Continuity and Differentiability

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

$e=\sqrt[]{1+\frac{b^2}{a^2}}$ $e\text{'}=\sqrt[]{1+\frac{a^2}{b^2}}$

$\mathcal{l}=\frac{2b^2}{a}$ $\mathcal{l}\text{'}=\frac{2a^2}{b}$

$\left(1+\frac{b^2}{a^2}\right)=\frac{11}{14}*\frac{2b^2}{a}$ $\left(1+\frac{a^2}{b^2}\right)=\frac{11}{8}*\frac{2a^2}{b}$

$7*\left\{{\left(\frac{7b}{4}\right)}^2+b^2\right\}=11*b^2*\frac{7b}{4}\Rightarrow a*77=65$

65b² = 44b³

65 = b * 44

$\therefore 77a+44b=65*2=130$

f (x) is an even function

$f\left(-\frac{1}{4}\right)=f\left(-\frac{1}{2}\right)=f\left(\frac{1}{2}\right)=f\left(\frac{1}{4}\right)=0$

So, f (x) has at least four roots in (-2, 2)

$g\left(\frac{-3}{4}\right)=g\left(\frac{3}{4}\right)=0$

So, g (x) has at least two roots in (2, 2)

now number of roots of f (x) $\cdot g"\left(x\right)=f\text{'}\left(x\right)\cdot g\text{'}\left(x\right)=0$

It is same as number of roots of $\frac{d}{dx}\left(f\left(x\right)\cdot g\text{'}\left(x\right)\right)=0$ will have atleast 4 roots in (2, 2)

$f\text{'}\left(x\right)=\frac{n_1\cdot f\left(x\right)}{x-3}+n_2\cdot \frac{f\left(x\right)}{x-5}$

$=\frac{f\left(x\right)\cdot \left(n_1+n_2\right)}{\left(x-3\right)\left(x-5\right)}\left(x-\frac{\left(5n_1+3n_2\right)}{n_1+n_2}\right)$

$f\text{'}\left(x\right)={\left(x-3\right)}^{n_1-1}\cdot {\left(x-5\right)}^{n_2-1}\cdot \left(n_1+n_2\right)\left(x-\frac{\left(5n_1+3n_2\right)}{n_1+n_{2l}}\right)$

option (C) is incorrect, there will be minima.

Let P (at², 2 at) where

$a=\frac{3}{2}$                

T : yt = x + at² so point Q is

$\left(-a,at-\frac{a}{t}\right)$                

N : y = -tx + 2at + at³ passes through (5, -8)

$-8=-5t+3t+\frac{3}{2}{t}^3$

⇒ $3t^3-4t+16=0$                

⇒ t = -2

So ordinate of point Q is $-\frac{9}{4}$  

Equation of L₁ = is

$\frac{xsec\theta }{4}-\frac{ytan\theta }{2}=1$ ….(i)

Equation of line L₂ is

$\frac{xtan\theta }{2}+\frac{ysec\theta }{4}=0$ ….(ii)

$?$ Required point of intersection of L₁ and L₂ is (x₁, y₁) then

$\frac{x_{1}sec\theta }{4}-\frac{y_{1}tan\theta }{2}-1=0$ ….(iii)

$and\frac{y_{1}sec\theta }{4}+\frac{x_{1}tan\theta }{2}=0$ ……(iv)

From equations (iii) and (iv)

$sec\theta =\frac{4x_1}{x_1^2+y_1^2}andtan\theta =\frac{-2y_1}{x_1^2+y_1^2}$        

$\therefore$ Required locus of (x₁, y₁) is

${\left(x^2+y^2\right)}^2=16x^2-4y^2$   

$\begin{array}{l}\therefore \alpha =16,\beta =-4\\ \therefore \alpha =\beta =12\end{array}$     

Given hyperbola :

$\frac{x^2}{a^2}-\frac{y^2}{9}=1$

$$ $?$  it passes through

$\left(8,3\sqrt[]{3}\right)$

$?\frac{64}{a^2}-\frac{27}{9}=1\Rightarrow a^2=16$

Now, equation of normal to hyperbola

$\frac{16x}{8}+\frac{9y}{3\sqrt[]{3}}=16+9$

$\text{exception 25:}$ $\left(-1,9\sqrt[]{3}\right)$  satisfied

Draw g(t) = t³ – 3t

g'(t) = 3(t² – 1)

g(1) is maximum in (-2, 2)

So, maximum (t³ – 3t) = $\{\begin{array}{cc}\hfill t^3-3t\hfill & \hfill ;-2<t<-1\hfill \\ \hfill 2\hfill & \hfill ;-1<t<2\hfill \end{array}$

$I={\displaystyle \underset{-2}{\overset{2}{\int }}f}\left(x\right)dx$

= ${\displaystyle \underset{-2}{\overset{-1}{\int }}\left(t^3-3t\right)dt}+{\displaystyle \underset{-1}{\overset{2}{\int }}2dt}$

I = $\frac{27}{4}$

again rewrite the f(x)

$f\left(x\right)=\left\{\begin{array}{ccc}\hfill \begin{array}{cc}\hfill x^3-3x\hfill & \hfill 2\hfill \end{array}\hfill & \hfill ;\hfill & \hfill \begin{array}{cc}\hfill x\le -1\hfill & \hfill -1<x\le 2\hfill \end{array}\hfill \\ \hfill \begin{array}{cc}\hfill x^2+2x-6\hfill & \hfill 9\hfill \end{array}\hfill & \hfill ;\hfill & \hfill \begin{array}{cc}\hfill 2<x<3\hfill & \hfill 3\le x<4\hfill \end{array}\hfill \\ \hfill \begin{array}{ccc}\hfill 10\hfill & \hfill 11\hfill & \hfill 2x+1\hfill \end{array}\hfill & \hfill ;\hfill & \hfill \begin{array}{ccc}\hfill 4\le x<5\hfill & \hfill x=5\hfill & \hfill x>5\hfill \end{array}\hfill \end{array}\right\}$

$f\text{'}\left(x\right)=\left\{\begin{array}{cccccc}\hfill 3x^2-3\hfill & \hfill ;\hfill & \hfill x<-1\hfill & \hfill 0\hfill & \hfill ;\hfill & \hfill -1<x<2\hfill \\ \hfill 2x+2\hfill & \hfill ;\hfill & \hfill 2<x<3\hfill & \hfill 0\hfill & \hfill ;\hfill & \hfill 3<x<4\hfill \\ \hfill 0\hfill & \hfill ;\hfill & \hfill 4<x<5\hfill & \hfill 2\hfill & \hfill ;\hfill & \hfill x>5\hfill \end{array}\right\}$

So f(x) is not differentiable at x = 2, 3, 4, 5

so m = 4

 $f\left(x\right)=\{\begin{array}{l}\frac{lo{g}_e\left(1-x+x^2\right)+lo{g}_e\left(1+x+x^2\right)}{secx-cosx},x\in \left(\frac{-\pi }{2},\frac{\pi }{2}\right)-\left\{0\right\}\\ k,x=0\end{array}$ for continuity at x = 0

$\underset{x\to 0}{lim}f\left(x\right)=k\therefore k=\underset{x\to 0}{lim}\frac{lo{g}_e\left(1+x^2+x^4\right)}{secx-cosx}\left(\frac{0}{0}form\right)=\underset{x\to 0}{lim}\frac{cosxlo{g}_e\left(1+x^2+x^4\right)}{si{n}^{2}x}=1$

f (3x)- f (x) = x

Replace $x\to \frac{x}{3}\Rightarrow f\left(x\right)-f\left(\frac{x}{3}\right)=\frac{x}{3}$

Again replace $x\to \frac{x}{3}\Rightarrow f\left(\frac{x}{3}\right)-f\left(\frac{x}{3^2}\right)-f\left(\frac{x}{3^2}\right)=\frac{x}{3^2}$

$\Rightarrow f\left(3x\right)-f\left(0\right)=\frac{3x}{2}puttingx=\frac{8}{3}\Rightarrow f\left(8\right)-f\left(0\right)=4\therefore f\left(0\right)=3$

Also putting x = $\frac{14}{3}$ in f (3x) – 3 = $\frac{3x}{2}\Rightarrow$ F (14) – 3 = 7 f (14) = 10

 $\beta =\frac{\alpha x-\left(e^{3x}-1\right)}{\alpha x\left(e^{3x}-1\right)},\alpha \in R\underset{x\to 0}{lim}\frac{\frac{\alpha }{3}-\left(\frac{e^{3x}-1}{3x}\right)}{\alpha x\left(\frac{e^{3x}-1}{3x}\right)}$

$=\underset{x\to 0}{lim}\frac{1-\frac{\left(1+3x+\frac{9x^2}{2}+..........-1\right)}{3x}}{1+3x+\frac{9x^2}{2}+........-1}=-\frac{1}{2}\therefore \alpha +\beta =\frac{5}{2}$