Continuity and Differentiability

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

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

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

Note : n should be given as a natural number:

$f\left(x\right)=\{\begin{array}{l}\begin{array}{ccc}\hfill \frac{-sin\left(x-1\right)}{x-1}\hfill & \hfill ,\hfill & \hfill x<-1\hfill \\ \hfill -\left(sin2+1\right)\hfill & \hfill ,\hfill & \hfill x=-1\hfill \\ \hfill cos2\pi x\hfill & \hfill ,\hfill & \hfill -1<x<1\hfill \end{array}\\ 1,x=1\end{array}$

$\frac{-sin\left(x-1\right)}{x-1},x>1$

f (x) is discontinuous at x = 1 and x = 1

Since,  ${\mathrm{l}\mathrm{i}\mathrm{m}}_{x\to 0}?\frac{f\left(x\right)}{x}$ exist $\Rightarrow f\left(0\right)=0$

Now,  $f^{\mathrm{\text{'}}}\left(x\right)={\mathrm{l}\mathrm{i}\mathrm{m}}_{h\to 0}?\frac{f(x+h)-f\left(x\right)}{h}={\mathrm{l}\mathrm{i}\mathrm{m}}_{h\to 0}?\frac{f\left(h\right)+x{h}^2+x^{2}h}{h}($ take $y=h)$

$={\mathrm{l}\mathrm{i}\mathrm{m}}_{h\to 0}?\frac{f\left(h\right)}{h}+{\mathrm{l}\mathrm{i}\mathrm{m}}_{h\to .0}?\left(xh\right)+x^2$

$\Rightarrow {\mathrm{f}}^{\mathrm{\text{'}}}\left(\mathrm{x}\right)=1+0+{\mathrm{x}}^2$

$\Rightarrow f^{\mathrm{\text{'}}}\left(3\right)=10$

$e^{4x}+4e^{3x}-58e^{2x}+4e^x+1=0$

$\left(e^{2x}+\frac{1}{e^{2x}}\right)+4\left(e^x+\frac{1}{e^x}\right)-58=0$

$\Rightarrow {\left(e^x+\frac{1}{e^x}+2\right)}^2=64$

$e^x$$=$$\frac{6\pm \text{exception 25:}}{2}$$=$$3$$\pm$$2$