80205
Let \(f(x)=\left\{\begin{array}{cc}(x-1) \sin \frac{1}{x-1} \text { if } x \neq 1 \\ 0 \text { if } x=1\end{array}\right.\) then which one of the following is true?
1 \(f\) is differentiable at \(x=0\) and \(x=1\)
2 \(f\) is differentiable at \(x=0\) but not at \(x=1\)
3 \(f\) is differentiable at \(x=1\) but not at \(x=0\)
4 \(f\) is neither differentiable at \(x=0\) nor at \(x=1\)
Explanation:
(B) Given, \(f(x)=\left\{\begin{array}{cc}(x-1) \cdot \sin \frac{1}{x-1}, & \text { if } x \neq 1 \\ 0, & \text { if } x=1\end{array}\right.\) \(f^{\prime}(1) =\lim _{x \rightarrow 0} \frac{f(1+h)-f(1)}{h}\) \(=\lim _{h \rightarrow 0} \frac{(1+h-1) \sin \frac{1}{(1+h-1)}-0}{h}\) \(=\lim _{h \rightarrow 0} \frac{h \cdot \sin \frac{1}{h}}{h}\) \(=\lim _{h \rightarrow 0} \sin \frac{1}{h}\) Which does not exist. Therefore, \(\mathrm{f}\) is not differentiable at \(\mathrm{x}=1\) Also, \(\mathrm{f}^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}\), if the limit exists But this limit doesn't exist. Hence, \(\mathrm{f}\) is differentiable at \(\mathrm{x}=0\) but not at \(\mathrm{x}=1\)
BITSAT-2014
Limits, Continuity and Differentiability
80206
The function \(f(x)=x-\left|x-x^{2}\right|, \quad-1 \leq x \leq 1\) continuous on the interval
1 \([-1,1](b)\)
2 \((-1,1)\)
3 \(\{-1,1]-\{0\}\)
4 \((-1,1)-\{0\}\)
Explanation:
(A) : Given, \(f(x)=x-\left|x-x^{2}\right|=x-|x(1-x)|\) \(f(x)=x-|x||1-x|\), Continuity is to be checked at, \(x =0 \text { and } x=1 . \text { At } x=0\) \(\text { LHL }=\lim _{h \rightarrow 0} f(0-h) =\lim _{h \rightarrow 0}-h-|-h||1+h|\) \(=\lim _{h \rightarrow 0}-h-h(1+h)=0\) RHL \(=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} h-|h||1-h|\) \(=\lim _{h \rightarrow 0} h-h(1-h)=0\) And, \(\mathrm{f}(0)=0\) Since, \(L H L=R H L=f(0)\), \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=0\). At \(x=1\); \(=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0}(1-h)-|1-h||1-(1-h)|\) \(=\lim _{\mathrm{h} \rightarrow 0}(1-\mathrm{h})-\mathrm{h}(1-\mathrm{h})=1\) Similarly, RHL \(=\lim _{h \rightarrow 0} \mathrm{f}(1+\mathrm{h})=1\) And, \(\mathrm{f}(1)=1-|1| .|1-1|=1\) \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=1\) Hence, \(\mathrm{f}(\mathrm{x})\) is continuous for all \(\mathrm{x} \in[-1,1]\).
BITSAT-2017
Limits, Continuity and Differentiability
80208
If \(f(x)=\left\{\begin{array}{l}\frac{x \log \cos x}{\log \left(1+x^{2}\right)}, x \neq 0 \\ 0,\end{array}\right.\) then \(f(x)\) is
1 continuous as well as differentiable at \(x=0\)
2 continuous but not differentiable at \(x=0\)
3 differentiable but not continuous at \(\mathrm{x}=0\)
4 neither continuous nor differentiable at \(x=0\)
Explanation:
(A) : We have, \(f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h}=\lim _{h \rightarrow 0} \frac{-h \log \cos h}{-h \log \left(1+h^{2}\right)}\) \(=\lim _{h \rightarrow 0} \frac{\log \cos h}{\log \left(1+h^{2}\right)} \quad\left(\frac{0}{0}\right.\) form \()\) \(=\lim _{h \rightarrow 0} \frac{-\tan h}{2 h /\left(1+h^{2}\right)}=-1 / 2\) \(R f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{h \log \cos h}{h \log \left(1+h^{2}\right)}\) \(=\lim _{h \rightarrow 0} \frac{\log \cos h}{\log \left(1+h^{2}\right)} \quad\left(\frac{0}{0}\right.\) form \()\) \(=\lim _{h \rightarrow 0} \frac{-\tan h}{2 h /\left(1+h^{2}\right)}=\frac{-1}{2}\) Since, \(\operatorname{Lf}^{\prime}(0)=\operatorname{Rf}^{\prime}(0)\) Therefore, \(\mathrm{f}(\mathrm{x})\) is differentiable at \(\mathrm{x}=0\) Since, any function which is differentiable at a point must be continuous. So, \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=0\).
BITSAT-2017
Limits, Continuity and Differentiability
80209
If \(f(x)=\cos ^{-1}\left[\frac{1-(\log x)^{2}}{1+(\log x)^{2}}\right]\) then the value of \(\mathbf{f}^{\prime}(\mathbf{e})\) is equal to
80205
Let \(f(x)=\left\{\begin{array}{cc}(x-1) \sin \frac{1}{x-1} \text { if } x \neq 1 \\ 0 \text { if } x=1\end{array}\right.\) then which one of the following is true?
1 \(f\) is differentiable at \(x=0\) and \(x=1\)
2 \(f\) is differentiable at \(x=0\) but not at \(x=1\)
3 \(f\) is differentiable at \(x=1\) but not at \(x=0\)
4 \(f\) is neither differentiable at \(x=0\) nor at \(x=1\)
Explanation:
(B) Given, \(f(x)=\left\{\begin{array}{cc}(x-1) \cdot \sin \frac{1}{x-1}, & \text { if } x \neq 1 \\ 0, & \text { if } x=1\end{array}\right.\) \(f^{\prime}(1) =\lim _{x \rightarrow 0} \frac{f(1+h)-f(1)}{h}\) \(=\lim _{h \rightarrow 0} \frac{(1+h-1) \sin \frac{1}{(1+h-1)}-0}{h}\) \(=\lim _{h \rightarrow 0} \frac{h \cdot \sin \frac{1}{h}}{h}\) \(=\lim _{h \rightarrow 0} \sin \frac{1}{h}\) Which does not exist. Therefore, \(\mathrm{f}\) is not differentiable at \(\mathrm{x}=1\) Also, \(\mathrm{f}^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}\), if the limit exists But this limit doesn't exist. Hence, \(\mathrm{f}\) is differentiable at \(\mathrm{x}=0\) but not at \(\mathrm{x}=1\)
BITSAT-2014
Limits, Continuity and Differentiability
80206
The function \(f(x)=x-\left|x-x^{2}\right|, \quad-1 \leq x \leq 1\) continuous on the interval
1 \([-1,1](b)\)
2 \((-1,1)\)
3 \(\{-1,1]-\{0\}\)
4 \((-1,1)-\{0\}\)
Explanation:
(A) : Given, \(f(x)=x-\left|x-x^{2}\right|=x-|x(1-x)|\) \(f(x)=x-|x||1-x|\), Continuity is to be checked at, \(x =0 \text { and } x=1 . \text { At } x=0\) \(\text { LHL }=\lim _{h \rightarrow 0} f(0-h) =\lim _{h \rightarrow 0}-h-|-h||1+h|\) \(=\lim _{h \rightarrow 0}-h-h(1+h)=0\) RHL \(=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} h-|h||1-h|\) \(=\lim _{h \rightarrow 0} h-h(1-h)=0\) And, \(\mathrm{f}(0)=0\) Since, \(L H L=R H L=f(0)\), \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=0\). At \(x=1\); \(=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0}(1-h)-|1-h||1-(1-h)|\) \(=\lim _{\mathrm{h} \rightarrow 0}(1-\mathrm{h})-\mathrm{h}(1-\mathrm{h})=1\) Similarly, RHL \(=\lim _{h \rightarrow 0} \mathrm{f}(1+\mathrm{h})=1\) And, \(\mathrm{f}(1)=1-|1| .|1-1|=1\) \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=1\) Hence, \(\mathrm{f}(\mathrm{x})\) is continuous for all \(\mathrm{x} \in[-1,1]\).
BITSAT-2017
Limits, Continuity and Differentiability
80208
If \(f(x)=\left\{\begin{array}{l}\frac{x \log \cos x}{\log \left(1+x^{2}\right)}, x \neq 0 \\ 0,\end{array}\right.\) then \(f(x)\) is
1 continuous as well as differentiable at \(x=0\)
2 continuous but not differentiable at \(x=0\)
3 differentiable but not continuous at \(\mathrm{x}=0\)
4 neither continuous nor differentiable at \(x=0\)
Explanation:
(A) : We have, \(f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h}=\lim _{h \rightarrow 0} \frac{-h \log \cos h}{-h \log \left(1+h^{2}\right)}\) \(=\lim _{h \rightarrow 0} \frac{\log \cos h}{\log \left(1+h^{2}\right)} \quad\left(\frac{0}{0}\right.\) form \()\) \(=\lim _{h \rightarrow 0} \frac{-\tan h}{2 h /\left(1+h^{2}\right)}=-1 / 2\) \(R f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{h \log \cos h}{h \log \left(1+h^{2}\right)}\) \(=\lim _{h \rightarrow 0} \frac{\log \cos h}{\log \left(1+h^{2}\right)} \quad\left(\frac{0}{0}\right.\) form \()\) \(=\lim _{h \rightarrow 0} \frac{-\tan h}{2 h /\left(1+h^{2}\right)}=\frac{-1}{2}\) Since, \(\operatorname{Lf}^{\prime}(0)=\operatorname{Rf}^{\prime}(0)\) Therefore, \(\mathrm{f}(\mathrm{x})\) is differentiable at \(\mathrm{x}=0\) Since, any function which is differentiable at a point must be continuous. So, \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=0\).
BITSAT-2017
Limits, Continuity and Differentiability
80209
If \(f(x)=\cos ^{-1}\left[\frac{1-(\log x)^{2}}{1+(\log x)^{2}}\right]\) then the value of \(\mathbf{f}^{\prime}(\mathbf{e})\) is equal to
80205
Let \(f(x)=\left\{\begin{array}{cc}(x-1) \sin \frac{1}{x-1} \text { if } x \neq 1 \\ 0 \text { if } x=1\end{array}\right.\) then which one of the following is true?
1 \(f\) is differentiable at \(x=0\) and \(x=1\)
2 \(f\) is differentiable at \(x=0\) but not at \(x=1\)
3 \(f\) is differentiable at \(x=1\) but not at \(x=0\)
4 \(f\) is neither differentiable at \(x=0\) nor at \(x=1\)
Explanation:
(B) Given, \(f(x)=\left\{\begin{array}{cc}(x-1) \cdot \sin \frac{1}{x-1}, & \text { if } x \neq 1 \\ 0, & \text { if } x=1\end{array}\right.\) \(f^{\prime}(1) =\lim _{x \rightarrow 0} \frac{f(1+h)-f(1)}{h}\) \(=\lim _{h \rightarrow 0} \frac{(1+h-1) \sin \frac{1}{(1+h-1)}-0}{h}\) \(=\lim _{h \rightarrow 0} \frac{h \cdot \sin \frac{1}{h}}{h}\) \(=\lim _{h \rightarrow 0} \sin \frac{1}{h}\) Which does not exist. Therefore, \(\mathrm{f}\) is not differentiable at \(\mathrm{x}=1\) Also, \(\mathrm{f}^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}\), if the limit exists But this limit doesn't exist. Hence, \(\mathrm{f}\) is differentiable at \(\mathrm{x}=0\) but not at \(\mathrm{x}=1\)
BITSAT-2014
Limits, Continuity and Differentiability
80206
The function \(f(x)=x-\left|x-x^{2}\right|, \quad-1 \leq x \leq 1\) continuous on the interval
1 \([-1,1](b)\)
2 \((-1,1)\)
3 \(\{-1,1]-\{0\}\)
4 \((-1,1)-\{0\}\)
Explanation:
(A) : Given, \(f(x)=x-\left|x-x^{2}\right|=x-|x(1-x)|\) \(f(x)=x-|x||1-x|\), Continuity is to be checked at, \(x =0 \text { and } x=1 . \text { At } x=0\) \(\text { LHL }=\lim _{h \rightarrow 0} f(0-h) =\lim _{h \rightarrow 0}-h-|-h||1+h|\) \(=\lim _{h \rightarrow 0}-h-h(1+h)=0\) RHL \(=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} h-|h||1-h|\) \(=\lim _{h \rightarrow 0} h-h(1-h)=0\) And, \(\mathrm{f}(0)=0\) Since, \(L H L=R H L=f(0)\), \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=0\). At \(x=1\); \(=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0}(1-h)-|1-h||1-(1-h)|\) \(=\lim _{\mathrm{h} \rightarrow 0}(1-\mathrm{h})-\mathrm{h}(1-\mathrm{h})=1\) Similarly, RHL \(=\lim _{h \rightarrow 0} \mathrm{f}(1+\mathrm{h})=1\) And, \(\mathrm{f}(1)=1-|1| .|1-1|=1\) \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=1\) Hence, \(\mathrm{f}(\mathrm{x})\) is continuous for all \(\mathrm{x} \in[-1,1]\).
BITSAT-2017
Limits, Continuity and Differentiability
80208
If \(f(x)=\left\{\begin{array}{l}\frac{x \log \cos x}{\log \left(1+x^{2}\right)}, x \neq 0 \\ 0,\end{array}\right.\) then \(f(x)\) is
1 continuous as well as differentiable at \(x=0\)
2 continuous but not differentiable at \(x=0\)
3 differentiable but not continuous at \(\mathrm{x}=0\)
4 neither continuous nor differentiable at \(x=0\)
Explanation:
(A) : We have, \(f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h}=\lim _{h \rightarrow 0} \frac{-h \log \cos h}{-h \log \left(1+h^{2}\right)}\) \(=\lim _{h \rightarrow 0} \frac{\log \cos h}{\log \left(1+h^{2}\right)} \quad\left(\frac{0}{0}\right.\) form \()\) \(=\lim _{h \rightarrow 0} \frac{-\tan h}{2 h /\left(1+h^{2}\right)}=-1 / 2\) \(R f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{h \log \cos h}{h \log \left(1+h^{2}\right)}\) \(=\lim _{h \rightarrow 0} \frac{\log \cos h}{\log \left(1+h^{2}\right)} \quad\left(\frac{0}{0}\right.\) form \()\) \(=\lim _{h \rightarrow 0} \frac{-\tan h}{2 h /\left(1+h^{2}\right)}=\frac{-1}{2}\) Since, \(\operatorname{Lf}^{\prime}(0)=\operatorname{Rf}^{\prime}(0)\) Therefore, \(\mathrm{f}(\mathrm{x})\) is differentiable at \(\mathrm{x}=0\) Since, any function which is differentiable at a point must be continuous. So, \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=0\).
BITSAT-2017
Limits, Continuity and Differentiability
80209
If \(f(x)=\cos ^{-1}\left[\frac{1-(\log x)^{2}}{1+(\log x)^{2}}\right]\) then the value of \(\mathbf{f}^{\prime}(\mathbf{e})\) is equal to
NEET Test Series from KOTA - 10 Papers In MS WORD
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Limits, Continuity and Differentiability
80205
Let \(f(x)=\left\{\begin{array}{cc}(x-1) \sin \frac{1}{x-1} \text { if } x \neq 1 \\ 0 \text { if } x=1\end{array}\right.\) then which one of the following is true?
1 \(f\) is differentiable at \(x=0\) and \(x=1\)
2 \(f\) is differentiable at \(x=0\) but not at \(x=1\)
3 \(f\) is differentiable at \(x=1\) but not at \(x=0\)
4 \(f\) is neither differentiable at \(x=0\) nor at \(x=1\)
Explanation:
(B) Given, \(f(x)=\left\{\begin{array}{cc}(x-1) \cdot \sin \frac{1}{x-1}, & \text { if } x \neq 1 \\ 0, & \text { if } x=1\end{array}\right.\) \(f^{\prime}(1) =\lim _{x \rightarrow 0} \frac{f(1+h)-f(1)}{h}\) \(=\lim _{h \rightarrow 0} \frac{(1+h-1) \sin \frac{1}{(1+h-1)}-0}{h}\) \(=\lim _{h \rightarrow 0} \frac{h \cdot \sin \frac{1}{h}}{h}\) \(=\lim _{h \rightarrow 0} \sin \frac{1}{h}\) Which does not exist. Therefore, \(\mathrm{f}\) is not differentiable at \(\mathrm{x}=1\) Also, \(\mathrm{f}^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}\), if the limit exists But this limit doesn't exist. Hence, \(\mathrm{f}\) is differentiable at \(\mathrm{x}=0\) but not at \(\mathrm{x}=1\)
BITSAT-2014
Limits, Continuity and Differentiability
80206
The function \(f(x)=x-\left|x-x^{2}\right|, \quad-1 \leq x \leq 1\) continuous on the interval
1 \([-1,1](b)\)
2 \((-1,1)\)
3 \(\{-1,1]-\{0\}\)
4 \((-1,1)-\{0\}\)
Explanation:
(A) : Given, \(f(x)=x-\left|x-x^{2}\right|=x-|x(1-x)|\) \(f(x)=x-|x||1-x|\), Continuity is to be checked at, \(x =0 \text { and } x=1 . \text { At } x=0\) \(\text { LHL }=\lim _{h \rightarrow 0} f(0-h) =\lim _{h \rightarrow 0}-h-|-h||1+h|\) \(=\lim _{h \rightarrow 0}-h-h(1+h)=0\) RHL \(=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} h-|h||1-h|\) \(=\lim _{h \rightarrow 0} h-h(1-h)=0\) And, \(\mathrm{f}(0)=0\) Since, \(L H L=R H L=f(0)\), \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=0\). At \(x=1\); \(=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0}(1-h)-|1-h||1-(1-h)|\) \(=\lim _{\mathrm{h} \rightarrow 0}(1-\mathrm{h})-\mathrm{h}(1-\mathrm{h})=1\) Similarly, RHL \(=\lim _{h \rightarrow 0} \mathrm{f}(1+\mathrm{h})=1\) And, \(\mathrm{f}(1)=1-|1| .|1-1|=1\) \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=1\) Hence, \(\mathrm{f}(\mathrm{x})\) is continuous for all \(\mathrm{x} \in[-1,1]\).
BITSAT-2017
Limits, Continuity and Differentiability
80208
If \(f(x)=\left\{\begin{array}{l}\frac{x \log \cos x}{\log \left(1+x^{2}\right)}, x \neq 0 \\ 0,\end{array}\right.\) then \(f(x)\) is
1 continuous as well as differentiable at \(x=0\)
2 continuous but not differentiable at \(x=0\)
3 differentiable but not continuous at \(\mathrm{x}=0\)
4 neither continuous nor differentiable at \(x=0\)
Explanation:
(A) : We have, \(f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h}=\lim _{h \rightarrow 0} \frac{-h \log \cos h}{-h \log \left(1+h^{2}\right)}\) \(=\lim _{h \rightarrow 0} \frac{\log \cos h}{\log \left(1+h^{2}\right)} \quad\left(\frac{0}{0}\right.\) form \()\) \(=\lim _{h \rightarrow 0} \frac{-\tan h}{2 h /\left(1+h^{2}\right)}=-1 / 2\) \(R f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{h \log \cos h}{h \log \left(1+h^{2}\right)}\) \(=\lim _{h \rightarrow 0} \frac{\log \cos h}{\log \left(1+h^{2}\right)} \quad\left(\frac{0}{0}\right.\) form \()\) \(=\lim _{h \rightarrow 0} \frac{-\tan h}{2 h /\left(1+h^{2}\right)}=\frac{-1}{2}\) Since, \(\operatorname{Lf}^{\prime}(0)=\operatorname{Rf}^{\prime}(0)\) Therefore, \(\mathrm{f}(\mathrm{x})\) is differentiable at \(\mathrm{x}=0\) Since, any function which is differentiable at a point must be continuous. So, \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=0\).
BITSAT-2017
Limits, Continuity and Differentiability
80209
If \(f(x)=\cos ^{-1}\left[\frac{1-(\log x)^{2}}{1+(\log x)^{2}}\right]\) then the value of \(\mathbf{f}^{\prime}(\mathbf{e})\) is equal to
