Question

Let $\alpha ,\beta ,\gamma$ and $\delta$ be 4 distinct roots of the equation $x^4-4x+3=x\left(x^3-f^{\prime }(1)x^2+f^{\prime \prime }(1)x-4\right)+f(1)$ and $f(x)$ is a monic polynomial of degree 3 .
If $\int \frac{dx}{\frac{f(x)-3}{x-1}+4}=\frac{1}{2}g(x)+C$, where $C$ is constant of integration and $g(3)=\frac{\pi }{4}$, then the value of $g(5)+g(7)$ is

• A. $\frac{\pi }{2}$
• B. $\frac{3\pi }{4}$
• C. $\pi$
• D. $\frac{5\pi }{4}$

Answer 1

Answer: (B)

$x^4-4x+3=x^4-f^{\prime }(1)x^3+f^{\prime \prime }(1)x^2-4x+f(1)$
which is a cubic equation and it has 4 distinct roots
$\therefore$ It is an identity
$f^{\prime }(1)=0,f^{\prime \prime }(1)=0,f(1)=3$
$\therefore f(x)=(x-1{)}^3+3$
(ii)$\int \frac{dx}{(x-1{)}^2+4}=\frac{1}{2}{\tan }^{-1}\left(\frac{x-1}{2}\right)+c$
$\therefore g(x)={\tan }^{-1}\left(\frac{x-1}{2}\right)$
$\Rightarrow g(5)+g(7)=\left({\tan }^{-1}2+{\tan }^{-1}3\right)=\frac{3\pi }{4}$.