Let [a,b R] be such that the function f given by [f( x ) = In| x | + bx^2 + ax,,x 0] has e…

Question

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

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

Accepted Answer

Given, $$f\left( x \right) = \ln \left| x \right| + b{x^2} + ax$$

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

At $x=-1,$  $$f'\left( x \right) =  - 1 - 2b + a = 0$$

$$ \Rightarrow a - 2b = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

At $x=2,$ $$\,\,f'\left( x \right) = {1 \over 2} + 4b + a = 0$$

$$ \Rightarrow a + 4b =  - {1 \over 2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

On solving $(i)$ and $(ii)$ we get $$a = {1 \over 2},b =  - {1 \over 4}$$

Thus, $$f'\left( x \right) = {1 \over x} - {x \over 2} + {1 \over 2} = {{2 - {x^2} + x} \over {2x}}$$

$$ = {{ - {x^2} + x + 2} \over {2x}} = {{ - \left( {{x^2} - x - 2} \right)} \over {2x}} = {{ - \left( {x + 1} \right)\left( {x - 2} \right)} \over {2x}}$$

So maximum at $x=-1,2$

Hence both the statements are true and statement $2$ is a correct explanation for $1.$

Solutions · 1