The united ages of a man and his wife are six times the united ages of their children. Two years ago their united ages were ten times the united ages of their children and six year hence their ages will be three times the united ages of their children. How many children have they?

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

  $\overrightarrow{a}+5\overrightarrow{b}$  is collinear with $\overrightarrow{c}$  

⇒   $\overrightarrow{a}+5\overrightarrow{b}$ = $\overrightarrow{c}$           …(1)

$\overrightarrow{b}+6\overrightarrow{c}$ is collinear with $\overrightarrow{a}$  

⇒   $\overrightarrow{b}+6\overrightarrow{c}=\mu \overrightarrow{a}$               …(2)

From (1) and (2)

  $\overrightarrow{b}+6\overrightarrow{c}=\mu \left(\lambda \overrightarrow{c}-5\overrightarrow{b}\right)$          

-> $\left(1+5\mu \right)\overrightarrow{b}+\left(6-\lambda \mu \right)\overrightarrow{c}=0$

$?\overrightarrow{b}$ and $\overrightarrow{c}$

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

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

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

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

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

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

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

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

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

$=\sqrt[]{\pi }$