Question

One root of the equation $\cos x-x+\frac{1}{2}=0$ lies in the interval:

• A. $(0,\frac{\pi }{2})$
• B. $(-\frac{\pi }{2},0)$
• C. $(\frac{\pi }{2},\pi )$
• D. $(\pi ,\frac{3\pi }{2})$

Answer 1

The correct option is A $(0,\frac{\pi }{2})$

$\begin{array}{c}Letf\left(x\right)=\cos x-x+\frac{1}{2}\\ f\left(0\right)=\cos 0-0+\frac{1}{2}\\ =1+\frac{1}{2}\\ =\frac{3}{2}>0\\ f\left(\frac{\pi }{2}\right)=\cos \frac{\pi }{2}-\frac{\pi }{2}+\frac{1}{2}\\ =0-\frac{\pi }{2}+\frac{1}{2}\\ =\frac{1-\pi }{2}<0\end{array}$

Since $f\left(x\right)$ changes its sign from $0$ to $\pi$

therefore there exists atleast one root in $\left(0,\frac{\pi }{2}\right)$