Question

Let P(x) be a polynomial of least degree whose graph has three points of inflection as (–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 ${60}^{\circ }$. 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

$\frac{3\sqrt{3}+4}{14}$

Required function is a polynomial, the abscissa of the points of inflection can only be among the roots of the second derivative

$p^{\prime \prime }\left(x\right)=ax(x-1)(x+1)=a(x^3-x)$

at the point x=0, $p^{\prime }\left(0\right)=tan{60}^{\circ }=\sqrt{3}$

$p^{\prime }\left(x\right)={\int }_0^x{p}^{\prime \prime }\left(x\right)dx+\sqrt{3}=a(\frac{x^4}{4}-\frac{x^2}{2})+\sqrt{3}$

Since $p\left(1\right)=1$, we get
$p\left(x\right)={\int }_1^x{p}^{\prime }\left(x\right)dx+1$

$=a(\frac{x^5}{20}-\frac{x^3}{6}+\frac{7}{60})+\sqrt{3}(x-1)+1$

$p(-1)=-1,\text{so}a=\frac{60(\sqrt{3}-1)}{7}$

${\int }_0^{1}p\left(x\right)dx={\int }_0^1\frac{\sqrt{3}-1}{7}(3x^5-10x^3)+x\sqrt{3}]dx$

$=\frac{\sqrt{3}-1}{7}(\frac{x^6}{2}-\frac{5}{2}{x}^4)+\frac{x^2}{2}{\begin{array}{c}\sqrt{3}\end{array}|}_0^1$

$=\frac{\sqrt{3}-1}{7}(\frac{1}{2}-\frac{5}{2})+\frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{14}+\frac{2}{7}$