Question

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

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

Answer 1

The correct option is C

$\frac{\mathrm{\pi }}{2}$

Explanation for the correct option:

Step 1. Find the value of $k$:

Let $y=\cos (x+y)$ ....(i)

Differentiate it with respect to $x$

$\frac{dy}{dx}=-\sin (x+y)\left(1+\frac{dy}{dx}\right)$ ....(i)

Equation of tangent is $x+2y=k$

$\therefore$Slope of tangent $=-\frac{1}{2}$

$\Rightarrow$ $\frac{dy}{dx}=-\frac{1}{2}$

$\Rightarrow$ $-\frac{1}{2}=\sin (x+y)\left(1–\frac{1}{2}\right)$

$\Rightarrow$ $-\frac{1}{2}=-\frac{1}{2}\sin (x+y)$

$\Rightarrow$ $\sin (x+y)=1$

$\Rightarrow$ $x+y=\frac{\pi }{2}$ ….(ii)

$\Rightarrow$ $y=\frac{\pi }{2}–x$

Step 2. Put the value of $y$ in equation (i), we get

$\frac{\pi }{2}–x=\cos (x+y)$

$\Rightarrow$$\frac{\pi }{2}–x=\cos \left(\frac{\pi }{2}\right)$ (from (ii))

$\Rightarrow$$\frac{\pi }{2}–x=0$

$\Rightarrow$ $x=\frac{\pi }{2}$

$\Rightarrow$ $y=0$

Step 3. Put $y=0$ in the equation of tangent, we get

$x+2y=k$

$\Rightarrow$$\frac{\pi }{2}+0=k$

$\mathbf{\therefore }\mathit{k}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{}$

Hence, Option ‘C’ is Correct.