Question

Assertion :Let a, b $\in$ R be such that the function f given by $f\left(x\right)=ln|x|+b{x}^2+ax$, $x\ne 0$ has extreme values at $x=-1$ and $x=2$. f has local maximum at $x=-1$ and at $x=2$. Reason: $a=\frac{1}{2}$ and $b=\frac{-1}{4}$.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is B Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

$f\left(x\right)=\ln \left|x\right|+b{x}^2+ax,x\ne 0$ has extreme values at $x=-1,x=2$

$\Rightarrow f^{\prime }\left(x\right)=\frac{1}{x}+2bx+a$

$f^{\prime }(-1)=0$ and $f^{\prime }\left(2\right)=0$ (given)

$\Rightarrow -1-2b+a=0$ and $\frac{1}{2}+2b\times 2+a=0$

$\Rightarrow a=2b+1$ and $a=-4b-\frac{1}{2}$

$\Rightarrow 2b+1=-4b-\frac{1}{2}$

$\Rightarrow 2b+4b=-\frac{1}{2}-1$

$\Rightarrow 6b=\frac{-1-2}{2}$

$\Rightarrow b=\frac{-3}{2\times 6}=\frac{-1}{4}$

Put $b=\frac{-1}{4}$ in $a=2b+1=2\times \frac{-1}{4}+1=\frac{-1}{2}+1=\frac{1}{2}$

$\therefore b=\frac{-1}{4}a=\frac{1}{2}$

$f^{\prime \prime }\left(x\right)=\frac{-1}{x^2}+2b=\frac{-1}{x^2}-\frac{1}{2}=-(\frac{1}{x^2}+\frac{1}{2})<0$ for all $x\in R-\left\{0\right\}$

$\Rightarrow f$ has a local maximum at $x=-1,x=2$

$\therefore f$ has local maxima at $x=-1,x=2$

$\therefore a=\frac{1}{2},b=\frac{-1}{4}$