In the following, determine the value(s) of constant(s) involved in the definition so that the given function is continuous:

Solution:

(i) Given: $f(x)=\left\{\begin{array}{l}\frac{\sin 2 x}{5 x}, \text { if } x \neq 0 \\ 3 k, \text { if } x=0\end{array}\right.$

If $f(x)$ is continuous at $x=0$, then

$\lim _{x \rightarrow 0} f(x)=f(0)$

$\Rightarrow \lim _{x \rightarrow 0} \frac{\sin 2 x}{5 x}=f(0)$

$\Rightarrow \lim _{x \rightarrow 0} \frac{2 \sin 2 x}{2 \times 5 x}=f(0)$

$\Rightarrow \frac{2}{5} \lim _{x \rightarrow 0} \frac{\sin 2 x}{2 x}=f(0)$

$\Rightarrow \frac{2}{5}=3 k$

$\Rightarrow k=\frac{2}{15}$

(ii) Given: $f(x)=\left\{\begin{array}{l}k x+5, \text { if } x \leq 2 \\ x-1, \text { if } x>2\end{array}\right.$

If $f(x)$ is continuous at $x=2$, then

$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x)$

$\Rightarrow \lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0} f(2+h)$

$\Rightarrow \lim _{h \rightarrow 0}(k(2-h)+5)=\lim _{h \rightarrow 0}(2+h-1)$

$\Rightarrow 2 k+5=1$

$\Rightarrow 2 k=-4$

$\Rightarrow k=-2$

(iii) Given: $f(x)=\left\{\begin{array}{l}k\left(x^{2}+3 x\right), \text { if } x<0 \\ \cos 2 x, \text { if } x \geq 0\end{array}\right.$

If $f(x)$ is continuous at $x=0$, then

$\Rightarrow \lim _{h \rightarrow 0} f(-h)=\lim _{h \rightarrow 0} f(h)$ $\Rightarrow \lim _{h \rightarrow 0}\left(k\left((-h)^{2}-3 h\right)\right)=\lim _{h \rightarrow 0}(\cos 2 h$ $\Rightarrow 0=1 \quad[$ It is not possible $]$

Hence, there does not exist any value of k, which can make the given function continuous.

(iv) Given: $f(x)=\left\{\begin{aligned} 2, & \text { if } x \leq 3 \\ a x+b, & \text { if } 3

If $f(x)$ is continuous at $x=3$ and 5 , then

$\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{+}} f(x)$ and $\lim _{x \rightarrow 5^{-}} f(x)=\lim _{x \rightarrow 5^{+}} f(x)$

$\Rightarrow \lim _{h \rightarrow 0} f(3-h)=\lim _{h \rightarrow 0} f(3+h)$ and $\lim _{h \rightarrow 0} f(5-h)=\lim _{h \rightarrow 0} f(5+h)$

$\Rightarrow \lim _{h \rightarrow 0}(2)=\lim _{h \rightarrow 0}(a(3+h)+b)$ and $\lim _{h \rightarrow 0}(a(5-h)+b)=\lim _{h \rightarrow 0}(9)$

$\Rightarrow 2=3 a+b$ and $5 a+b=9$

$\Rightarrow 2=3 a+b$ and $5 a+b=9$

$\Rightarrow a=\frac{7}{2}$ and $b=\frac{-17}{2}$

(v)

Given: $f(x)=\left\{\begin{array}{c}4, \text { if } x \leq-1 \\ a x^{2}+b, \text { if }-1

If $f(x)$ is continuous at $x=-1$ and 0 , then

$\lim _{x \rightarrow-1^{-}} f(x)=\lim _{x \rightarrow-1^{+}} f(x)$ and $\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)$

$\Rightarrow \lim _{h \rightarrow 0} f(-1-h)=\lim _{h \rightarrow 0} f(-1+h)$ and $\lim _{h \rightarrow 0} f(-h)=\lim _{h \rightarrow 0} f(h)$

$\Rightarrow \lim _{h \rightarrow 0}(4)=\lim _{h \rightarrow 0}\left(a(-1+h)^{2}+b\right)$ and $\lim _{h \rightarrow 0}\left(a(-h)^{2}+b\right)=\lim _{h \rightarrow 0}(\cos h)$

$\Rightarrow 4=a+b$ and $b=1$

$\Rightarrow a=3$ and $b=1$

(vi)

Given: $f(x)=\left\{\begin{array}{c}\frac{\sqrt{1+p x}-\sqrt{1-p x}}{x}, \text { if }-1 \leq x<0 \\ \frac{2 x+1}{x-2}, \text { if } 0 \leq x \leq 1\end{array}\right.$

If $f(x)$ is continuous at $x=0$, then

$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)$

$\Rightarrow \lim _{h \rightarrow 0} f(-h)=\lim _{h \rightarrow 0} f(h)$

$\Rightarrow \lim _{h \rightarrow 0}\left(\frac{\sqrt{1-p h}-\sqrt{1+p h}}{-h}\right)=\lim _{h \rightarrow 0}\left(\frac{2 h+1}{h-2}\right)$

$\Rightarrow \lim _{h \rightarrow 0}\left(\frac{(\sqrt{1-p h}-\sqrt{1+p h})(\sqrt{1-p h}+\sqrt{1+p h})}{-h(\sqrt{1-p h}+\sqrt{1+p h})}\right)=\lim _{h \rightarrow 0}\left(\frac{2 h+1}{h-2}\right)$

$\Rightarrow \lim _{h \rightarrow 0}\left(\frac{(1-p h-1-p h)}{-h(\sqrt{1-p h}+\sqrt{1+p h})}\right)=\lim _{h \rightarrow 0}\left(\frac{2 h+1}{h-2}\right)$

$\Rightarrow \lim _{h \rightarrow 0}\left(\frac{(-2 p h)}{-h(\sqrt{1-p h}+\sqrt{1+p h})}\right)=\lim _{h \rightarrow 0}\left(\frac{2 h+1}{h-2}\right)$

$\Rightarrow \lim _{h \rightarrow 0}\left(\frac{(2 p)}{(\sqrt{1-p h}+\sqrt{1+p h})}\right)=\lim _{h \rightarrow 0}\left(\frac{2 h+1}{h-2}\right)$

$\Rightarrow\left(\frac{(2 p)}{(2)}\right)=\left(\frac{1}{-2}\right)$

$\Rightarrow p=\frac{-1}{2}$

(vii)

Given: $f(x)=\left\{\begin{aligned} 5, & \text { if } x \leq 2 \\ a x+b, & \text { if } 2

If $f(x)$ is continuous at $x=2$ and 10 , then

$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x)$ and $\lim _{x \rightarrow 10^{-}} f(x)=\lim _{x \rightarrow 10^{+}} f(x)$

$\Rightarrow \lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0} f(2+h)$ and $\lim _{h \rightarrow 0} f(10-h)=\lim _{h \rightarrow 0} f(10+h)$

$\Rightarrow \lim _{h \rightarrow 0}(5)=\lim _{h \rightarrow 0}[a(2+h)+b]$ and $\lim _{h \rightarrow 0}[a(10-h)+b]=\lim _{h \rightarrow 0}(21)$

$\Rightarrow 5=2 a+b \quad \ldots(1) \quad$ and $\quad 10 a+b=21 \quad \ldots(2)$

On solving eqs. (1) and (2), we get

$a=2$ and $b=1$

(viii)

Given: $f(x)=\left\{\begin{array}{cc}\frac{k \cos x}{\pi-2 \mathrm{x}}, & x<\frac{\pi}{2} \\ 3, & x=\frac{\pi}{2} \\ \frac{3 \tan 2 x}{2 x-\pi}, & x>\frac{\pi}{2}\end{array}\right.$

If $f(x)$ is continuous at $x=\frac{\pi}{2}$, then

$\lim _{x \rightarrow \frac{\pi}{2}^{-}} f(x)=f\left(\frac{\pi}{2}\right)$

$\Rightarrow \lim _{h \rightarrow 0} f\left(\frac{\pi}{2}-h\right)=f\left(\frac{\pi}{2}\right)$

$\Rightarrow \lim _{h \rightarrow 0} f\left(\frac{\pi}{2}-h\right)=3$

$\Rightarrow \lim _{h \rightarrow 0}\left[\frac{k \cos \left(\frac{x}{2}-h\right)}{\pi-2\left(\frac{\pi}{2}-\mathrm{h}\right)}\right]=1$

$\Rightarrow \lim _{h \rightarrow 0}\left[\frac{k \sin h}{\pi-\pi+2 \mathrm{~h}}\right]=1$

$\Rightarrow \lim _{h \rightarrow 0}\left[\frac{k \sin h}{2 \mathrm{~h}}\right]=1$

$\Rightarrow \frac{k}{2} \lim _{h \rightarrow 0}\left[\frac{\sin h}{\mathrm{~h}}\right]=1$

$\Rightarrow \frac{k}{2}=1$

$\Rightarrow k=2$