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Continuity — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity1 Mark
Question
The values of the constants a, b and for which the function $\text{f(x)}=\begin{cases}(1+\text{ax})^{\frac{1}{\text{x}}},&\text{x}>0\\\text{b},&\text{x}=0\\\frac{(\text{x}+\text{c})^{\frac{1}{2}}-1}{(\text{x}+1)^{\frac{1}{2}}-1},&\text{x}>0\end{cases}$ may be continuous at x = 0, are:
  1. $\text{a}=\log_{\text{e}}\Big(\frac{2}{3}\Big),\text{ b}=-\frac{2}{3},\text{ c}=1$
  2. $\text{a}=\log_{\text{e}}\Big(\frac{2}{3}\Big),\text{ b}=\frac{2}{3},\text{ c}=-1$
  3. $\text{a}=\log_{\text{e}}\Big(\frac{2}{3}\Big),\text{ b}=\Big(\frac{2}{3}\Big),\text{ c}=1$
  4. none of these
✓

Answer

  1. $\text{a}=\log_{\text{e}}\Big(\frac{2}{3}\Big),\text{ b}=\Big(\frac{2}{3}\Big),\text{ c}=1$

Solution:

$\text{f}(0)=\lim\limits_{\text{x}\rightarrow0}(1+\text{ax})^{\frac{1}{\text{x}}}$

$\text{b}=\lim\limits_{\text{x}\rightarrow{\text{a}}}(1+\text{ax})^{\frac{1}{\text{ax}}\times\text{a}}$

$\text{b}=\text{e}^{\text{a}}$

$\text{a}=\log_{\text{e}}\text{b}$

$\text{f}(0)=\lim\limits_{\text{x}\rightarrow\text{a}^+}\frac{(\text{x}+\text{c})^{\frac{1}{3}}-1}{(\text{x}+1)^{\frac{1}{2}}-1}$

Here, $\text{c}=1$

$\text{x}+1=\text{y}$

$\text{x}\rightarrow0\Rightarrow\text{y}\rightarrow1$

$\text{f}(0)=\lim\limits_{\text{y}\rightarrow1}\frac{\text{y}^{\frac{1}{3}}-1}{\text{y}^{\frac{1}{2}}-1}$

$\text{b}=\lim\limits_{\text{y}\rightarrow1}\frac{\frac{\text{y}^{\frac{1}{3}}-1}{\text{y}-1}}{\frac{\text{y}^{\frac{1}{2}}-1}{\text{y}-1}}=\frac{\frac{1}{3}}{\frac{1}{2}}=\frac{2}{3}$

$\text{a}=\log\text{b}=\log\frac{2}{3}$

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