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JUNE 2024 — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsJUNE 20243 Marks
Question
Prove that $y=\frac{4 \sin \theta}{(2+\cos \theta)}-\theta$ is an increasing function in $\left[0, \frac{\pi}{2}\right]$.
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Answer


$\begin{array}{l}y=\frac{4 \sin \theta}{(2+\cos \theta)}-\theta, \theta \in\left[0, \frac{\pi}{2}\right] \\ \therefore \frac{d y}{d \theta}=\frac{(2+\cos \theta)(4 \cos \theta)-4 \sin \theta(-\sin \theta)}{(2+\cos \theta)^2}-1\end{array}$
$\begin{array}{l}=\frac{8 \cos \theta+4 \cos ^2 \theta+4 \sin ^2 \theta}{(2+\cos \theta)^2}-1 \\ =\frac{8 \cos \theta+4\left(\cos ^2 \theta+\sin ^2 \theta\right)-(2+\cos \theta)^2}{(2+\cos \theta)^2} \\ =\frac{8 \cos \theta+4-4-4 \cos \theta-\cos ^2 \theta}{(2+\cos \theta)^2} \\ =\frac{4 \cos \theta-\cos ^2 \theta}{(2+\cos \theta)^2}\end{array}$
$\frac{d y}{d \theta}=\frac{\cos \theta(4-\cos \theta)}{(2+\cos \theta)^2}$
Here, $\theta \in\left[0, \frac{\pi}{2}\right] \Rightarrow \cos \theta \geq 0$
$\begin{array}{l}
\Rightarrow(4-\cos \theta)>0 \\
\Rightarrow(2+\cos \theta)^2>0 \\
\Rightarrow \frac{\cos \theta(4-\cos \theta)}{(2+\cos \theta)^2} \geq 0 \\
\Rightarrow \frac{d y}{d \theta} \geq 0
\end{array}$
Therefore, $f(\theta)$ is increasing function in the interval of $\left[0, \frac{\pi}{2}\right]$.

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