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Continuity — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsContinuity2 Marks
MCQ
For what value of $k$, the function defined by $f(x)=\left\{\begin{array}{cc}\frac{\log (1+2 x) \sin x^{\circ}}{x^2} & \text { for } x \neq 0 \\ k & \text { for } x \neq 0\end{array}\right.$ is continuous at $x=0$ ?
  • A
    2
  • B
    $\frac{1}{2}$
  • ✓
    $\frac{\pi}{90}$
  • D
    $\frac{90}{\pi}$
✓

Answer

Correct option: C.
$\frac{\pi}{90}$
(c) : $f(x)= \begin{cases}\frac{\log (1+2 x) \sin x^0}{x^2} & , x \neq 0 \\ k & , x=0\end{cases}$
Since, $f(x)$ is continuous at $x=0$
$
\begin{aligned}
& \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0) \\
& \therefore \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}\left[2 \cdot \frac{\log (1+2 x)}{2 x} \frac{\sin \left(\frac{\pi x}{180}\right)}{x}\right] \\
& =2 \times \lim _{x \rightarrow 0} \frac{\log (1+2 x)}{2 x} \frac{\sin \left(\frac{\pi x}{180}\right)}{\left(\frac{\pi x}{180}\right)} \times \frac{\pi}{180} \\
& \pi
\end{aligned}
$
$
=2 \times \frac{\pi}{180}=\frac{\pi}{90}\left[\because \lim _{x \rightarrow 0} \frac{\log (1+2 x)}{2 x}=1 \text { and } \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]
$
Also $f(0)=k$
$\therefore k=\frac{\pi}{90}$

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