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Model Paper 4 — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSModel Paper 42 Marks
Question
Show that $f(x)=\sin x-\cos x$ is an increasing function on $\left(\frac{-\pi}{4}, \frac{\pi}{4}\right)$.
✓

Answer

Given : $f(x) = \sin x - \cos x$
$f (x) = \cos x + \sin x$
$=\sqrt{2}\left(\frac{1}{\sqrt{2}} \cos x+\frac{1}{\sqrt{2}} \sin x\right)$
$=\sqrt{2}\left(\frac{\sin \pi}{4} \cos x+\frac{\cos \pi}{4} \sin x\right)$
$=\sqrt{2} \sin \left(\frac{\pi}{4}+x\right)$
Now,
$x \in\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$
$\Rightarrow-\frac{\pi}{4}< x < \frac{\pi}{4}$
$\Rightarrow 0 < \frac{\pi}{4}+x < \frac{\pi}{2}$
$\Rightarrow \sin 0^{\circ}<\sin \left(\frac{\pi}{4}+x\right)<\sin \frac{\pi}{2}$
$\Rightarrow 0<\sin \left(\frac{\pi}{4}+x\right)<1$
$\Rightarrow \sqrt{2} \sin \left(\frac{\pi}{4}+x\right);0$
$=f^{\prime}(x);0$
Hence $, f(x)$ is an increasing function on $\left(\frac{-\pi}{4}, \frac{\pi}{4}\right)$

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