MCQ
If $f (x)=\left\{\begin{array}{ll} ax +1 & , x \leq \frac{\pi}{2} \\ \sin x+ b & , x>\frac{\pi}{2}\end{array}\right.$ is continuous at $x=\frac{\pi}{2}$, then
  • A
    $a=1, b=0$
  • B
    $a=b \frac{\pi}{2}+1$
  • $b =\frac{ a \pi}{2}$
  • D
    $a = b =\frac{\pi}{2}$

Answer

Correct option: C.
$b =\frac{ a \pi}{2}$
(C)
Since $f (x)$ is continuous at $x=\frac{\pi}{2}$.
$\therefore \quad \lim _{x \rightarrow \frac{\pi^{-}}{2}} f (x)=\lim _{x \rightarrow \frac{\pi^{+}}{2}} f (x)$
$\Rightarrow \lim _{x \rightarrow \frac{\pi}{2}}(a x+1)=\lim _{x \rightarrow \frac{\pi}{2}}(\sin x+b)$
$\Rightarrow a \cdot \frac{\pi}{2}+1=1+ b \quad \Rightarrow b =\frac{ a \pi}{2}$

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