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Continuity — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity1 Mark
Question
Let $\text{f(x)}=\frac{\tan\Big(\frac{\pi}{4}-\text{x}\Big)}{\cot2\text{x}},\text{ x}\neq\frac{\pi}{4}$ The value which should be assigned to f(x) at $\text{x}=\frac{\pi}{4},$ so that it is continuous everywhere is:
  1. 1
  2. $\frac{1}{2}$
  3. 2
  4. None of these.
✓

Answer

  1. $\frac{1}{2}$

Solution:

$\text{f}\big(\frac{\pi}{2}\big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{\tan\Big(\frac{\pi}{4}-\text{x}\Big)}{\cot2\text{x}}$

$\text{f}\Big(\frac{\pi}{4}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{\frac{1-\tan\text{x}}{1+\tan\text{x}}}{\cot2\text{x}}$

$\text{f}\Big(\frac{\pi}{4}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\tan2\text{x}\Big(\frac{1-\tan\text{x}}{1+\tan\text{x}}\Big)$

$\text{f}\Big(\frac{\pi}{4}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{2\tan\text{x}}{1-\tan^2\text{x}}\Big(\frac{1-\tan\text{x}}{1+\tan\text{x}}\Big)$

$\text{f}\Big(\frac{\pi}{4}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{2\tan\text{x}}{(1-\tan\text{x})(1+\tan\text{x})}\Big(\frac{1-\tan\text{x}}{1+\tan\text{x}}\Big)$ $\begin{pmatrix}\because\tan\frac{\pi}{4}\rightarrow1\\1-\tan\frac{\pi}{4}\neq0 \end{pmatrix}$

$\text{f}\Big(\frac{\pi}{4}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{2\tan\text{x}}{(1+\tan\text{x})^2}=\frac{2}{4}=\frac{1}{2}$

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