80233
The set of points, where \(f(x)=\frac{x}{1+|x|}\) is differentiable, is
1 \((-\infty,-1) \cup(-1, \infty)\)
2 \((-\infty, \infty)\)
3 \((0, \infty)\)
4 \((-\infty, 0) \cup(0, \infty)\)
Explanation:
(B) : Given than, \(\quad \mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{1+|\mathrm{x}|}\) we can write this fraction as - \(f(x)=\left\{\begin{array}{lll}\frac{x}{1+x} & \text { if } & x \geq 0 \\ \frac{x}{1-x} & \text { if } & x\lt 0\end{array}\right.\) Looking at \(\mathrm{f}(\mathrm{x})\), we can say that \(\mathrm{f}(\mathrm{x})\) is differentiable at everywhere except at 0 , Now, we need to check its differentiability at \(x=0\) LHD - \(=\lim _{h \rightarrow 0^{-}} \frac{f(0-h)-f(0)}{h}\) \(=\lim _{h \rightarrow 0^{-}} \frac{\frac{-h}{1+h}-0}{-h}=\lim _{h \rightarrow 0^{-}} \frac{1}{1+h}=1\) RHD at \(x=0\), \(=\lim _{h \rightarrow 0^{+}} \frac{f(0+h)-f(0)}{h}\) \(=\lim _{h \rightarrow 0^{+}} \frac{\frac{h}{1+h}-0}{h}=\lim _{h \rightarrow 0^{+}} \frac{1}{1+h}=1\) \(\because\) LHD \(=\) RHD at \(\mathrm{x}=0\) \(\therefore \mathrm{f}\) is differentiable at \(\mathrm{x}=0\) Hence, the set of points where \(\mathrm{f}(\mathrm{x})\) is differentiable is (\(\infty, \infty)\)
AIEEE-2006
Limits, Continuity and Differentiability
80234
Let \(f: R \rightarrow R\) be a function defined by \(f(x)=\) \(\min \{x+1,|x|+1\}\). Then, which one of the following is true?
1 \(f(x) \geq 1\) for all \(x \in R\)
2 \(f(x)\) is not differentiable at \(x=1\)
3 \(f(x)\) is differentiable everywhere
4 \(f(x)\) is not differentiable at \(x=0\)
Explanation:
(C) : Given that, \(\mathrm{f}(\mathrm{x})=\min \{\mathrm{x}+1,|\mathrm{x}|+1\}\) When, \(x\lt 0,(x+1)\) is minimum When, \(x \geq 0,(x+1)\) is minimum \(\mathrm{f}(\mathrm{x})=\) When, \(\mathrm{x}\lt 0,(\mathrm{x}+1)\) is minimum When, \(x \geq 0,(x+1)\) is minimum Overall, \(\mathrm{f}(\mathrm{x})=\mathrm{x}+1\) is differentiable in everywhere
AIEEE-2007
Limits, Continuity and Differentiability
80235
If \(\mathrm{f}:(-1,1) \rightarrow \mathrm{R}\) be a differentiable function with \(f(0)=-1\) and \(f^{\prime}(0)=1\). Let \(g(x)=[f(2 f(x)+\) \(21^{2}\). Then, \(\mathbf{g}^{\prime}(0)\) is equal to
80236
Let \(S\) be the set of all points in \((-\pi, \pi)\) at which the function, \(f(x)=\min \{\sin x, \cos x\}\) is not differentiable. Then, \(S\) is a subset of which of the following?
(C) : We have - \(f(x)=\min \{\sin x, \cos x\}\) Let us draw the graph of \(y=f(x)\), as shown below, clearly, the function - \(\mathrm{f}(\mathrm{x})=\min \{\sin \mathrm{x}, \cos \mathrm{x}\}\) is not differentiable at \(\mathrm{x}=-\frac{3 \pi}{4}\) and \(\frac{\pi}{4}\) So, \(\mathrm{S}=\left\{-\frac{3 \pi}{4}, \frac{\pi}{4}\right\}\) which is a subset of \(\left\{\frac{-3 \pi}{4}, \frac{-\pi}{4}, \frac{3 \pi}{4}, \frac{\pi}{4}\right\}\)
80233
The set of points, where \(f(x)=\frac{x}{1+|x|}\) is differentiable, is
1 \((-\infty,-1) \cup(-1, \infty)\)
2 \((-\infty, \infty)\)
3 \((0, \infty)\)
4 \((-\infty, 0) \cup(0, \infty)\)
Explanation:
(B) : Given than, \(\quad \mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{1+|\mathrm{x}|}\) we can write this fraction as - \(f(x)=\left\{\begin{array}{lll}\frac{x}{1+x} & \text { if } & x \geq 0 \\ \frac{x}{1-x} & \text { if } & x\lt 0\end{array}\right.\) Looking at \(\mathrm{f}(\mathrm{x})\), we can say that \(\mathrm{f}(\mathrm{x})\) is differentiable at everywhere except at 0 , Now, we need to check its differentiability at \(x=0\) LHD - \(=\lim _{h \rightarrow 0^{-}} \frac{f(0-h)-f(0)}{h}\) \(=\lim _{h \rightarrow 0^{-}} \frac{\frac{-h}{1+h}-0}{-h}=\lim _{h \rightarrow 0^{-}} \frac{1}{1+h}=1\) RHD at \(x=0\), \(=\lim _{h \rightarrow 0^{+}} \frac{f(0+h)-f(0)}{h}\) \(=\lim _{h \rightarrow 0^{+}} \frac{\frac{h}{1+h}-0}{h}=\lim _{h \rightarrow 0^{+}} \frac{1}{1+h}=1\) \(\because\) LHD \(=\) RHD at \(\mathrm{x}=0\) \(\therefore \mathrm{f}\) is differentiable at \(\mathrm{x}=0\) Hence, the set of points where \(\mathrm{f}(\mathrm{x})\) is differentiable is (\(\infty, \infty)\)
AIEEE-2006
Limits, Continuity and Differentiability
80234
Let \(f: R \rightarrow R\) be a function defined by \(f(x)=\) \(\min \{x+1,|x|+1\}\). Then, which one of the following is true?
1 \(f(x) \geq 1\) for all \(x \in R\)
2 \(f(x)\) is not differentiable at \(x=1\)
3 \(f(x)\) is differentiable everywhere
4 \(f(x)\) is not differentiable at \(x=0\)
Explanation:
(C) : Given that, \(\mathrm{f}(\mathrm{x})=\min \{\mathrm{x}+1,|\mathrm{x}|+1\}\) When, \(x\lt 0,(x+1)\) is minimum When, \(x \geq 0,(x+1)\) is minimum \(\mathrm{f}(\mathrm{x})=\) When, \(\mathrm{x}\lt 0,(\mathrm{x}+1)\) is minimum When, \(x \geq 0,(x+1)\) is minimum Overall, \(\mathrm{f}(\mathrm{x})=\mathrm{x}+1\) is differentiable in everywhere
AIEEE-2007
Limits, Continuity and Differentiability
80235
If \(\mathrm{f}:(-1,1) \rightarrow \mathrm{R}\) be a differentiable function with \(f(0)=-1\) and \(f^{\prime}(0)=1\). Let \(g(x)=[f(2 f(x)+\) \(21^{2}\). Then, \(\mathbf{g}^{\prime}(0)\) is equal to
80236
Let \(S\) be the set of all points in \((-\pi, \pi)\) at which the function, \(f(x)=\min \{\sin x, \cos x\}\) is not differentiable. Then, \(S\) is a subset of which of the following?
(C) : We have - \(f(x)=\min \{\sin x, \cos x\}\) Let us draw the graph of \(y=f(x)\), as shown below, clearly, the function - \(\mathrm{f}(\mathrm{x})=\min \{\sin \mathrm{x}, \cos \mathrm{x}\}\) is not differentiable at \(\mathrm{x}=-\frac{3 \pi}{4}\) and \(\frac{\pi}{4}\) So, \(\mathrm{S}=\left\{-\frac{3 \pi}{4}, \frac{\pi}{4}\right\}\) which is a subset of \(\left\{\frac{-3 \pi}{4}, \frac{-\pi}{4}, \frac{3 \pi}{4}, \frac{\pi}{4}\right\}\)
80233
The set of points, where \(f(x)=\frac{x}{1+|x|}\) is differentiable, is
1 \((-\infty,-1) \cup(-1, \infty)\)
2 \((-\infty, \infty)\)
3 \((0, \infty)\)
4 \((-\infty, 0) \cup(0, \infty)\)
Explanation:
(B) : Given than, \(\quad \mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{1+|\mathrm{x}|}\) we can write this fraction as - \(f(x)=\left\{\begin{array}{lll}\frac{x}{1+x} & \text { if } & x \geq 0 \\ \frac{x}{1-x} & \text { if } & x\lt 0\end{array}\right.\) Looking at \(\mathrm{f}(\mathrm{x})\), we can say that \(\mathrm{f}(\mathrm{x})\) is differentiable at everywhere except at 0 , Now, we need to check its differentiability at \(x=0\) LHD - \(=\lim _{h \rightarrow 0^{-}} \frac{f(0-h)-f(0)}{h}\) \(=\lim _{h \rightarrow 0^{-}} \frac{\frac{-h}{1+h}-0}{-h}=\lim _{h \rightarrow 0^{-}} \frac{1}{1+h}=1\) RHD at \(x=0\), \(=\lim _{h \rightarrow 0^{+}} \frac{f(0+h)-f(0)}{h}\) \(=\lim _{h \rightarrow 0^{+}} \frac{\frac{h}{1+h}-0}{h}=\lim _{h \rightarrow 0^{+}} \frac{1}{1+h}=1\) \(\because\) LHD \(=\) RHD at \(\mathrm{x}=0\) \(\therefore \mathrm{f}\) is differentiable at \(\mathrm{x}=0\) Hence, the set of points where \(\mathrm{f}(\mathrm{x})\) is differentiable is (\(\infty, \infty)\)
AIEEE-2006
Limits, Continuity and Differentiability
80234
Let \(f: R \rightarrow R\) be a function defined by \(f(x)=\) \(\min \{x+1,|x|+1\}\). Then, which one of the following is true?
1 \(f(x) \geq 1\) for all \(x \in R\)
2 \(f(x)\) is not differentiable at \(x=1\)
3 \(f(x)\) is differentiable everywhere
4 \(f(x)\) is not differentiable at \(x=0\)
Explanation:
(C) : Given that, \(\mathrm{f}(\mathrm{x})=\min \{\mathrm{x}+1,|\mathrm{x}|+1\}\) When, \(x\lt 0,(x+1)\) is minimum When, \(x \geq 0,(x+1)\) is minimum \(\mathrm{f}(\mathrm{x})=\) When, \(\mathrm{x}\lt 0,(\mathrm{x}+1)\) is minimum When, \(x \geq 0,(x+1)\) is minimum Overall, \(\mathrm{f}(\mathrm{x})=\mathrm{x}+1\) is differentiable in everywhere
AIEEE-2007
Limits, Continuity and Differentiability
80235
If \(\mathrm{f}:(-1,1) \rightarrow \mathrm{R}\) be a differentiable function with \(f(0)=-1\) and \(f^{\prime}(0)=1\). Let \(g(x)=[f(2 f(x)+\) \(21^{2}\). Then, \(\mathbf{g}^{\prime}(0)\) is equal to
80236
Let \(S\) be the set of all points in \((-\pi, \pi)\) at which the function, \(f(x)=\min \{\sin x, \cos x\}\) is not differentiable. Then, \(S\) is a subset of which of the following?
(C) : We have - \(f(x)=\min \{\sin x, \cos x\}\) Let us draw the graph of \(y=f(x)\), as shown below, clearly, the function - \(\mathrm{f}(\mathrm{x})=\min \{\sin \mathrm{x}, \cos \mathrm{x}\}\) is not differentiable at \(\mathrm{x}=-\frac{3 \pi}{4}\) and \(\frac{\pi}{4}\) So, \(\mathrm{S}=\left\{-\frac{3 \pi}{4}, \frac{\pi}{4}\right\}\) which is a subset of \(\left\{\frac{-3 \pi}{4}, \frac{-\pi}{4}, \frac{3 \pi}{4}, \frac{\pi}{4}\right\}\)
80233
The set of points, where \(f(x)=\frac{x}{1+|x|}\) is differentiable, is
1 \((-\infty,-1) \cup(-1, \infty)\)
2 \((-\infty, \infty)\)
3 \((0, \infty)\)
4 \((-\infty, 0) \cup(0, \infty)\)
Explanation:
(B) : Given than, \(\quad \mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{1+|\mathrm{x}|}\) we can write this fraction as - \(f(x)=\left\{\begin{array}{lll}\frac{x}{1+x} & \text { if } & x \geq 0 \\ \frac{x}{1-x} & \text { if } & x\lt 0\end{array}\right.\) Looking at \(\mathrm{f}(\mathrm{x})\), we can say that \(\mathrm{f}(\mathrm{x})\) is differentiable at everywhere except at 0 , Now, we need to check its differentiability at \(x=0\) LHD - \(=\lim _{h \rightarrow 0^{-}} \frac{f(0-h)-f(0)}{h}\) \(=\lim _{h \rightarrow 0^{-}} \frac{\frac{-h}{1+h}-0}{-h}=\lim _{h \rightarrow 0^{-}} \frac{1}{1+h}=1\) RHD at \(x=0\), \(=\lim _{h \rightarrow 0^{+}} \frac{f(0+h)-f(0)}{h}\) \(=\lim _{h \rightarrow 0^{+}} \frac{\frac{h}{1+h}-0}{h}=\lim _{h \rightarrow 0^{+}} \frac{1}{1+h}=1\) \(\because\) LHD \(=\) RHD at \(\mathrm{x}=0\) \(\therefore \mathrm{f}\) is differentiable at \(\mathrm{x}=0\) Hence, the set of points where \(\mathrm{f}(\mathrm{x})\) is differentiable is (\(\infty, \infty)\)
AIEEE-2006
Limits, Continuity and Differentiability
80234
Let \(f: R \rightarrow R\) be a function defined by \(f(x)=\) \(\min \{x+1,|x|+1\}\). Then, which one of the following is true?
1 \(f(x) \geq 1\) for all \(x \in R\)
2 \(f(x)\) is not differentiable at \(x=1\)
3 \(f(x)\) is differentiable everywhere
4 \(f(x)\) is not differentiable at \(x=0\)
Explanation:
(C) : Given that, \(\mathrm{f}(\mathrm{x})=\min \{\mathrm{x}+1,|\mathrm{x}|+1\}\) When, \(x\lt 0,(x+1)\) is minimum When, \(x \geq 0,(x+1)\) is minimum \(\mathrm{f}(\mathrm{x})=\) When, \(\mathrm{x}\lt 0,(\mathrm{x}+1)\) is minimum When, \(x \geq 0,(x+1)\) is minimum Overall, \(\mathrm{f}(\mathrm{x})=\mathrm{x}+1\) is differentiable in everywhere
AIEEE-2007
Limits, Continuity and Differentiability
80235
If \(\mathrm{f}:(-1,1) \rightarrow \mathrm{R}\) be a differentiable function with \(f(0)=-1\) and \(f^{\prime}(0)=1\). Let \(g(x)=[f(2 f(x)+\) \(21^{2}\). Then, \(\mathbf{g}^{\prime}(0)\) is equal to
80236
Let \(S\) be the set of all points in \((-\pi, \pi)\) at which the function, \(f(x)=\min \{\sin x, \cos x\}\) is not differentiable. Then, \(S\) is a subset of which of the following?
(C) : We have - \(f(x)=\min \{\sin x, \cos x\}\) Let us draw the graph of \(y=f(x)\), as shown below, clearly, the function - \(\mathrm{f}(\mathrm{x})=\min \{\sin \mathrm{x}, \cos \mathrm{x}\}\) is not differentiable at \(\mathrm{x}=-\frac{3 \pi}{4}\) and \(\frac{\pi}{4}\) So, \(\mathrm{S}=\left\{-\frac{3 \pi}{4}, \frac{\pi}{4}\right\}\) which is a subset of \(\left\{\frac{-3 \pi}{4}, \frac{-\pi}{4}, \frac{3 \pi}{4}, \frac{\pi}{4}\right\}\)
