Question

The tangents to the curve $y=\sin (x+y),-2\pi \le x\le 2\pi$ that are parallel to the line $x+2y=0$ is

• A. $x+2y-\pi =0$
• B. $x+2y+\pi =0$
• C. $x-2y+\pi =0$
• D. $x-2y-\pi =0$

Answer 1

Answer: (A), (B)

$x+2y+\pi =0$

Let $(x_1,y_1)$ be the point of contact of one of the tangents.
Tangent is parallel to $x+2y=0$
Slope of the tangent is, $m=-\frac{1}{2}$
$y=\sin (x+y)$
$\Rightarrow \frac{dy}{dx}=-\frac{1}{2}$
$\Rightarrow \cos (x+y)(1+\frac{dy}{dx})=-\frac{1}{2}$
At $(x_1,y_1)$, we get
$\cos (x_1+y_1)(1-\frac{1}{2})=-\frac{1}{2}$
$\Rightarrow \cos (x_1+y_1)=-1\Rightarrow x_1+y_1=\pi ,-\pi$

So, $y_1=\sin (x_1+y_1)=0$
$\Rightarrow x_1=\pi ,-\pi$
Equations of tangents are
$y-0=-\frac{1}{2}(x-\pi )$ and $y-0=-\frac{1}{2}(x+\pi )$

Hence, the required equation of the tangents are
$x+2y-\pi =0$ and $x+2y+\pi =0$