Question

The tangent to the graph of the function y = f(x) at the point with abscissa x = 1 from an angle of p/6 and at the point x = 2 an angle of p/3, at the point x = 3 an angle of p/4. The value of $\int_{1}^{3}{f'(x)f''(x)dx + \int_{2}^{3}{f"(x)dx}}$ (f"(x) is supposed to be

continuous) is :

• A. $\frac{4\sqrt{3} - 1}{3\sqrt{3}}$
• B. $\frac{3\sqrt{3} - 1}{2}$
• C. $\frac{4\sqrt{3}}{3}$
• D. None

Answer 1

Answer: (D)

None

Answer 2

f(1) = tan p/6 = $\frac{1}{\sqrt{3}}$

f '(2) = tan p/3 = $\sqrt{3}$

f ' (3) = tan p/4 = 1

None $\int_{1}^{3}{f'(x)f''(x)dx + \int_{2}^{3}{f"(x)dx}}$

= $\left( \frac{(f'(x)^{2}}{2} \right)_{1}^{3}$+ $\left( f'(x) \right)_{2}^{3}$

= $\frac{1}{2}$ [f '(3)² –(f '(1)²] + [f '(3) –f '(2)]

= $\frac{1}{2}$ [1–$\frac{1}{3}$] + [1–$\sqrt{3}$] =$\frac{1}{3}$+ 1 – $\sqrt{3}$

= $\frac{1}{3}$ [4 – 3$\sqrt{3}$]