Question

The equation of the curve is $y=f\left(x\right)$.

The tangents at $(1,f\left(1\right)),(2,f(2\left)\right)$, $(2,f(2\left)\right)$ and $(3,f(3\left)\right)$ make angles $\frac{\pi }{6}$, $\frac{\pi }{3}$ and $\frac{\pi }{4}$

respectively with the positive direction of the $x$-axis.

Then the value of ${\int }_2^3{f}^{{}^{\prime }}\left(x\right)f^{\prime \prime }\left(x\right)dx+{\int }_1^3{f}^{\prime \prime }\left(x\right)dx$ is equal to?

• A. $-\frac{1}{\sqrt{3}}$
• B. $\frac{1}{\sqrt{3}}$
• C. $0$
• D. None of these

Answer 1

The correct option is A $-\frac{1}{\sqrt{3}}$

Given $f^{{}^{\prime }}\left(1\right)=\frac{1}{\sqrt{3}},f^{{}^{\prime }}\left(2\right)=\sqrt{3},f^{{}^{\prime }}\left(3\right)=1$
Now ${\int }_2^3{f}^{{}^{\prime }}\left(x\right)f^{\prime \prime }\left(x\right)dx+{\int }_1^3{f}^{\prime \prime }\left(x\right)dx$
$=\frac{f^{\prime }(3{)}^2-f^{\prime }(2{)}_2}{2}+f^{\prime }\left(3\right)-f^{\prime }\left(1\right)$
$=-1+\frac{\sqrt{3}-1}{\sqrt{3}}=-\frac{1}{\sqrt{3}}$