Question

If the function $g\left(t\right)={\int }_{2t}^{t^2}co{t}^{-1}\left|\frac{1+x}{(1+t{)}^2-x}\right|dx,then\frac{g\left(5\right)}{g\left(3\right)}$ is equal to ___.

• A. 5

Answer 1

The correct option is A

5

$g\left(t\right)={\int }_{2t}^{t^2}co{t}^{-1}\left|\frac{1+x}{(1+t{)}^2-x}\right|dx\cdots \cdots \cdots \left(i\right)g\left(t\right)={\int }_{2t}^{t^2}co{t}^{-1}\left|\frac{(1+t{)}^2-x}{1+x}\right|dx\cdots \cdots \left(ii\right)$
(i) + (ii) gives
$2g\left(t\right)={\int }_{2t}^{t^2}\left(\frac{\pi }{2}\right)dx\Rightarrow g\left(t\right)=\frac{\pi }{4}\left(\right(t-1{)}^2-1)\Rightarrow \frac{g\left(5\right)}{g\left(3\right)}=\frac{\frac{\pi }{4}\left(15\right)}{\frac{\pi }{4}\left(3\right)}=5$