Question

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

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

Answer 1

The correct option is C $y-\sqrt{2}=-\sqrt{2}(x-\frac{\pi }{4})$

y=2 cos x
At $x=\frac{\pi }{4},y=\frac{2}{\sqrt{2}}=\sqrt{2}$ and $\frac{dy}{dx}=-2.sinx$
$\therefore {\left(\frac{dy}{dx}\right)}_{x-\pi /4}=-\sqrt{2}$
$\therefore$ Equation of tangent at $(\frac{\pi }{4},\sqrt{2})$ is $y-\sqrt{2}=-\sqrt{2}(x-\frac{\pi }{4})$