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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY1 Mark
Question
The value of a for which the function $\text{f(x)}=\begin{cases}\frac{(4^\text{x}-1)^3}{\sin\Big(\frac{\text{x}}{\text{a}}\Big)\log\Big\{\Big(1+\frac{\text{x}^2}{3}\Big)\Big\}},&\text{x}\neq0\\12(\log4)^3,&\text{x}=0\end{cases}$ may be continuous at x = 0 is:
  1. 1
  2. 2
  3. 3
  4. None of these.
✓

Answer

  1. None of these.

Solution:

$\lim\limits_{\text{x}\rightarrow0}\frac{(4^\text{x}-1)^3}{\sin\Big(\frac{\text{x}}{\text{a}}\Big)\log\Big\{\Big(1+\frac{\text{x}^2}{3}\Big)\Big\}}=12(\log4)^3$

$\lim\limits_{\text{x}\rightarrow0}\frac{\frac{(4^\text{x}-1)^2}{\text{x}^3}}{\frac{\sin\Big(\frac{\text{x}}{\text{a}}\Big)}{\text{x}}}\times\frac{1}{\frac{\log\Big\{\Big(1+\frac{\text{x}}{3}\Big)\Big\}}{\text{x}^2}}=12(\log4)^3$

$\lim\limits_{\text{x}\rightarrow0}\frac{\frac{(4^\text{x}-1)^3}{\text{x}^3}}{\frac{\sin\Big(\frac{\text{x}}{\text{a}}\Big)}{\text{x}}}\times\frac{1}{\frac{\log\Big\{\Big(1+\frac{\text{x}^2}{3}\Big)\Big\}}{\text{x}^2}}=12(\log4)^3$

$\lim\limits_{\text{x}\rightarrow0}\frac{\Big(\frac{(4^\text{x}-1}{\text{x}^3}\Big)}{\frac{\sin\Big(\frac{\text{x}}{\text{a}}\Big)}{\frac{\text{x}}{\text{a}}}}\text{a}\text{x}\ {\times}\frac{\frac{1}{\log\Big\{\Big(1+\frac{\text{x}^2}{3}\Big)\Big\}}}{\frac{\text{x}^3}{3}}\text{x}^3=12(\log4)^3$

$3(\log4)^3=12(\log4)^3$

$3\text{a}=12$

$\text{a}=12$

Note: The question is incorrect, so it has been modified.

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