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Chapter 2: Problem 30
For \(g(x)\) to be continuous, then \(\mathrm{a}>\) range of \(\mathrm{f}(\mathrm{x})\) \(\Rightarrow \mathrm{a}>\sqrt{26}\) least positive integral value of \(a=3\) Hence, B is correct.
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\(\lim _{n \rightarrow \infty} \lim _{x \rightarrow 0} \cos ^{2 n} x=0\) \(\lim _{n \rightarrow \infty} \lim _{x \rightarrow 0} \sqrt[n]{1+x^{n}}=1\) \(\lim _{n \rightarrow \infty} \lim _{x \rightarrow 1} \sqrt[n]{1+x^{n}}=1\) \(\lim _{n \rightarrow \infty} \lim _{x \rightarrow 1} \frac{1}{1+x^{n}}=0\)
\(\lim _{x \rightarrow 0} \frac{\left(3^{x}-1\right)^{2}}{(\sin x) \ln (1+x)}\) $=\lim _{x \rightarrow 0} \frac{\left(3^{x}-1\right)^{2}}{x^{2}} \times \frac{x}{\sin x} \times \frac{x}{\ln (1+x)}$ \(=(\ln 3)^{2}\) Hence, \(\mathrm{C}\) is correct.
$f(x)=\frac{\log _{\sin x} \cos ^{3} x}{\log _{\sin |3 x|} \cos \frac{x}{2}}, x \in\left(-\frac{\pi}{3}, \frac{\pi}{3}\right), \quad x \neq 0$ $=\frac{3 \log (1+\cos x-1)}{\log \left(1+\cos \frac{x}{2}-1\right)} \times \frac{\log (1+\sin |3 x|-1)}{\log (1+\sin |x|-1)}$ \(f(x)\) is discontinuous at \(x=\pm \frac{\pi}{6}\)
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.
$\lim _{x \rightarrow 0} \frac{\sin ^{4}\left(\frac{1}{x}\right)-\sin ^{2}\left(\frac{1}{x}\right)+1}{\cos ^{4}\left(\frac{1}{x}\right)-\cos ^{2}\left(\frac{1}{x}\right)+1}$ $\lim _{x \rightarrow 0} \frac{-\sin ^{2}\left(\frac{1}{x}\right) \cos ^{2}\left(\frac{1}{x}\right)+1}{-\cos ^{2}\left(\frac{1}{x}\right) \sin ^{2}\left(\frac{1}{x}\right)+1}=1$
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