Let f(x) = $\{\begin{array}{cc}\hfill x^3-x^2+10x-7,x\le 1\hfill & \hfill -2x+lo{g}_2\left(b^2-4\right),x>1\hfill \end{array}.$  Then the set of all values of b, for which f(x) has maximum value at x = 1, is:

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

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

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

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

$f\left(x\right)=\left|x^2-3x-2\right|-x$

$=\left|\left(x-\frac{3-\sqrt[]{17}}{2}\right)\left(x-\frac{3+\sqrt[]{17}}{2}\right)\right|-x$

$\Rightarrow f\left(x\right)=\left[\begin{array}{cc}\hfill x^2-4x-2;-1\le x\le \frac{3-\sqrt[]{17}}{2}\hfill & \hfill -x^2+2x+2;\frac{3-\sqrt[]{17}}{2}<x\le 2\hfill \end{array}\right]$

absolute minimum $f\left(\frac{3-\sqrt[]{17}}{2}\right)=\frac{-3+\sqrt[]{17}}{2}$  

 absolute maximum = 3

$\therefore sum3+\frac{-3+\sqrt[]{17}}{2}=\frac{3+\sqrt[]{17}}{2}$  

OP² = x² = y²

y = e^x, y’ = e^x,

slope of normal =  $-\frac{1}{e^x}$

$\frac{y}{x}=-\frac{1}{e^x}$    

$-\frac{1}{e^x}=\frac{e^x}{x}\Rightarrow -x=e^{2x}$      

By hit and trial we get  $x=-\frac{2}{5}$

$P\equiv \left(-\frac{2}{5},e^{-2/5}\right)$

$OP=\sqrt[]{\frac{4}{25}+e^{-4/5}}\Rightarrow O{P}^2=\frac{14}{25}=\frac{m}{n}$

c = -5, a = 2, b = 1

CD = √ (10+x²)² – (10–x²)² = 2√10|x|
Area
= 1/2 × CD × AB = 1/2 × 2√10|x| (20–2x²)
=> 10 – x² = 2x
3x² = 10
x = k
3k² = 10