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( ax ^2+ bx + c \right) $
$f ^{\prime}( x )=( x -1)^2(2 ax + b )+\left( ax ^2+ bx + c \right) 2( x -1)$
$f ^{\prime}(-1)= f ^{\prime}(3)=0 \& f (-1)= f (3)=-2$
$f ^{\prime}(3)=4(6 a + b )+4(9 a +3 b + c )=0 $
$\Rightarrow 15 a +4 b + c =0$....(1)
$f (3)=4(9 a +3 b + c )=-2$....(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{8 dx }{ x ^4-8 x ^2}=\int \frac{8 dx }{ x ^2\left( x ^2-8\right)} $
$=\int\left(\frac{1}{ x ^2-8}-\frac{1}{ x ^2}\right) dx =\frac{1}{2 \sqrt{2}} \ln \left|\frac{ x -2 \sqrt{2}}{ x +2 \sqrt{2}}\right|+\frac{1}{ x }+ k$