Question

Let $y=f(x)$ be a polynomial function of degree 4 , which is symmetrical about the line $x=1$ and touches the $x$-axis and straight line $y+2=0$ and distance between both absolute minima is 4 .
The value of $\int \frac{dx}{f(x+1)}$ is equal to
[Note: Where $k$ is the constant of integration.]

• A. $\frac{1}{2\sqrt{2}}\ln \left|\frac{x+2\sqrt{2}}{x-2\sqrt{2}}\right|-\frac{1}{x}+k$
• B. $\frac{1}{2\sqrt{2}}\ln \left|\frac{x-2\sqrt{2}}{x+2\sqrt{2}}\right|+\frac{1}{x}+k$
• C. $\frac{1}{2\sqrt{2}}{\tan }^{-1}\left(\frac{x}{2\sqrt{2}}\right)+\frac{1}{x}+k$
• D. $\frac{1}{2\sqrt{2}}{\tan }^{-1}\left(\frac{x}{2\sqrt{2}}\right)-\frac{1}{x}+k$

Answer 1

Answer: (B)

$f(x)=(x-1{)}^2\left(a{x}^2+bx+c\right)$
$f^{\prime }(x)=(x-1{)}^2(2ax+b)+\left(a{x}^2+bx+c\right)2(x-1)$
$f^{\prime }(-1)=f^{\prime }(3)=0\&f(-1)=f(3)=-2$
$f^{\prime }(3)=4(6a+b)+4(9a+3b+c)=0$
$\Rightarrow 15a+4b+c=0$....(1)
$f(3)=4(9a+3b+c)=-2$....(2)
$f(-1)=4(a-b+c)=-2$....(3)

From (1), (2) and (3)
$a=\frac{1}{8},b=-\frac{1}{4},c=-\frac{7}{8}$
$\therefore f(x)=(x-1{)}^2\left(\frac{1}{8}{x}^2-\frac{1}{4}x-\frac{7}{8}\right)$
$=\frac{1}{8}(x-1{)}^2\left((x-1{)}^2-8\right)$
$=\frac{1}{8}\left[(x-1{)}^4-8(x-1{)}^2\right]$
$\int \frac{dx}{f(x+1)}=\int \frac{8dx}{x^4-8x^2}=\int \frac{8dx}{x^2\left(x^2-8\right)}$
$=\int \left(\frac{1}{x^2-8}-\frac{1}{x^2}\right)dx=\frac{1}{2\sqrt{2}}\ln \left|\frac{x-2\sqrt{2}}{x+2\sqrt{2}}\right|+\frac{1}{x}+k$