Question

The tangent to the curve $y=f\left(x\right)$ at the point with abscissa $x=1$ form an angle of $\frac{\pi }{6}$ and at the point $x=2$ an angle of $\frac{\pi }{3}$ and at the point $x=3$ an angle of $\frac{\pi }{4}$. If $f^n\left(x\right)$ is continuous, then the value of ${\int }_1^3{f}^{\prime \prime }\left(x\right)f^{\prime }\left(x\right)dx+{\int }_2^3{f}^{\prime \prime }\left(x\right)dx$ is

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

Answer 1

Answer: (C)

$\frac{4-3\sqrt{3}}{3}$

According to question $f^{\prime }\left(1\right)=\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}},f^{\prime }\left(2\right)=\tan \frac{\pi }{3}=\sqrt{3}$ and $f^{\prime }\left(3\right)=\tan \frac{\pi }{4}=1$

So, ${\int }_1^3{f}^{\prime \prime }\left(x\right)f^{\prime }\left(x\right)dx+{\int }_2^3{f}^{\prime \prime }\left(x\right)dx={\left[\frac{{\left(f^{\prime }\left(x\right)\right)}^2}{2}\right]}_1^3+{\left[f^{\prime }\left(x\right)\right]}_2^3$

$=\frac{1}{2}[1-\frac{1}{3}]+[1-\sqrt{3}]=\frac{4}{3}-\sqrt{3}$