In each of the following, find the value of the constant k so that the given function is continuous at the indicated point;

(i) Given:

$f(x)= \begin{cases}\frac{1-\cos 2 k x}{x^{2}}, & \text { if } x \neq 0 \\ 8 & , \text { if } x=0\end{cases}$

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

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

$\Rightarrow \lim _{x \rightarrow 0} \frac{1-\cos 2 k x}{x^{2}}=8$

$\Rightarrow \lim _{x \rightarrow 0} \frac{2 k^{2} \sin ^{2} k x}{k^{2} x^{2}}=8$

$\Rightarrow 2 k^{2} \lim _{x \rightarrow 0}\left(\frac{\sin k x}{k x}\right)^{2}=8$

$\Rightarrow 2 k^{2} \times 1=8$

$\Rightarrow k^{2}=4$

$\Rightarrow k=\pm 2$

(ii) Given:

$f(x)=\left\{\begin{array}{l}(x-1) \tan \frac{\pi x}{2}, \text { if } x \neq 1 \\ k, \text { if } x=1\end{array}\right.$

If f(x) is continuous at x = 1, then

$\lim _{x \rightarrow 1} f(x)=f(1)$

$\Rightarrow \lim _{x \rightarrow 1}(x-1) \tan \frac{\pi x}{2}=k$

Putting $x-1=y$, we get

$\lim _{y \rightarrow 0} y \tan \frac{\pi(y+1)}{2}=k$

$\Rightarrow \lim _{y \rightarrow 0} y \tan \left(\frac{\pi y}{2}+\frac{\pi}{2}\right)=k$

$\Rightarrow \lim _{y \rightarrow 0} y \tan \left(\frac{\pi}{2}+\frac{\pi y}{2}\right)=k$

$\Rightarrow-\lim _{y \rightarrow 0} y \cot \left(\frac{\pi y}{2}\right)=k$

$\Rightarrow \frac{-2}{\pi} \lim _{y \rightarrow 0} \frac{\frac{\pi y}{2} \cos \left(\frac{x y}{2}\right)}{\sin \left(\frac{y y}{2}\right)}=k$

$\Rightarrow \frac{-2}{\pi} \frac{\lim _{y \rightarrow 0} \cos \left(\frac{x y}{2}\right)}{\lim _{y \rightarrow 0}\left(\frac{\sin \left(\frac{x y}{2}\right)}{\frac{x y}{2}}\right)}=k$

$\Rightarrow \frac{-2}{\pi} \times \frac{1}{1}=k$

$\Rightarrow k=\frac{-2}{\pi}$

(iii) Given:

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

We have

$(\mathrm{LHL}$ at $x=0)=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(-h)=\lim _{h \rightarrow 0} k\left(h^{2}+2 h\right)=0$

(RHL at $x=0)=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(h)=\lim _{h \rightarrow 0} \cos h=1$

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

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

(iv) Given:

$f(x)=\left\{\begin{array}{l}k x+1, \text { if } x \leq \pi \\ \cos x, \text { if } x>\pi\end{array}\right.$

We have

$($ LHL at $x=\pi)=\lim _{x \rightarrow \pi^{-}} f(x)=\lim _{h \rightarrow 0} f(\pi-h)=\lim _{h \rightarrow 0} k(\pi-\mathrm{h})+1=k \pi+1$

$(\mathrm{RHL}$ at $x=\pi)=\lim _{x \rightarrow \pi^{+}} f(x)=\lim _{h \rightarrow 0} f(\pi+h)=\lim _{h \rightarrow 0} \cos (\pi+h)=\cos \pi=-1$

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

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

$\Rightarrow k \pi+1=-1$

$\Rightarrow \mathrm{k}=\frac{-2}{\pi}$

(v) Given:

$f(x)= \begin{cases}k x+1, & \text { if } x \leq 5 \\ 3 x-5, & \text { if } x>5\end{cases}$

We have

$($ LHL at $x=5)=\lim _{x \rightarrow 5^{-}} f(x)=\lim _{h \rightarrow 0} f(5-h)=\lim _{h \rightarrow 0} k(5-\mathrm{h})+1=5 k+1$

(RHL at $x=5$ ) $=\lim _{x \rightarrow 5^{+}} f(x)=\lim _{h \rightarrow 0} f(5+h)=\lim _{h \rightarrow 0} 3(5+\mathrm{h})-5=10$

If f(x) is continuous at x = 5, then

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

$\Rightarrow 5 k+1=10$

$\Rightarrow k=\frac{9}{5}$

(vi) Given:

$f(x)=\left\{\begin{array}{l}\frac{x^{2}-25}{x-5}, \quad x \neq 5 \\ k, \quad x=5\end{array}\right.$

$\Rightarrow f(x)=\left\{\begin{array}{l}\frac{(x-5)(x+5)}{x-5}, \quad x \neq 5 \\ k, \quad x=5\end{array}\right.$

$\Rightarrow f(x)=\left\{\begin{array}{l}x+5, \quad x \neq 5 \\ k, \quad x=5\end{array}\right.$

If f(x) is continuous at x = 5, then

$\lim _{x \rightarrow 5} f(x)=f(5)$

$\Rightarrow \lim _{x \rightarrow 5}(x+5)=k$

$\Rightarrow k=5+5=10$

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

We have

(LHL at $x=1$ ) $=\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0} 4=4$

(RHL at $x=1$ ) $=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h)=\lim _{h \rightarrow 0} k(1+\mathrm{h})^{2}=k$

If f(x) is continuous at x = 1, then

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

$\Rightarrow k=4$

(viii) Given:

$f(x)=\left\{\begin{array}{l}k\left(x^{2}+2\right), \text { if } x \leq 0 \\ 3 x+1, \text { if } x>0\end{array}\right.$

We have

$($ LHL at $x=0)=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} k\left((-h)^{2}+2\right)=2 k$

$($ RHL at $x=0)=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} 3 h+1=1$

If f(x) is continuous at x = 0, then

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

$\Rightarrow 2 k=1$

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

(ix) Given:

$f(x)=\left\{\begin{array}{l}\frac{x^{3}+x^{2}-16 x+20}{(x-2)^{2}}, \quad x \neq 2 \\ k, \quad x=2\end{array}\right.$

$\Rightarrow f(x)=\left\{\begin{array}{l}\frac{x^{3}+x^{2}-16 x+20}{x^{2}-4 x+4}, \quad x \neq 2 \\ k, x=2\end{array}\right.$

$\Rightarrow f(x)= \begin{cases}x+5, & x \neq 2 \\ k, & x=2\end{cases}$

If f(x) is continuous at x = 2, then

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

$\Rightarrow \lim _{x \rightarrow 2}(x+5)=k$

$\Rightarrow k=2+5=7$