If $f(x)=a \log |x|+b x^2+x$ has extreme values at $x=-1$ and $x=2$, then the ordered pair $(a, b)=$

• A

  (2 ,-1 )
• Correct answer
  B

  $\left(2,-\frac{1}{2}\right)$
• C

  $(-1,2)$
• D

  $\left(-\frac{1}{2}, 2\right)$

Solution

Given, $f(x)=a \cdot \log |x|+b x^2+x$ $$ \begin{aligned} & f^{\prime}(x)=a \cdot \frac{1}{|x|} \cdot \frac{x}{|x|}+b(2 x)+1 \\ & f^{\prime}(x)=\frac{a x}{|x|^2}+2 b x+1 \end{aligned} $$ Given that $x=-1$ is one of extremity of $f(x)$. $$ \begin{aligned} \Rightarrow f^{\prime}(-1)= & 0 \\ & \frac{a(-1)}{1}+2 b(-1)+1=0 \end{aligned} $$
Also, given $x=2$ is the other extremity of $f(x)$. $$ \begin{aligned} f^{\prime}(2) & =0 \\ \frac{a(2)}{4}+2 b(2)+1 & =0 \\ \frac{a}{2}+4 b+1 & =0 \end{aligned} $$
Adding on Eqs. (i) and (ii), we get $$ \begin{array}{rlrl} \Rightarrow & & 6 b+3 & =0 \\ \Rightarrow & b & =-\frac{1}{2} \end{array} $$ Substituting $b=-\frac{1}{2}$ in Eq. (i), we get $$ \begin{aligned} a & =2 \\ \therefore(a, b) & =\left(2,-\frac{1}{2}\right) \end{aligned} $$ $\therefore$ Hence option (b) is correct.

Hence, the correct answer is option: 'B'.