Mahesh and Anand together can polish the floors of a building in 60 days. Mahesh alone can complete (1\4) part of the total work in 45 days. In how many days Anand alone can polish the floor of the same building?

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

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

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

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

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

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

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

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

$=\sqrt[]{\pi }$