Thank you for reporting, we will resolve it shortly
Q.
If an equation of a tangent to the curve, $y=\cos\left(x + y\right),-1\leq x\leq 1+\pi $ , is $x+2y=k$ then $k$ is equal to
NTA AbhyasNTA Abhyas 2022
Solution:
Let $y=\cos x+y$
$\Rightarrow \frac{d y}{d x}=-\sin x+y 1+\frac{d y}{d x} \ldots .1$
Now, given equation of tangent is
$x+2 y=k$
$\Rightarrow$ Slope $=\frac{-1}{2}$
So, $\frac{d y}{d x}=\frac{-1}{2}$ put this value in 1 , we get
$\frac{-1}{2}=-\sin x+y 1-\frac{1}{2}$
$\Rightarrow \sin x+y=1$
$\Rightarrow x+y=\frac{\pi}{2} \Rightarrow y=\frac{\pi}{2}-x$
Now, $\frac{\pi}{2}-x=\cos x+y$
$\Rightarrow x=\frac{\pi}{2}$ and $y=0$
Thus $x+2 y=k \Rightarrow \frac{\pi}{2}=k$