COMEDK · Maths · 25. Continuity and Differentiability
If \(f(x) = \left\{ {\matrix{
{2\sin x} & ; & { - \pi \le x \le {{ - \pi } \over 2}} \cr
{a\sin x + b} & ; & { - {\pi \over 2} < x < {\pi \over 2}} \cr
{\cos x} & ; & {{\pi \over 2} \le x \le \pi } \cr
} } \right.\) and it is continuous on \([-\pi, \pi]\), then
- A \(a=-1\) and \(b=-1\)
- B \(a=1\) and \(b=-1\)
- C \(a=-1\) and \(b=1\)
- D \(a=1\) and \(b=1\)
Answer & Solution
Correct Answer
(B) \(a=1\) and \(b=-1\)
Step-by-step Solution
Detailed explanation
For \(f(x)\) to be continuous on \([-\pi, \pi]\), it must be continuous at \(x = -\dfrac{\pi}{2}\) and \(x = \dfrac{\pi}{2}\). At \(x = -\dfrac{\pi}{2}\), the left-hand limit equals the right-hand limit: \(\lim_{x \to -\pi/2^{-}} 2\sin x = \lim_{x \to -\pi/2^{+}} (a\sin x + b)\)…
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