Q62 / 150 · MHT CET 2024 (11 May Shift 2) · Mathematics

If $y=\mathrm{a} \log x+\mathrm{b} x^2+x$ has its extreme values at $x=-1$ and $x=2$, then the value of $\left(\frac{\mathrm{a}}{\mathrm{b}}+\frac{\mathrm{b}}{\mathrm{a}}\right)$ is

  • A
    $-\frac{7}{4}$
  • B
    $-\frac{15}{4}$
  • Correct answer
    C
    $-\frac{17}{4}$
  • D
    $-\frac{5}{4}$

Solution

$\begin{aligned} & \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{a}}{x}+2 \mathrm{~b} x+1 \Rightarrow\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)_{x=-1}=-\mathrm{a}-2 \mathrm{~b}+1=0 \\ & \Rightarrow \mathrm{a}=-2 \mathrm{~b}+1 \\ & \text { and }\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{x=2}=\frac{\mathrm{a}}{2}+4 \mathrm{~b}+1=0 \\ & \Rightarrow \frac{-2 \mathrm{~b}+1}{2}+4 \mathrm{~b}+1=0 \\ & \Rightarrow-\mathrm{b}+4 \mathrm{~b}+\frac{3}{2}=0 \\ & \Rightarrow 3 \mathrm{~b}=\frac{-3}{2} \Rightarrow \mathrm{~b}=\frac{-1}{2} \text { and } \mathrm{a}=2 \\ & \Rightarrow\left(\frac{\mathrm{a}}{\mathrm{b}}+\frac{\mathrm{b}}{\mathrm{a}}\right)=\frac{-17}{4}\end{aligned}$

Hence, the correct answer is option: 'C'.

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