(3)

$f\left(x\right)$$=\{\begin{array}{cc}\frac{1-\cos 2x}{x^2},& x<0\\ \alpha ,& x=0,\\ \frac{\beta \sqrt{1-\cos x}}{x},& x>0\end{array}$

$f\left(x\right)$ is continuous at $x=0$

$\Rightarrow f\left(0\right)=\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }f\left(x\right)$

Now, $\underset{x\to 0^{-}}{\lim }f\left(x\right)=f\left(0\right)$

$\therefore$     $\underset{x\to 0^{-}}{\lim }f\left(x\right)=\alpha \Rightarrow \underset{x\to 0^{-}}{\lim }\left(\frac{1-\cos 2x}{x^2}\right)=\alpha$

$\Rightarrow \underset{x\to 0^{-}}{\lim }\frac{2{\sin }^{2}x}{x^2}=\alpha \Rightarrow \underset{h\to 0}{\lim }\frac{2{\sin }^{2}h}{h^2}=\alpha \Rightarrow \alpha =2$

Also, $\underset{x\to 0^{+}}{\lim }f\left(x\right)=f\left(0\right)$

$\Rightarrow \underset{x\to 0^{+}}{\lim }\frac{\beta \sqrt{1-\cos x}}{x}=2\Rightarrow \underset{h\to 0}{\lim }\frac{\beta \sqrt{1-\cos h}}{h}=2$

$\Rightarrow \underset{h\to 0}{\lim }\frac{\beta \sqrt{2}\sin \frac{h}{2}}{h}=2\Rightarrow \underset{h\to 0}{\lim }\frac{\beta \sqrt{2}\sin \frac{h}{2}}{2\times \frac{h}{2}}=2$

$\Rightarrow \frac{\beta }{\sqrt{2}}=2\Rightarrow \beta =2\sqrt{2}$

Hence, ${\alpha }^2+{\beta }^2=4+8=12$

(4)

$f\left(x\right)=\{\begin{array}{cc}\frac{{72}^x-9^x-8^x+1}{\sqrt{2}-\sqrt{1+\cos x}},& x\ne 0\\ a{\log }_e 2{\log }_{e}3,& x=0\end{array}$

$\because$     $f\left(x\right)$ is continuous at $x=0$

$f\left(0\right)=\underset{x\to 0}{\lim }\frac{{72}^x-9^x-8^x+1}{\sqrt{2}-\sqrt{1+\cos x}}$

$=\underset{x\to 0}{\lim }\frac{(9^x-1)(8^x-1)(\sqrt{2}+\sqrt{1+\cos x})}{\frac{(1-\cos x)}{x^2}{\displaystyle \times }{x}^2}$

$=\underset{x\to 0}{\lim }\left(\frac{{\displaystyle 9^x-1}}{{\displaystyle x}}\right)\left(\frac{{\displaystyle 8^x-1}}{{\displaystyle x}}\right)\frac{(\sqrt{2}+\sqrt{1+\cos x})}{\left(\frac{{\displaystyle 2{\sin }^2\frac{{\displaystyle x}}{{\displaystyle 2}}}}{{\displaystyle 4\times \frac{{\displaystyle x^2}}{{\displaystyle 4}}}}\right)}$

$=\underset{x\to 0}{\lim }\frac{2\left(\frac{9^x-1}{x}\right)\left(\frac{8^x-1}{x}\right)(\sqrt{2}+\sqrt{1+\cos x})}{{{\displaystyle (}{\displaystyle \frac{{\displaystyle \sin \frac{{\displaystyle x}}{{\displaystyle 2}}}}{{\displaystyle \frac{{\displaystyle x}}{{\displaystyle 2}}}}}{\displaystyle )}}^2}$

$=2\times \left({\log }_{e}9{\log }_{e}8\right(2\sqrt{2}\left)\right)=4\sqrt{2}\times 2\times 3{\log }_{e}2{\log }_{e}3$

$=24\sqrt{2}{\log }_{e}2{\log }_{e}3=a{\log }_{e}2{\log }_{e}3$                      $(\because \mathrm{Given})$

$\Rightarrow a=24\sqrt{2}$

Hence, $a^2=1152$

(1)

$f\left(x\right)=\frac{\sin 3x+\alpha \sin x-\beta \cos 3x}{x^3}$

$f\left(0\right)=\underset{x\to 0}{\lim }\frac{(3x-\frac{27x^3}{6}+\dots )+\alpha (x-\frac{x^3}{6}+\dots )-\beta (1-\frac{9x^2}{2}+\dots )}{x^3}$

$=\underset{x\to 0}{\lim }\frac{-\beta +x(3+\alpha )+x^2\frac{9}{2}\beta +x^3(\frac{-9}{2}-\frac{\alpha }{6})}{x^3}$          (Rest of the term will be zero)

Since this limit exists so, $\beta =0$

$3+\alpha =0$ and $\frac{9}{2}\beta =0$

$\Rightarrow \alpha =-3$

$\therefore$    $f\left(0\right)=\underset{x\to 0}{\lim }\frac{(\frac{-9}{2}-\frac{\alpha }{6})x^3}{x^3}=\frac{-9}{2}+\frac{3}{6}$

$=\frac{-24}{6}=-4$

So, $f\left(0\right)=-4$ 

(4)

$\text{We have, }g\left(f\right(x\left)\right)=x$

$\text{On differentiating w.r.t. }x,\text{ we get}$ $g\text{'}\left(f\right(x\left)\right)·f\text{'}\left(x\right)=1$

$\Rightarrow g\text{'}\left(f\right(x\left)\right)=\frac{1}{f\text{'}\left(x\right)}$                                   ...(i)

$\text{Now, put }f\left(x\right)=7,\text{ we get}$ $x^5+2x^3+3x+1=7$

$\Rightarrow x^5+2x^3+3x-6=0$

$x=1$ is the only solution because this function is increasing

$\Rightarrow g\left(7\right)=1$                                                    ...(ii)

$\text{Now, }g\text{'}\left(7\right)=\frac{1}{f\text{'}\left(1\right)}$                              [By (i) and (ii)]

$=\frac{1}{5+6+3}=\frac{1}{14}$

$\therefore \frac{g\left(7\right)}{g\text{'}\left(7\right)}=14$

(3)

We have, $y=\frac{2\cos \theta +\cos 2\theta }{\cos 3\theta +4\cos 2\theta +5\cos \theta +2}$

$=\frac{2\cos \theta +2{\cos }^2\theta -1}{4{\cos }^3\theta -3\cos \theta +8{\cos }^2\theta -4+5\cos \theta +2}$

$=\frac{2{\cos }^2\theta +2\cos \theta -1}{4{\cos }^3\theta +8{\cos }^2\theta +2\cos \theta -2}$

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

$=\frac{2{\cos }^2\theta +2\cos \theta -1}{2\cos \theta (2{\cos }^2\theta +2\cos \theta -1)+2(2{\cos }^2\theta +2\cos \theta -1)}$

$=\frac{1}{2\cos \theta +2}$

$=\frac{1}{2}\left(\frac{{\displaystyle 1}}{{\displaystyle \cos \theta +1}}\right)$

$y\text{'}\left(\theta \right)=\frac{1}{2}\left[\frac{{\displaystyle \sin \theta }}{{\displaystyle {(1+\cos \theta )}^2}}\right]$

$y\text{'}\text{'}\left(\theta \right)=\frac{1}{2}\left[\frac{{\displaystyle \cos \theta {(1+\cos \theta )}^2+{\sin }^2\theta ·2(1+\cos \theta )}}{{\displaystyle {(1+\cos \theta )}^4}}\right]$

$=\frac{1}{2}\left[\frac{{\displaystyle \cos \theta (1+\cos \theta )+2{\sin }^2\theta }}{{\displaystyle {(1+\cos \theta )}^3}}\right]$

at $\theta =\frac{\pi }{2}, y\text{'}=\frac{1}{2}, y\text{'}\text{'}=1$ and $y=\frac{1}{2}$

So, $y\text{'}\text{'}+y\text{'}+y=1+\frac{1}{2}+\frac{1}{2}=2$

(1)

$f\left(x\right)=2x^2+x+\left[x^2\right]-\left[x\right]$

We check continuity at

$x=0,1,2,-1,\sqrt{2},\sqrt{3}$

At $x=0$

$\text{LHL}=\underset{x\to 0^{-}}{\lim }f\left(x\right)=0+0+0-(-1)=1$

$\text{RHL}=\underset{x\to 0^{+}}{\lim }f\left(x\right)=0+0+0-\left(0\right)=-0$

$\text{∵ LHL}\ne \text{RHL at }x=0$

$\therefore$   $f\left(x\right)$ is not continuous at $x=0$

        At $x=1$

$\text{LHL}=\underset{x\to 1^{-}}{\lim }f\left(x\right)=2+1+0-0=3$

$\text{RHL}=\underset{x\to 1^{+}}{\lim }f\left(x\right)=2+1+1-1=3$

$f\left(1\right)=2+1+1-1=3$

$\text{∵ LHL}=\text{RHL}=f\left(x\right)\text{ at }x=1$

$\therefore f\left(x\right)\text{ is continuous at }x=1$

At   $x=2$

$\text{LHL}=\underset{x\to 2^{-}}{\lim }f\left(x\right)=8+2+3-1=12$

$\text{RHL}=\underset{x\to 2^{+}}{\lim }f\left(x\right)=8+2+4-2=12$

$f\left(2\right)=8+2+4-2=12$

$\text{∵ LHL}=\text{RHL}=f\left(x\right)\text{ at }x=2$

$\therefore$  $f\left(x\right)\text{ is continuous at }x=2$

At  $x=\sqrt{2}$

$\text{LHL}=\underset{x\to {\sqrt{2}}^{-}}{\lim }f\left(x\right)=4+\sqrt{2}+1-1=4+\sqrt{2}$

$\text{RHL}=\underset{x\to {\sqrt{2}}^{+}}{\lim }f\left(x\right)=4+\sqrt{2}+2-1=5+\sqrt{2}$

$\text{∵ LHL}\ne \text{RHL at }x=\sqrt{2}$

$\therefore f\left(x\right)\text{ is not continuous at }x=\sqrt{2}$

Similarly, $\mathrm{LHL}\ne \mathrm{RHL}$ at $x=-1$ and $\sqrt{3}$

So there are 4 points of discontinuity.

(4)

$f\left(x\right)$$=\{\begin{array}{cc}{x}^3\sin \frac{1}{x},& x\ne 0\\ 0,& x=0\end{array},$

It is a differentiable function.

$\therefore$   $f\text{'}\left(x\right)$$=3x^2\sin \frac{1}{x}+x^3\cos \frac{1}{x}\left(\frac{-1}{x^2}\right)$

            $=3x^2\sin \frac{1}{x}-x\cos \frac{1}{x}$

           $f\text{'}\text{'}\left(x\right)$$=6x\sin \frac{1}{x}-3\cos \frac{1}{x}-\cos \frac{1}{x}-\frac{\sin \frac{1}{x}}{x}$

          $f\text{'}\text{'}\left(\frac{{\displaystyle 2}}{{\displaystyle \pi }}\right)=\frac{6\times 2}{\pi }-\frac{\pi }{2}=\frac{24-{\pi }^2}{2\pi }$

(4)

$\text{We have, }g\left(x\right)=h\left(e^x\right)e^{h\left(x\right)}$

$g\text{'}\left(x\right)=h\left(e^x\right)e^{h\left(x\right)}h\text{'}\left(x\right)+e^{h\left(x\right)}h\text{'}\left(e^x\right)·e^x$

$g\text{'}\left(0\right)=h\left(1\right)e^{h\left(0\right)}h\text{'}\left(0\right)+e^{h\left(0\right)}h\text{'}\left(1\right)=1\times 2+2=4$

(2)

$\text{We have, }f\left(x\right)\text{ is continuous at }x=0.$

$\therefore$     $f\left(0^{-}\right)=f\left(0\right)=f\left(0^{+}\right)$

$\therefore$   $\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{-}}{\lim }\frac{\tan \left[\right(a+1\left)x\right]+b\tan x}{x}$

                  $=\underset{x\to 0^{-}}{\lim }\frac{\tan \left[\right(a+1\left)x\right]}{(a+1)x}\times (a+1)+b\underset{x\to 0^{-}}{\lim }\frac{\tan x}{x}$

                  $=a+1+b=3$                            $[\because f(0)=3]$

$\text{and }\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }\frac{\sqrt{ax+b^2{x}^2}-\sqrt{ax}}{b\sqrt{a} x\sqrt{x}}$

$=\underset{x\to 0^{+}}{\lim }\frac{\sqrt{a+b^{2}x}-\sqrt{a}}{b\sqrt{a}x}\times \frac{\sqrt{a+b^{2}x}+\sqrt{a}}{\sqrt{a+b^{2}x}+\sqrt{a}}$

$=\underset{x\to 0^{+}}{\lim }\frac{b^{2}x}{b\sqrt{a} x(\sqrt{a+b^{2}x}+\sqrt{a})}$

$=\underset{x\to 0^{+}}{\lim }\frac{b^2}{b\sqrt{a}(\sqrt{a+b^{2}x}+\sqrt{a})}=\frac{b^2}{b\sqrt{a}(\sqrt{a+0}+\sqrt{a})}=\frac{b}{2a}=3$

$\therefore$   $a+1+b=3 \text{and} \frac{b}{2a}=3\Rightarrow \frac{b}{a}=6$

(4)

$\text{We have, }f\left(x\right)=a{x}^3+b{x}^2+cx+41$

$f\left(1\right)=40\Rightarrow a+b+c+41=40\Rightarrow a+b+c=-1$              ...(i)

$f\text{'}\left(1\right)=2\Rightarrow 3a+2b+c=2$                                                          ...(ii)

$\text{Also, }f\text{'}\text{'}\left(x\right)=6ax+2b$

$f\text{'}\text{'}\left(1\right)=4\Rightarrow 6a+2b=4$                                                                ...(iii)

$\text{On solving (i), (ii), and (iii), we get }a=-1, b=5, c=-5$

$\therefore$    $a^2+b^2+c^2=1+25+25=51$

(2)

${\log }_{e}y=3{\sin }^{-1}x$

Differentiating w.r.t. $\text{'}x\text{'}$ we get

$\frac{1}{y}y\text{'}=\frac{3}{\sqrt{1-x^2}}$

$y\text{'}=\frac{3y}{\sqrt{1-x^2}}$                                                                    ...(i)

Again differentiating w.r.t. $\text{'}x\text{'}$ we get

$y\text{'}\text{'}=3\left[\frac{{\displaystyle \sqrt{1-x^2} y\text{'}-y \frac{{\displaystyle 1}}{{\displaystyle 2\sqrt{1-x^2}}}(-2x)}}{{\displaystyle (1-x^2)}}\right]$

$\Rightarrow (1-x^2)y\text{'}\text{'}=3[\sqrt{1-x^2}·\frac{{\displaystyle 3y}}{{\displaystyle \sqrt{1-x^2}}}+\frac{{\displaystyle xy}}{{\displaystyle \sqrt{1-x^2}}}]$         [Using (i)]

                              $=3[3y+\frac{{\displaystyle xy}}{{\displaystyle \sqrt{1-x^2}}}]$

Now, $y\text{'}\left(\frac{{\displaystyle 1}}{{\displaystyle 2}}\right)=\frac{3e^{3{\sin }^{-1}\left(\frac{1}{2}\right)}}{\sqrt{1-\frac{1}{4}}}=\frac{3e^{\pi /2}}{{\displaystyle \frac{\sqrt{3}}{2}}}=2\sqrt{3} e^{\pi /2}$

$(1-x^2)y\text{'}\text{'}\left(x\right)\text{ at (}x=\frac{{\displaystyle 1}}{{\displaystyle 2}}\text{)}=3[3e^{\pi /2}+\frac{{\displaystyle \frac{{\displaystyle 1}}{{\displaystyle 2}}{e}^{\pi /2}}}{{\displaystyle \frac{\sqrt{3}}{2}}}]$

$=3[3e^{\pi /2}+\frac{{\displaystyle 1}}{{\displaystyle \sqrt{3}}}{e}^{\pi /2}]=3e^{\pi /2}[3+\frac{{\displaystyle 1}}{{\displaystyle \sqrt{3}}}]$

So, $(1-x^2)y\text{'}\text{'}-xy\text{'}$ at $x=\frac{1}{2}$ is given by

$3e^{\pi /2}[3+\frac{{\displaystyle 1}}{{\displaystyle \sqrt{3}}}]-\frac{1}{2}(2\sqrt{3} e^{\pi /2})=9e^{\pi /2}$

(4)

$\text{Given, }f\left(x\right)\text{ is continuous everywhere}$

$f\left(0^{-}\right)=f\left(0\right)\Rightarrow 2b=2\Rightarrow b=1$

$f\left(1\right)=f\left(1^{+}\right)\Rightarrow 3+c=3\Rightarrow c=0$

$\text{Since, }f\left(0^{-}\right)=2$

$=\underset{h\to 0}{\lim }\frac{a-b\cos 2h}{h^2}=\underset{h\to 0}{\lim }\frac{a-b\{1-\frac{4h^2}{2!}+\frac{16h^4}{4!}-\dots \}}{h^2}$

$=\underset{h\to 0}{\lim }\frac{(a-b)+b\{2h^2-\frac{2}{3}{h}^4\dots \}}{h^2}$

$\text{=for limit to exist }a-b=0\text{ and limit is }2b$         $\therefore a=b=1$

$\text{Now, }Lf\text{'}\left(0\right)=\underset{x\to 0}{\lim }\frac{\frac{1-\cos 2h}{h^2}-2}{-h}$

$=\underset{x\to 0}{\lim }\frac{1-(1-\frac{4h^2}{2!}+\frac{16h^2}{4!}\dots )-2h^2}{-h^3}=0$

$\text{Also, }Rf\text{'}\left(0\right)=\underset{x\to 0}{\lim }\frac{{(0+h)}^2+2-2}{h}=0$

$\therefore m=0$

$\therefore m+a+b+c=0+1+1+0=2$

(3)

$\text{As }f\left(x\right)\text{ is continuous at }x=3.$

$\text{So, LHL=RHL = }f\left(a\right)$

$\Rightarrow \underset{x\to 3^{-}}{\lim }\frac{a(7x-12-x^2)}{b|x^2-7x+12|}=\underset{x\to 3^{+}}{\lim }{2}^{\frac{\sin (x-3)}{x-\left[x\right]}}=b$

$\Rightarrow \frac{-a}{b}=2^1=b$                                     $[\because \underset{n\to 0}{\lim }\frac{\sin n}{n}=1]$

 $\Rightarrow b=2, a=-4 \therefore \text{Number of elements in }S=1$

(3)

(4)

We have, $f\left(x\right)$$=\frac{(2^x+2^{-x})\tan x\sqrt{{\tan }^{-1}(x^2-x+1)}}{{(7x^2+3x+1)}^3}$

Let $A\left(x\right)=(2^x+2^{-x}), B\left(x\right)=\tan x$ and $C\left(x\right)$$=\sqrt{{\tan }^{-1}(x^2-x+1)}$

$A\text{'}\left(x\right)=2^x\log 2-2^{-x}\log 2=(2^x-2^{-x})\log 2,$

$B\text{'}\left(x\right)={\sec }^{2}x$

and $C\text{'}\left(x\right)$$=\frac{1}{2\sqrt{{\tan }^{-1}(x^2-x+1)}}\times \frac{1}{1+{(x^2-x+1)}^2}\times$$(2x-1)$

$=\frac{(2x-1)}{2(1+{(x^2-x+1)}^2)\sqrt{{\tan }^{-1}(x^2-x+1)}}$

$A\left(0\right)=2, B\left(0\right)=0, \text{and }C\left(0\right)=\sqrt{\frac{\pi }{4}}=\frac{\sqrt{\pi }}{2}$

Also, $A\text{'}\left(0\right)=0, B\text{'}\left(0\right)=1, \text{and }C\text{'}\left(0\right)=-\frac{1}{2\sqrt{\pi }}$

Now, $f\text{'}\left(x\right)$$=\frac{[A\text{'}(x\left)B\right(x\left)C\right(x)+A(x)B\text{'}(x\left)C\right(x)+A(x\left)B\right(x)C\text{'}(x\left)\right]-A\left(x\right)B\left(x\right)C\left(x\right)\left[3{(7x^2+3x+1)}^2\right(14x+3\left)\right]}{{(7x^2+3x+1)}^6}$

$f\text{'}\left(0\right)$$=\frac{[A\text{'}(0\left)B\right(0\left)C\right(0)+A(0)B\text{'}(0\left)C\right(0)+A(0\left)B\right(0)C\text{'}(0\left)\right]-A\left(0\right)B\left(0\right)C\left(0\right)\left[3{\left(1\right)}^2\right(3\left)\right]}{{\left(1\right)}^6}$

$=(0+2\times 1\times \frac{{\displaystyle \sqrt{\pi }}}{{\displaystyle 2}}+0)-0=\sqrt{\pi }$

(2)

We have, $y={\log }_e\left(\frac{{\displaystyle 1-x^2}}{{\displaystyle 1+x^2}}\right)$

$\Rightarrow y={\log }_e(1-x^2)-{\log }_e(1+x^2)$

$\Rightarrow y\text{'}=\frac{1}{1-x^2}\times (-2x)-\frac{1}{1+x^2}\times \left(2x\right)$

$\Rightarrow y\text{'}=\frac{-4x}{1-x^4}\Rightarrow y\text{'}\text{'}=\frac{(1-x^4)(-4)-(-4x)(-4x^3)}{{(1-x^4)}^2}$

$\Rightarrow y\text{'}\text{'}=\frac{-4(1+3x^4)}{{(1-x^4)}^2}$

Now, $225 (y\text{'}-y\text{'}\text{'})=225[\frac{{\displaystyle -4x}}{{\displaystyle 1-x^4}}+\frac{{\displaystyle 4(1+3x^4)}}{{\displaystyle {(1-x^4)}^2}}]$

$\Rightarrow 225 (y\text{'}-y\text{'}\text{'})\text{ at }x=\frac{1}{2}\text{ is equal to}$ $225[\frac{{\displaystyle -4\times \frac{{\displaystyle 1}}{{\displaystyle 2}}}}{{\displaystyle 1-\frac{{\displaystyle 1}}{{\displaystyle 16}}}}+\frac{{\displaystyle 4(1+\frac{{\displaystyle 3}}{{\displaystyle 16}})}}{{\displaystyle {(1-\frac{{\displaystyle 1}}{{\displaystyle 16}})}^2}}]=736$

(3)

$\text{We have, }f\left(x\right)=\frac{1}{2}\left[g\right(x)+g(2-x\left)\right]$

$\text{So }f\text{'}\left(x\right)=\frac{1}{2}(g\text{'}(x)+g\text{'}(2-x\left)\right(-1\left)\right)=\frac{1}{2}(g\text{'}(x)-g\text{'}(2-x\left)\right)$

      $\Rightarrow f\text{'}\left(\frac{1}{2}\right)=\frac{1}{2}(g\text{'}(\frac{1}{2})-g\text{'}(2-\frac{1}{2}\left)\right)$

       $=\frac{1}{2}(g\text{'}(\frac{1}{2})-g\text{'}(\frac{3}{2}\left)\right)=0$                               $[\because g\text{'}(\frac{1}{2})=g\text{'}(\frac{3}{2}\left)\right]$

$\text{Similarly }f\text{'}\left(\frac{3}{2}\right)=\frac{1}{2}(g\text{'}(\frac{3}{2})-g\text{'}(\frac{1}{2}\left)\right)=0$

$\text{Now, }f\text{'}\left(x\right)=0\text{ when }x=\frac{1}{2}\text{ and }\frac{3}{2}$

$\text{So, }f\text{'}\text{'}\left(x\right)\text{ has at least two roots in }(0,2).$           $\text{[By Rolle's theorem]}$

(2)

$\text{We have,} f\left(\frac{x}{y}\right)=\frac{f\left(x\right)}{f\left(y\right)} \text{and} f\text{'}\left(1\right)=2024$

$\text{On putting }x=y=1,\text{ we get }$

$f\left(1\right)=1$                                                                                      ...(i)

$\text{Also, on putting }x=1,\text{ we get }$

$f\left(\frac{1}{y}\right)=\frac{f\left(1\right)}{f\left(y\right)}=\frac{1}{f\left(y\right)} \text{(Using (i))}$

$\Rightarrow f\left(y\right)=\pm y^n (\because If f(x\left)f\right(\frac{1}{x})=1,\text{ then }f(x)=\pm x^n)$