Question

If $y=aln|x|+b{x}^2+x$ has its extreme values at $x=-1$ and $x=2$ then P (a , b) is

• A. $(2,-1)$
• B. $(2,-1/2)$
• C. $(-2,1/2)$
• D. none of these

Answer 1

Answer: (B)

$(2,-1/2)$

$y=aln|x|+b{x}^2+x$

$\frac{dy}{dx}=a\frac{1}{|x|}\frac{|x|}{x}+2bx+1$

$=a\frac{1}{x}+2bx+1$

The function has extreme values at $x=-1$ and $x=2$

$⟹-a-2b+1=0$ and

$⟹\frac{a}{2}+4b+1=0$

Solving these equations we get

$a=2,b=-\frac{1}{2}$

So, the answer is option (B).