circ} f_1(x)= \begin{cases}x^2, & x<0 \\ e^{2 x}, & x \geq 0\end{cases}$

Draw the graph of $y=f_2{ }^{\circ} f_1(x)$

It is clear that $f_2{ }^{\circ} f_1(x)$ is neither continuous nor one-one.

Hence, R match with 2.

Given, $f_4: \mathrm{R} \rightarrow[0, \infty)$ and

$f_4(x)=\left\{\begin{array}{l} f_2\left(f_1(x)\right), \text { if } x<0 \\ f_2\left(f_1(x)\right)-1, \text { if } x \geq 0 \end{array}\right.$

Draw the graph of $y=f_4(x)$

Range of $f_4(x)=$ co-domain of $f_4(x)=[0, \infty)$

$\therefore \quad f_4(x)$ is an onto function

$f_4(x)$ is continuous, on to but not one-one.

Hence, P match with 1.

Hint:

(i) If $f(x)$ is continuous at $x=a$, if and only if $\lim _\limits{x \rightarrow a^{+}} f(x)=\lim _\limits{x \rightarrow a^{-}} f(x)=f(a)$.

(ii) If $f(x)$ is differentiable at $x=a$ if and only if $\lim _\limits{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}=\lim _\limits{h \rightarrow 0} \frac{f(a-h)-f(a)}{-h}$.

(iii) If Range of $f(x)=\operatorname{co-domain}$ of $f(x)$, then $f(x)$ is called an onto function.

(iv) If $f\left(x_1\right)=f\left(x_2\right)$ is possible only for $x_1=x_2$, then $f(x)$ is called an one-one function.

$f(\cos 4\theta ) = {2 \over {2 - {{\sec }^2}\theta }}$

Let $\cos 4\theta = t$

$ \Rightarrow 2{\cos ^2}2\theta - 1 = t \Rightarrow {\cos ^2}2\theta = {2 \over 3}$

For $t = {1 \over 3}$ we have ${\cos ^2}2\theta = {2 \over 3}$

$\cos 2\theta = \sqrt {{2 \over 3}} $ or $\cos 2\theta = - \sqrt {{2 \over 3}} $

$f(\cos 4\theta ) = {2 \over {2 - {1 \over {{{\cos }^2}\theta }}}} = {{2{{\cos }^2}\theta } \over {2{{\cos }^2}\theta - 1}} = {{1 + \cos 2\theta } \over {\cos 2\theta }} = 1 + {1 \over {\cos 2\theta }}$

Hence, $f\left( {{1 \over 3}} \right) = 1 + \sqrt {{3 \over 2}} $ or $1 - \sqrt {{3 \over 2}} $

The function $f:[0,3] \to [1,29]$, defined by $f(x) = 2x^3 - 15x^2 + 36x + 1$, is :

First, calculate the derivative of $ f $ :

$ f'(x) = 6x^2 - 30x + 36 $

Which simplifies to :

$ f'(x) = 6(x^2 - 5x + 6) = 6(x - 2)(x - 3) $

For the given domain $[0, 3]$, the function $ f(x) $ is both increasing and decreasing. Therefore, $ f(x) $ is many-to-one.

Setting $ f'(x) = 0 $ gives the critical points :

$ x = 2, 3 $

We now evaluate the function at these points within the domain to determine the range :

$ f(0) = 1 $

$ f(2) = 29 $

$ f(3) = 28 $

Thus, the range of $ f $ is $[1, 29]$, indicating that the function is onto.

Here, $f(x) = {{b - x} \over {1 - bx}}$

where, 0 < b < 1, 0 < x < 1

For function to be invertible it should be one-one onto.

$\therefore$ Check range :

Let $f(x) = y \Rightarrow y = {{b - x} \over {1 - bx}}$

$ \Rightarrow y - bxy = b - x \Rightarrow x(1 - by) = b - y$

$ \Rightarrow x = {{b - y} \over {1 - by}}$

where, 0 < x < 1

$\therefore$ $0 < {{b - y} \over {1 - by}} < 1$

${{b - y} \over {1 - by}} > 0$ and ${{b - y} \over {1 - by}} < 1$

$ \Rightarrow y < b$ or $y > {1 \over b}$ .... (i)

${{(b - 1)(y + 1)} \over {1 - by}} < - 1 < y < {1 \over b}$ ..... (ii)

From Eqs. (i) and (ii), we get

$y \in \left( { - 1,{1 \over b}} \right) \subset $ Codomain

Thus, f(x) is not invertible.

$f(x) = {x^2}$, $g(x) = \sin x$

$(g \circ f)(x) = \sin {x^2}$

$g \circ (g \circ f)(x) = \sin (\sin {x^2})$

$(f \circ g \circ g \circ f)(x) = {(\sin (\sin {x^2}))^2}$ ..... (i)

Again, $(g \circ f)(x) = \sin {x^2}$

$(g \circ g \circ f)(x) = \sin (\sin {x^2})$ ..... (ii)

Given, $(f \circ g \circ g \circ f)(x) = (g \circ g \circ f)(x)$

$ \Rightarrow {(\sin (\sin {x^2}))^2} = \sin (\sin {x^2})$

$ \Rightarrow \sin (\sin {x^2})\{ \sin (\sin {x^2}) - 1\} = 0$

$ \Rightarrow \sin (\sin {x^2}) = 0$ or $\sin (\sin {x^2}) = 1$

$ \Rightarrow \sin {x^2} = 0$ or $\sin {x^2} = {\pi \over 2}$

$\therefore$ ${x^2} = n\pi $ (i.e. not possible as $ - 1 \le \sin \theta \le 1$)

$x = \pm \sqrt {n\pi } $