Q85 / 150 · MHT CET 2021 (22 Sep Shift 2) · Mathematics

The function $f(x)=\frac{\lambda \sin x+6 \cos x}{2 \sin x+3 \cos x}$ is increasing, if

  • Correct answer
    A
    $\lambda>2$
  • B
    $\lambda < 4$
  • C
    $\lambda \geq 4$
  • D
    $\lambda>1$

Solution

$$ \begin{aligned} & f(x)=\frac{\lambda \sin x+6 \cos x}{2 \sin x+3 \cos x} \\ & f^{\prime}(x)=\frac{\{[(2 \sin x+3 \cos x)(\lambda \cos x-6 \sin x]-[\lambda \sin x+6 \cos x)(2 \cos x-3 \sin x)]\}}{(2 \sin x+3 \cos x)^2} \end{aligned} $$ When $\mathrm{f}^{\prime}(\mathrm{x}) \geq 0$, we get $$ \begin{aligned} & {\left[\left(2 \lambda \sin \mathrm{x} \cos \mathrm{x}+3 \lambda \cos ^2 \mathrm{x}-12 \sin ^2 \mathrm{x}-18 \sin \mathrm{x} \cos \mathrm{x}\right)\right.} \\ & \left.-\left(2 \lambda \sin \mathrm{x} \cos \mathrm{x}+12 \cos ^2 \mathrm{x}-3 \lambda \sin ^2 \mathrm{x}-18 \operatorname{sun} \mathrm{x} \cos \mathrm{x}\right)\right] \geq 0 \\ & \therefore 3 \lambda\left(\sin ^2 \mathrm{x}+\cos ^2 \mathrm{x}\right)-12\left(\sin ^2 \mathrm{x}+\cos ^2 \mathrm{x}\right) \geq 0 \\ & \therefore 3 \lambda-12 \geq 0 \Rightarrow \lambda \geq 4 \end{aligned} $$

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

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