Question

If $y=a\log x+b{x}^2+x$ has its extremum values at $x=-1$ and $x=2,$ then

• A. $a=2,b=-1$
• B. $a=2,b=\frac{1}{2}$
• C. $a=-2,b=\frac{1}{2}$
• D. $a=2,b=-\frac{1}{2}$

Answer 1

Answer: (D)

We have, $y=a\log x+b{x}^2+x$
$\Rightarrow \frac{dy}{dx}=\frac{a}{x}+2bx+1$
For extremum, $\frac{dy}{dx}=0$ at $x=-1$ and $x=2$
$\therefore -a-2b+1=0$
$\Rightarrow a+2b=1$...(i)
and $\frac{a}{2}+4b+1=0$
$\Rightarrow a+8b=-2$...(ii)
From Eqs. (i) and (ii), we have
$a=2\text{ and }b=-\frac{1}{2}$