Question

In the following, determine the value(s) of constant(s) involved in the definition so that the given function is continuous:
(i) $f\left(x\right)=\left\{\begin{array}{ll}\frac{\sin 2x}{5x},& \mathrm{if}x\ne 0\\ 3k,& \mathrm{if}x=0\end{array}\right.$

(ii) $f\left(x\right)=\left\{\begin{array}{ll}kx+5,& \mathrm{if}x\le 2\\ x-1,& \mathrm{if}x>2\end{array}\right.$

(iii) $f\left(x\right)=\left\{\begin{array}{ll}k(x^2+3x),& \mathrm{if}x<0\\ \cos 2x,& \mathrm{if}x\ge 0\end{array}\right.$

(iv) $f\left(x\right)=\left\{\begin{array}{cc}2,& \mathrm{if}x\le 3\\ ax+b,& \mathrm{if}3<x<5\\ 9,& \mathrm{if}x\ge 5\end{array}\right.$

(v) $f\left(x\right)=\left\{\begin{array}{cc}4,& \mathrm{if}x\le -1\\ a{x}^2+b,& if-1<x<0\\ \cos x,& ifx\ge 0\end{array}\right.$

(vi) $f\left(x\right)=\left\{\begin{array}{ll}\frac{\sqrt{1+px}-\sqrt{1-px}}{x},& \mathrm{if}-1\le x<0\\ \frac{2x+1}{x-2},& \mathrm{if}0\le x\le 1\end{array}\right.$

(vii) $f\left(x\right)=\left\{\begin{array}{ccc}5,& \mathrm{if}& x\le 2\\ ax+b,& \mathrm{if}& 2<x<10\\ 21,& \mathrm{if}& x\ge 10\end{array}\right.$

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

Answer 1

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

If $f\left(x\right)$ is continuous at x = 0, then
$\underset{x\to 0}{\lim }f\left(x\right)=f\left(0\right)$

$$\begin{gathered} \Rightarrow \underset{x\to 0}{\lim }\frac{\sin 2x}{5x}=f\left(0\right) \\ \Rightarrow \underset{x\to 0}{\lim }\frac{2\sin 2x}{2\times 5x}=f\left(0\right) \\ \Rightarrow \frac{2}{5}\underset{x\to 0}{\lim }\frac{\sin 2x}{2x}=f\left(0\right) \\ \Rightarrow \frac{2}{5}=3k \\ \Rightarrow k=\frac{2}{15} \end{gathered}$$

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

If $f\left(x\right)$ is continuous at x = 2, then
$\underset{x\to 2^{-}}{\lim }f\left(x\right)=\underset{x\to 2^{+}}{\lim }f\left(x\right)$

$$\begin{gathered} \Rightarrow \underset{h\to 0}{\lim }f\left(2-h\right)=\underset{h\to 0}{\lim }f\left(2+h\right) \\ \Rightarrow \underset{h\to 0}{\lim }\left(k\left(2-h\right)+5\right)=\underset{h\to 0}{\lim }\left(2+h-1\right) \\ \Rightarrow 2k+5=1 \\ \Rightarrow 2k=-4 \\ \Rightarrow k=-2 \end{gathered}$$

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

If $f\left(x\right)$ is continuous at x = 0, then
$\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }f\left(x\right)$

$$\begin{gathered} \Rightarrow \underset{h\to 0}{\lim }f\left(-h\right)=\underset{h\to 0}{\lim }f\left(h\right) \\ \Rightarrow \underset{h\to 0}{\lim }\left(k\left({\left(-h\right)}^2-3h\right)\right)=\underset{h\to 0}{\lim }\left(\cos 2h\right) \\ \Rightarrow 0=1\left[\mathrm{It}\mathrm{is}\mathrm{not}\mathrm{possible}\right] \end{gathered}$$

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


(iv) Given: $f\left(x\right)=\left\{\begin{array}{c}2,\mathrm{if}x\le 3\\ ax+b,\mathrm{if}3<x<5\\ 9,\mathrm{if}x\ge 5\end{array}\right.$

If $f\left(x\right)$ is continuous at x = 3 and 5, then
$\underset{x\to 3^{-}}{\lim }f\left(x\right)=\underset{x\to 3^{+}}{\lim }f\left(x\right)\mathrm{and}\underset{x\to 5^{-}}{\lim }f\left(x\right)=\underset{x\to 5^{+}}{\lim }f\left(x\right)$

$$\begin{gathered} \Rightarrow \underset{h\to 0}{\lim }f\left(3-h\right)=\underset{h\to 0}{\lim }f\left(3+h\right)\mathrm{and}\underset{h\to 0}{\lim }f\left(5-h\right)=\underset{h\to 0}{\lim }f\left(5+h\right) \\ \Rightarrow \underset{h\to 0}{\lim }\left(2\right)=\underset{h\to 0}{\lim }\left(a\left(3+h\right)+b\right)\mathrm{and}\underset{h\to 0}{\lim }\left(a\left(5-h\right)+b\right)=\underset{h\to 0}{\lim }\left(9\right) \\ \Rightarrow 2=3a+b\mathrm{and}5a+b=9 \\ \Rightarrow 2=3a+b\mathrm{and}5a+b=9 \\ \Rightarrow a=\frac{7}{2}\mathrm{and}b=\frac{-17}{2} \end{gathered}$$

(v)

Given: $f\left(x\right)=\left\{\begin{array}{c}4,\mathrm{if}x\le -1\\ a{x}^2+b,\mathrm{if}-1<x<0\\ \cos x,\mathrm{if}x\ge 0\end{array}\right.$

If $f\left(x\right)$ is continuous at x = −1 and 0, then
$\underset{x\to -1^{-}}{\lim }f\left(x\right)=\underset{x\to -1^{+}}{\lim }f\left(x\right)\mathrm{and}\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }f\left(x\right)$

$$\begin{gathered} \Rightarrow \underset{h\to 0}{\lim }f\left(-1-h\right)=\underset{h\to 0}{\lim }f\left(-1+h\right)\mathrm{and}\underset{h\to 0}{\lim }f\left(-h\right)=\underset{h\to 0}{\lim }f\left(h\right) \\ \Rightarrow \underset{h\to 0}{\lim }\left(4\right)=\underset{h\to 0}{\lim }\left(a{\left(-1+h\right)}^2+b\right)\mathrm{and}\underset{h\to 0}{\lim }\left(a{\left(-h\right)}^2+b\right)=\underset{\mathrm{h}\to 0}{\lim }\left(\cos h\right) \\ \Rightarrow 4=a+b\mathrm{and}b=1 \\ \Rightarrow a=3\mathrm{and}b=1 \end{gathered}$$

(vi)

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

If $f\left(x\right)$ is continuous at x = 0, then
$\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }f\left(x\right)$

$$\begin{gathered} \Rightarrow \underset{h\to 0}{\lim }f\left(-h\right)=\underset{h\to 0}{\lim }f\left(h\right) \\ \Rightarrow \underset{h\to 0}{\lim }\left(\frac{\sqrt{1-ph}-\sqrt{1+ph}}{-h}\right)=\underset{h\to 0}{\lim }\left(\frac{2h+1}{h-2}\right) \\ \Rightarrow \underset{h\to 0}{\lim }\left(\frac{\left(\sqrt{1-ph}-\sqrt{1+ph}\right)\left(\sqrt{1-ph}+\sqrt{1+ph}\right)}{-h\left(\sqrt{1-ph}+\sqrt{1+ph}\right)}\right)=\underset{h\to 0}{\lim }\left(\frac{2h+1}{h-2}\right) \\ \Rightarrow \underset{h\to 0}{\lim }\left(\frac{\left(1-ph-1-ph\right)}{-h\left(\sqrt{1-ph}+\sqrt{1+ph}\right)}\right)=\underset{h\to 0}{\lim }\left(\frac{2h+1}{h-2}\right) \\ \Rightarrow \underset{h\to 0}{\lim }\left(\frac{\left(-2ph\right)}{-h\left(\sqrt{1-ph}+\sqrt{1+ph}\right)}\right)=\underset{h\to 0}{\lim }\left(\frac{2h+1}{h-2}\right) \\ \Rightarrow \underset{h\to 0}{\lim }\left(\frac{\left(2p\right)}{\left(\sqrt{1-ph}+\sqrt{1+ph}\right)}\right)=\underset{h\to 0}{\lim }\left(\frac{2h+1}{h-2}\right) \\ \Rightarrow \left(\frac{\left(2p\right)}{\left(2\right)}\right)=\left(\frac{1}{-2}\right) \\ \Rightarrow p=\frac{-1}{2} \end{gathered}$$

(vii)

Given: $f\left(x\right)=\left\{\begin{array}{c}5,\mathrm{if}x\le 2\\ ax+b,\mathrm{if}2<x<10\\ 21,\mathrm{if}x\ge 10\end{array}\right.$

If $f\left(x\right)$ is continuous at x = 2 and 10, then
$\underset{x\to 2^{-}}{\lim }f\left(x\right)=\underset{x\to 2^{+}}{\lim }f\left(x\right)\mathrm{and}\underset{x\to {10}^{-}}{\lim }f\left(x\right)=\underset{x\to {10}^{+}}{\lim }f\left(x\right)$

$$\begin{gathered} \Rightarrow \underset{h\to 0}{\lim }f\left(2-h\right)=\underset{h\to 0}{\lim }f\left(2+h\right)\mathrm{and}\underset{h\to 0}{\lim }f\left(10-h\right)=\underset{h\to 0}{\lim }f\left(10+h\right) \\ \Rightarrow \underset{h\to 0}{\lim }\left(5\right)=\underset{h\to 0}{\lim }\left[a\left(2+h\right)+b\right]\mathrm{and}\underset{h\to 0}{\lim }\left[a\left(10-h\right)+b\right]=\underset{h\to 0}{\lim }\left(21\right) \\ \Rightarrow 5=2a+b...\left(1\right)\mathrm{and}10a+b=21...\left(2\right) \\ \mathrm{On}\mathrm{solving}\mathrm{eqs}.\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get} \\ a=2\mathrm{and}b=1 \end{gathered}$$


(viii)

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

If $f\left(x\right)$ is continuous at x = $\frac{\mathrm{\pi }}{2}$, then
$\underset{x\to {\frac{\mathrm{\pi }}{2}}^{-}}{\lim }f\left(x\right)=f\left(\frac{\mathrm{\pi }}{2}\right)$

$$\begin{gathered} \Rightarrow \underset{h\to 0}{\lim }f\left(\frac{\mathrm{\pi }}{2}-h\right)=f\left(\frac{\mathrm{\pi }}{2}\right) \\ \Rightarrow \underset{h\to 0}{\lim }f\left(\frac{\mathrm{\pi }}{2}-h\right)=3 \\ \Rightarrow \underset{h\to 0}{\lim }\left[\frac{k\cos \left({\displaystyle \frac{\mathrm{\pi }}{2}}-h\right)}{\mathrm{\pi }-2\left({\displaystyle \frac{\mathrm{\pi }}{2}}-\mathrm{h}\right)}\right]=1 \\ \Rightarrow \underset{h\to 0}{\lim }\left[\frac{k\sin h}{\mathrm{\pi }-\mathrm{\pi }+2\mathrm{h}}\right]=1 \\ \Rightarrow \underset{h\to 0}{\lim }\left[\frac{k\sin h}{2\mathrm{h}}\right]=1 \\ \Rightarrow \frac{k}{2}\underset{h\to 0}{\lim }\left[\frac{\sin h}{\mathrm{h}}\right]=1 \\ \Rightarrow \frac{k}{2}=1 \\ \Rightarrow k=2 \end{gathered}$$