Question

Let $a,b\in R$ be such that the function $f$ given by $f\left(x\right)=\ln \left(\left|x\right|\right)+b{x}^2+ax$, $x\ne 0$ 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 $b=-\frac{1}{4}$

• A. Statement $1$ is true, Statement $2$ is true; statement $2$ is a correct explanation for statement $1$.
• B. Statement $1$ is true, statement $2$ is true; statement $2$ is not a correct explanation for statement $1$
• C. Statement $1$ is true, statement $2$ is false
• D. Statement $1$ is false, statement $2$ is true

Answer 1

The correct option is A

Statement $1$ is true, Statement $2$ is true; statement $2$ is a correct explanation for statement $1$.

Explanation for Correct Answer:

Statement $1$: $f$ has local maximum at $x=-1$ and at $x=2$$f\left(x\right)=\ln \left(\left|x\right|\right)+b{x}^2+ax$

Finding the local maximum and the value of $a$ and $b$:

For finding local maximum equate $f\text{'}\left(x\right)=0$ and find the values of $x$. and substitute the value $f\text{'}\text{'}\left(x\right)$ where $f\text{'}\text{'}\left(x\right)>0$ then that particular $x$ gives the maximum value for $f\left(x\right)$.

$f\text{'}\left(x\right)=\frac{1}{x}+2bx+a$

equating $f\text{'}\left(x\right)=0$

$0=\frac{1}{x}+2bx+a$

using Statement $1$

where $x=-1$$\Rightarrow 0=\frac{1}{-1}+2b\left(-1\right)+a\Rightarrow a-2b=0\to \left(i\right)$

where $x=2$ $\Rightarrow 0=\frac{1}{2}+2b\left(2\right)+a\Rightarrow a+4b=-\frac{1}{2}\to \left(ii\right)$

solving equation $\left(i\right)\text{and}\left(ii\right)$

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

sub $b=-\frac{1}{4}$ in equation $\left(i\right)$

$a-2\left(\frac{-1}{4}\right)=0\Rightarrow a=\frac{1}{2}$

Therefore Statement $2$ is correct. and correct explanation for a statement $1$

Statement $2$: $a=\frac{1}{2}$ and $b=-\frac{1}{4}$

$$\begin{gathered} f\left(x\right)=\ln \left(\left|x\right|\right)+\left(\frac{-1}{4}\right)x^2+\left(\frac{1}{2}\right)x \\ f\text{'}\left(x\right)=\frac{1}{x}+2\left(\frac{-1}{4}\right)x^2+\left(\frac{1}{2}\right) \end{gathered}$$

$f\text{'}\left(x\right)=0$$\Rightarrow 0=\frac{1}{x}+2\left(\frac{-1}{4}\right)x^2+\left(\frac{1}{2}\right)$

$$\begin{gathered} 0=\frac{1}{x}-\left(\frac{1}{2}\right)x+\left(\frac{1}{2}\right) \\ 0=\frac{2-2x^2+x}{2x} \\ 0=2-x^2-x^2+x \\ 0=\left(x+1\right)(x-2) \end{gathered}$$

$x=-1,+2$

So maxima at $x=-1,+2$.

Therefore Statement $1$ is correct.

Hence, option (A) is correct.