Question

If $y\cos x+x\cos y=\pi ,$ then $y^{\prime \prime }\left(0\right)$ is

• A. $1$
• B. $\pi$
• C. $0$
• D. $-\pi$

Answer 1

The correct option is C $0$

$y\cos x+x\cos y=\pi$ ...(1)

Differentiating (1) w.r.t x

$-y\sin x+y^{\prime }\cos x+\cos y-\left(x\sin y\right)y^{\prime }=0$

or $-y\sin x+\cos y+y^{\prime }(\cos x-x\sin y)=0$ ...(2)

Again differentiate (2) w.r.t. $x$

$-y\cos x-y^{\prime }\sin x-(-\sin y)y^{\prime }+y^{\prime \prime }[\cos x-x\sin y]+y^{\prime }(-\sin x-1.\sin y-x\cos y.y^{\prime }]=0$ ...(3)

Putting $x=0$ in (1), we get, $y=\pi$

Putting $x=0$ and $y=\pi$ in (2), we get , $0+\cos \pi +y^{\prime }(-1-\pi \sin \pi )$

or $-1-y^{\prime }=0$

$\therefore y^{\prime }=-1$

Putting $x=0$ and $y=\pi$ and $y^{\prime }=1$ in (3), we get,

$y^{\prime \prime }\left(0\right)=0$.