Question

If an equation of a tangent to the curve, $y-cos\left(x+f\right),-1-1\le x\le 1+\pi ,isx+2y=k$ then $k$ is equal to :

• A. 1
• B. 2
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (D)

Let $y=cos(x+y)$
$\Rightarrow \frac{dy}{dx}=-sin\left(x+y\right)\left(1+\frac{dy}{dx}\right)\dots \left(1\right)$
Now, given equation of tangent is
$x+2y=k$
$\Rightarrow Slope=\frac{-1}{2}$
$So,\frac{dy}{dx}=\frac{-1}{2}putthisvaluein\left(1\right),weget$
$\frac{-1}{2}=-sin\left(x+y\right)\left(1-\frac{1}{2}\right)$
$\Rightarrow sin\left(x+y\right)=1$
$\Rightarrow x+y=\frac{\pi }{2}\Rightarrow y=\frac{\pi }{2}-x$
$Now,\frac{\pi }{2}-x=cos\left(x+y\right)$
$\Rightarrow x=\frac{\pi }{2}andy=0$
$Thusx+2y=k\Rightarrow \frac{\pi }{2}=k$