Q68 / 150 · MHT CET 2024 (16 May Shift 1) · Mathematics
If $y=\mathrm{a} \log x+\mathrm{b} x^2+x$ has its extremum values at $x=-1$ and $x=2$, then
- A$\mathrm{a}=2, \mathrm{~b}=-1$
- Correct answerB$\mathrm{a}=2, \mathrm{~b}=-\frac{1}{2}$
- C$\mathrm{a}=-2, \mathrm{~b}=\frac{1}{2}$
- D$\mathrm{a}=2, \mathrm{~b}=\frac{1}{2}$
Solution
$\begin{array}{ll}
& y=a \log x+b x^2+x \\
\therefore \quad & \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{\mathrm{a}}{x}+2 \mathrm{~b} x+1 \\
& \Rightarrow\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)_{x=-1}=-\mathrm{a}-2 \mathrm{~b}+1=0 \\
& \Rightarrow \mathrm{a}+2 \mathrm{~b}=1...(i)
\end{array}$
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 $a=2, b=-\frac{1}{2}$
Solving (i) and (ii), we get $a=2, b=-\frac{1}{2}$
Hence, the correct answer is option: 'B'.