Assume $e^{-\frac{4}{5}}=\frac{2}{5}.$  If x, y satisfy, y = e^x and the minimum value of (x² + y²) is expressed in the form of $\frac{m}{n}$ tehn $\frac{\left(2m-n\right)}{5}$  equals (where m & n are coprime natural numbers)

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

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

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

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

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

$f\left(x\right)=\{\begin{array}{cc}\hfill x^3-x^2+10x-7,\hfill & \hfill x\le 1\hfill \\ \hfill -2x+lo{g}_2\left(b^2-4\right),\hfill & \hfill x>1\hfill \end{array}$  

If f(x) has maximum value at x = 1 then

$f\left(1\right)\le f\left(1\right)\Rightarrow -2+lo{g}_2\left(b^2-4\right)\le 1-1+10-7$

$\Rightarrow lo{g}_2\left(b^2-4\right)\le 5\Rightarrow 0<b^2-4\le 32$

$\Rightarrow b^2-4>0\Rightarrow b\in \left(-\infty ,-2\right)\cup \left(2,\infty \right)$         ……..(i)

$And{b}^2-4\le 32\Rightarrow b\in \left[-6,6\right]$                      ……..(ii)

From (i) and (ii) we get  $b\in [-6,-2)\cup (2,6]$  

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