Question

The tangent to the graph of a continuous function $y=f\left(x\right)$ at the point with abscissa $x=a$ forms with the $x$ axis an angle of $\frac{\pi }{3}$ and at the point with abscissa $x=b$ an angle of $\frac{\pi }{4}$, then what is the value of the integral $\underset{a}{\overset{b}{\int }}{e}^x\left\{f^{\prime }\right(x)+f^{\prime \prime }(x\left)\right\}dx$ ?

(where $f^{\prime }\left(x\right)$ the derivative of $f$ w.r.t. $x$ which is assumed to be continuous and similarly $f^{\prime \prime }\left(x\right)$ the double derivative of $f$ w.r.t. $x$)

• A. $-e^b+\sqrt{3e^a}$
• B. $e^b+\sqrt{3}{e}^a$
• C. $e^b-\sqrt{3}{e}^a$
• D. $e^b+\sqrt{3e^a}$

Answer 1

The correct option is C $e^b-\sqrt{3}{e}^a$

We know that
$f^{\prime }\left(a\right)=\tan \frac{\pi }{3}=\sqrt{3}$ and
$f^{\prime }\left(b\right)=\tan \frac{\pi }{4}=1$
$\underset{a}{\overset{b}{\int }}{e}^x\left\{f^{\prime }\right(x)+f^{\prime \prime }(x\left)\right\}dx=[e^x{f}^{\prime }\left(x\right){]}_a^b=e^b-e^a\sqrt{3}=e^b-\sqrt{3}{e}^a$