80204
Let \(h(x)=\min \left\{x, x^{2}\right\}\), for every real number of \(x\), then
1 \(h\) is continuous for all \(x\)
2 \(h\) is differentiable for all \(x\)
3 \(\mathrm{h}^{\prime}(\mathrm{x})=2\), for all \(\mathrm{x}>1\)
4 \(h\) is not differentiable at three values of \(x\)
Explanation:
(A) : From the figure it is clear that, \(\mathrm{h}(\mathrm{x})= \begin{cases}\mathrm{x}, \text { if } \mathrm{x} \leq 0 \\ \mathrm{x}^{2}, \text { if } 0\lt \mathrm{x}\lt 1 \\ \mathrm{x}, \text { if } \mathrm{x} \geq 1\end{cases}\) From the graph it is clear that \(\mathrm{h}\) is continuous for all \(\mathrm{x} \in \mathrm{R}, \mathrm{h}^{\prime}(\mathrm{x})=1\) for all \(\mathrm{x}>1\) and \(\mathrm{h}\) is not differentiable at \(\mathrm{x}=0\) and 1 .
BITSAT-2006
Limits, Continuity and Differentiability
80207
The value of \(f(0)\), so that the function \(f(x)=\frac{2 x-\sin ^{-1} x}{2 x+\tan ^{-1} x}\) is continuous at each point in its domain, is
1 \(\frac{1}{3}\)
2 \(-\frac{1}{3}\)
3 \(\frac{2}{3}\)
4 \(-\frac{2}{3}\)
Explanation:
(A) : Since, \(x \sin ^{-1} x, \tan ^{-1} x\) are continuous function, so the function (f) is clearly continuous at each point of its domain except possibly at \(\mathrm{x}=0\). So, for \(\mathrm{f}\) to be continuous at \(\mathrm{x}=0\) \(f(0)=\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{2-\frac{\sin ^{-1} x}{x}}{2+\frac{\tan ^{-1} x}{x}}=\frac{1}{3}\) \(\left[\because \lim _{x \rightarrow 0} \frac{\sin ^{-1} x}{x}=1, \lim _{x \rightarrow 0} \frac{\tan ^{-1} x}{x}=1\right]\)
BITSAT-2017
Limits, Continuity and Differentiability
80210
If \(\left.\begin{array}{rl}\mathbf{f}(\mathbf{x}) & =\sin x, \text { when } x \text { is rational } \\ & =\cos x, \text { when } x \text { is irrational }\end{array}\right\}\) Then the function is
1 discontinuous at \(\mathrm{x}=\pi \mathrm{n}+\pi / 4\)
2 continuous at \(\mathrm{x}=\mathrm{n} \pi+\pi / 4\)
3 discontinuous at all \(\mathrm{x}\)
4 none of these
Explanation:
(B) : The function can be continuous only at those points for which \(\sin \mathrm{x}=\cos \mathrm{x}=\sin \left(\frac{\pi}{2}-\mathrm{x}\right)\) \(\left(\begin{array}{l}\because \sin (\theta)=\sin (\alpha) \\ \Rightarrow \theta=\mathrm{n} \pi \pm \alpha\end{array}\right)\) \(\Rightarrow \mathrm{x}=\mathrm{n} \pi+\frac{\pi}{4}\)
80204
Let \(h(x)=\min \left\{x, x^{2}\right\}\), for every real number of \(x\), then
1 \(h\) is continuous for all \(x\)
2 \(h\) is differentiable for all \(x\)
3 \(\mathrm{h}^{\prime}(\mathrm{x})=2\), for all \(\mathrm{x}>1\)
4 \(h\) is not differentiable at three values of \(x\)
Explanation:
(A) : From the figure it is clear that, \(\mathrm{h}(\mathrm{x})= \begin{cases}\mathrm{x}, \text { if } \mathrm{x} \leq 0 \\ \mathrm{x}^{2}, \text { if } 0\lt \mathrm{x}\lt 1 \\ \mathrm{x}, \text { if } \mathrm{x} \geq 1\end{cases}\) From the graph it is clear that \(\mathrm{h}\) is continuous for all \(\mathrm{x} \in \mathrm{R}, \mathrm{h}^{\prime}(\mathrm{x})=1\) for all \(\mathrm{x}>1\) and \(\mathrm{h}\) is not differentiable at \(\mathrm{x}=0\) and 1 .
BITSAT-2006
Limits, Continuity and Differentiability
80207
The value of \(f(0)\), so that the function \(f(x)=\frac{2 x-\sin ^{-1} x}{2 x+\tan ^{-1} x}\) is continuous at each point in its domain, is
1 \(\frac{1}{3}\)
2 \(-\frac{1}{3}\)
3 \(\frac{2}{3}\)
4 \(-\frac{2}{3}\)
Explanation:
(A) : Since, \(x \sin ^{-1} x, \tan ^{-1} x\) are continuous function, so the function (f) is clearly continuous at each point of its domain except possibly at \(\mathrm{x}=0\). So, for \(\mathrm{f}\) to be continuous at \(\mathrm{x}=0\) \(f(0)=\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{2-\frac{\sin ^{-1} x}{x}}{2+\frac{\tan ^{-1} x}{x}}=\frac{1}{3}\) \(\left[\because \lim _{x \rightarrow 0} \frac{\sin ^{-1} x}{x}=1, \lim _{x \rightarrow 0} \frac{\tan ^{-1} x}{x}=1\right]\)
BITSAT-2017
Limits, Continuity and Differentiability
80210
If \(\left.\begin{array}{rl}\mathbf{f}(\mathbf{x}) & =\sin x, \text { when } x \text { is rational } \\ & =\cos x, \text { when } x \text { is irrational }\end{array}\right\}\) Then the function is
1 discontinuous at \(\mathrm{x}=\pi \mathrm{n}+\pi / 4\)
2 continuous at \(\mathrm{x}=\mathrm{n} \pi+\pi / 4\)
3 discontinuous at all \(\mathrm{x}\)
4 none of these
Explanation:
(B) : The function can be continuous only at those points for which \(\sin \mathrm{x}=\cos \mathrm{x}=\sin \left(\frac{\pi}{2}-\mathrm{x}\right)\) \(\left(\begin{array}{l}\because \sin (\theta)=\sin (\alpha) \\ \Rightarrow \theta=\mathrm{n} \pi \pm \alpha\end{array}\right)\) \(\Rightarrow \mathrm{x}=\mathrm{n} \pi+\frac{\pi}{4}\)
80204
Let \(h(x)=\min \left\{x, x^{2}\right\}\), for every real number of \(x\), then
1 \(h\) is continuous for all \(x\)
2 \(h\) is differentiable for all \(x\)
3 \(\mathrm{h}^{\prime}(\mathrm{x})=2\), for all \(\mathrm{x}>1\)
4 \(h\) is not differentiable at three values of \(x\)
Explanation:
(A) : From the figure it is clear that, \(\mathrm{h}(\mathrm{x})= \begin{cases}\mathrm{x}, \text { if } \mathrm{x} \leq 0 \\ \mathrm{x}^{2}, \text { if } 0\lt \mathrm{x}\lt 1 \\ \mathrm{x}, \text { if } \mathrm{x} \geq 1\end{cases}\) From the graph it is clear that \(\mathrm{h}\) is continuous for all \(\mathrm{x} \in \mathrm{R}, \mathrm{h}^{\prime}(\mathrm{x})=1\) for all \(\mathrm{x}>1\) and \(\mathrm{h}\) is not differentiable at \(\mathrm{x}=0\) and 1 .
BITSAT-2006
Limits, Continuity and Differentiability
80207
The value of \(f(0)\), so that the function \(f(x)=\frac{2 x-\sin ^{-1} x}{2 x+\tan ^{-1} x}\) is continuous at each point in its domain, is
1 \(\frac{1}{3}\)
2 \(-\frac{1}{3}\)
3 \(\frac{2}{3}\)
4 \(-\frac{2}{3}\)
Explanation:
(A) : Since, \(x \sin ^{-1} x, \tan ^{-1} x\) are continuous function, so the function (f) is clearly continuous at each point of its domain except possibly at \(\mathrm{x}=0\). So, for \(\mathrm{f}\) to be continuous at \(\mathrm{x}=0\) \(f(0)=\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{2-\frac{\sin ^{-1} x}{x}}{2+\frac{\tan ^{-1} x}{x}}=\frac{1}{3}\) \(\left[\because \lim _{x \rightarrow 0} \frac{\sin ^{-1} x}{x}=1, \lim _{x \rightarrow 0} \frac{\tan ^{-1} x}{x}=1\right]\)
BITSAT-2017
Limits, Continuity and Differentiability
80210
If \(\left.\begin{array}{rl}\mathbf{f}(\mathbf{x}) & =\sin x, \text { when } x \text { is rational } \\ & =\cos x, \text { when } x \text { is irrational }\end{array}\right\}\) Then the function is
1 discontinuous at \(\mathrm{x}=\pi \mathrm{n}+\pi / 4\)
2 continuous at \(\mathrm{x}=\mathrm{n} \pi+\pi / 4\)
3 discontinuous at all \(\mathrm{x}\)
4 none of these
Explanation:
(B) : The function can be continuous only at those points for which \(\sin \mathrm{x}=\cos \mathrm{x}=\sin \left(\frac{\pi}{2}-\mathrm{x}\right)\) \(\left(\begin{array}{l}\because \sin (\theta)=\sin (\alpha) \\ \Rightarrow \theta=\mathrm{n} \pi \pm \alpha\end{array}\right)\) \(\Rightarrow \mathrm{x}=\mathrm{n} \pi+\frac{\pi}{4}\)
