Question

# The angle of intersection of the curves y = 2 sin2 x and y = cos 2 x at $x=\frac{\mathrm{\pi }}{6}$ is (a) π/4 (b) π/2 (c) π/3 (d) none of these

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Solution

## (c) π/3 $\text{Given:}\phantom{\rule{0ex}{0ex}}x=\frac{\mathrm{\pi }}{6}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}y=2{\mathrm{sin}}^{2}x\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}=4\mathrm{sin}x\mathrm{cos}x\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}=2\mathrm{sin}2x\phantom{\rule{0ex}{0ex}}⇒{m}_{1}={\left(\frac{dy}{dx}\right)}_{x=\frac{\mathrm{\pi }}{6}}=2×\frac{\sqrt{3}}{2}=\sqrt{3}\phantom{\rule{0ex}{0ex}}\mathrm{Also},\phantom{\rule{0ex}{0ex}}y=\mathrm{cos}2x\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}=-2\mathrm{sin}2x\phantom{\rule{0ex}{0ex}}⇒{m}_{2}={\left(\frac{dy}{dx}\right)}_{x=\frac{\mathrm{\pi }}{6}}=-2×\frac{\sqrt{3}}{2}=-\sqrt{3}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{tan}\theta =\left|\frac{{m}_{1}-{m}_{2}}{1+{m}_{1}{m}_{2}}\right|=\left|\frac{\sqrt{3}+\sqrt{3}}{1-\sqrt{3}\sqrt{3}}\right|=\left|\frac{2\sqrt{3}}{-2}\right|=\sqrt{3}\phantom{\rule{0ex}{0ex}}⇒\theta ={\mathrm{tan}}^{-1}\left(\sqrt{3}\right)=\frac{\mathrm{\pi }}{3}$

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