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 ${\int }_a^b{e}^x\left\{f^{\prime }\right(x)+f^{\prime \prime }(x\left)\right\}dx$?
(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. $e^b+\sqrt{3}{e}^a$
• B. $e^b-\sqrt{3}{e}^a$
• C. $e^b+\sqrt{3e^a}$
• D. $-e^b+\sqrt{3e^a}$

Answer 1

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

Given, tangent to the graph of function $y=f\left(x\right)$ at the point $x=a$ forms with the X-axis an angle of $\frac{\pi }{3}$

$\therefore {\left|\frac{dy}{dx}\right|}_{x=a}=\tan \frac{\pi }{3}=\sqrt{3}$

The tangent to the graph of function $y=f\left(x\right)$ at the point $x=b$ forms with the X-axis an angle of $\frac{\pi }{4}$

$\therefore {\left|\frac{dy}{dx}\right|}_{x=b}=\tan \frac{\pi }{4}=1$

$\therefore f^{\prime }\left(a\right)=\sqrt{3};f^{\prime }\left(b\right)=1$

$I={\int }_a^b{e}^x\left\{f^{\prime }\right(x)+f^{\prime \prime }(x\left)\right\}dx$

$={\int }_a^b\left\{e^x{f}^{\prime }\right(x)+e^x{f}^{\prime \prime }(x\left)\right\}dx={\int }_a^b\frac{d}{dx}\left\{e^x{f}^{\prime }\right(x\left)\right\}dx={\left[e^x{f}^{\prime }\right(x\left)\right]}_a^b$

$=e^b{f}^{\prime }\left(b\right)-e^a{f}^{\prime }\left(a\right)=e^b.1-e^a.\sqrt{3}=e^b-\sqrt{3}{e}^a$