Question

Let $P(x)$ be a polynomial of degree 4 having a relative maximum at $x=2$ and $\underset{x\to 0}{\mathrm{Lim}}\left(3-\frac{P(x)}{x}\right)=27$. Also $P(1)=-9$ and $P^{\prime \prime }(x)$ has a local minimum at $x=2$.
The value of definite integral ${\int }_0^1(P(-x)-P(x))dx$ equals

• A. 12
• B. 16
• C. 28
• D. 24

Answer 1

Answer: (C)

We have
$\underset{x\to 0}{\mathrm{Lim}}\left(3-\frac{P(x)}{x}\right)=27\Rightarrow \underset{x\to 0}{\mathrm{Lim}}\frac{P(x)}{x}=-24$
$\therefore$ Let $P(x)=a{x}^4+b{x}^3+c{x}^2-24x$....(1)
Now, $P(1)=-9\Rightarrow a+b+c=15$....(2)
Also, $P^{\prime }(2)=0\Rightarrow 8a+3b+c=6$....(3)
and $P^{\prime \prime \prime }(2)=0\Rightarrow b=-8a$....(4)
$\therefore$ On solving (2), (3), (4), we get
$a=1,b=-8,c=22$
Hence, $P(x)=x^4-8x^3+22x^2-24x$....(5)

$\therefore P^{\prime }(x)=4x^3-24x^2+44x-24=4\left(x^3-6x^2+11x-6\right)=4(x-1)(x-2)(x-3)$
$\Rightarrow P^{\prime \prime }(x)=4\left(3x^2-12x+11\right)$
Clearly, $P^{\prime \prime }(x)>0\forall x\in [3,4]$ (As $P^{\prime }(x)$ is increasing function on $[3,4]$ )
Consider,
$\underset{0}{\overset{1}{\int }}(P(-x)-P(x))dx=\underset{0}{\overset{1}{\int }}\left(\left(x^4+8x^3+22x^2+24x\right)-\left(x^4-8x^3+22x^2-24x\right)\right)dx$
$=\underset{0}{\overset{1}{\int }}\left(16x^3+48x\right)dx=16\underset{0}{\overset{1}{\int }}\left(x^3+3x\right)dx=16{\left(\frac{x^4}{4}+\frac{3x^2}{2}\right)}_0^1=16\left(\frac{1}{4}+\frac{3}{2}\right)=4+24=28.$