Question

The least positive root of the function $\sin x-\frac{\pi }{2}+1=0$ lies in the interval

• A. $[0,\frac{\pi }{2}]$
• B. $[\frac{\pi }{2},\pi ]$
• C. $[\frac{\pi }{2},\frac{3\pi }{2}]$
• D. None

Answer 1

The correct option is A $[0,\frac{\pi }{2}]$

$\sin x-\frac{\pi }{2}+1=f\left(x\right)$

$\because \sin x$ and $1-\frac{\pi }{2}$ both are continuous funtions, $\therefore f\left(x\right)$ is also continuous

$\sin x+1-\frac{\pi }{2}$

$f^{\prime }\left(x\right)=\cos x$

$\therefore f^{\prime }\left(x\right)>0$ for $xϵ[0,\frac{\pi }{2}]$ &

$xϵ[\frac{3\pi }{2},2\pi ]$

Hence $f\left(x\right)$ is an increasing function only in the above range i.e.,

$xϵ[0,\frac{\pi }{2}]\bigcup [\frac{3\pi }{2},2\pi ]$

$f\left(0\right)=0+1-\frac{\pi }{2}=1-\frac{\pi }{2}$

$=1-\frac{3.14}{2}=1-1.57=-ve$

$\because f\left(0\right)$ is negative

$f\left(\frac{\pi }{2}\right)=1+1-\frac{\pi }{2}=2-1.57=+ve$

$\therefore f\left(\frac{\pi }{2}\right)$ is positive

and $\because f^{\prime }\left(x\right)>0$ in $xϵ[0,\frac{\pi }{2}]$

$\therefore f\left(x\right)$ is an increasing continuous function for $xϵ[0,\frac{\pi }{2}]$

$\because f\left(0\right)=$ negative

and $f\left(\frac{\pi }{2}\right)=$positive

$\therefore$There must be one root of $f\left(x\right)$ present in $xϵ[0,\frac{\pi }{2}]$

We can also say this using intermediate value theorem

Hence $xϵ[0,\frac{\pi }{2}]$ is correct