Question

Find $a$ such that function is continuous at $x=1$; Where f(x) =$\left\{\begin{array}{cc}sin\frac{\pi x}{2},& x\le 1\\ 3x-a& x>1\end{array}\right\}$

• A. 0
• B. 1
• C. 2
• D. Do not exisrt

Answer 1

Answer: (C)

2

For continuity of function $RHL=LHL$ at $x=1$. Hence

$\underset{x\to 1^{+}}{lim}3x-a=\underset{x\to 1^1}{lim}sin\frac{\pi x}{2}$

$⟹3-a=sin\frac{\pi }{2}$

$⟹a=3-1=2$