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

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Aicharya · Accepted Answer

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

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

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