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Chapter 2: Problem 15
For \(\mathrm{f}(\mathrm{x})\) to be continuous, \(2 x+1=x^{2}-2 x+5\) \(\Rightarrow x^{2}-4 x+4=0\) \(\Rightarrow(x-2)^{2}=0\) \(\Rightarrow \mathrm{x}=2\). Hence, \(f(x)\) is continuous at only \(x=2\).
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$\mathrm{f}(\mathrm{x})= \begin{cases}\sin \mathrm{x}, & \mathrm{x} \in \text { Rational } \\ 1-2 \cos \mathrm{x}, & \mathrm{x} \in \text { Irrational }\end{cases}$ For continuity, \(\sin x=1-2 \cos x\) \(\sin x+2 \cos x=1\) \(\frac{1}{\sqrt{5}} \sin x+\frac{2}{\sqrt{5}} \cos x=\frac{1}{\sqrt{5}}\) \(\sin (x+\alpha)=\frac{1}{\sqrt{5}}\), where \(\cos \alpha=\frac{1}{\sqrt{5}}\) There are infinite values of ' \(x\) ' which satisfies the eqn.
$\lim _{x \rightarrow 1^{\prime}} g(x)=\lim _{x \rightarrow 1^{-} n \rightarrow \infty} \lim _{x \rightarrow \infty} \frac{x+x^{2 n} \sin x}{1+x^{4 n}}$ $\quad=\lim _{x \rightarrow 1^{-}} \lim _{n \rightarrow \infty} \frac{x^{-4 n+1}+x^{-2 n} \sin x}{x^{-4 n}+1}$ \(=0\) $\lim _{x \rightarrow 1} g(x)=\lim _{x \rightarrow 1} \lim _{n \rightarrow \infty} \frac{x+x^{2 n} \sin x}{1+x^{4 n}}=1$
\(f(x)=\cos \left[\frac{\pi}{x}\right] \cos \left(\frac{\pi}{2}(x-1)\right)\) \(L H L=\operatorname{RHL}_{x \rightarrow 0^{-}}=f(0)=0\) $\lim _{x \rightarrow 1^{+}} f(x)=\cos 3, \quad \lim _{x \rightarrow 1^{-}} f(x)=\cos 3, \quad f(1)=\cos 3$ $\lim _{x \rightarrow 2^{\prime}} f(x)=0, \quad \lim _{x \rightarrow 2} f(x)=0, \quad f(2)=0$
$\lim _{x \rightarrow 2} \frac{x-[x]}{x-2}=\lim _{x \rightarrow 2} \frac{\\{x\\}}{\\{x\\}}=1$ So, \(b=1\) $\lim _{x \rightarrow 2} \frac{a\left|x^{2}-x-2\right|}{2+x-x^{2}}=\lim _{x \rightarrow 2} \frac{a\left(2+x-x^{2}\right)}{2+x-x^{2}}$ \(=a\) So, \(a=1\)
If \(x \geq 0\) \(x^{2}-x+1-2=0\) \(x^{2}-x-1=0\) \(x=1 \pm \sqrt{5} \Rightarrow x=\frac{1+\sqrt{5}}{2}\) If \(x<0\) \(x^{2}-x-2=0\) \((x-2)(x+1)=0\) \(\mathrm{x}=2\) or \(-1\) so, \(x=-1\) No options matches.
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