Find a and b if the following function is continuous at the point…

Question

Find $a$ and $b$ if the following function is continuous at the point or on the interval indicated against them: $\begin{aligned} & =\frac{4 \tan x+5 \sin x}{a^2-1}, & & \text { for } x<0 \\ f(x) & =\frac{9}{\log 2}, & & \text { for } x=0 \\ & =\frac{11 x+7 x-\cos x}{b^2-1}, & & \text { for } x>0 \end{aligned}$

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Answer

\begin{aligned}
& f(x) \text { is continuous at } x=0 \\
& \therefore \lim _{x \rightarrow 0} f(x)=f(0) \\
& \therefore \lim _{x \rightarrow 0}\left[\frac{4 \tan x+5 \sin x}{a^x-1}\right]=\frac{9}{\log 2} \\
& \therefore \lim _{x \rightarrow 0}\left[\frac{\frac{4 \tan x+5 \sin x}{x}}{\frac{a^x-1}{x}}\right] \ldots[\because x \rightarrow 0, x \neq 0] \\
& \left.=\frac{9}{\log 2}\right] \frac{\lim _{x \rightarrow 0}\left(\frac{4 \tan x}{x}+\frac{5 \sin x}{x}\right)}{\lim _{x \rightarrow 0} \frac{a^x-1}{x}}=\frac{9}{\log 2} \\
& \therefore \frac{4 \lim _{x \rightarrow 0} \frac{\tan x}{x}+5 \lim _{x \rightarrow 0} \frac{\sin x}{x}}{\lim _{x \rightarrow 0} \frac{a^x-1}{x}}=\frac{9}{\log 2} \\
& \therefore \frac{4(1)+5(1)}{\log a}=\frac{9}{\log 2} \cdots\left[\because \lim _{x \rightarrow 0} \frac{a^x-1}{x}=\log a\right] \\
& \therefore \frac{9}{\log a}=\frac{9}{\log 2} \\
& \therefore \log a=\log 2
\end{aligned}
\begin{aligned}
& \therefore \mathrm{a}=2 \\
& \text { Also } \lim _{x \rightarrow 0^{+}} f(x)=\mathrm{f}(0) \\
& \therefore \lim _{x \rightarrow 0} \frac{11 x+7 x \cdot \cos x}{\mathrm{~b}^x-1}=\frac{9}{\log 2} \\
& \therefore \lim _{x \rightarrow 0} \frac{\frac{11 x+7 x \cos x}{x}}{\frac{\mathrm{b}^x-1}{x}}=\frac{9}{\log 2} \ldots[\because x \rightarrow 0, x \neq 0] \\
& \therefore \frac{\lim _{x \rightarrow 0}(11+7 \cos x)}{\lim _{x \rightarrow 0}\left(\frac{\mathrm{b}^x-1}{x}\right)}=\frac{9}{\log 2} \\
& \therefore \frac{11+7 \cos 0}{\log \mathrm{b}}=\frac{9}{\log 2} \ldots\left[\because \lim _{x \rightarrow 0} \frac{\mathrm{a}^x-1}{x}=\log \mathrm{a}\right] \\
& \therefore \frac{11+7(1)}{\log \mathrm{b}}=\frac{9}{\log 2} \\
& \therefore 9 \log \mathrm{b}=18 \log 2 \\
& \therefore \log \mathrm{b}=2 \log 2 \\
& =\log (2)^2 \\
& \therefore \log \mathrm{b}=\log 4 \\
& \therefore \mathrm{b}=4 \\
& \therefore a=2 and b=4
\end{aligned}