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Differentiation of Derivatives — Applied Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceApplied MathsDifferentiation of Derivatives5 Marks
Question
If
$f(x)=\left\{\begin{array}{cl}\frac{\sin (a+1) x+2 \sin x}{x}, & x<0 \\ 2, & x=0 \\ \frac{\sqrt{1+b x}-1}{x}, & x>0\end{array}\right.$
is continuous at $x=0$, then find the values of $a$ and $b$.
✓

Answer

We have given that $f(x)$ is continuous at $x=0$
$
\therefore \quad \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)
$
$=f(0)=2\quad\ldots(i) $
Now,
$
\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)
$
Considering $x=0-h$ or $x \rightarrow 0^{-}$or $h \rightarrow 0$ 
Thus,
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(h)$
$\begin{array}{l}=\lim _{h \rightarrow 0} \frac{\sin (a+1)(-h)+2 \sin (-h)}{-h} \\ =\lim _{h \rightarrow 0} \frac{-\sin (a+1) h-2 \sin h}{-h} \\ =\lim _{h \rightarrow 0} \frac{\sin (a+1) h}{h}+\lim _{h \rightarrow 0} \frac{2 \sin h}{h}\end{array}$
$\begin{array}{r}=(a+1) \lim _{h \rightarrow 0} \frac{\sin (a+1) h}{(a+1) h}+2 \lim _{h \rightarrow 0} \frac{\sin h}{h}\ (x \neq 0)\end{array}$
$\begin{array}{l}=a+1+2 \\ =a+3\end{array}$
Now using eqn. (ii).
$
\begin{array}{l}
\lim _{x \rightarrow 0^{-}} f(x)=2 \\
\text { or } \quad a+3=2 \\
\text { or } \quad a=-1 \\
\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h) \\
=\lim _{h \rightarrow 0} f(h) \\
\text { Considering } x=0+h, x \rightarrow 0^{+} \text {or } h \rightarrow 0
\end{array}
$
Now,
$
\lim _{h \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} \frac{\sqrt{1+b h}-1}{h}
$
$\begin{array}{l}=\lim _{h \rightarrow 0} \frac{\sqrt{1+b h}-1}{h} \times \frac{\sqrt{1+b h}+1}{\sqrt{1+b h}+1} \\ =\lim _{h \rightarrow 0}\left[\frac{(\sqrt{1+b h})^2-(1)^2}{h(\sqrt{1+b h}+1)}\right] \\ =\lim _{h \rightarrow 0} \frac{1+b h-1}{h(\sqrt{1+b h}+1)}\end{array}$
$=\lim _{h \rightarrow 0} \frac{b h}{h(\sqrt{1+b h}+1)} \quad (h \neq 0)$
$\begin{array}{l}=\lim _{h \rightarrow 0} \frac{b}{\sqrt{1+b h}+1} \\ =\frac{b}{\sqrt{1+b \times 0+1}} \\ =\frac{b}{1+1}=\frac{b}{2}\end{array}$
Now, using eqn. (i),
$
\lim _{x \rightarrow 0^{+}} f(x)=f(0)=2
$
or $\frac{b}{2}=2$
or $\quad b=4$.

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