Q. If $y = a\log{x + bx^{2} + x}$ has its extremum at $x = - 1$ and $x = 2$, then
A) $a = 2,\ b = \frac{1}{2}$
B) $a = 2,\ b = - \frac{1}{2}$
C) $a = \frac{1}{2},\ b = 2$
D) $a = - \frac{1}{2},\ b = 2$
Explanation
(b)
Given, $y = a\log x + bx^{2} + x$
$\Rightarrow \ \ \ \ \frac{dy}{dx} = \frac{a}{x} + 2bx + 1$
At $x = - 1$ and $x = 2,\ y$ has its extrema.
$\therefore\ \ \ - a - 2b + 1 = 0\ \ \ \Rightarrow \ \ \ a + 2b = 1$ ...(i)
And $\frac{a}{2} + 4b + 1 = 0\ \ \ \ \Rightarrow \ \ a + 8b = - 2$ ...(ii)
On solving Eqs. (i) and (ii), we get
$a = 2,\ b = - \frac{1}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.