Functions

Given,

$f(x)=l^{\log (\sin x)}+(\tan x)^3-\operatorname{cosec}(3 x-5)$

$ f(x)=\sin x+(\tan x)^3-\operatorname{cosec}(3 x-5) $

The period of $\sin x$ is $2 \pi$.

The period of $(\tan x)^3$ is $\pi$.

The period of $\operatorname{cosec}(3 x-5)$ is $\frac{2 \pi}{3}$.

The LCM of $2 \pi, \pi$ and $\frac{2 \pi}{3}$ is $2 \pi$.

So, the period of

$ f(x)=e^{\log (\sin x)}+(\tan x)^3-\operatorname{cosec}(3 x-5) $

be $2 \pi$

Consider the function $f(x)=\sqrt{16-x^2}$

Clearly, $f:[0,4] \rightarrow[0,4]$ defined by $f(x)=\sqrt{16-x^2}$ is a bijection.

Given, $f(x)=\frac{\sqrt{|x|-x}}{\sqrt{x-[x]}}$

Case I $|x|-x \geq 0$

$ \therefore \quad x \in R \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Case II $x-[x]>0$

$ \therefore \quad\{x\}>0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

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

Domain $=R-Z$

For interval $x \in(-\infty,-1)$

Range of ( $2 x-3$ )

$ \Rightarrow \quad(-\infty,-5) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

For interval $x \in[-1,1]$

Range of $1-x^2$

$ \Rightarrow \quad[0,1] \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

For interval $x \in(1, \infty)$

Range of $3 x^2+2$

$ \begin{aligned} & \Rightarrow \quad(5, \infty) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\\ & (-\infty,-5) \cup[0,1] \cup(5, \infty) \end{aligned} $

Given, $\sinh x=-\frac{4}{3}$

We know that, $\cosh ^2 x-\sinh ^2 x=1$,

$ \begin{aligned} & \cosh x>0 \forall x \in R \\ & \sinh 2 x=2 \sinh x \cosh x \end{aligned} $

$ \begin{aligned} &\text { and } \cosh 2 x=\cosh ^2 x+\sinh ^2 x\\ &\begin{aligned} \therefore \sinh 2 x+\cosh 2 x= & \sinh ^2 x+\cosh ^2 x +2 \sinh x \cosh x \\ = & (\sinh x+\cosh x)^2 \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \because \cosh ^2 x-\sinh ^2 x=1, \sinh x=-\frac{4}{3} \\ & \quad \cosh x>0 \forall x \in R \\ & \therefore \cosh x=\sqrt{1+\sinh ^2 x}=\sqrt{1+\left(-\frac{4}{3}\right)^2}=\frac{5}{3} \end{aligned}\\ &\text { Now, } \sinh 2 x+\cosh 2 x\\ &\begin{aligned} & =(\sinh x+\cosh x)^2 \\ & =\left(-\frac{4}{3}+\frac{5}{3}\right)^2=\left(\frac{1}{3}\right)^2=\frac{1}{9} \end{aligned} \end{aligned} $

$f: R \rightarrow R$ defined by

$ f(x)=\left\{\begin{array}{lc} 2 x-3, & x<-2 \\ x^2-1, & -2 \leq x \leq 2 \\ 3 x+2, & x>2 \end{array}\right. $

For $x=2, f(2)=2^2-1=3$

For $x=-2, f(-2)=(-2)^2-1=3$

$\therefore f$ is not one-one (not injective).

Now, for $x<-2 \Rightarrow 2 x-3<-7$

For $-2 \leq x \leq 2$

$ \Rightarrow \quad 0 \leq x^2 \leq 4 \Rightarrow-1 \leq x^2-1 \leq 3 $

For   $ x>2 $

$ 3 x+2>8 $

$\therefore f(x) \in(-\infty,-7) \cup[-1,3] \cup(8, \infty)$

⇒ Range of $f \subset$ co-domain of $f$.

$\therefore f$ is not onto and not surjective.

Hence, $f(x)$ is neither injection nor surjection.

$f(x)=\frac{\sqrt{\log _{10}\left(\frac{x}{x-2}\right)}}{\sqrt{[x]^2-5[x]+6}}$

Function is defined, when

$ \begin{aligned} & {[x]^2-5[x]+6>0} \\ & \Rightarrow \quad([x]-2)([x]-3)>0 \\ & \Rightarrow \quad[x] \in(-\infty, 2) \cup(3, \infty) \\ & \Rightarrow \quad x \in(-\infty, 2) \cup[4, \infty) \end{aligned} $

and $\log _{10}\left(\frac{x}{x-2}\right) \geq 0$

$ \begin{array}{lr} \Rightarrow & \frac{x}{x-2} \geq 1 \Rightarrow \frac{2}{x-2} \geq 0 \\ \Rightarrow & x-2 \geq 0 \Rightarrow x \geq 2 \\ \Rightarrow & x \in[2, \infty) \end{array} $

∴ Domain of $f(x)$ is $[4, \infty)$.

$ \begin{aligned} & \text { } f(x)=\frac{1}{x-|x|} \\ & \because \quad|x|>x \quad[\because x \neq|x|] \\ & \Rightarrow \quad x-|x|<0 \Rightarrow \frac{1}{x-|x|}<0 \\ & \Rightarrow \quad f(x) \in(-\infty, 0) \end{aligned} $

$ \begin{aligned} & \frac{6 x^4+13 x^3+2 x^2-x+3}{2 x^2+3 x-2} \\ & =3 x^2+2 x+1+\frac{5}{2 x^2+3 x-2} \end{aligned} $

$ \begin{aligned} &\begin{aligned} f(x) & =\frac{A}{a x-1}+\frac{B}{x+b} \\ & =3 x^2+2 x+1+\frac{5}{(2 x-1)(x+2)} \\ & =3 x^2+2 x+1+\frac{2}{2 x-1}+\frac{(-1)}{x+2} \end{aligned}\\ &\text { On comparing both sides, we get }\\ &\begin{gathered} f(x)=3 x^2+2 x+1 \\ A=2, B=-1 \\ a=2, b=2 \end{gathered}\\ &f(1)+a \cdot B+b \cdot A=6-2+4=8 \end{aligned} $

The given function is

$ f(x)=\sin ^{-1}\left(\log _2\left(\frac{x^2}{2}\right)\right) $

To find the domain of $f(x)$

Domain $(f(x))=$ Domain of

$ \sin ^{-1}\left(\log _2\left(\frac{x^2}{2}\right)\right) $

where, $\log _2\left(\frac{x^2}{2}\right) \in[-1,1]$

$ \begin{aligned} & \Rightarrow \quad-1 \leq \log _2\left(\frac{x^2}{2}\right) \leq 1 \\ & \Rightarrow \quad \frac{1}{2} \leq \frac{x^2}{2} \leq 2 \\ & \Rightarrow \quad 1 \leq x^2 \leq 4 \end{aligned} $

For, $x^2 \leq 4 \Rightarrow-2 \leq x \leq 2$

$ \Rightarrow x \in[-2,2] \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

and also, $x^2 \geq 1$

$ \Rightarrow x \in(-\infty,-1] \cup[1, \infty) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Now, from Eqs. (i) and (ii), taking intersection, we get

$ x \in[-2,-1] \cup[1,2] $

Given function,

$ \begin{aligned} f(x)= & -\sqrt{-x^2-6 x-5} \\ & =-\sqrt{-\left(x^2+6 x+5\right)} \end{aligned} $

Let $y=f(x)=-\sqrt{-(x+5)(x+1)}$

Now, squaring on both sides, we get

$ \begin{aligned} & y^2=-\left(x^2+6 x+5\right) \\ \Rightarrow & x^2+6 x+\left(5+y^2\right)=0 \end{aligned} $

Applying quadratic formula, we get

$ \begin{aligned} & x=\frac{-6 \pm \sqrt{36-4\left(5+y^2\right)}}{2} \\ & =\frac{-6 \pm \sqrt{16-4 y^2}}{2} \end{aligned} $

Clearly, $16-4 y^2 \geq 0 \Rightarrow y^2 \leq \frac{16}{4}=4$

$ \begin{array}{lc} \Rightarrow & y^2 \leq 4 \\ \Rightarrow & -2 \leq y \leq 2 \end{array} $

Clearly, $y<0$ because $\sqrt{-\left(x^2-6 x-5\right)}$ always positive.

So, $y \in[-2,0]$

Thus, range of given function is $[-2,0]$

Given function is, $f(x)=2 x+\sin x$, $x \in R$

For one one $f\left(x_1\right)=f\left(x_2\right)$

$ \Rightarrow 2 x_1+\sin x_1=2 x_2+\sin x_2 $

This can be possible only, when $x_1=x_2$

So, $f(x)$ is one-one.

For onto Range of

$ f(x)=R=\text { Co-domain } $

So, Range $=$ Co-domain

Thus, $f(x)$ is onto.

To find the locus of the centroid of triangle $ \triangle OAB $ where the vertices are $ A = (a \sec t, b \tan t) $, $ B = (-a \tan t, b \sec t) $, and $ O = (0,0) $, we calculate the centroid $(x, y)$ using the formula for the centroid of a triangle:

$ x = \frac{0 + a \sec t - a \tan t}{3} = \frac{a (\sec t - \tan t)}{3} $

Similarly,

$ y = \frac{0 + b \tan t + b \sec t}{3} = \frac{b (\sec t + \tan t)}{3} $

Next, we compute the product $ (3x)(3y) $:

$ 3x = a (\sec t - \tan t) \quad \text{and} \quad 3y = b (\sec t + \tan t) $

So,

$ (3x)(3y) = [a (\sec t - \tan t)][b (\sec t + \tan t)] = ab (\sec^2 t - \tan^2 t) $

Since $ \sec^2 t - \tan^2 t = 1 $ (using the identity $\sec^2 t - \tan^2 t = 1$), we have:

$ 9xy = ab \quad \text{or} \quad xy = \frac{ab}{9} $

Thus, the locus of the centroid of $ \triangle OAB $ satisfies the equation $ 9xy = ab $.

Let's explore the function $ f(x) = x^2 - 4x + 5 $, which is defined on the interval $[2, \infty)$.

First, consider the discriminant $\Delta$ of the quadratic equation:

$ \Delta = (-4)^2 - 4 \times 1 \times 5 = 16 - 20 = -4 $

Since the discriminant $\Delta$ is negative, the quadratic equation does not have real roots, which means it doesn't intersect the x-axis and is always positive.

Next, find the vertex of the parabola, as this will help us determine its minimum value. In a quadratic function of the form $ax^2 + bx + c$, the x-coordinate of the vertex is given by:

$ x_{\min} = -\frac{b}{2a} = \frac{4}{2} = 2 $

Since the vertex lies at $x = 2$ and the parabola is opening upwards (as the coefficient of $x^2$ is positive), $f(x)$ is increasing for $x \geq 2$.

Calculate $f(x)$ at $x = 2$:

$ f(2) = 2^2 - 4 \times 2 + 5 = 4 - 8 + 5 = 1 $

Therefore, $f(2) = 1$, which is the minimum value of the function on the given interval. As $x$ approaches infinity, $f(x)$ also approaches infinity.

Thus, the range of $f(x)$ is $[1, \infty)$.

To solve the problem, let's analyze the function $ f(x) = -|x| $.

First, let's evaluate $ f(-x) $:

$ f(-x) = -|-x| = -|x| $

Next, consider the composition $ f(f(x)) $:

$ f(f(x)) = -|f(x)| = -|-|x|| = -|x| = f(x) $

Now evaluate $ f(f(f(x))) $:

$ f(f(f(x))) = f(f(x)) = f(x) $

Thus, $ f(f(f(x))) = f(x) $.

Similarly, since $ f(x) = -|x| $, we have:

$ f(f(f(-x))) = f(f(-|x|)) = f(-|x|) = -|-|x|| = -|x| = f(x) $

Therefore, the expression $ (f \circ f \circ f)(x) + (f \circ f \circ f)(-x) $ simplifies to:

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

So, the result is:

$ 2f(x) $

$f(1) = a + b + c = 3$ ..... (i)

$f(3) = 9a + 3b + c = 4$ .... (ii)

$f(0) + f(1) + f( - 2) + f(3) = 14$

OR $c + 3 + (4a - 2b + c) + 4 = 14$

OR $4a - 2b + 2c = 7$ ..... (iii)

From (i) and (ii) $8a + 2b = 1$ ..... (iv)

From (iii) $ - (2) \times $ (i)

$ \Rightarrow 2a - 4b = 1$ ..... (v)

From (iv) and (v) $a = {1 \over 6},\,b = {{ - 1} \over 6}$ and $c = 3$

$f( - 2) = 4a - 2b + c$

$ = {4 \over 6} + {2 \over 6} + 3 = 4$

$f\left( {g\left( {{1 \over n}\sum\limits_{i = 1}^n {f({a_i})} } \right)} \right)$

${{{a_1} + {a_2} + {a_3}\, + \,......\, + \,{a_n}} \over n} = 0$

$\therefore$ First and last term, second and second last and so on are equal in magnitude but opposite in sign.

$f(x) = \alpha {x^5} + \beta {x^3} + \gamma x$

$\sum\limits_{i = 1}^n {f({a_i}) = \alpha \left( {a_1^5 + a_2^5 + a_3^5\, + \,.....\, + \,a_n^5} \right) + \beta \left( {a_1^3 + a_2^3\, + \,....\, + \,a_n^3} \right) + \gamma \left( {{a_1} + {a_2}\, + \,....\, + \,{a_n}} \right)} $

$ = 0\alpha + 0\beta + 0\gamma $

$ = 0$

$\therefore$ $f\left( {g\left( {{1 \over n}\sum\limits_{i = 1}^n {f({a_i})} } \right)} \right) = {1 \over n}\sum\limits_{i = 1}^n {f({a_i}) = 0} $

$f,g:N - \{ 1\} \to N$ defined as

$f(a) = \alpha $, where $\alpha$ is the maximum power of those primes p such that p^$\alpha$ divides a.

$g(a) = a + 1$,

Now,

$\matrix{ {f(2) = 1,} & {g(2) = 3} & \Rightarrow & {(f + g)\,(2) = 4} \cr {f(3) = 1,} & {g(3) = 4} & \Rightarrow & {(f + g)\,(3) = 5} \cr {f(4) = 2,} & {g(4) = 5} & \Rightarrow & {(f + g)\,(4) = 7} \cr {f(5) = 1,} & {g(5) = 6} & \Rightarrow & {(f + g)\,(5) = 7} \cr } $

$\because$ $(f + g)\,(5) = (f + g)\,(4)$

$\therefore$ $f + g$ is not one-one

Now, $\because$ ${f_{\min }} = 1,\,{g_{\min }} = 3$

So, there does not exist any $x \in N - \{ 1\} $ such that $(f + g)(x) = 1,2,3$

$\therefore$ $f + g$ is not onto

As function is one-one and onto, out of 50 elements of domain set 17 elements are following restriction $f(3) > f(9) > f(15)\,.......\, > f(99)$

So number of ways $ = {}^{50}{C_{17}}\,.\,1\,.\,33!$

$ = {}^{50}{P_{33}}$

Given, $f(1) + f(2) = f(3)$

It means $f(1),f(2)$ and $f(3)$ are dependent on each other. But there is no condition on $f(4)$, so $f(4)$ can be $f(4) = 1,2,3,4,5,6$.

For $f(1),f(2)$ and we have to find how many functions possible which will satisfy the condition $f(1) + f(2) = f(3)$

Case 1 :

When $f(3) = 2$ then possible values of $f(1)$ and $f(2)$ which satisfy $f(1) + f(2) = f(3)$ is $f(1) = 1$ and $f(2) = 1$.

And $f(4)$ can be = 1, 2, 3, 4, 5, 6

$\therefore$ Total possible functions $=1\times6=6$

Case 2 :

When $f(3) = 3$ then possible values

(1) $f(1) = 1$ and $f(2) = 2$

(2) $f(1) = 2$ and $f(2) = 1$

And $f(4)$ can be = 1, 2, 3, 4, 5, 6.

$\therefore$ Total functions $ = 2 \times 6 = 12$

Case 3 :

When $f(3) = 4$ then

(1) $f(1) = 1$ and $f(2) = 3$

(2) $f(1) = 2$ and $f(2) = 2$

(3) $f(1) = 3$ and $f(2) = 1$

And $f(4)$ can be = 1, 2, 3, 4, 5, 6

$\therefore$ Total functions $ = 3 \times 6 = 18$

Case 4 :

When $f(3) = 5$ then

(1) $f(1) = 1$ and $f(4) = 4$

(2) $f(1) = 2$ and $f(4) = 3$

(3) $f(1) = 3$ and $f(4) = 2$

(4) $f(1) = 4$ and $f(4) = 1$

And $f(4)$ can be = 1, 2, 3, 4, 5 and 6

$\therefore$ Total functions $ = 4 \times 6 = 24$

Case 5 :

When $f(3) = 6$ then

(1) $f(1) = 1$ and $f(2) = 5$

(2) $f(1) = 2$ and $f(2) = 4$

(3) $f(1) = 3$ and $f(2) = 3$

(4) $f(1) = 4$ and $f(2) = 2$

(5) $f(1) = 5$ and $f(2) = 1$

And $f(4)$ can be = 1, 2, 3, 4, 5 and 6

$\therefore$ Total possible functions $ = 5 \times 6 = 30$

$\therefore$ Total functions from those 5 cases we get

$ = 6 + 12 + 18 + 24 + 30$

$ = 90$

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.

$f:R \to R$ defined as

$f(x) = x - 1$ and $g:R \to \{ 1, - 1\} \to R,\,g(x) = {{{x^2}} \over {{x^2} - 1}}$

Now $fog(x) = {{{x^2}} \over {{x^2} - 1}} - 1 = {1 \over {{x^2} - 1}}$

$\therefore$ Domain of $fog(x) = R - \{ - 1,1\} $

And range of $fog(x) = ( - \infty , - 1] \cup (0,\infty )$

Now, ${d \over {dx}}(fog(x)) = {{ - 1} \over {{x^2} - 1}}\,.\,2x = {{2x} \over {1 - {x^2}}}$

$\therefore$ ${d \over {dx}}(fog(x)) > 0$ for ${{2x} \over {(1 - x)(1 + x)}} > 0$

$ \Rightarrow {x \over {(x - 1)(x + 1)}} < 0$

$\therefore$ $x \in ( - \infty , - 1) \cup (0,1)$

and ${d \over {dx}}(fog(x)) < 0$ for $x \in ( - 1,0) \cup (1,\infty )$

$\therefore$ $fog(x)$ is neither one-one nor onto.

Given,

$f(x) = {{x - 1} \over {x + 1}}$

Also given,

${f^{n + 1}}(x) = f({f^n}(x))$ ..... (1)

$\therefore$ For $n = 1$

${f^{1 + 1}}(x) = f({f^1}(x))$

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

$ = f\left( {{{x - 1} \over {x + 1}}} \right)$

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

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

$ = {{ - 2} \over {2x}}$

$ = - {1 \over x}$

From equation (1), when n = 2

${f^{2 + 1}}(x) = f({f^2}(x))$

$ \Rightarrow {f^3}(x) = f({f^2}(x))$

$ = f\left( { - {1 \over x}} \right)$

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

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

$ = {{ - 1 - x} \over { - 1 + x}} = {{ - (x + 1)} \over {x - 1}}$

Similarly,

${f^4}(x) = f({f^3}(x))$

$ = f\left( {{{ - x + 1} \over {x - 1}}} \right)$

$ = {{{{ - (x + 1)} \over {x - 1}} - 1} \over {{{ - (x + 1)} \over {x - 1}} + 1}}$

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

$ = {{ - 2x} \over { - 2}} = x$

$\therefore$ ${f^5}(x) = f({f^4}(x))$

$ = f(x)$

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

${f^6}(x) = f({f^5}(x))$

$ = f\left( {{{x - 1} \over {x + 1}}} \right)$

$ = - {1 \over x}$ (Already calculated earlier)

${f^7}(x) = f({f^6}(x))$

$ = f\left( { - {1 \over x}} \right)$

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

$ = {{ - (x + 1)} \over {x - 1}}$

$\therefore$ ${f^6}(6) = - {1 \over 6}$

and ${f^7}(7) = {{ - (7 + 1)} \over {7 - 1}} = - {8 \over 6}$

So, ${f^6}(6) + {f^7}(7)$

$ = - {1 \over 6} - {8 \over 6}$

$ = - {3 \over 2}$

Given,

$f(x + y) = 2f(x)f(y)$

and $f(1) = 2$

For x = 1 and y = 1,

$f(1 + 1) = 2f(1)f(1)$

$ \Rightarrow f(2) = 2{\left( {f(1)} \right)^2} = 2{(2)^2} = {2^3}$

For x = 1, y = 2,

$f(1 + 2) = 2f(1)y(2)$

$ \Rightarrow f(3) = 2\,.\,2\,.\,{2^3} = {2^5}$

For x = 1, y = 3,

$f(1 + 3) = 2f(1)f(3)$

$ \Rightarrow f(4) = 2\,.\,2\,.\,{2^5} = {2^7}$

For x = 1, y = 4,

$f(1 + 4) = 2f(1)f(4)$

$ \Rightarrow f(5) = 2\,.\,2\,.\,{2^7} = {2^9}$ ..... (1)

Also given

$\sum\limits_{k = 1}^{10} {f(\alpha + k) = {{512} \over 3}({2^{20}} - 1)} $

$ \Rightarrow f(\alpha + 1) + f(\alpha + 2) + f(\alpha + 3)\, + \,\,...\,\, + \,\,f(\alpha + 10) = {{512} \over 3}({2^{20}} - 1)$

$ \Rightarrow f(\alpha + 1) + f(\alpha + 2) + f(\alpha + 3)\, + \,\,....\,\, + f(\alpha + 10) = {{{2^9}\left( {{{({2^2})}^{10}} - 1} \right)} \over {{2^2} - 1}}$

This represent a G.P with first term = 2⁹ and common ratio = 2²

$\therefore$ First term $ = f(\alpha + 1) = {2^9}$ ..... (2)

From equation (1), $f(5) = {2^9}$

$\therefore$ From (1) and (2), we get

$f(\alpha + 1) = {2^9} = f(5)$

$ \Rightarrow f(\alpha + 1) = f(5)$

$ \Rightarrow f(\alpha + 1) = f(4 + 1)$

Comparing both sides we get,

$\alpha = 4$

$f(x) = {\log _e}({x^2} + 1) - {e^{ - x}} + 1$

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

$ = {2 \over {x + {1 \over x}}} + {e^{ - x}} > 0\,\,\forall x \in R$

$g(x) = {e^{ - x}} - 2{e^x}$

$g'(x) - - {e^{ - x}} - 2{e^x} < 0\,\,\,\,\forall x \in R$

$\Rightarrow$ f(x) is increasing and g(x) is decreasing function.

$f\left( {g\left( {{{{{(\alpha - 1)}^2}} \over 3}} \right)} \right) > f\left( {g\left( {\alpha - {5 \over 3}} \right)} \right)$

$ \Rightarrow {{{{(\alpha - 1)}^2}} \over 3} < \alpha - {5 \over 3}$

$ = {\alpha ^2} - 5\alpha + 6 < 0$

$ = (\alpha - 2)(\alpha - 3) < 0$

$ = \alpha \in (2,\,3)$

$ f(\theta)=\frac{1}{2}\left|\begin{array}{ccc} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{array}\right|+\left|\begin{array}{ccc} \sin \pi & \cos \left(\theta+\frac{\pi}{4}\right) & \tan \left(\theta-\frac{\pi}{4}\right) \\ \sin \left(\theta-\frac{\pi}{4}\right) & -\cos \frac{\pi}{2} & \log _{e}\left(\frac{4}{\pi}\right) \\ \cot \left(\theta+\frac{\pi}{4}\right) & \log _{e}\left(\frac{\pi}{4}\right) & \tan \pi \end{array}\right| $

Here $\cos \left(\theta+\frac{\pi}{4}\right)=-\sin \left(\theta-\frac{\pi}{4}\right)$

and $\tan \left(\theta-\frac{\pi}{4}\right)=-\cot \left(\theta+\frac{\pi}{4}\right)$

and $\log _e\left(\frac{4}{\pi}\right)=-\log _e\left(\frac{\pi}{4}\right)$

Also $\sin \pi=-\cos \frac{\pi}{2}=\tan \pi=0$

$ \Rightarrow $ $ f(\theta)=\frac{1}{2}\left|\begin{array}{ccc} 2 & \sin \theta & 1 \\ 0 & 1 & \sin \theta \\ 0 & -\sin \theta & 1 \end{array}\right|+\left|\begin{array}{ccc} 0 & -\sin \left(\theta-\frac{\pi}{4}\right) & \tan \left(\theta-\frac{\pi}{4}\right) \\ \sin \left(\theta-\frac{\pi}{4}\right) & 0 & \log _{\mathrm{e}}\left(\frac{4}{\pi}\right) \\ -\tan \left(\theta-\frac{\pi}{4}\right) & -\log _{\mathrm{e}}\left(\frac{4}{\pi}\right) & 0 \end{array}\right| $

$ \begin{aligned} & f(\theta)=\left(1+\sin ^2 \theta\right)+0 \text { (skew symmetric) } \\\\ & g(\theta)=\sqrt{f(\theta)-1}+\sqrt{f\left(\frac{\pi}{2}-\theta\right)-1} \\\\ & =|\sin \theta|+|\cos \theta| \quad \\\\ & g(\theta) \in[1, \sqrt{2}] \text { for } \theta \in\left[0, \frac{\pi}{2}\right] \end{aligned} $

Again let $\mathrm{P}(\mathrm{x})=\mathrm{k}(\mathrm{x}-\sqrt{2})(\mathrm{x}-1)$

$ \begin{aligned} & 2-\sqrt{2}=\mathrm{k}(2-\sqrt{2})(2-1) \\\\ & \Rightarrow \mathrm{k}=1 \quad(\mathrm{P}(2)=2-\sqrt{2} \text { given }) \\\\ & \therefore \mathrm{P}(\mathrm{x})=(\mathrm{x}-\sqrt{2})(\mathrm{x}-1) \end{aligned} $

For option (A) $P\left(\frac{3+\sqrt{2}}{4}\right)<0$ Correct.

For option (B) $P\left(\frac{1+3 \sqrt{2}}{4}\right)<0$ Incorrect.

For option (C) $P\left(\frac{5 \sqrt{2}-1}{4}\right)>0$ Correct.

For option (D) $P\left(\frac{5-\sqrt{2}}{4}\right)>0$ Incorrect.

$f(x)=\sqrt{\frac{[x]-x}{x-[x]}}$

$x \notin I$ : $\quad[\because$ for $x \in I$, function is indeterminate form]

For $x \in$ non integer values, we have

$ f(x)=\sqrt{\frac{-(x-[x])}{x-[x]}}=\sqrt{-1} $

For $x \in$ non integer values, $f(x)$ is not real.

Therefore, $x \in \phi$.

$f(x)=\frac{1}{\sqrt{x-[x]}}$

We know, $x-[x]=\{x\}$ and $0 \leq\{x\}<1$

$ \begin{aligned} f(x) & =\frac{1}{\sqrt{\{x\}}} \\ \because \quad 0 \leq\{x\}<1 & \Rightarrow 0 \leq \sqrt{\{x\}}<1 \end{aligned} $

$ \begin{array}{ll} \Rightarrow & \frac{1}{0}>\frac{1}{\sqrt{\{x\}}}>\frac{1}{1} \\ \Rightarrow & \infty>\frac{1}{\sqrt{\{x\}}}>1 \\ \Rightarrow & \infty>f(x)>1 \\ \Rightarrow & f(x) \in(1, \infty) \end{array} $

Assertion

$ \begin{aligned} \operatorname{coth} x & =\frac{1-k}{1+k}(0 < k<2) \\ \because \quad \operatorname{coth} x & =\frac{e^x+e^{-x}}{e^x-e^{-x}} \\ & =\frac{1+e^{-2 x}}{1-e^{-2 x}}=\frac{1-\left(-e^{-2 x}\right)}{1+\left(-e^{-2 x}\right)} \end{aligned} $

Here, $k=-e^{-2 x}$

$ -e^{-2 x} \in(-\infty, 0) \Rightarrow k \in(-\infty, 0) $

Therefore, Assertion (A) is false.

Reason $y=\tanh x$

Here, $y$ lies between -1 and 1 , Reason is true.

$ \begin{aligned} & \text { Given, } f(x)=\sqrt{\frac{2 x^2-7 x+5}{3 x^2-5 x-2}} \\ & \quad \frac{2 x^2-7 x+5}{3 x^2-5 x-2} \geq 0 \\ & \Rightarrow \quad \frac{2 x^2-2 x-5 x+5}{3 x^2-6 x+x-2} \geq 0 \\ & \Rightarrow \frac{2 x(x-1)-5(x-1)}{3 x(x-2)+1(x-2)} \geq 0 \\ & \Rightarrow \quad \frac{(2 x-5)(x-1)}{(3 x+1)(x-2)} \geq 0 \\ & \Rightarrow \quad x \neq-\frac{1}{3}, x \neq 2 \end{aligned} $

$ \begin{aligned} &\text { Domain : }\\ &x \in\left(-\infty,-\frac{1}{3}\right) \cup[1,2) \cup\left[\frac{5}{2}, \infty\right) \end{aligned} $

Given, $f(x)=|x-2|+|x-3|$

We know that for $f(x)=|x-a|+|x-b|$, the minimum value of $f(x)$,

we get either at $x=a$

or at $x=b$

$ \begin{aligned}So\,\, for\,\,\,\, & f(x)=|x-2|+|x-3| \\ & f(2)=0+1=1 \\ & f(3)=1+0=1 \end{aligned} $

So, $f(x)_{\text {min }}=1$ and maximum value will be $\infty$ at $x=\infty$.

$ \because \quad f_{\max }=\infty+\infty=\infty $

So, range $=[1, \infty)$

$ \begin{aligned} &\text { Given, } f(x)=\frac{1}{2}-\tan \left(\frac{\pi x}{2}\right)\\ &\begin{aligned} & \text { and } g(x)=\sqrt{3+4 x-4 x^2} \\ & \Rightarrow \quad 3+4 x-4 x^2 \geq 0 \\ & \Rightarrow \quad 3-4\left(x^2-x\right) \geq 0 \\ & \Rightarrow 3-4\left(x^2-x+\frac{1}{4}\right)+1 \geq 0 \\ & \Rightarrow \quad 4-4\left(x-\frac{1}{2}\right)^2 \geq 0 \\ & \Rightarrow \quad\left(x-\frac{1}{2}\right)^2 \leq 1 \\ & \Rightarrow \quad-1 \leq x-\frac{1}{2} \leq 1 \\ & \Rightarrow \quad-1+\frac{1}{2} \leq x \leq 1+\frac{1}{2} \\ & \Rightarrow \quad-\frac{1}{2} \leq x \leq \frac{3}{2} \Rightarrow x \in\left[-\frac{1}{2}, \frac{3}{2}\right] \end{aligned} \end{aligned} $

Now, $f(x)=\frac{1}{2}-\tan \left(\frac{\pi x}{2}\right)$

As $f: A \rightarrow B$ and $g: B \rightarrow C$, so domain of $g(x)=$ range of $f(x)$

∴ Range of $f(x)=\left[-\frac{1}{2}, \frac{3}{2}\right]$.

$ \begin{aligned} & \text { Given, } f(x)=\sqrt{\frac{1-x^2}{1+x^2}} \\ & \qquad \begin{array}{ll} & \frac{1-x^2}{1+x^2} \geq 0 \\ \Rightarrow & 1-x^2 \geq 0 \quad\left[\because 1+x^2>0\right] \\ \Rightarrow & x^2 \leq 1 \\ \Rightarrow & x \in[-1,1] \\ \therefore & D=[-1,1] \\ \text { Let } y=\frac{1-x^2}{1+x^2} \\ & y+x^2 y=1-x^2 \end{array} \end{aligned} $

          $ \begin{aligned} &\begin{array}{ll} \Rightarrow & x^2(y+1)=1-y \\ \Rightarrow & x^2=\frac{1-y}{1+y} \\ \Rightarrow & x= \pm \sqrt{\frac{1-y}{1+y}} \\ \Rightarrow & \frac{1-y}{1+y} \geq 0 \end{array}\\ &\text { As we know that } y \geq 0,1+y>0\\ &\begin{aligned} \Rightarrow \quad 1-y & \geq 0 \\ y & \leq 1 \\ y & \in[0,1] \\ G & =[0,1] \\ D \cap G & =[0,1] \end{aligned} \end{aligned} $

Given,

$ \begin{array}{rc} & f(x)=\log _2\left(2^x-2\right)+\sqrt{1-x} \\ & 1-x \geq 0 \\ \Rightarrow &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, x \leq 1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{array} $

Also, $\quad 2^x-2>0$

$ \begin{aligned}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ 2^x & >2 \\ x & >1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

From Eqs. (i) and (ii), it can be concluded that $x$ contains no real values.

$ \begin{aligned} & \text { } \begin{aligned} Given,\,\,\,\,\, f(x) & =1-x, g(x)=\frac{1}{1-x} \\ h(x) & =1 / x \\ f(F(h(x))) & =g(x) \\ \Rightarrow \quad f\left(F\left(\frac{1}{x}\right)\right) & =\frac{1}{1-x} \end{aligned} \end{aligned} $

$ \begin{aligned} \Rightarrow & & 1-F\left(\frac{1}{x}\right) & =\frac{1}{1-x} \\ \Rightarrow & & F\left(\frac{1}{x}\right) & =1-\frac{1}{1-x} \\ \Rightarrow & & F\left(\frac{1}{x}\right) & =\frac{-x}{1-x} \end{aligned} $

Substitute $x=\frac{1}{2022}$ to determine $F(2022)$

$ \begin{aligned} F(2022) & =\frac{-\frac{1}{2022}}{1-\frac{1}{2022}} \\ & =-\frac{1}{2021}=\frac{1}{1-2022} \\ & =g(2022) \end{aligned} $

$ \begin{aligned} &\text { }\\ &\begin{aligned} & \cos [\sinh (\log x)+\cosh (\log x)] \\ & =\cos \left[\frac{e^{\log x}-e^{-\log x}}{2}+\frac{e^{\log x}+e^{-\log x}}{2}\right] \\ & {\left[\because \sinh x=\frac{e^x-e^{-x}}{2}\right. \text { and }} \left.\cosh x=\frac{e^x+e^{-x}}{2}\right] \end{aligned} \end{aligned} $

$ \begin{aligned} &=\cos \left[\frac{2 x}{2}\right]=\cos x\\ &\text { The minimum value of } \cos x=-1\\ &\begin{aligned} \therefore &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, k=-1 \\ \text { And, } \cosh (k+1) & =\frac{e^{k+1}+e^{-(k+1)}}{2} \\ & =\frac{e^0+e^0}{2}=1 \end{aligned} \end{aligned} $

$ \text { } \begin{aligned} & f:\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \rightarrow R \\ & f(x)=\sec x+\tan x \\ &=\frac{1+\tan ^2 x / 2}{1-\tan ^2 x / 2}+\frac{2 \tan x / 2}{1-\tan ^2 x / 2} \\ &=\frac{(1+\tan x / 2)^2}{1-\tan ^2 x / 2}=\frac{1+\tan x / 2}{1-\tan x / 2} \\ &=\tan \left(\frac{\pi}{4}+\frac{x}{2}\right) \\ & f^{\prime}(x)=\sec ^2\left(\frac{\pi}{4}+\frac{x}{2}\right) \cdot \frac{1}{2}>0 \forall \\ & x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \end{aligned} $

$\therefore f(x)$ is an increasing function.

Hence, $f(x)$ is one-one function.

$ \begin{aligned} & f:[0, \infty) \rightarrow R, f(x)=x^2 \\ & f^{\prime}(x)=2 x \geq 0 \forall x \in[0, \infty) \end{aligned} $

$\therefore f(x)$ is one-one function.

Both Statements I and II are true.

$f(x)=\left\{\begin{array}{cc}2 x-5, & x<-3 \\ x+2, & -3 \leq x<5 \\ 3 x+1, & x \geq 5\end{array}\right.$

$ \begin{aligned}(A) f(-5)+f(0)+ & f(-1)=(2(-5)-5) +(0+2)+(-1+2) \\ = & -15+2+1=-12 \end{aligned} $

$ \begin{aligned}(B) & f((f(5))+10 f(-3)) \\ & \quad=f((3 \times 5+1)+10 \cdot(-3+2)) \\ & \quad=f(6)=3 \times 6+1=19 \end{aligned} $

(C) $f(|f(-4)|)=f(|-13|)=3 \times 13+1=40$

$ \begin{aligned}(D) f(f(f(1))) & =f(f(3))=f(5) \\ & =3 \times 5+1=16 \end{aligned} $

$ \begin{aligned} & \text { } f(x)=\frac{\sqrt{6 x^2+5 x-6}}{\sqrt{4-x}-\sqrt{x+4}} \\ & 6 x^2+5 x-6 \geq 0 \\ & \Rightarrow \quad 6 x^2+9 x-4 x-6 \geq 0 \\ & \Rightarrow \quad 3 x(2 x+3)-2(2 x+3) \geq 0 \\ & \Rightarrow \quad(3 x-2)(2 x+3) \geq 0 \\ & \end{aligned} $

$ \begin{array}{lc} & x \in\left(-\infty, \frac{-3}{2}\right] \cup\left[\frac{2}{3}, \infty\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \Rightarrow & 4-x \geq 0 \Rightarrow x \leq 4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\\ \Rightarrow & x \in(-\infty, 4] \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ \Rightarrow & x+4 \geq 0 \Rightarrow x \geq-4 \end{array} $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, x \in[-4, \infty) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

$ \therefore \quad x \in\left[-4, \frac{-3}{2}\right] \cup\left[\frac{2}{3}, 4\right] $