Question

Let P(x) 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 60ŗ. Then $\int_{0}^{1}{}$P(x) 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

Answer: (A)

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

Answer 2

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

p''(x) = ax (x – 1) (x + 1) = a(x³ – x)

at the point x = 0, p'(0) = tan 60ŗ = $\sqrt{3}$

p'(x) = $\int_{0}^{x}{}$p''(x) dx + $\sqrt{3}$ = a$\left( \frac{x^{4}}{4} - \frac{x^{2}}{2} \right)$+$\sqrt{3}$

Since p(1) = 1, we get

p(x) = $\int_{1}^{x}{}$p'(x) dx + 1

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

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

$\int_{1}^{x}{}$p(x) dx = $\frac{\sqrt{3} - 1}{7}$ (3x⁵ – 10x³) + x$\sqrt{3}$]dx

= $\frac { \sqrt { 3 } - 1 } { 7 }$ $\left( \frac{x^{6}}{2} - \frac{5}{2}x^{4} \right)$ + $\left. \ \sqrt{3} \right|_{0}^{1}$

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