Question

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