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Continuity — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsContinuity3 Marks
Question
In the following, find the value of the constant k so that the given function is continuous at the indicated point:
$\text{f(x)}=\begin{cases}\frac{1-\cos2\text{kx}}{\text{x}^2},&\text{if}\text{ x}\neq0\\8,&\text{if}\text{ x}=0\end{cases}\text{at x}=0$
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Answer

Given,
$\text{f(x)}=\begin{cases}\frac{1-\cos2\text{kx}}{\text{x}^2},&\text{if}\text{ x}\neq0\\8,&\text{if}\text{ x}=0\end{cases}$
If f(x) is continuous at x = 0, then
$\lim_\limits{\text{x}\rightarrow 0}\text{f(x)}=\text{f}(0)$
$\Rightarrow\lim_\limits{\text{x}\rightarrow 0}\frac{1-\cos2\text{kx}}{\text{x}^2}=8$
$\Rightarrow\lim_\limits{\text{x}\rightarrow 0}\frac{2\text{k}^2\sin^2\text{kx}}{\text{k}^2\text{x}^2}=8$
$\Rightarrow2\text{k}^2\lim_\limits{\text{x}\rightarrow 0}\Big(\frac{\sin\text{kx}}{\text{kx}}\Big)^2=8$
$\Rightarrow2\text{k}^2\times1=8$
$\Rightarrow\text{k}^2=4$
$\Rightarrow\text{k}=\pm2$

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