Let us define a relation $R$ in $R$ as $aRb$ if $a\ge b$ . Then $R$ is:

(A) An equivalence relation

(B) Reflexive, transitive but not symmetric

(C) Symmetric, transitive but not reflexive

(D) Neither transitive nor reflexive but symmetric

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

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

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

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

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

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

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

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

$=\sqrt[]{\pi }$