Cos this is simple, solve it in 10 seconds!

$y=-x\cos x$ is an odd function, so its graph is symmetrical about $(0,0)$ , therefore A and C are impossible.

Furthermore, when $x\in (0,\frac{\pi}{2})$ , $y=-x\cos x< 0$ .

When you zoom in close to zero, y= -x. (d) is the only one with this behavior.

Since there are options given., my solution is going to be very short.

When $\dfrac{-\pi}{2} <x < 0 , -x > 0 , \cos x > 0$

Hence the graph where the $y$ values are $< 0$ , for $x \in (-\dfrac{\pi}{2} , 0)$ , are excluded.

When $0 < x < \dfrac{\pi}{2}, -x < 0 , \cos x > 0$

Hence the graph where the $y$ values are $> 0$ , for $x \in (0,\dfrac{\pi}{2})$ , are excluded.

Option D is all that is left.

So let us analyse the graph.

At x=0, y=0 and all the graphs satisfy this. At x= $\Pi /2$ , y=- $\Pi/2$ cos( $\Pi/2$ ) = 0 and the graphs seems so satisfy this even though the scale is not mentioned. Now for the critical analysis let us take $\Pi /4$ .

At x= $\Pi /4$ , y=- $\Pi /4$ cos( $\Pi /4$ ) = - $\Pi /(4\sqrt { 2 }$ ), which is a negative value. so, A and B are eliminated.

Now at x=- $\Pi /4$ , y = $\Pi /4$ cos(- $\Pi /4$ ) = $\Pi /4$ cos( $\Pi /4$ ) as cos(-x) = cos(x).

so y is positive in this case and thus C is eliminated and answer is D. And as it cos is a periodic function, -xcosx will also be periodic.