Question

In each of the following, find the value of the constant k so that the given function is continuous at the indicated point;
(i) $f\left(x\right)=\left\{\begin{array}{ll}\frac{1-\cos 2kx}{x^2},\mathrm{if}& x\ne 0\\ 8,\mathrm{if}& x=0\end{array}\right.$at x = 0

(ii) $f\left(x\right)=\left\{\begin{array}{ll}(x-1)\tan \frac{\mathrm{\pi }x}{2},\mathrm{if}& x\ne 1\\ k,\mathrm{if}& x=1\end{array}\right.$at x = 1

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

(iv) $f\left(x\right)=\left\{\begin{array}{ll}kx+1,\mathrm{if}& x\le \mathrm{\pi }\\ \cos x,\mathrm{if}& x>\mathrm{\pi }\end{array}\right.$at x = π

(v) $f\left(x\right)=\left\{\begin{array}{ll}kx+1,\mathrm{if}& x\le 5\\ 3x-5,\mathrm{if}& x>5\end{array}\right.$at x = 5

(vi) $f\left(x\right)=\left\{\begin{array}{ll}\frac{x^2-25}{x-5},& x\ne 5\\ k,& x=5\end{array}\right.$at x = 5

(vii) $f\left(x\right)=\left\{\begin{array}{ll}k{x}^2,& x\ge 1\\ 4,& x<1\end{array}\right.$at x = 1

(viii) $f\left(x\right)=\left\{\begin{array}{ll}k(x^2+2),\mathrm{if}& x\le 0\\ 3x+1,\mathrm{if}& x>0\end{array}\right.$

Answer 1

(i) Given:
$f\left(x\right)=\left\{\begin{array}{l}\frac{1-\cos 2kx}{x^2},\mathrm{if}x\ne 0\\ 8,\mathrm{if}x=0\end{array}\right.$

If f(x) is continuous at x = 0, then
$$\begin{gathered} \underset{x\to 0}{\lim }f\left(x\right)=f\left(0\right) \\ \Rightarrow \underset{x\to 0}{\lim }\frac{1-\cos 2kx}{x^2}=8 \\ \Rightarrow \underset{x\to 0}{\lim }\frac{2k^2{\sin }^{2}kx}{k^2{x}^2}=8 \\ \Rightarrow 2k^2\underset{x\to 0}{\lim }{\left(\frac{\sin kx}{kx}\right)}^2=8 \\ \Rightarrow 2k^2\times 1=8 \\ \Rightarrow k^2=4 \\ \Rightarrow k=\pm 2 \end{gathered}$$

(ii) Given:
$f\left(x\right)=\left\{\begin{array}{l}\left(x-1\right)\tan \frac{\mathrm{\pi }x}{2},\mathrm{if}x\ne 1\\ k,\mathrm{if}x=1\end{array}\right.$

If f(x) is continuous at x = 1, then
$$\begin{gathered} \underset{x\to 1}{\lim }f\left(x\right)=f\left(1\right) \\ \Rightarrow \underset{x\to 1}{\lim }\left(x-1\right)\tan \frac{\mathrm{\pi }x}{2}=k \end{gathered}$$

$\mathrm{Putting}x-1=y,\mathrm{we}\mathrm{get}$

$$\begin{gathered} \underset{y\to 0}{\lim }y\tan \frac{\mathrm{\pi }\left(\mathrm{y}+1\right)}{2}=k \\ \Rightarrow \underset{y\to 0}{\lim }y\tan \left(\frac{\mathrm{\pi y}}{2}+\frac{\mathrm{\pi }}{2}\right)=k \\ \Rightarrow \underset{y\to 0}{\lim }y\tan \left(\frac{\mathrm{\pi }}{2}+\frac{\mathrm{\pi y}}{2}\right)=k \\ \Rightarrow -\underset{y\to 0}{\lim }y\cot \left(\frac{\mathrm{\pi y}}{2}\right)=k \\ \Rightarrow \frac{-2}{\mathrm{\pi }}\underset{y\to 0}{\lim }\frac{{\displaystyle \frac{\mathrm{\pi y}}{2}}\cos \left({\displaystyle \frac{\mathrm{\pi y}}{2}}\right)}{\sin \left({\displaystyle \frac{\mathrm{\pi y}}{2}}\right)}=k \\ \Rightarrow \frac{-2}{\mathrm{\pi }}\frac{\underset{y\to 0}{\lim }\cos \left({\displaystyle \frac{\mathrm{\pi y}}{2}}\right)}{\underset{y\to 0}{\lim }{\displaystyle \left(\frac{\sin \left({\displaystyle \frac{\mathrm{\pi y}}{2}}\right)}{{\displaystyle \frac{\mathrm{\pi y}}{2}}}\right)}}=k \\ \Rightarrow \frac{-2}{\mathrm{\pi }}\times \frac{1}{{\displaystyle 1}}=k \\ \Rightarrow k=\frac{-2}{\mathrm{\pi }} \end{gathered}$$

(iii) Given:
$f\left(x\right)=\left\{\begin{array}{l}k\left(x^2-2x\right),\mathrm{if}x<0\\ \cos x,\mathrm{if}x\ge 0\end{array}\right.$
We have
(LHL at x = 0) = $\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(0-h\right)=\underset{h\to 0}{\lim }f\left(-h\right)=\underset{h\to 0}{\lim }k\left(h^2+2h\right)=0$
(RHL at x = 0) = $\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(0+h\right)=\underset{h\to 0}{\lim }f\left(h\right)=\underset{h\to 0}{\lim }\cos h=1$

$\therefore \underset{x\to 0^{-}}{\lim }f\left(x\right)\ne \underset{x\to 0^{+}}{\lim }f\left(x\right)$

$\mathrm{Thus},\mathrm{no}\mathrm{value}\mathrm{of}\mathrm{k}\mathrm{exists}\mathrm{for}\mathrm{which}f\left(x\right)\mathrm{is}\mathrm{continuous}\mathrm{at}x=0.$


(iv) Given:
$f\left(x\right)=\left\{\begin{array}{l}kx+1,\mathrm{if}x\le \mathrm{\pi }\\ \cos x,\mathrm{if}x>\mathrm{\pi }\end{array}\right.$
We have
(LHL at x = $\mathrm{\pi }$) = $\underset{x\to {\mathrm{\pi }}^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(\mathrm{\pi }-h\right)=\underset{h\to 0}{\lim }k\left(\mathrm{\pi }-\mathrm{h}\right)+1=k\mathrm{\pi }+1$
(RHL at x = $\mathrm{\pi }$) = $\underset{x\to {\mathrm{\pi }}^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(\mathrm{\pi }+h\right)=\underset{h\to 0}{\lim }\cos \left(\mathrm{\pi }+h\right)=\mathrm{cos\pi }=-1$

If f(x) is continuous at x = $\mathrm{\pi }$, then
$\underset{x\to {\mathrm{\pi }}^{-}}{\lim }f\left(x\right)=\underset{x\to {\mathrm{\pi }}^{+}}{\lim }f\left(x\right)$
$$\begin{gathered} \Rightarrow k\mathrm{\pi }+1=-1 \\ \Rightarrow \mathrm{k}=\frac{-2}{\mathrm{\pi }} \end{gathered}$$

(v) Given:
$f\left(x\right)=\left\{\begin{array}{l}kx+1,\mathrm{if}x\le 5\\ 3x-5,\mathrm{if}x>5\end{array}\right.$

We have
(LHL at x = 5) = $\underset{x\to 5^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(5-h\right)=\underset{h\to 0}{\lim }k\left(5-\mathrm{h}\right)+1=5k+1$
(RHL at x = 5) = $\underset{x\to 5^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(5+h\right)=\underset{h\to 0}{\lim }3\left(5+\mathrm{h}\right)-5=10$

If f(x) is continuous at x = 5, then
$$\begin{gathered} \underset{x\to 5^{-}}{\lim }f\left(x\right)=\underset{x\to 5^{+}}{\lim }f\left(x\right) \\ \Rightarrow 5k+1=10 \\ \Rightarrow k=\frac{9}{5} \end{gathered}$$

(vi) Given:
$f\left(x\right)=\left\{\begin{array}{l}\frac{x^2-25}{x-5},x\ne 5\\ k,x=5\end{array}\right.$
$\Rightarrow f\left(x\right)=\left\{\begin{array}{l}\frac{\left(x-5\right)\left(x+5\right)}{x-5},x\ne 5\\ k,x=5\end{array}\right.$
$\Rightarrow f\left(x\right)=\left\{\begin{array}{l}x+5,x\ne 5\\ k,x=5\end{array}\right.$

If f(x) is continuous at x = 5, then
$$\begin{gathered} \underset{x\to 5}{\lim }f\left(x\right)=f\left(5\right) \\ \Rightarrow \underset{x\to 5}{\lim }\left(x+5\right)=k \\ \Rightarrow k=5+5=10 \end{gathered}$$

(vii) Given: $f\left(x\right)=\left\{\begin{array}{l}k{x}^2,x\ge 1\\ 4,x<1\end{array}\right.$

We have
(LHL at x = 1) = $\underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(1-h\right)=\underset{h\to 0}{\lim }4=4$
(RHL at x = 1) = $\underset{x\to 1^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(1+h\right)=\underset{h\to 0}{\lim }k{\left(1+\mathrm{h}\right)}^2=k$

If f(x) is continuous at x = 1, then
$$\begin{gathered} \underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }f\left(x\right) \\ \Rightarrow k=4 \end{gathered}$$

(viii) Given:
$f\left(x\right)=\left\{\begin{array}{l}k\left(x^2+2\right),\mathrm{if}x\le 0\\ 3x+1,\mathrm{if}x>0\end{array}\right.$
We have
(LHL at x = 0) = $\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(0-h\right)=\underset{h\to 0}{\lim }k\left({\left(-h\right)}^2+2\right)=2k$
(RHL at x = 0) = $\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(0+h\right)=\underset{h\to 0}{\lim }3h+1=1$

If f(x) is continuous at x = 0, then
$$\begin{gathered} \underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }f\left(x\right) \\ \Rightarrow 2k=1 \\ \Rightarrow k=\frac{1}{2} \end{gathered}$$