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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability4 Marks
Question
A function f(x) is said to be continuous in an open interval (a, b), if it is continuous at every point in this interval.
A function f(x) is said to be continuous in the closed interval [a, b), if f(x) is continuous in (a, b) and $\lim\limits_{\text{x}\rightarrow0}\text{f}(\text{a}+\text{h})=\text{f}(\text{a})$ and $\lim\limits_{\text{x}\rightarrow0}\text{f}(\text{b}-\text{h})=\text{f}(\text{b})$
If function $\text{f}(\text{x})=\begin{cases}\frac{\sin(\text{a}+1)\text{x}+\sin\text{x}}{\text{x}}&,\text{x}<0\\\text{c}&,\text{x}=0\\\frac{\sqrt{\text{x}+\text{bx}^2}-\sqrt{\text{x}}}{\text{bx}^{\frac{3}{2}}}&,\text{x}>0\end{cases}$ is continuous at x = 0, then answer the following questions.
  1. The value of a is:
  1. $-\frac{3}{2}$
  2. $0$
  3. $\frac{1}{2}$
  4. $-\frac{1}{2}$
  1. The value of b is:
  1. 1
  2. -1
  3. 0
  4. Any real number.
  1. The value of c is:
  1. $1$
  2. $\frac{1}{2}$
  3. $-1$
  4. $-\frac{1}{2}$
  1. The value of a + c is:
  1. 1
  2. 0
  3. -1
  4. -2
  1. The value of c - a is:
  1. 1
  2. 0
  3. -1
  4. 2
✓

Answer

$\text{L.H.L.}(\text{at x})=\lim\limits_{\text{x}\rightarrow0}\frac{\sin(\text{a}+1)\text{x}+\sin\text{x}}{\text{x}}\Big(\frac{0}{0}\text{ form}\Big)$
Using L' Hospital rule, we get
$\text{L.H.L.} (\text{at x} = 0)$
$=\lim\limits_{\text{x}\rightarrow0}(\text{a}+1)\cos(\text{a}+1)\text{x}+\cos\text{x}=\text{a}+2$
$\text{R.H.L.} (\text{at x} = 0)=\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{\text{x}+\text{bx}^2}-\sqrt{\text{x}}}{\text{bx}^\frac{3}{2}}=\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+\text{bx}}-1}{\text{bx}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{1}{\sqrt{1+\text{bx}}+1}=\frac{1}{2}$
Since,f(x) is continuous at x = 0.
$\therefore$ From (i) and (ii), we get
$\text{a}+2=\text{c}=\frac{1}{2}\Rightarrow\text{a}=-\frac{3}{2},\text{c}=\frac{1}{2}$
Also, value of b does not affect the continuity of f(x), so b can be any real number.
  1. (a) $-\frac{3}{2}$
  1. (d) Any real number.
  1. (b) $\frac{1}{2}$
  1. (c) -1
Solution:

$\text{a}+\text{c}=-\frac{3}{2}+\frac{1}{2}=-1$
  1. (d) 2
Solution:

$\text{c}-\text{a}=\frac{1}{2}+\frac{3}{2}=2$

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