Question

The number of solutions of the equation ${\int }_{-2}^x\left|\cos x\right|dx=0$ is?

where $0<x<\frac{x}{2}$

• A. $0$
• B. $1$
• C. $2$
• D. $4$

Answer 1

The correct option is A $0$

Number of solutions of the above equations $0<x<\frac{\pi }{2}$

$\left(\cos x\right)$ is negative for $-2<x<-\frac{\pi }{2}$

And positive for $-\frac{\pi }{2}<x<\frac{\pi }{2}$

${\int }_{-2}^x\left|\cos x\right|dx={\int }_{-2}^{-\frac{\pi }{2}}-\cos xdx+{\int }_{-\frac{\pi }{2}}^x\cos xdx\Rightarrow {\left[\sin x\right]}_{-2}^{-\frac{\pi }{2}}-{\left[\sin x\right]}_{-\frac{\pi }{2}}^x=0\Rightarrow -1+\sin \left(2\right)-\sin x-1=0\Rightarrow \sin x=\sin \left(2\right)-2$

$-1<\sin \left(t\right)<1-3<\sin \left(t\right)-2<-1$

$\sin \left(2\right)-2$ is negative

But $\sin x$ is positive ; since $0<x<\frac{\pi }{2}$

Hence, correct answer is $0$ solution