Maxima and Minima

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

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

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

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  $b\in [-6,-2)\cup (2,6]$  

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

P' (x) = a (x-1) (x+1) = a (x²-1).
P (x) = ∫ P' (x) dx = a (x³/3 - x) + b.
Given P (-3) = 0 ⇒ a (-9+3) + b = 0 ⇒ b = 6a.
Given ∫ P (x)dx = 18. Assuming the integration is over a symmetric interval like [-c, c] and using the fact that a (x³/3-x) is an odd function, ∫ (a (x³/3 - x)dx = 0. Then ∫ b dx = 1

Equation x²/5 + y²/4 = 1 then P (√5cosθ, 2sinθ)
(PQ)² = 5cos²θ + 4 (sinθ+2)² = cos²θ + 16sinθ + 20
= -sin²θ + 16sinθ + 21
= 85 - (sinθ - 8)²
∴ (PQ)²max = 85 - 49 = 36
? (sinθ - 8)² ∈

C? → C? - C?
f (θ) = | -sin²θ -1 |
| -cos²θ -1 |
| 12 -2 -2|
= 4 (cos²θ - sin²θ) = 4 (cos2θ), θ ∈ [π/4, π/2]
f (θ)max = M = 0
f (θ)min = m = -4

A (α, 0), B (-α, 0)
⇒ D (α, α² − 1)
Area (ABCD) = (AB) (AD)
⇒ S = (2α) (1 − α²) = 2α – 2α³
dS/dα = 2 - 6α²
= 0 ⇒ α² = 1/3

⇒ α = 1/√3
Area = 2α – 2α³ = 2/√3 - 2/ (3√3)
= 4/ (3√3)