If f ( x ) = { m x + 1 , if x 2 sin x + n , if x 2 , is continuous at x = 2 , then

Question

If $f (x) =$ ${\begin{matrix} m x + 1, & if x \leq \frac{π}{2} \\ \sin x + n, & if x > \frac{π}{2} \end{matrix},$ is continuous at $x = \frac{π}{2}$ , then

Smart Achievers · Accepted Answer

(3)

Ans. $n = \frac{m π}{2}$

Explanation:

We have, $f (x) =$ ${\begin{matrix} m x + 1, & if x \leq \frac{π}{2} \\ \sin x + n, & if x > \frac{π}{2} \end{matrix}$ is continuous at $x = \frac{π}{2}$

$∴$ $L.H.L. = \lim_{x \to {\frac{π}{2}}^{-}} (m x + 1) = \lim_{h \to 0} [m (\frac{π}{2} - h) + 1] = \frac{m π}{2} + 1$

and $R.H.L. = \lim_{x \to {\frac{π}{2}}^{+}} (\sin x + n) = \lim_{h \to 0} [\sin (\frac{π}{2} + h) + n]$

$= \lim_{h \to 0} (\cos h + n) = 1 + n$

We must have $L.H.L. = R.H.L.$

$\Rightarrow \frac{m π}{2} + 1 = n + 1$

$∴$ $n = \frac{m π}{2}$

Solution

Q.
If $f (x) =$ ${\begin{matrix} m x + 1, & if x \leq \frac{π}{2} \\ \sin x + n, & if x > \frac{π}{2} \end{matrix},$ is continuous at $x = \frac{π}{2}$ , then

1 $m = 1, n = 0$

2 $m = \frac{n π}{2} + 1$

3 $n = \frac{m π}{2}$

4 $m = n = \frac{π}{2}$

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Solution

Q. If f(x)={mx+1,if x≤π2sinx+n,if x>π2, is continuous at x=π2, then

1 m=1, n=0

2 m=nπ2+1

3 n=mπ2

4 m=n=π2