question_answer

Let a, \[b\in R\] be such that the function f given by \[f(x)=\ell n\left| x \right|+b{{x}^{2}}+ax,\,x\ne 0\] has extreme values at \[x=-1\] and \[x=2\]. Statement-1: f has local maximum at \[x=-1\] and at \[x=2\]. Statement-2: \[a=\frac{1}{2}\] and \[b=\frac{-1}{4}\].   AIEEE  Solved  Paper-2012

A) Statement-1 is false, Statement-2 is true.

B) Statement-1 is true, statement-2 is true; statement-2 is a correct explanation for Statement-1.

C) Statement-1 is true, statement-2 is true; statement-2 is not a correct explanation for Statement-1.

D) Statement-1 is true, statement-2 is false.

Correct Answer: B

Solution :

             \[f'(x)=\frac{1}{x}+2bx+a\]              at \[x=-1\]                              \[-1-1b+a=0\]              \[a-2b=1\]                                              ... (i) at \[x=2\]                                \[\frac{1}{2}+4b+a=0\]                                    \[a+4b=-\frac{1}{2}\]     ... (ii) On solving (i) and (ii) \[a=\frac{1}{2},\,\,b=-\frac{1}{4}\] \[f'(x)=\frac{1}{x}-\frac{x}{2}+\frac{1}{2}=\frac{2-{{x}^{2}}+x}{2x}=\frac{-(x+1)(x-2)}{2x}\]                                   So maxima at \[x=-1,2\]