Solve this
Question:

If $f(x)=\left\{\begin{array}{ll}\frac{x^{2}}{2}, & \text { if } 0 \leq x \leq 1 \\ 2 x^{2}-3 x+\frac{3}{2}, & \text { if } 1<x \leq 2\end{array} .\right.$ Show that $f$ is continuous at $x=1 .$

Solution:

Given: $f(x)=\left\{\begin{array}{l}\frac{x^{2}}{2}, \text { if } 0 \leq \mathrm{x} \leq 1 \\ 2 x^{2}-3 x+\frac{3}{2}, \text { if } 1<\mathrm{x} \leq 2\end{array}\right.$

We have

$(\mathrm{LHL}$ at $x=1)=\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0} \frac{(1-h)^{2}}{2}=\frac{1}{2}$

$(\mathrm{RHL}$ at $x=1)=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h)=\lim _{h \rightarrow 0}\left[2(1+h)^{2}-3(1+h)+\frac{3}{2}\right]=2-3+\frac{3}{2}=\frac{1}{2}$

Also, $f(1)=\frac{(1)^{2}}{2}=\frac{1}{2}$

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

Hence, the given function is continuous at $x=1$.

Administrator
Solve this
Question:

(i) $\sin ^{-1} \frac{63}{65}=\sin ^{-1} \frac{5}{13}+\cos ^{-1} \frac{3}{5}$

(ii) $\sin ^{-1} \frac{\frac{55}{5}}{13}+\cos ^{-1} \frac{13}{5}=\tan ^{-1} \frac{\frac{63}{16}}{16}$

(iii) $\frac{9 \pi}{8}-\frac{9}{4} \sin ^{-1} \frac{1}{3}=\frac{9}{4} \sin ^{-1} \frac{2 \sqrt{2}}{3}$

Solution:

(i) 

RHS

$\sin ^{-1} \frac{5}{13}+\cos ^{-1} \frac{3}{5}$

$=\sin ^{-1} \frac{5}{13}+\sin ^{-1} \frac{4}{5} \quad\left[\because \cos ^{-1} x=\sin ^{-1} \sqrt{1-x^{2}}\right]$

$=\sin ^{-1}\left\{\frac{5}{13} \sqrt{1-\left(\frac{4}{5}\right)^{2}}+\frac{4}{5} \sqrt{1-\left(\frac{5}{13}\right)^{2}}\right\}$

$=\sin ^{-1}\left\{\frac{5}{13} \times \frac{3}{5}+\frac{4}{5} \times \frac{12}{13}\right\}$

$=\sin ^{-1}\left\{\frac{15}{65}+\frac{48}{65}\right\}$

$=\sin ^{-1} \frac{63}{65}=$ LHS


(ii)

LHS $=\sin ^{-1} \frac{5}{13}+\cos ^{-1} \frac{3}{5}$

$=\sin ^{-1} \frac{5}{13}+\cos ^{-1} \frac{3}{5}$

$=\sin ^{-1} \frac{5}{13}+\sin ^{-1} \sqrt{1-\left(\frac{3}{5}\right)^{2}} \quad\left[\because \sin ^{-1} x=\cos ^{-1} \sqrt{1-x^{2}}\right]$

$=\sin ^{-1} \frac{5}{13}+\sin ^{-1} \frac{4}{5}$

$=\sin ^{-1}\left[\frac{5}{13} \sqrt{1-\left(\frac{4}{5}\right)^{2}}+\frac{4}{5} \sqrt{1-\left(\frac{5}{13}\right)^{2}}\right]$                              $\left[\because \sin ^{-1} x+\sin ^{-1} y=\sin ^{-1}\left(x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}\right)\right]$

$=\sin ^{-1}\left(\frac{5}{13} \times \frac{3}{5}+\frac{4}{5} \times \frac{12}{13}\right)$

$=\sin ^{-1}\left(\frac{3}{13}+\frac{48}{65}\right)$

$=\sin ^{-1}\left(\frac{63}{65}\right)$

$=\tan ^{-1}\left(\frac{\frac{63}{65}}{\sqrt{1-\frac{63^{2}}{65}}}\right) \quad\left[\because \sin ^{-1} x=\tan ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right)\right]$

$=\tan ^{-1}\left(\frac{\frac{63}{85}}{\frac{10}{65}}\right)$

$=\tan ^{-1}\left(\frac{63}{16}\right)=$ RHS


(iii)

$\frac{9 \pi}{8}-\frac{9}{4} \sin ^{-1} \frac{1}{3}=\frac{9}{4} \sin ^{-1} \frac{2 \sqrt{2}}{3}$

LHS $=\frac{9 \pi}{8}-\frac{9}{4} \sin ^{-1} \frac{1}{3}$

$=\frac{9}{4}\left(\frac{\pi}{2}-\sin ^{-1} \frac{1}{3}\right)$

$=\frac{9}{4}\left(\cos ^{-1} \frac{1}{3}\right)$

$=\frac{9}{4}\left(\sin ^{-1} \sqrt{1-\frac{1}{9}}\right)$

$=\frac{9}{4}\left(\sin ^{-1} \frac{2 \sqrt{2}}{3}\right)=\mathrm{RHS}$

 

 

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