Question

If \[\mathrm{f}(x)= \begin{cases} \mathrm{m}x+1, & x\leqslant\frac{\pi}{2} \\ \\ \mathrm{sin}x+\mathrm{n}, & x>\frac{\pi}{2} & \end{cases}\], is continuous at \[x=\frac{\pi}{2},( \begin{array} {c}\mathrm{m,n\in\mathbb{Z}} \end{array})\] then

विकल्प

• m = 1, n = 0

• \[\mathrm{m=\frac{n\pi}{2}}\]

• \[\mathrm{m}=\mathrm{n}=\frac{\pi}{2}\]

• \[\mathrm{n}=\frac{\mathrm{m}\pi}{2}\]

Answer 1

\[\mathrm{n}=\frac{\mathrm{m}\pi}{2}\]

Explanation:

Since the function is continuous at x = \[\frac{\pi}{2}\]

\[\therefore\quad\lim_{x\to\frac{\pi-}{2}}\mathrm{f}\left(x\right)=\lim_{x\to\frac{\pi^{+}}{2}}\mathrm{f}\left(x\right)\]

\[\therefore\quad\lim_{x\to\frac{\pi}{2}}\mathrm{m}x+1=\lim_{x\to\frac{\pi}{2}}\sin x+\mathrm{n}\]

\[\therefore\quad\frac{\pi}{2}\mathrm{m}+1=\mathrm{sin}\frac{\pi}{2}+\mathrm{n}\]

\[\therefore\quad\mathrm{n}=\frac{\mathrm{m}\pi}{2}\]