Functions

$f(x)=\frac{1}{\sqrt{|x|-x^2}}$

For $f(x)$ to be defined $|x|-x^2>0$

$\operatorname{Case} \mathbf{I}|x|=x, x>0$

$ \begin{array}{lr} \Rightarrow & x-x^2>0 \\ \Rightarrow & x(1-x) > 0 \\ \Rightarrow & 0 < x < 1 \end{array} $

Case II $|x|=-x, x<0$

$ \begin{aligned} & \Rightarrow \quad-x-x^2 > 0 \\ & \Rightarrow x(1+x) < 0 \\ & \Rightarrow \quad-1 < x < 0 \end{aligned} $

So, $A=(-1,1)-\{0\}$

Let $y=\frac{1}{\sqrt{|x|-x^2}}$

When $0 < x < 1$

$ y=\frac{1}{\sqrt{x-x^2}} $

As, $x \in(0,1)$

$ \begin{array}{ll} \Rightarrow & 0 < x-x^2 \leq \frac{1}{4} \\ \Rightarrow & 0<\sqrt{x-x^2} \leq \frac{1}{2} \\ \Rightarrow & 2 \leq \frac{1}{\sqrt{x-x^2}}<\infty \end{array} $

So, $y \in[2, \infty)$ for $x \in(0,1)$

$ y=\frac{1}{\sqrt{-x-x^2}} $

$ y \in[2, \infty) \text { for } x \in(-1,0) $

So, $B=[2, \infty)$

So, $A \cup B$

$ \begin{aligned} & (-1,1)-\{0\} \cup[2, \infty) \\ & (-1,0) \cup(0,1) \cup[2, \infty) \end{aligned} $

$f(x)= \begin{cases}2 x+3, & x \leq \frac{4}{3} \\ -3 x^2+8 x, & x>\frac{4}{3}\end{cases}$

$ \Rightarrow f^{\prime}(x)= \begin{cases}2 & x<\frac{4}{3} \\ -6 x+8, & x>\frac{4}{3}\end{cases} $

$ f^{\prime}(x)>0, \forall x<\frac{4}{3} \text { and } f^{\prime}(x)<0, \forall x>\frac{4}{3} $

So, $f$ is many-one.

Sign scheme for $f^{\prime}(x)$

$ \begin{aligned} & f(-\infty) \rightarrow-\infty \\ & f(\infty) \rightarrow-\infty \\ & \qquad f\left(\frac{4}{3}\right)=2 \times \frac{4}{3}+3 \end{aligned} $

So, range $=\left(-\infty, \frac{17}{3}\right]$

Range is subset of codomain.

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

$ \begin{aligned} & \text { Let } f(x)=2^{4 n+3}+3^{3 n+1} \\ & \Rightarrow P \text { divides } f(1), f(2), f(3), \ldots, f(n) \\ & \qquad \begin{array}{l} f(1)=2^7+3^4=128+81 \\ f(1)=209 \end{array} \end{aligned} $

$f(1)$ is divisible by 11 .

11 is an odd prime integer.

$ \begin{aligned} &\\ &\begin{aligned} & \cosh (x)=\frac{e^x+e^{-x}}{2} \\ & \tanh (x)=\frac{e^x-e^{-x}}{e^x+e^{-x}} \\ & x=\frac{e^{\cosh ^{-1}(x)}+e^{-\cosh ^{-1}(x)}}{2} \\ & x=\frac{y+\frac{1}{y}}{2} \\ & y^2-2 x y+1=0 \\ & y=\frac{2 x \pm \sqrt{4 x^2-4}}{2} \\ & e^{\cosh ^{-1}(x)}=x+\sqrt{x^2-1} \\ & \cosh ^{-1}(x)=\ln \left(x+\sqrt{x^2-1}\right) \end{aligned} \end{aligned} $

Again, $x=\frac{e^{\tanh ^{-1}(x)}-e^{-\tanh ^{-1}(x)}}{e^{\tanh ^{-1}(x)}+e^{-\tanh ^{-1}(x)}}$

Let $y=e^{\tanh ^{-1}(x)}$

$ \begin{array}{ll} \Rightarrow & x=\frac{y^2-1}{y^2+1} \\ \Rightarrow & x y^2+x=y^2-1 \\ \Rightarrow & \frac{x+1}{1-x}=y^2 \\ \Rightarrow & e^{\tanh ^{-1}(x)}=\sqrt{\frac{1+x}{1-x}} \\ \Rightarrow & \tanh ^{-1}(x)=\frac{1}{2} \ln \left(\frac{1+x}{1-x}\right) \end{array} $

Now, $\cosh ^{-1}(x)=\tanh ^{-1}(x)$

$ \ln \left(x+\sqrt{x^2-1}\right)=\frac{1}{2} \ln \left(\frac{1+x}{1-x}\right) $

No solution

Aas $x \geq 1$ for $\ln \left(x+\sqrt{x^2-1}\right)$ to be real.

And for $\ln \left(\frac{1+x}{1-x}\right)$ to be defined

$ -1

Now, $\cosh ^{-1}(x)=\operatorname{coth}^{-1}(x)$

$ \ln \left(x+\sqrt{x^2-1}\right)=\frac{1}{2} \ln \left(\frac{x+1}{x-1}\right) $

Now, $\ln \left(x+\sqrt{x^2-1}\right)$ is strictly increasing function, $\forall x \geq 1$ and $\frac{1}{2} \ln \left(\frac{x+1}{x-1}\right)$ is strictly decreasing function for $x>1$.

⇒ Both the graph will cut and one point exactly.

For $f(x)$ to be defined.

$ \begin{aligned} & \sqrt{x^2+x}+\sqrt{x^2-x}>0 \\ & \Rightarrow \quad x^2+x \geq 0 \text { and } x^2-x \geq 0 \\ & \Rightarrow \quad x(x+1) \geq 0 \text { and } x(x-1) \geq 0 \\ & \Rightarrow \quad(-\infty,-1] \cup[0, \infty) \end{aligned} $

and $(-\infty, 0] \cup[1, \infty)$

Domain

$ \begin{aligned} & =(-\infty,-1] \cup[0, \infty) \cap(-\infty, 0] \cup[1, \infty)-\{0\} \\ & =(-\infty,-1] \cup[1, \infty) \end{aligned} $

$ \begin{aligned} & \text { } \frac{x+1}{x^3(x-1)}=\frac{a}{x}+\frac{b}{x^2}+\frac{c}{x^3}+\frac{d}{x-1} \\ & =\frac{a x^2(x-1)+b x(x-1)+c(x-1)+d x^3}{x^3(x-1)} \\ & \Rightarrow x+1=x^3(a+d)+x^2(b-a)+x(c-b)-c \\ & \Rightarrow a+d=0, b-a=0 \end{aligned} $

$ \begin{aligned} & c-b=1,-c=1 \\ & \text { So, } c=-1, b=-2 a=-2 \\ & \Rightarrow a=b=2 c=-d \end{aligned} $

Given, $f(x)=5^{|x|}+\operatorname{Sgn}\left(5^x\right)$

$ \begin{array}{ll} \because & 5^x>0 \forall x \in R \\ \therefore & \operatorname{sgn}\left(5^x\right)=1 \\ \text { And } & |x| \geq 0 \Rightarrow-|x| \leq 0 \\ \Rightarrow & 5^x \leq 5^0 \Rightarrow 5^x \leq 1 \\ & 5^{|x|} \leq 5^0 \Rightarrow 5^{|x|} \leq 1 \\ \therefore & f(x)=5^{|x|}+1 \leq 1+1 \leq 2 \\ \because & f(1)=f(-1) \end{array} $

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

$f(x)=\frac{x^2+x+k}{x^2-x+k}$

$ \text { ∵ Range }=\left[\frac{1}{3}, 3\right] $

For range let $f(x)=y$

$ \begin{aligned} & y=\frac{x^2+x+k}{x^2-x+k} \\ & \Rightarrow \quad x^2(y-1)-x(y+1)+k y-k=0 \\ & \therefore \quad D \geq 0 \\ & \Rightarrow \quad(y+1)^2-4 k(y-1)(y-1) \geq 0 \\ & \Rightarrow \quad(y+1)^2-(2 \sqrt{k}(y-1))^2 \geq 0 \\ & \Rightarrow \quad(y+1+2 \sqrt{k}(y-1)) \\ & \quad \quad(y+1-2 \sqrt{k}(y-1) \geq 0 \\ & \Rightarrow \quad(y(2 \sqrt{k}+1)+1-2 \sqrt{k}) \\ & \quad(y(2 \sqrt{k}-1)-2 \sqrt{k}-1) \leq 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \\\because \,\,\,\, & y \in\left[\frac{1}{3}, 3\right] \\ &(3 y-1)(y-3) \leq 0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Comparing Eqs. (i) and (ii),

$ 2 \sqrt{k}+1=3 \Rightarrow 2 \sqrt{k}=2 \Rightarrow k=1 $

We have,

$ \begin{aligned} & f(x)=4 x^3+a x^2+3 x-2 \\ & f^{\prime}(x)=12 x^2+2 a x+3 \end{aligned} $

For monotonic $f^{\prime}(x)>0$

$ \begin{aligned} & 12 x^2+2 a x+3>0 \\ & D<0 \\ \Rightarrow \quad & (2 a)^2-4(12)(3)<0 \Rightarrow a^2-36<0 \\ \Rightarrow \quad & (a-6)(a+6)<0 \Rightarrow a \in(-6,6) \end{aligned} $

$f(x)=\frac{x^2+x+a}{x^2-x+a}$

$\because f(x)$ is surjection, then range $=$ Codomain

Range $=R$

Now for range, put $f(x)=y$.

$ \begin{gathered} y=\frac{x^2+x+a}{x^2-x+a} \\ x^2(y-1)-x(y+1)+a y-a=0 \end{gathered} $

Now $x \in R \Rightarrow D \geq 0$

$ \begin{aligned} & (y+1)^2-4(y-1)(a y-a) \geq 0 \\ & \Rightarrow(y+1)^2-4 a(y-1)^2 \geq 0, \forall y \in R \\ & \Rightarrow y^2(1-4 a)+y(2+8 a)+1-4 a \geq 0, \forall y \in R\end{aligned} $

$ \begin{aligned} & \therefore 1-4 a \geq 0 \text { and } D \leq 0 \\ & \qquad a \leq \frac{1}{4} \text { and }(2+8 a)^2-4(1-4 a)^2 \leq 0 \\ & \Rightarrow(1+4 a)^2-(1-4 a)^2 \leq 0 \\ & \Rightarrow(1+4 a+1-4 a)(1+4 a-1+4 a) \leq 0 \\ & \Rightarrow 2(8 a) \leq 0 \\ & \Rightarrow a \leq 0 \\ & \therefore a \in(-\infty, 0) . \end{aligned} $

Given, $f(x)=\frac{1}{\sqrt{\log _{1 / 3}\left(\frac{x-1}{2-x}\right)}}$

∴ Domain

$ \begin{aligned} & \log _{1 / 3} \frac{x-1}{2-x}>0 \text { and } \frac{x-1}{2-x}>0 \\ & \Rightarrow \quad \frac{x-1}{2-x}<\left(\frac{1}{3}\right)^0 \Rightarrow \frac{x-1}{2-x}<1 \\ & \Rightarrow \quad \frac{x-1}{2-x}-1<0 \\ & \Rightarrow \quad \frac{x-1-2+x}{2-x}<0 \Rightarrow \frac{2 x-3}{2-x}<0 \end{aligned} $

Now,

Now, $x \in\left(1, \frac{3}{2}\right)$

$ \therefore \quad a=1, b=\frac{3}{2} $

Now, $2 b=3=a+2$

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

Let $g(x)=x^2+x+1, \quad h(x)=x^2-3 x-4$

$ \begin{aligned} & g^{\prime}(x)=2 x+1, \quad h^{\prime}(x)=2 x-3 \\ & \therefore \quad x>4 \\ & \quad g^{\prime}(x)>0 \text { and } h^{\prime}(x)>0 \end{aligned} $

∴ both $g(x)$ and $h(x)$ are increasing

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

We have, $3 f(x)+4 f\left(\frac{1}{x}\right)=\frac{2}{x}-1$

$ \therefore \quad 3 f(3)+4 f\left(\frac{1}{3}\right)=\frac{2}{3}-1=\frac{-1}{3} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Also, $3 f\left(\frac{1}{3}\right)+4 f(3)=6-1=5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

On solving Eqs. (i) and (ii), we get

$ f(3)=3 $

$ \begin{aligned} & \text { We have, } y=\frac{10^x-10^{-x}}{10^x+10^{-x}}+1 \\ & \Rightarrow \quad y-1=\frac{10^x-10^{-x}}{10^x+10^{-x}} \\ & \text { Let } \quad a=10^x \\ & \therefore \quad 10^{-x}=\frac{1}{a} \\ & \therefore \quad y-1=\frac{a-\frac{1}{a}}{a+\frac{1}{a}}=\frac{a^2-1}{a^2+1} \\ & \Rightarrow \quad(y-1)\left(a^2+1\right)=a^2-1 \Rightarrow(y-1) a^2+(y-1)=a^2-1 \\ & \Rightarrow \quad a^2(y-1-1)=-y \Rightarrow a^2(y-2)=-y \\ & \Rightarrow \quad a^2=\frac{-y}{y-2} \Rightarrow 10^{2 x}=\frac{-y}{y-2} \\ & \Rightarrow \quad 2 x=\log _{10}\left(\frac{-y}{y-2}\right) \Rightarrow 2 x=\log _{10}\left(\frac{y}{2-y}\right) \\ & \Rightarrow \quad x=\frac{1}{2} \log _{10}\left(\frac{y}{2-y}\right) \end{aligned} $

Given, $f(x)=\tan \left(\frac{\pi}{\sqrt{x+1}+4}\right)$

For $\sqrt{x+1}$ to be defined,

$ \begin{array}{ll} \therefore & \sqrt{x+1}+4 \geq 0+4=4 \\ \Rightarrow & 0<\frac{1}{\sqrt{x+1}+4} \leq \frac{1}{4} \\ \Rightarrow & 0<\frac{\pi}{\sqrt{x+1}+4} \leq \frac{\pi}{4} \\ \therefore & f(x)=\tan \left(\frac{\pi}{\sqrt{x+1}}+4\right), \text { where } \\ & 0<\frac{\pi}{\sqrt{x+1}+4} \leq \frac{\pi}{4} \end{array} $

So, the function $f(x)$ is strictly increasing in the interval $\left(0, \frac{\pi}{4}\right]$

And the range of the given function is $\left(\tan (0), \tan \left(\frac{\pi}{4}\right)\right]$, which is $(0,1]$

$\therefore$ The range of $f(x)$ is $(0,1]$

Given,

$ \frac{x^3+3}{(x-3)^3}=a+\frac{b}{x-3}+\frac{c}{(x-3)^2}+\frac{d}{(x-3)^3} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Let $y=x-3$

$ \Rightarrow x=y+3 $

Then, $\frac{x^3+3}{(x-3)^3}=\frac{(y+3)^3+3}{y^3}$

$ \begin{aligned} & =\frac{y^3+9 y^2+27 y+27+3}{y^3} \\ & =\frac{y^3+9 y^2+27 y+30}{y^3} \\ & =1+\frac{9}{y}+\frac{27}{y^2}+\frac{30}{y^3} \\ & =1+\frac{9}{x-3}+\frac{27}{(x-3)^2}+\frac{30}{(x-3)^3} \end{aligned} $

Comparing this with Eq. (i), we get

$ a=1, b=9, c=27, d=30 $

So, $(a+d)-(b+c)=(1+30)-(9+27)$

$ =31-36=-5 $

Given two functions, $ f(x) = ax^2 + bx + c $, which is even, and $ g(x) = px^3 + qx^2 + rx $, which is odd, we need to analyze the combined function $ h(x) = f(x) + g(x) $. We know that $ h(-2) = 0 $.

Properties of Even and Odd Functions:

Even Function: $ f(x) = f(-x) $.

Odd Function: $ g(x) = -g(-x) $.

Since $ f(x) $ is even:

$ f(x) = f(-x) \quad \Rightarrow \quad f(2) = f(-2) $

For the odd function $ g(x) $:

$ g(-x) = -g(x) \quad \Rightarrow \quad g(-2) = -g(2) $

Using the Given Information:

Given $ h(-2) = 0 $, we have:

$ h(-2) = f(-2) + g(-2) = 0 $

From the properties mentioned:

$ f(-2) = f(2) \quad \text{and} \quad g(-2) = -g(2) $

Substitute these into $ h(-2) = 0 $:

$ f(2) - g(2) = 0 $

This implies:

$ f(2) = g(2) $

Expressing $ f(2) $ and $ g(2) $:

Calculate $ f(2) $:

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

And $ g(2) $:

$ g(2) = p(2)^3 + q(2)^2 + r(2) = 8p + 4q + 2r $

Equating the two expressions from $ f(2) = g(2) $:

$ 4a + 2b + c = 8p + 4q + 2r $

Thus, the expression for $ 8p + 4q + 2r $ is:

$ 8p + 4q + 2r = 4a + 2b + c $

To determine the range of the function $ f(x) = \log_3(5 + 4x - x^2) $, we need to analyze the behavior of the quadratic expression $ 5 + 4x - x^2 $.

Analysis of $ 5 + 4x - x^2 $

The expression $ 5 + 4x - x^2 $ is a quadratic function of the form $ ax^2 + bx + c $, where $ a = -1 $, $ b = 4 $, and $ c = 5 $.

Determining the Range of the Quadratic

Since $ a = -1 < 0 $, the parabola opens downwards, and the quadratic function achieves its maximum value at its vertex. The maximum (or minimum for upward-opening parabolas) value of a quadratic function $ ax^2 + bx + c $ is given by $ -\frac{D}{4a} $, where $ D = b^2 - 4ac $ is the discriminant.

Calculate the discriminant:

$ D = 4^2 - 4 \times (-1) \times 5 = 16 + 20 = 36 $

The maximum value of the quadratic $ 5 + 4x - x^2 $ is:

$ -\frac{D}{4a} = -\frac{36}{4 \times (-1)} = 9 $

Hence, the range of $ 5 + 4x - x^2 $ is $ (-\infty, 9] $.

Translating to the Logarithmic Function

Considering the function $ f(x) = \log_3(5 + 4x - x^2) $, the possible values of $ 5 + 4x - x^2 $ determine the input values for the logarithm.

Since $ 5 + 4x - x^2 \leq 9 $, we take the logarithm base 3 of each side:

$ \log_3(5 + 4x - x^2) \leq \log_3(9) $

Given that $ \log_3(9) = 2 $, we conclude:

$ f(x) = \log_3(5 + 4x - x^2) \leq 2 $

Thus, the range of $ f(x) $ is $ (-\infty, 2] $.

To find the sum of the maximum and minimum values of the function $ f(x) = \frac{x^2 - x + 1}{x^2 + x + 1} $, we start by setting:

$ y = \frac{x^2 - x + 1}{x^2 + x + 1} $

Rewriting, we have:

$ y(x^2 + x + 1) = x^2 - x + 1 $

This simplifies to:

$ (y - 1)x^2 + (y + 1)x + (y - 1) = 0 $

To ensure real values of $ x $, the discriminant $ D $ of this quadratic in $ x $ must be non-negative. The discriminant is given by:

$ (y + 1)^2 - 4(y - 1)^2 \geq 0 $

Simplifying the inequality:

$ (y + 1 - 2(y - 1))(y + 1 + 2(y - 1)) \geq 0 $

$ (3 - y)(3y - 1) \geq 0 $

Reversing the inequality:

$ (y - 3)(3y - 1) \leq 0 $

This implies $ y $ lies within the interval:

$ y \in \left[\frac{1}{3}, 3\right] $

Therefore, the sum of the maximum and minimum values of $ y $ is:

$ \frac{1}{3} + 3 = \frac{10}{3} $

Given,

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

For

$ \begin{array}{r} x>0,|x|=x \\\\ x-|x|=0 \\\\ |x-|x||=0 \end{array} $

$\therefore f(x)$ is not possible for $x>0$.

For

$ \begin{aligned} x<0,|x| & =-x \\\\ x-|x| & =-2 x \\\\ |x-|x|| & =2 x \\\\ f(x) & =\frac{1}{\sqrt{-2 x}} \end{aligned} $

$\therefore A$ is set of negative numbers and $B$ is a set of positive numbers.

$ \therefore A \cap B=\phi $

We have,

$ \begin{aligned} & f(x)=\sqrt[3]{\frac{x-2}{2 x^2-7 x+5}}+\log \left(x^2-x-2\right) \\ \therefore & \quad 2 x^2-7 x+5 \neq 0 \text { and } x^2-x-2>0 \\ \Rightarrow & (2 x-5)(x-1) \neq 0 \text { and }(x+1)(x-2)>0 \\ \Rightarrow & x \neq 1, \frac{5}{2} \text { and } x \in(-\infty,-1) \cup(2, \infty) \\ \Rightarrow & x \in(-\infty,-1) \cup\left(2, \frac{5}{2}\right) \cup\left(\frac{5}{2}, \infty\right) \end{aligned} $

$\therefore$ Domain of $f$ is

$ (-\infty,-1) \cup\left(2, \frac{5}{2}\right) \cup\left(\frac{5}{2}, \infty\right) $

We have, $f\left(3 x+\frac{1}{2 x}\right)=9 x^2+\frac{1}{4 x^2}$

$ \therefore \quad f\left(3 x+\frac{1}{2 x}\right)=\left(3 x+\frac{1}{2 x}\right)^2-3 $

Now, $f\left(x+\frac{1}{x}\right)=1$

$ \begin{array}{cc} \Rightarrow & \left(x+\frac{1}{x}\right)^2-3=1 \Rightarrow\left(x+\frac{1}{x}\right)^2=4 \\ \Rightarrow & x^2+\frac{1}{x^2}+2=4 \Rightarrow x^2+\frac{1}{x^2}-2=0 \\ \Rightarrow & \left(x-\frac{1}{x}\right)^2=0 \Rightarrow x-\frac{1}{x}=0 \\ \Rightarrow & x^2-1=0 \Rightarrow x^2=1 \\ \Rightarrow & x= \pm 1 \end{array} $

We have,

$ \begin{aligned} & f(x)=\sqrt{\cos (\sin x)}+\cos ^{-1}\left(\frac{1+x^2}{2 x}\right) \\\\& \Rightarrow \quad \cos (\sin x)>0 \\\\ & \because \quad \sin x \in[-1,1] \\\\ & \therefore \quad \cos (\sin x)>0, \forall x \in R \\\\ & \text { and } \quad-1 \leq \frac{1+x^2}{2 x} \leq 1 \\\\ & \quad-1 \leq \frac{1+x^2}{2 x} \text { and } \frac{1+x^2}{2 x} \leq 1 \\\\ & \Rightarrow \quad x \in\{-1,1\} \end{aligned} $

$ \text { Here, } f(x)=\left\{\begin{array}{cc} x^2-4 x+3 & , x<2 \\ x-3 & , x \geq 2 \end{array}\right. $

$ \text { Number of solutions }=2 $

$ \begin{aligned} & \text { Here, } f(x)=3 x-2, g(x)=x^2+2 \\ & g o f(x)=(3 x-2)^2+2=9 x^2-12 x+6 \\ & f og(x)=3\left(x^2+2\right)-2=3 x^2+4 \\ & \text { Now, }[(g \circ f)+(f \circ g)](x)=12 x^2-12 x+10 \\ & =12\left(x^2+2\right)-4(f(x))-22 \\ & =12 g(x)-4 f(x)-22 \end{aligned} $

$ \begin{array}{r} f(x)=a x^9+b x^7+c x^5+d x^3 \\ +e x+\frac{15}{x}+12 \end{array} $

Let $g(x)=a x^9+b x^7+c x^5+d x^3+e x+\frac{15}{x}$ which is an odd function odd function, let $g(x)$ We have,

$ \begin{aligned} \Rightarrow \quad g(-x) & =-g(x) \\ -4 & =f(4)=g(4)+12 \end{aligned} $

$ \begin{array}{llrl} \Rightarrow g(4) =-16 \\ \Rightarrow g(-4) =16 \\ f(-4)=g(-4)+12 & =16+12=28 \end{array} $

As, we know in ${ }^n C_r, n \geq r$ and $n \in I$, $r \geq 0$

So, to define, ${ }^{(16-x)} C_{(2 x-1)}$

$ \begin{array}{ll} & 16-x \geq 2 x-1 \\ \Rightarrow & 3 x \leq 17 \\ \Rightarrow & x \leq \frac{17}{3} \text { and } 2 x-1 \geq 0 \\ \Rightarrow & x \geq \frac{1}{2} \end{array} $

So, $\quad \frac{1}{2} \leq x \leq \frac{17}{3}$

Hence, $x=\{1,2,3,4,5\}$

$ \begin{aligned} & X=\left\{\left.\left[\begin{array}{ll} a & b \\ c & d \end{array}\right] \right\rvert\, a, b, c, d \in R\right\} \\ & f: X \rightarrow R \\ & f(A)=\operatorname{det}(A), \forall A \in X \end{aligned} $

Now, by taking examples

$ \begin{aligned} & {\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right],\left[\begin{array}{ll} 2 & 2 \\ 2 & 2 \end{array}\right] \in X} \\ & \operatorname{det}\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right]=\operatorname{det}\left[\begin{array}{ll} 2 & 2 \\ 2 & 2 \end{array}\right]=0 \\ & f\left(\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right]\right)=f\left(\left[\begin{array}{ll} 2 & 2 \\ 2 & 2 \end{array}\right]\right) \\ & \Rightarrow\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right] \neq\left[\begin{array}{ll} 2 & 2 \\ 2 & 2 \end{array}\right] \end{aligned} $

So, this function is not one-one.

Let $f: X \rightarrow Y$

to show onto for each such that $f(x)=y_0$; $y_0 \in Y$

there exists $x \in X$

$ \begin{aligned} & y_0 \in R \\ & {\left[\begin{array}{cc} y_0 & 0 \\ 0 & 1 \end{array}\right] \text { such that }\left|\begin{array}{cc} y_0 & 0 \\ 0 & 1 \end{array}\right|=y_0} \\ & f\left(\left[\begin{array}{cc} y_0 & 0 \\ 0 & 1 \end{array}\right]\right)=y_0 \end{aligned} $

This function is onto function.

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) $

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

$x \notin I$ : $\quad[\because$ 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 $f(x)=\left[-\frac{1}{2}, \frac{3}{2}\right]$.