MHT CET · Maths · Continuity and Differentiability
The value of \(a\) and \(b\) such that the function
\(f(x)=\left\{\begin{array}{ll}-2 \sin x, & -\pi \leq x \leq-\frac{\pi}{2} \ a \sin x+b, \end{array}\right.\) \(-\frac{\pi}{2} < x < \frac{\pi}{2} \text { is continuous } \ \cos x, \frac{\pi}{2} \leq x \leq \pi\)
in \([-\pi, \pi]\), are
- A \(-1,0\)
- B 1,0
- C 1,1
- D \(-1,1\)
Answer & Solution
Correct Answer
(D) \(-1,1\)
Step-by-step Solution
Detailed explanation
For continuity in \([-\pi, \pi]\), we must have
\(\text { At } x=-\frac{\pi}{2}, f\left(-\frac{\pi}{2}\right)=\lim _{x \rightarrow\left(-\frac{\pi}{2}\right)^{-}}(-2 \sin x) \)
\( =\lim _{x \rightarrow\left(-\frac{\pi}{2}\right)^{+}}(a \sin x+b) \)
\( \Rightarrow 2=-a+b\)
At \(x=\frac{\pi}{2}\),
\(
f\left(\frac{\pi}{2}\right)=\lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{-}}(a \sin x+b)=\lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{+}} \cos x
\)
\(
\Rightarrow 0=a+b
\)
\(\cdots\)
On solving Eqs. (i) and (ii) we get, \(a=-1, b=1\)
\(\text { At } x=-\frac{\pi}{2}, f\left(-\frac{\pi}{2}\right)=\lim _{x \rightarrow\left(-\frac{\pi}{2}\right)^{-}}(-2 \sin x) \)
\( =\lim _{x \rightarrow\left(-\frac{\pi}{2}\right)^{+}}(a \sin x+b) \)
\( \Rightarrow 2=-a+b\)
At \(x=\frac{\pi}{2}\),
\(
f\left(\frac{\pi}{2}\right)=\lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{-}}(a \sin x+b)=\lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{+}} \cos x
\)
\(
\Rightarrow 0=a+b
\)
\(\cdots\)
On solving Eqs. (i) and (ii) we get, \(a=-1, b=1\)
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