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Increasing and Decreasing Functions — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIncreasing and Decreasing Functions5 Marks
Question
Show that $\text{f}(\text{x})=\tan^{-1}(\sin\text{x}+\cos\text{x})$ is a decreasing function on the interval $\Big(\frac{\pi}{4},\frac{\pi}{2}\Big).$
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Answer

$\text{f}(\text{x})=\tan^{-1}(\sin\text{x}+\cos\text{x})$ $\therefore\ \text{f}'(\text{x})=\frac{1}{1+(\sin\text{x}+\cos\text{x})^2}(\cos\text{x}-\sin\text{x})$ $=\frac{1}{1+1+2\sin\text{x}\cos\text{x}}(\cos\text{x}-\sin\text{x})$ $=\frac{(\cos\text{x}-\sin\text{x})}{2+\sin2\text{x}}$ Here, $\frac{\pi}{4}<\text{x}<\frac{\pi}{2}$ $\Rightarrow\frac{\pi}{2}<2\text{x}<\pi$ $\Rightarrow\sin2\text{x}>0$ $\Rightarrow2+\sin2\text{x}>0\ ...(1)$Also,
$\frac{\pi}{4}<\text{x}<\frac{\pi}{2}$ $\cos\text{x}<\sin\text{x}$ $\Rightarrow\cos\text{x}-\sin\text{x}<0\ ....(2)$ $\text{f}'(\text{x})=\frac{(\cos\text{x}-\sin\text{x})}{2+\sin2\text{x}}<0,\ \forall\ \text{x}\in\Big(\frac{\pi}{4},\frac{\pi}{2}\Big).$ [From eqs. (1) and (2)] Hence, f(x) is decreasing on $\Big(\frac{\pi}{4},\frac{\pi}{2}\Big).$

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