Question

The tangent to the graph of a continuous function $y=f(x)$ 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 $\int_\limits{a}^{b} e^{x}\left\{f'(x)+f''(x)\right\} d x ?$
(where $f'(x)$ the derivative of $f$ w.r.t. $x$ which is assumed to be continuous and similarly $f''(x)$ 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. $e^{b} \sqrt{3 e^{a}}$

Answer 1

Answer: (C)

Given, the tangent to the graph of function $y=f(x)$ at the point $x=a$ forms with the $X$ -axis an angle of $\frac{\pi}{3}$.
$\left.\therefore \, \frac{d y}{d x}\right|_{x-a}=\tan \frac{\pi}{3}=\sqrt{3}$
and the tangent to the graph of function $y=f(x)$ at the point $x=b$ forms with the $X$ -axis an angle of $\frac{\pi}{4}$.
$\therefore \,\frac{d y}{d x}|_{x=b}=\tan \frac{\pi}{4}=1$
$\therefore \, f'(a)=\sqrt{3}$ and $f'(b)=1$
Now, $ I =\int_{a}^{b} e^{x} {f'(x)+f''(x)} d x $
$=\int_{a}^{b}\left[e^{x} f'(x)+e^{x} f''(x)\right] d x $
$=\int_{a}^{b} \frac{d}{d x}\left\{e^{x} f'(x)\right\} d x$
$=\left[e^{x} f'(x)\right]_{a}^{b} $
$=e^{b} f'(b)-e^{a} f'(a)$
$=e^{b} \cdot 1-e^{a} \sqrt{3} $
$[\because f'(b)=1$ and $f'(a)=\sqrt{3}] $
$=e^{b}-\sqrt{3} \,e^{a}$