Question
Let $$a,b \in R$$ be such that the function $$f$$ given by $$f\left( x \right) = \ln \left| x \right| + b{x^2} + ax,x \ne 0$$ has extreme values at $$x = - 1$$ and $$x = 2$$
Let $$a,b \in R$$ be such that the function $$f$$ given by $$f\left( x \right) = \ln \left| x \right| + b{x^2} + ax,x \ne 0$$ has extreme values at $$x = - 1$$ and $$x = 2$$
Statement-1 : $$f$$ has local maximum at $$x = - 1$$ and at $$x = 2.$$
Statement-2 : $$a = \frac{1}{2}$$ and $$b = \frac{{ - 1}}{4}$$
A.
Statement-1 is false, Statement-2 is true.
B.
Statement-1 is true, statement-2 is true; statement-2 is a correct explanation for Statement-1.
C.
Statement-1 is true, statement-2 is true; statement-2 is not a correct explanation for Statement-1.
D.
Statement-1 is true, statement-2 is false.
Answer :
Statement-1 is true, statement-2 is true; statement-2 is a correct explanation for Statement-1.
Solution :
$$\eqalign{ & {\text{Given, }}f\left( x \right) = \ln |x| + b{x^2} + ax \cr & \therefore f'\left( x \right) = \frac{1}{x} + 2bx + a \cr & {\text{At }}x = - 1,f'\left( x \right) = - 1 - 2{\text{b}} + a = 0 \cr & \Rightarrow a - 2b = 1\,......\left( {\text{i}} \right) \cr & {\text{At }}x = 2,\,f'\left( x \right) = \frac{1}{2} + 4b + a = 0 \cr & \Rightarrow a + 4b = - \frac{1}{2}\,......\left( {{\text{ii}}} \right) \cr & {\text{On solving }}\left( {\text{i}} \right){\text{ and}}\,\left( {{\text{ii}}} \right){\text{ we get }}a = \frac{1}{2},b = - \frac{1}{4} \cr & {\text{Thus, }}f'\left( x \right) = \frac{1}{x} - \frac{x}{2} + \frac{1}{2} = \frac{{2 - {x^2} + x}}{{2x}} \cr & = \frac{{ - {x^2} + x + 2}}{{2x}} = \frac{{ - \left( {{x^2} - x - 2} \right)}}{{2x}} = \frac{{ - \left( {x + 1} \right)\left( {x - 2} \right)}}{{2x}} \cr} $$
So maxima at $$x = - 1,2$$
Hence both the statements are true and statement 2 is a correct explanation for 1.
$$\eqalign{ & {\text{Given, }}f\left( x \right) = \ln |x| + b{x^2} + ax \cr & \therefore f'\left( x \right) = \frac{1}{x} + 2bx + a \cr & {\text{At }}x = - 1,f'\left( x \right) = - 1 - 2{\text{b}} + a = 0 \cr & \Rightarrow a - 2b = 1\,......\left( {\text{i}} \right) \cr & {\text{At }}x = 2,\,f'\left( x \right) = \frac{1}{2} + 4b + a = 0 \cr & \Rightarrow a + 4b = - \frac{1}{2}\,......\left( {{\text{ii}}} \right) \cr & {\text{On solving }}\left( {\text{i}} \right){\text{ and}}\,\left( {{\text{ii}}} \right){\text{ we get }}a = \frac{1}{2},b = - \frac{1}{4} \cr & {\text{Thus, }}f'\left( x \right) = \frac{1}{x} - \frac{x}{2} + \frac{1}{2} = \frac{{2 - {x^2} + x}}{{2x}} \cr & = \frac{{ - {x^2} + x + 2}}{{2x}} = \frac{{ - \left( {{x^2} - x - 2} \right)}}{{2x}} = \frac{{ - \left( {x + 1} \right)\left( {x - 2} \right)}}{{2x}} \cr} $$
So maxima at $$x = - 1,2$$
Hence both the statements are true and statement 2 is a correct explanation for 1.