Question

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

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

Answer 1

Answer: (D)

Let $y = cos (x+y)$
$\Rightarrow \quad\frac{d y}{d x}=-sin\, \left(x+y\right) \left(1+\frac{d y}{d x}\right)\quad\ldots\left(1\right)$
Now, given equation of tangent is
$x + 2y = k$
$\Rightarrow \quad Slope\, =\frac{-1}{2}$
$So, \frac{d y}{d x}=\frac{-1}{2} put \,this\, value\, in \left(1\right), we\, get$
$\frac{-1}{2}=-sin \left(x+y\right) \left(1-\frac{1}{2}\right)$
$\Rightarrow sin \, \left(x+y\right)=1$
$\Rightarrow x+y=\frac{\pi}{2} \Rightarrow y=\frac{\pi}{2}-x$
$Now, \frac{\pi}{2}-x=cos\left(x+y\right)$
$\Rightarrow x=\frac{\pi}{2} and \,y=0$
$Thus\, x+2y=k \Rightarrow \frac{\pi}{2}=k$