79960
Let \(f(x)=\left\{\begin{array}{rr}x^{n} \sin \frac{1}{x}, x \neq 0 \\ 0, x=0\end{array}\right\}\), Then, \(f(x)\) is continuous but not differentiable at \(\mathrm{x}=\mathbf{0}\), if
1 \(\mathrm{n} \in(0,1)\)
2 \(\mathrm{n} \in[1, \infty)\)
3 \(\mathrm{n} \in(-\infty, 0)\)
4 \(\mathrm{n}=0\)
Explanation:
(A) : Given, \(\mathrm{f}(\mathrm{x})=\mathrm{x}^{\mathrm{n}} \sin \frac{1}{\mathrm{x}}\) \(\mathrm{f}(0)=0\) \(\because \mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=0\), Hence, \(\mathrm{f}(\mathrm{x})\) will not be differentiable if \(\mathrm{n} \leq 1\) but \(\lim _{x \rightarrow 0} f(x)=f(0)\) \(\lim _{x \rightarrow 0} x^{n} \sin \frac{1}{x}=0\) \(f^{\prime}(0)=\frac{f(x)-f(0)}{x-0}\) \(f^{\prime}(0)=\frac{x^{n} \sin \frac{1}{x}-0}{x}\) \(f^{\prime}(0)=x^{n-1} \sin \frac{1}{x}\) continuous at \(\mathrm{x}=0\) \(\therefore\) Possible values of \(\mathrm{n}\) is \(\mathrm{n} \in(0,1)\)
UPSEE-2014
Limits, Continuity and Differentiability
79961
\(\lim _{x \rightarrow 0} \frac{x}{|x|+x^{2}}\) is equal to
79963
If \(f(x)=\left\{\begin{array}{l}-x^{2}, \quad \text { When } x \leq 0 \\ 5 x-4, \text { When } 0\lt x \leq 1 \\ 4 x^{2}-3 x, \text { When } 1\lt x\lt 2 \\ 3 x+4, \text { When } x \geq 2\end{array}\right.\) then
79960
Let \(f(x)=\left\{\begin{array}{rr}x^{n} \sin \frac{1}{x}, x \neq 0 \\ 0, x=0\end{array}\right\}\), Then, \(f(x)\) is continuous but not differentiable at \(\mathrm{x}=\mathbf{0}\), if
1 \(\mathrm{n} \in(0,1)\)
2 \(\mathrm{n} \in[1, \infty)\)
3 \(\mathrm{n} \in(-\infty, 0)\)
4 \(\mathrm{n}=0\)
Explanation:
(A) : Given, \(\mathrm{f}(\mathrm{x})=\mathrm{x}^{\mathrm{n}} \sin \frac{1}{\mathrm{x}}\) \(\mathrm{f}(0)=0\) \(\because \mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=0\), Hence, \(\mathrm{f}(\mathrm{x})\) will not be differentiable if \(\mathrm{n} \leq 1\) but \(\lim _{x \rightarrow 0} f(x)=f(0)\) \(\lim _{x \rightarrow 0} x^{n} \sin \frac{1}{x}=0\) \(f^{\prime}(0)=\frac{f(x)-f(0)}{x-0}\) \(f^{\prime}(0)=\frac{x^{n} \sin \frac{1}{x}-0}{x}\) \(f^{\prime}(0)=x^{n-1} \sin \frac{1}{x}\) continuous at \(\mathrm{x}=0\) \(\therefore\) Possible values of \(\mathrm{n}\) is \(\mathrm{n} \in(0,1)\)
UPSEE-2014
Limits, Continuity and Differentiability
79961
\(\lim _{x \rightarrow 0} \frac{x}{|x|+x^{2}}\) is equal to
79963
If \(f(x)=\left\{\begin{array}{l}-x^{2}, \quad \text { When } x \leq 0 \\ 5 x-4, \text { When } 0\lt x \leq 1 \\ 4 x^{2}-3 x, \text { When } 1\lt x\lt 2 \\ 3 x+4, \text { When } x \geq 2\end{array}\right.\) then
79960
Let \(f(x)=\left\{\begin{array}{rr}x^{n} \sin \frac{1}{x}, x \neq 0 \\ 0, x=0\end{array}\right\}\), Then, \(f(x)\) is continuous but not differentiable at \(\mathrm{x}=\mathbf{0}\), if
1 \(\mathrm{n} \in(0,1)\)
2 \(\mathrm{n} \in[1, \infty)\)
3 \(\mathrm{n} \in(-\infty, 0)\)
4 \(\mathrm{n}=0\)
Explanation:
(A) : Given, \(\mathrm{f}(\mathrm{x})=\mathrm{x}^{\mathrm{n}} \sin \frac{1}{\mathrm{x}}\) \(\mathrm{f}(0)=0\) \(\because \mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=0\), Hence, \(\mathrm{f}(\mathrm{x})\) will not be differentiable if \(\mathrm{n} \leq 1\) but \(\lim _{x \rightarrow 0} f(x)=f(0)\) \(\lim _{x \rightarrow 0} x^{n} \sin \frac{1}{x}=0\) \(f^{\prime}(0)=\frac{f(x)-f(0)}{x-0}\) \(f^{\prime}(0)=\frac{x^{n} \sin \frac{1}{x}-0}{x}\) \(f^{\prime}(0)=x^{n-1} \sin \frac{1}{x}\) continuous at \(\mathrm{x}=0\) \(\therefore\) Possible values of \(\mathrm{n}\) is \(\mathrm{n} \in(0,1)\)
UPSEE-2014
Limits, Continuity and Differentiability
79961
\(\lim _{x \rightarrow 0} \frac{x}{|x|+x^{2}}\) is equal to
79963
If \(f(x)=\left\{\begin{array}{l}-x^{2}, \quad \text { When } x \leq 0 \\ 5 x-4, \text { When } 0\lt x \leq 1 \\ 4 x^{2}-3 x, \text { When } 1\lt x\lt 2 \\ 3 x+4, \text { When } x \geq 2\end{array}\right.\) then
79960
Let \(f(x)=\left\{\begin{array}{rr}x^{n} \sin \frac{1}{x}, x \neq 0 \\ 0, x=0\end{array}\right\}\), Then, \(f(x)\) is continuous but not differentiable at \(\mathrm{x}=\mathbf{0}\), if
1 \(\mathrm{n} \in(0,1)\)
2 \(\mathrm{n} \in[1, \infty)\)
3 \(\mathrm{n} \in(-\infty, 0)\)
4 \(\mathrm{n}=0\)
Explanation:
(A) : Given, \(\mathrm{f}(\mathrm{x})=\mathrm{x}^{\mathrm{n}} \sin \frac{1}{\mathrm{x}}\) \(\mathrm{f}(0)=0\) \(\because \mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=0\), Hence, \(\mathrm{f}(\mathrm{x})\) will not be differentiable if \(\mathrm{n} \leq 1\) but \(\lim _{x \rightarrow 0} f(x)=f(0)\) \(\lim _{x \rightarrow 0} x^{n} \sin \frac{1}{x}=0\) \(f^{\prime}(0)=\frac{f(x)-f(0)}{x-0}\) \(f^{\prime}(0)=\frac{x^{n} \sin \frac{1}{x}-0}{x}\) \(f^{\prime}(0)=x^{n-1} \sin \frac{1}{x}\) continuous at \(\mathrm{x}=0\) \(\therefore\) Possible values of \(\mathrm{n}\) is \(\mathrm{n} \in(0,1)\)
UPSEE-2014
Limits, Continuity and Differentiability
79961
\(\lim _{x \rightarrow 0} \frac{x}{|x|+x^{2}}\) is equal to
79963
If \(f(x)=\left\{\begin{array}{l}-x^{2}, \quad \text { When } x \leq 0 \\ 5 x-4, \text { When } 0\lt x \leq 1 \\ 4 x^{2}-3 x, \text { When } 1\lt x\lt 2 \\ 3 x+4, \text { When } x \geq 2\end{array}\right.\) then
79960
Let \(f(x)=\left\{\begin{array}{rr}x^{n} \sin \frac{1}{x}, x \neq 0 \\ 0, x=0\end{array}\right\}\), Then, \(f(x)\) is continuous but not differentiable at \(\mathrm{x}=\mathbf{0}\), if
1 \(\mathrm{n} \in(0,1)\)
2 \(\mathrm{n} \in[1, \infty)\)
3 \(\mathrm{n} \in(-\infty, 0)\)
4 \(\mathrm{n}=0\)
Explanation:
(A) : Given, \(\mathrm{f}(\mathrm{x})=\mathrm{x}^{\mathrm{n}} \sin \frac{1}{\mathrm{x}}\) \(\mathrm{f}(0)=0\) \(\because \mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=0\), Hence, \(\mathrm{f}(\mathrm{x})\) will not be differentiable if \(\mathrm{n} \leq 1\) but \(\lim _{x \rightarrow 0} f(x)=f(0)\) \(\lim _{x \rightarrow 0} x^{n} \sin \frac{1}{x}=0\) \(f^{\prime}(0)=\frac{f(x)-f(0)}{x-0}\) \(f^{\prime}(0)=\frac{x^{n} \sin \frac{1}{x}-0}{x}\) \(f^{\prime}(0)=x^{n-1} \sin \frac{1}{x}\) continuous at \(\mathrm{x}=0\) \(\therefore\) Possible values of \(\mathrm{n}\) is \(\mathrm{n} \in(0,1)\)
UPSEE-2014
Limits, Continuity and Differentiability
79961
\(\lim _{x \rightarrow 0} \frac{x}{|x|+x^{2}}\) is equal to
79963
If \(f(x)=\left\{\begin{array}{l}-x^{2}, \quad \text { When } x \leq 0 \\ 5 x-4, \text { When } 0\lt x \leq 1 \\ 4 x^{2}-3 x, \text { When } 1\lt x\lt 2 \\ 3 x+4, \text { When } x \geq 2\end{array}\right.\) then