Question

The equation of tangent to the curve $y=2 \cos x$ at $x=\frac{\pi}{4}$ is

• A. $y-\sqrt{2}=2 \sqrt{2}\left(x-\frac{\pi}{4}\right)$
• B. $y+\sqrt{2}=\sqrt{2}\left(x+\frac{\pi}{4}\right)$
• C. $y-\sqrt{2}=-\sqrt{2}\left(x-\frac{\pi}{4}\right)$
• D. $y-\sqrt{2}=\sqrt{2}\left(x-\frac{\pi}{4}\right)$

Answer 1

Answer: (C)

$y=2 \cos x A t x=\frac{\pi}{4} i j, y=\frac{2}{\sqrt{2}}=\sqrt{2}$
and $\frac{ dy }{ dx }=-2 \cdot \sin x \therefore\left(\frac{ dy }{ dx }\right)_{ x -\pi / 4}=-\sqrt{2}$
$\therefore$ Equation of tangent at $\left(\frac{\pi}{2}, \sqrt{2}\right)$ is
$y-\sqrt{2}=-\sqrt{2}\left(x-\frac{\pi}{4}\right)$