If $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$  are non-zero and $\overrightarrow{b}$  and $\overrightarrow{c}$  are non-collinear. $\overrightarrow{a}+5\overrightarrow{b}$  is collinear with $\overrightarrow{c}$  and $\overrightarrow{b}+6\overrightarrow{c}$  is collinear with $\overrightarrow{a}$ . If $\overrightarrow{a}+\alpha \overrightarrow{b}+\beta \overrightarrow{b}=0$ , then find α + β.

It's difficult but in some colleges you may can get 

y (x) = 2^x – x²

y? (x) = 2x log 2 – 2x

M = 3

N = 2

M + N = 5

${\displaystyle \int \frac{\left(sinx-cosx\right)si{n}^{2}x}{tanx\left(si{n}^{3}x+co{s}^{3}x\right)}dx}$

${\displaystyle \int \frac{\left(sinx-cosx\right)sinxcosx}{si{n}^{3}x+co{s}^{3}x}dx}$ , put sin³x + cos³x = t(3 sin²x×cosx – 3cos²xsinx) dx = dt

-> $\frac{1}{3}{\displaystyle \int \frac{dt}{t}}$

$=\frac{lnt}{3}+c$

$=\frac{ln\left|si{n}^{3}x+co{s}^{3}x\right|}{3}+c$

$f\left(x\right)=\frac{\left(2^x+2^{-x}\right)tanx\sqrt[]{ta{n}^{-1}\left(2x^2-3x+1\right)}}{{\left(7x^2-3x+1\right)}^3}$

$f\left(x\right)=\left(2^x+2^{-x}\right).tanx.\sqrt[]{ta{n}^{-1}\left(2x^2-3x+1\right)}.{\left(7x^2-3x+1\right)}^{-3}$

$f\text{'}\left(x\right)=\left(2^x+2^{-x}\right).se{c}^{2}x.\sqrt[]{ta{n}^{-1}\left(2x^2-3x+1\right)}.{\left(7x^2-3x+1\right)}^{-3}+tanx.\left(Q\left(x\right)\right)$

$f\text{'}\left(0\right)=2.1\sqrt[]{\frac{\pi }{4}}.1$

$=\sqrt[]{\pi }$

$Area=\frac{\pi {\left(13\right)}^2}{2}-\left[\frac{1}{2}\times 25\times 5+{\displaystyle \underset{12}{\overset{13}{\int }}\sqrt[]{\left(169-y^2\right)}}\cdot dy\right]$

$=\frac{169\pi }{2}-\left[\frac{125}{2}+{\left[\frac{y}{2}\sqrt[]{169-y^2}+\frac{169}{2}si{n}^{-1}\frac{y}{13}\right]}_{12}^{13}\right]$

$=\frac{169}{2}\pi -\frac{125}{2}-\left[\frac{169}{2}\times \frac{\pi }{2}-6\times 5-\frac{169}{2}si{n}^{-1}\frac{12}{13}\right]$

$=\frac{169\pi }{4}-\frac{65}{2}+\frac{169}{2}si{n}^{-1}\frac{12}{13}$