Question

Let $P\left(x\right)$ be a polynomial of least degree whose graph has three points of inflection $(-1,-1),(1,1)$ and a point with abscissa $0$ at which the curve is inclined to the axis of abscissa at an angle of ${60}^0$. Then ${\int }_0^{1}P\left(x\right)dx$ equals to:

• A. $\frac{3\sqrt{3}+4}{14}$
• B. $\frac{3\sqrt{3}}{7}$
• C. $\frac{\sqrt{3}+\sqrt{7}}{14}$
• D. $\frac{\sqrt{3}+2}{7}$

Answer 1

The correct option is A $\frac{3\sqrt{3}+4}{14}$

For any polynomial $P\left(x\right)$, the second derivative must be equal to zero at the point of inflexion.
As per the question, $P^{\prime \prime }\left(x\right)$ has three roots: $-1,1$ and $0$
Hence, $P^{\prime \prime }\left(x\right)$ is a third degree polynomial, which can be written as
$P^{\prime \prime }\left(x\right)=ax(x-1)(x+1)=a(x^3-x)$, where 'a' is a constant.
at the point $x=0$, it is given that $P^{\prime }\left(0\right)=tan{60}^0=\sqrt{3}$
$P^{\prime }\left(x\right)-P^{\prime }\left(0\right)={\int }_0^x{P}^{\prime \prime }\left(x\right)dx$
$\Rightarrow P^{\prime }\left(x\right)=a(\frac{x^4}{4}-\frac{x^2}{2})+\sqrt{3}$
Since $P\left(1\right)=1$, we get
$P\left(x\right)-P\left(1\right)={\int }_1^x{P}^{\prime }\left(x\right)dx$
$\Rightarrow P\left(x\right)=a(\frac{x^5}{20}-\frac{x^3}{6}+\frac{7}{60})+\sqrt{3}(x-1)+1$
Using $P(-1)=-1$, we get $a=\frac{60(\sqrt{3}-1)}{7}$
Using the value of 'a', we get $P\left(x\right)=\frac{60(\sqrt{3}-1)}{7}\left(\frac{3x^5-10x^3+7}{60}\right)+\sqrt{3}x+1-\sqrt{3}$

$=\frac{\sqrt{3}-1}{7}(3x^5-10x^3)+\sqrt{3}-1+\sqrt{3}x+1-\sqrt{3}=\frac{\sqrt{3}-1}{7}(3x^5-10x^3)+\sqrt{3}x$
Now ${\int }_0^{1}P\left(x\right)dx={\int }_0^1\frac{\sqrt{3}-1}{7}(3x^5-10x^3)+x\sqrt{3}]dx$
$={|\frac{\sqrt{3}-1}{7}(\frac{x^6}{2}-\frac{5}{2}{x}^4)+\frac{x^2}{2}\sqrt{3}|}_0^1$
$=\frac{\sqrt{3}-1}{7}(\frac{1}{2}-\frac{5}{2})+\frac{\sqrt{3}}{2}=\frac{\sqrt{3}-1}{7}(-2)+\frac{\sqrt{3}}{2}=\frac{3\sqrt{3}+4}{14}$