one root of the equation cosx- x+1/2=0 lies in the interval 0 to π/2.prove

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$$\begin{gathered} \mathrm{We} \mathrm{have}: \\ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{cosx}-\mathrm{x}+\frac{1}{2} \\ \mathrm{Now} \\ \mathrm{f}\left(0\right)=\cos 0-0+\frac{1}{2}=1+\frac{1}{2}=\frac{3}{2} \\ \Rightarrow \mathrm{f}\left(0\right)>0 \\ \mathrm{f}\left(\frac{\mathrm{\pi }}{2}\right)=\cos \frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{2}+\frac{1}{2}=0+\frac{1-\mathrm{\pi }}{2}=\frac{1-\mathrm{\pi }}{2} \\ \mathrm{f}\left(\frac{\mathrm{\pi }}{2}\right)<0 \\ \mathrm{As} \mathrm{f}\left(0\right)>0 \mathrm{and} \mathrm{f}\left(\frac{\mathrm{\pi }}{2}\right)<0, \mathrm{hence} \mathrm{there} \mathrm{will} \mathrm{exist} \mathrm{atleast} \mathrm{one} \mathrm{root} \mathrm{of} \\ \mathrm{f}\left(\mathrm{x}\right) \mathrm{in} \mathrm{x}\in \left(0,\frac{\mathrm{\pi }}{2}\right) \mathrm{as} \mathrm{f}\left(\mathrm{x}\right) \mathrm{will} \mathrm{cross} \mathrm{X}-\mathrm{axis} \mathrm{when} \mathrm{it} \mathrm{changes} \mathrm{from} \mathrm{positive} \mathrm{to} \mathrm{negative}. \end{gathered}$$

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