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Continuity — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsContinuity5 Marks
Question
Find the value of 'a' for which the function f defined by
$\text{f}\text{(x)}=\begin{cases}\text{a}\sin\frac{\pi}{2}(\text{x}+1),& \text{x}\leq0 \\\frac{\tan\text{x-sin}\text{x}}{\text{x}^3} &\text{x} > 0\end{cases}$ is discontinuous at x = 0.
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Answer

Since f(x) is continuous at x = 0, L.H.L = R.H.L.
Thus, we have
$\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{(x)}$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0^-}\text{a}\sin\frac{\pi}{2}(\text{x}+1)=\lim\limits_{\text{x} \rightarrow 0^+}\frac{\tan\text{x}-\sin\text{x}}{\text{x}^3}$
$\Rightarrow\text{a}\times1=\lim\limits_{\text{x} \rightarrow 0}\frac{\tan\text{x}-\sin\text{x}}{\text{x}^3}$
$\Rightarrow\text{a}=\lim\limits_{\text{x} \rightarrow 0}\frac{\frac{\sin\text{x}}{\cos\text{x}}-\sin\text{x}}{\text{x}^3}$
$\Rightarrow\text{a}=\lim\limits_{\text{x} \rightarrow 0}\frac{\frac{\sin\text{x}}{\text{x}}\Big(\frac{1}{\cos\text{x}}-1\Big)}{\text{x}^2}$
$\Rightarrow\text{a}=\lim\limits_{\text{x} \rightarrow 0}\frac{\frac{\sin\text{x}}{\text{x}}\Big(\frac{1-\cos\text{x}}{\cos\text{x}}\Big)}{\text{x}^2}$
$\Rightarrow\text{a}=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\text{x}}{\text{x}}\times\lim\limits_{\text{x} \rightarrow 0}\frac{1}{\cos\text{x}}\times\lim\limits_{\text{x} \rightarrow 0}\frac{1-\cos\text{x}}{\text{x}^2}$
$\Rightarrow\text{a}=1\times1\times\lim\limits_{\text{x} \rightarrow 0}\frac{1-\cos\text{x}}{\text{x}^2}$
$\Rightarrow\text{a}=\lim\limits_{\text{x} \rightarrow 0}\frac{1-\cos\text{x}}{\text{x}^2}\times\frac{1+\cos\text{x}}{1+\cos\text{x}}$
$\Rightarrow\text{a}=\lim\limits_{\text{x} \rightarrow 0}\frac{1-\cos^2\text{x}}{\text{x}^2(1+\cos\text{x)}}$
$\Rightarrow\text{a}=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin^2\text{x}}{\text{x}^2(1+\cos\text{x)}}$
$\Rightarrow\text{a}=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin^2\text{x}}{\text{x}^2}\times\lim\limits_{\text{x} \rightarrow 0}\frac{1}{1+\cos\text{x}}$
$\Rightarrow\text{a}=1\times\lim\limits_{\text{x} \rightarrow 0}\frac{1}{1+\cos\text{x}}$
$\Rightarrow\text{a}=1\times\frac{1}{1+1}$
$\Rightarrow\text{a}=\frac{1}{2}$

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