(B)

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

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

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

(D)

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

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

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

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

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

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

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

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

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

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

$\therefore m=0$

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

(C)

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

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

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

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

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

(C)

(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)

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

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

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

$\Rightarrow f\left(y\right)=y^n \text{(∵ }f\left(1\right)=1\text{)}$

$\Rightarrow f\text{'}\left(y\right)=n{y}^{n-1} \Rightarrow f\text{'}\left(1\right)=n=2024$

$\text{Now, }f\left(x\right)=x^{2024}$

$\Rightarrow f\text{'}\left(x\right)=2024x^{2023} \Rightarrow xf\text{'}\left(x\right)=2024x^{2024}$

$\therefore xf\text{'}\left(x\right)-2024x^{2024}=0$

(4)

$f\left(x\right)$$=\{\begin{array}{cc}g\left(x\right),& x<0\\ {\left(\frac{1+x}{2+x}\right)}^{\frac{1}{x}},& x>0\end{array}$ is continuous at $x=0$

Let $g\left(x\right)=px+q$

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

$\Rightarrow g\left(x\right)=p\left(x\right) (q=0)$

$\text{Now, }f\text{'}\left(1\right)=f(-1), y={\left(\frac{{\displaystyle 1+x}}{{\displaystyle 2+x}}\right)}^{1/x}$

$\Rightarrow \log y=\log {\left(\frac{{\displaystyle 1+x}}{{\displaystyle 2+x}}\right)}^{1/x}\Rightarrow \log y=\frac{1}{x}\log \left(\frac{{\displaystyle 1+x}}{{\displaystyle 2+x}}\right)$

$\Rightarrow \frac{1}{y}\frac{dy}{dx}=-\frac{1}{x^2}\log \left(\frac{{\displaystyle 1+x}}{{\displaystyle 2+x}}\right)+\frac{1}{x}\times \left(\frac{1}{{\displaystyle \frac{1+x}{2+x}}}\right)\times \frac{(x+2)-(x+1)}{{(2+x)}^2}$

at $x=1$

     $f\text{'}\left(1\right)=\frac{2}{3}[-\log (\frac{{\displaystyle 2}}{{\displaystyle 3}})+\frac{{\displaystyle 3}}{{\displaystyle 2}}(\frac{{\displaystyle 1}}{{\displaystyle 9}}\left)\right]\Rightarrow -\frac{2}{3}\log \left(\frac{{\displaystyle 2}}{{\displaystyle 3}}\right)+\frac{1}{9}$

and at $x=-1$

$f(-1)=-p=-\frac{2}{3}\log \left(\frac{{\displaystyle 2}}{{\displaystyle 3}}\right)+\frac{1}{9}\Rightarrow p=\frac{2}{3}\log \left(\frac{{\displaystyle 2}}{{\displaystyle 3}}\right)-\frac{1}{9}$

$g\left(3\right)=3p=2\log \left(\frac{{\displaystyle 2}}{{\displaystyle 3}}\right)-\frac{1}{3}=\log \left(\frac{{\displaystyle 4}}{{\displaystyle 9}}\right)+{\log }_e{ }^{-1/3}$

$=\log \left(\frac{{\displaystyle 4}}{{\displaystyle 9e^{1/3}}}\right)$

(2)

$f\left(x\right)$$=$$\left|\begin{array}{ccc}{x}^3& 2x^2+1& 1+3x\\ 3x^2+2& 2x& x^3+6\\ x^3-x& 4& x^2-2\end{array}\right|$

$f\left(0\right)$$=$$\left|\begin{array}{ccc}0& 1& 1\\ 2& 0& 6\\ 0& 4& -2\end{array}\right|$$=12$

$f\text{'}\left(0\right)$$=$$\left|\begin{array}{ccc}0& 0& 3\\ 2& 0& 6\\ 0& 4& -2\end{array}\right|$$+$$\left|\begin{array}{ccc}0& 1& 1\\ 0& 2& 0\\ 0& 4& -2\end{array}\right|$$+$$\left|\begin{array}{ccc}0& 1& 1\\ 2& 0& 6\\ -1& 0& 0\end{array}\right|$

            $=24+0-6=18$

$\Rightarrow 2f\left(0\right)+f\text{'}\left(0\right)=2\times 12+\left(18\right)=24+18\Rightarrow 42$