If $y=\mathrm{a} \log x+\mathrm{b} x^2+x$ has its extremum values at $x=-1$ and $x=2$, then

Question

Aicharya · Accepted Answer

$\begin{array}{ll} & y=a \log x+b x^2+x \\ \therefore \quad & \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{\mathrm{a}}{x}+2 \mathrm{~b} x+1 \\ & \Rightarrow\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)_{x=-1}=-\mathrm{a}-2 \mathrm{~b}+1=0 \\ & \Rightarrow \mathrm{a}+2 \mathrm{~b}=1...(i) \end{array}$ and $\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{x=2}=\frac{\mathrm{a}}{2}+4 \mathrm{~b}+1=0$ $\Rightarrow a+8 b+2=0...(ii)$
Solving (i) and (ii), we get $a=2, b=-\frac{1}{2}$

If $y=\mathrm{a} \log x+\mathrm{b} x^2+x$ has its extremum values at $x=-1$ and $x=2$, then

Solution