Question

The equation of one of the tangents to the curve $y=\cos (x+y),-2\pi \le x\le 2\pi$ that is parallel to the line $x+2y=0$:

• A. $x+2y=1$
• B. $x+2y=\frac{\pi }{2}$
• C. $x+2y=\frac{\pi }{4}$
• D. None of these

Answer 1

The correct option is B

$x+2y=\frac{\pi }{2}$

Explanation for the correct answer:

Given $y=\cos (x+y)$

Now on differentiating with respect to $x$

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

Now we calculate slope of the line

$x+2y=0$ is $\frac{-1}{2}$.

Put the value $\frac{dy}{dx}=\frac{-1}{2}$ in (i)

$$\begin{gathered} \frac{-1}{2}=-\sin (x+y)(1–\frac{1}{2}) \\ \sin (x+y)=1 \\ (x+y)=\pi /2 \end{gathered}$$

Now putting $(x+y)=\pi /2$ in $y=\cos (x+y)$

$$\begin{gathered} y=\cos \frac{\pi }{2} \\ \Rightarrow y=0 \end{gathered}$$

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

Now applying $\sin$ both sides we get

$$\begin{gathered} \sin x=\sin \frac{\pi }{2} \\ \Rightarrow \sin x=1 \end{gathered}$$

Also it is given $-2\pi \le x\le 2\pi$

So $x=\frac{-3\pi }{2},\frac{\pi }{2}.$

Thus the points are $(\frac{-3\pi }{2},0)and(\frac{\pi }{2},0)$.

Equation of tangent is $y-y_1=m(x-x_1)$

At Point $(\frac{-3\pi }{2},0)$where slope $\frac{-1}{2}$ so equation of tangent

$$\begin{gathered} y–0=\frac{-1}{2}(x+\frac{3\pi }{2}) \\ y=\frac{-1}{2}x+\frac{-3\pi }{4} \\ 4y=-2x-3\pi \\ 2x+4y=-3\pi \end{gathered}$$

At $(\frac{\pi }{2},0)$ where slope is $\frac{-1}{2}$ equation of tangent is

$$\begin{gathered} y–0=-\frac{1}{2}(x–\frac{\pi }{2}) \\ y=\frac{-1}{2}x+\frac{\pi }{4} \\ 4y=-2x+\pi \\ 2x+4y=\pi \end{gathered}$$

Hence option (B) is the answer.