Two statements are given, one labelled Assertion (A) and the other labelled Reason (R).
Let $a, b ∈ R$ be such that the function $f$ given by $f (x) = \log |x| + bx^2 + ax, x ≠ 0$ has extreme values at $x = -1$ and $x= 2$.
Assertion (A): f has local maximum at $x = -1$ and at $x = 2$
Reason (R): $a =\frac{1}{2}$ and $b =-\frac{1}{4}$.
Select the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
$f(x) = \log |x| + bx^2 + ax, x ≠0$
$⇒ f'(x) =\frac{1}{x}+ 2bx^2+a, x ≠0$.
Given $x = −1$ and $x = 2$ are extreme values of $f(x)$.
So, $f'(-1) = 0$ and $f'(2) = 0$
$⇒ -1-2b+ a = 0$ and $\frac{1}{2}+ 4b + a = 0$.
Solving these equations, we get $a =\frac{1}{2}, b =-\frac{1}{4}$
∴ Reason is true.
Now, $f'(x)=\frac{1}{x}-\frac{1}{2}x+\frac{1}{2}=\frac{2-x^2+x}{2x}$
$⇒f'(x)=\frac{(2-x)(x+1)}{2x}$.
So, $f'(x)=0⇒x=-1$ and $x=2$
$f''(x)=-\frac{1}{x^2}-\frac{1}{2}⇒f''(-1)=-\frac{1}{1}-\frac{1}{2}=-\frac{3}{2}<0$
$⇒x=-1$ is a point of local maxima.
Also, $f''(2) =-\frac{1}{4}-\frac{1}{2}=-\frac{3}{4}<0$
$⇒x = 2$ is a point of local maxima.
Hence, Assertion and Reason both are true and Reason is the correct explanation of Assertion.