79947
\(\mathbf{f}(\mathbf{x})=\left\{\begin{array}{ccc}x \sin 1 / x & , & x \neq 0 \\ 0 & , & x=0\end{array}\right.\) at \(x=0\) is
1 continuous as well as differentiable
2 differentiable but not continuous
3 continuous but not differentiable
4 neither continuous nor differentiable
Explanation:
(C) : Given, \(f(x)=\left\{\begin{array}{cc}x \sin 1 / x , x \neq 0 \\ 0 , \quad x=0\end{array}\right.\) at \(x=0\) For function to be continuous, \(\mathrm{f}(0+\mathrm{h})=\mathrm{f}(0-\mathrm{h})=\mathrm{f}(0)\) \(\because \quad \mathrm{f}(\mathrm{x})=\mathrm{x} \sin \frac{1}{\mathrm{x}}\) \(\mathrm{f}(0+\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0} \mathrm{~h} \sin \frac{1}{\mathrm{~h}}\) \(=0 \times(\text { a finite quantity })=0\) \(\mathrm{f}(0-\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0}-\mathrm{h} \sin \frac{1}{-\mathrm{h}}\) \(=0 \times(\text { a finite quantity })=0\) Also, \(\quad f(0)=\lim _{x \rightarrow 0} x \sin \frac{1}{x}\) \(=0 \times(\text { a finite quantity })=0\) \(\Rightarrow\) function is continuous at \(x=0\) For function to be differentiable : \(f^{\prime}(0+h) =f^{\prime}(0-h)\) \(f^{\prime}(0+h) =\frac{f(0+h)-f(0)}{h}\) \(=\lim _{h \rightarrow 0} \frac{h \sin \frac{1}{h}-0}{h}=\lim _{h \rightarrow 0} \sin \frac{1}{h}\) Which does not exist. So, continuous but not differentiable at \(\mathrm{x}=0\).
VITEEE-2018
Limits, Continuity and Differentiability
79949
The value of \(\lim (\cos x)^{\cot ^{2} x}\) is
79950
Let \(f(x)=\left\{\begin{array}{cl}5^{1 / x}, x\lt 0 \\ \lambda[x], x \geq 0\end{array}\right.\) and \(\lambda \in R\), then at \(\mathbf{x}=\mathbf{0}\)
1 \(f\) is discontinuous
2 \(f\) is continuous only, if \(\lambda=0\)
3 \(f\) is continuous only, whatever \(\lambda\) may be
79947
\(\mathbf{f}(\mathbf{x})=\left\{\begin{array}{ccc}x \sin 1 / x & , & x \neq 0 \\ 0 & , & x=0\end{array}\right.\) at \(x=0\) is
1 continuous as well as differentiable
2 differentiable but not continuous
3 continuous but not differentiable
4 neither continuous nor differentiable
Explanation:
(C) : Given, \(f(x)=\left\{\begin{array}{cc}x \sin 1 / x , x \neq 0 \\ 0 , \quad x=0\end{array}\right.\) at \(x=0\) For function to be continuous, \(\mathrm{f}(0+\mathrm{h})=\mathrm{f}(0-\mathrm{h})=\mathrm{f}(0)\) \(\because \quad \mathrm{f}(\mathrm{x})=\mathrm{x} \sin \frac{1}{\mathrm{x}}\) \(\mathrm{f}(0+\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0} \mathrm{~h} \sin \frac{1}{\mathrm{~h}}\) \(=0 \times(\text { a finite quantity })=0\) \(\mathrm{f}(0-\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0}-\mathrm{h} \sin \frac{1}{-\mathrm{h}}\) \(=0 \times(\text { a finite quantity })=0\) Also, \(\quad f(0)=\lim _{x \rightarrow 0} x \sin \frac{1}{x}\) \(=0 \times(\text { a finite quantity })=0\) \(\Rightarrow\) function is continuous at \(x=0\) For function to be differentiable : \(f^{\prime}(0+h) =f^{\prime}(0-h)\) \(f^{\prime}(0+h) =\frac{f(0+h)-f(0)}{h}\) \(=\lim _{h \rightarrow 0} \frac{h \sin \frac{1}{h}-0}{h}=\lim _{h \rightarrow 0} \sin \frac{1}{h}\) Which does not exist. So, continuous but not differentiable at \(\mathrm{x}=0\).
VITEEE-2018
Limits, Continuity and Differentiability
79949
The value of \(\lim (\cos x)^{\cot ^{2} x}\) is
79950
Let \(f(x)=\left\{\begin{array}{cl}5^{1 / x}, x\lt 0 \\ \lambda[x], x \geq 0\end{array}\right.\) and \(\lambda \in R\), then at \(\mathbf{x}=\mathbf{0}\)
1 \(f\) is discontinuous
2 \(f\) is continuous only, if \(\lambda=0\)
3 \(f\) is continuous only, whatever \(\lambda\) may be
79947
\(\mathbf{f}(\mathbf{x})=\left\{\begin{array}{ccc}x \sin 1 / x & , & x \neq 0 \\ 0 & , & x=0\end{array}\right.\) at \(x=0\) is
1 continuous as well as differentiable
2 differentiable but not continuous
3 continuous but not differentiable
4 neither continuous nor differentiable
Explanation:
(C) : Given, \(f(x)=\left\{\begin{array}{cc}x \sin 1 / x , x \neq 0 \\ 0 , \quad x=0\end{array}\right.\) at \(x=0\) For function to be continuous, \(\mathrm{f}(0+\mathrm{h})=\mathrm{f}(0-\mathrm{h})=\mathrm{f}(0)\) \(\because \quad \mathrm{f}(\mathrm{x})=\mathrm{x} \sin \frac{1}{\mathrm{x}}\) \(\mathrm{f}(0+\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0} \mathrm{~h} \sin \frac{1}{\mathrm{~h}}\) \(=0 \times(\text { a finite quantity })=0\) \(\mathrm{f}(0-\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0}-\mathrm{h} \sin \frac{1}{-\mathrm{h}}\) \(=0 \times(\text { a finite quantity })=0\) Also, \(\quad f(0)=\lim _{x \rightarrow 0} x \sin \frac{1}{x}\) \(=0 \times(\text { a finite quantity })=0\) \(\Rightarrow\) function is continuous at \(x=0\) For function to be differentiable : \(f^{\prime}(0+h) =f^{\prime}(0-h)\) \(f^{\prime}(0+h) =\frac{f(0+h)-f(0)}{h}\) \(=\lim _{h \rightarrow 0} \frac{h \sin \frac{1}{h}-0}{h}=\lim _{h \rightarrow 0} \sin \frac{1}{h}\) Which does not exist. So, continuous but not differentiable at \(\mathrm{x}=0\).
VITEEE-2018
Limits, Continuity and Differentiability
79949
The value of \(\lim (\cos x)^{\cot ^{2} x}\) is
79950
Let \(f(x)=\left\{\begin{array}{cl}5^{1 / x}, x\lt 0 \\ \lambda[x], x \geq 0\end{array}\right.\) and \(\lambda \in R\), then at \(\mathbf{x}=\mathbf{0}\)
1 \(f\) is discontinuous
2 \(f\) is continuous only, if \(\lambda=0\)
3 \(f\) is continuous only, whatever \(\lambda\) may be
79947
\(\mathbf{f}(\mathbf{x})=\left\{\begin{array}{ccc}x \sin 1 / x & , & x \neq 0 \\ 0 & , & x=0\end{array}\right.\) at \(x=0\) is
1 continuous as well as differentiable
2 differentiable but not continuous
3 continuous but not differentiable
4 neither continuous nor differentiable
Explanation:
(C) : Given, \(f(x)=\left\{\begin{array}{cc}x \sin 1 / x , x \neq 0 \\ 0 , \quad x=0\end{array}\right.\) at \(x=0\) For function to be continuous, \(\mathrm{f}(0+\mathrm{h})=\mathrm{f}(0-\mathrm{h})=\mathrm{f}(0)\) \(\because \quad \mathrm{f}(\mathrm{x})=\mathrm{x} \sin \frac{1}{\mathrm{x}}\) \(\mathrm{f}(0+\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0} \mathrm{~h} \sin \frac{1}{\mathrm{~h}}\) \(=0 \times(\text { a finite quantity })=0\) \(\mathrm{f}(0-\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0}-\mathrm{h} \sin \frac{1}{-\mathrm{h}}\) \(=0 \times(\text { a finite quantity })=0\) Also, \(\quad f(0)=\lim _{x \rightarrow 0} x \sin \frac{1}{x}\) \(=0 \times(\text { a finite quantity })=0\) \(\Rightarrow\) function is continuous at \(x=0\) For function to be differentiable : \(f^{\prime}(0+h) =f^{\prime}(0-h)\) \(f^{\prime}(0+h) =\frac{f(0+h)-f(0)}{h}\) \(=\lim _{h \rightarrow 0} \frac{h \sin \frac{1}{h}-0}{h}=\lim _{h \rightarrow 0} \sin \frac{1}{h}\) Which does not exist. So, continuous but not differentiable at \(\mathrm{x}=0\).
VITEEE-2018
Limits, Continuity and Differentiability
79949
The value of \(\lim (\cos x)^{\cot ^{2} x}\) is
79950
Let \(f(x)=\left\{\begin{array}{cl}5^{1 / x}, x\lt 0 \\ \lambda[x], x \geq 0\end{array}\right.\) and \(\lambda \in R\), then at \(\mathbf{x}=\mathbf{0}\)
1 \(f\) is discontinuous
2 \(f\) is continuous only, if \(\lambda=0\)
3 \(f\) is continuous only, whatever \(\lambda\) may be
