Question

If $\int \frac{2dx}{\left(\right(x-5)+(x-7\left)\right)\sqrt{(x-5)(x-7)}}=f\left(g\right(x\left)\right)+c$, then

• A. $f\left(x\right)=se{c}^{-1}x$
  $g\left(x\right)=\sqrt{(x-5)+(x-7)}$
• B. $f\left(x\right)=se{c}^{-1}x$
  $g\left(x\right)=\sqrt{(x-5)(x-7)}$
• C. $f\left(x\right)=ta{n}^{-1}x$
  $g\left(x\right)=\sqrt{(x-5)+(x-7)}$
• D. $f\left(x\right)=ta{n}^{-1}x$
  $g\left(x\right)=\sqrt{(x-5)(x-7)}$

Answer 1

The correct option is D

$f\left(x\right)=ta{n}^{-1}x$
$g\left(x\right)=\sqrt{(x-5)(x-7)}$

$(x-5)+(x-7)=2(x-6)$
and $(x-5)(x-7)=x^2-12x+35=(x-6{)}^2-1....(1)$
$\therefore I=\int \frac{dx}{(x-6)\sqrt{(x-6{)}^2-1}}=se{c}^{-1}(x-6)=ta{n}^{-1}\left[\right(x-6{)}^2-1{]}^{\frac{1}{2}}\because se{c}^{-1}z=ta{n}^{-1}\sqrt{z^2-1}=ta{n}^{-1}\sqrt{(x-5)(x-7)}$
=f(g(x))
$\therefore f\left(x\right)=ta{n}^{-1}x$ and $g\left(x\right)=\sqrt{(x-5)(x-7)}$