Question

Let $f:R\to R$ be a function defined as $f\left(x\right)=(x-1)(x+2)(x-3)(x-6)-100.$ If $g\left(x\right)$ is a polynomial of degree at most $3$ such that $\int \frac{g\left(x\right)}{f\left(x\right)}dx$ does not contain any logarithmic function and $g(-2)=-10,$ then

• A. Minimum value of $f\left(x\right)$ is $-84$ when $x=2$
• B. Minimum value of $f\left(x\right)$ is $-42$ when $x=2$
• C. $\underset{0}{\overset{4}{\int }}\frac{g\left(x\right)}{f\left(x\right)}dx=\frac{\pi }{2}$
• D. $\underset{0}{\overset{4}{\int }}\frac{g\left(x\right)}{f\left(x\right)}dx=\pi$

Answer 1

Answer: (A), (C)

The correct options are
A Minimum value of $f\left(x\right)$ is $-84$ when $x=2$
C $\underset{0}{\overset{4}{\int }}\frac{g\left(x\right)}{f\left(x\right)}dx=\frac{\pi }{2}$

$f\left(x\right)=(x-1)(x+2)(x-3)(x-6)-100$
$=(x^2-4x-17)(x^2-4x+8)=\left(\right(x-2{)}^2-21)\left(\right(x-2{)}^2+4)$

Minimum value of $f\left(x\right)=-84$ at $x=2$

$\frac{g\left(x\right)}{f\left(x\right)}=\frac{g\left(x\right)}{(x^2-4x-17)(x^2-4x+8)}=\frac{Ax+B}{x^2-4x+8}+\frac{Cx+D}{x^2-4x-17}$

Since, $\int \frac{g\left(x\right)}{f\left(x\right)}dx$ does not contain any logarithmic function.
$\therefore A=C=0$

$\Rightarrow \frac{g\left(x\right)}{f\left(x\right)}=\frac{B}{(x-2{)}^2+2^2}+\frac{D}{(x-2{)}^2-(\sqrt{21}{)}^2}$

Since, second integral is in the form of $\int \frac{dx}{x^2-a^2},$ which contain logarithmic function.
$\therefore D=0$

$g\left(x\right)=B(x^2-4x-17),g(-2)=-10$
$\Rightarrow B=2$

$\therefore \underset{0}{\overset{4}{\int }}\frac{g\left(x\right)}{f\left(x\right)}dx=\underset{0}{\overset{4}{\int }}\frac{2}{(x-2{)}^2+2^2}dx$
$=\frac{2}{2}{\left[{\tan }^{-1}\left(\frac{x-2}{2}\right)\right]}_0^4$
$=\frac{\pi }{2}$