80200
If \(f(x)\) and \(g(x)\) are two functions with \(g(x)=x-\frac{1}{x}\) and \(f o g(x)=x^{3}-\frac{1}{x^{3}}\), then \(f^{\prime}(x)=\)
1 \(3 x^{2}+3\)
2 \(x^{2}-\frac{1}{x^{2}}\)
3 \(1+\frac{1}{\mathrm{x}^{2}}\)
4 \(3 x^{2}+\frac{3}{x^{4}}\)
Explanation:
(A) : Given, \(g(x)=x-\frac{1}{x}\) And, \(f \circ g(x)=x^{3}-\frac{1}{x^{3}}\) Using \((a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)\) Then, \(\left(x-\frac{1}{x}\right)^{3}=x^{3}-\frac{1}{x^{3}}-3 x \cdot \frac{1}{x}\left(x-\frac{1}{x}\right)\) \(\mathrm{x}^{3}-\frac{1}{\mathrm{x}^{3}}=\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)^{3}+3\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)\) We have, \(f o g(x)=x^{3}-\frac{1}{x^{3}}\) \(f o g(x)=\left(x-\frac{1}{x}\right)^{3}+3\left(x-\frac{1}{x}\right)\) As \(g(x)=x-\frac{1}{x}\), this yields \(f\left(x-\frac{1}{x}\right)=\left(x-\frac{1}{x}\right)^{3}+3\left(x-\frac{1}{x}\right)\) Putting, \(x-\frac{1}{x}=\mathrm{t}\) \(f(t)=t^{3}+3 t\) Thus, \(f(x)=x^{3}+3 x\) \(f^{\prime}(x)=3 x^{2}+3\)
Karnataka CET-2006
Limits, Continuity and Differentiability
80201
If \(f(x)=\left\{\begin{array}{l}\frac{e^{3 x}-1}{4 x} \text { for } x \neq 0 \\ \frac{k+x}{4} \text { for } x=0\end{array}\right.\) is continuous at \(x=0\),
1 5
2 3
3 2
4 0
Explanation:
(B)\((x)=\left\{\begin{array}{ll}\frac{e^{3 x}-1}{4 x}, & \text { for } x \neq 0 \\ \frac{k+x}{4}, & \text { for } x=0\end{array}\right\}\) \(f(x) \text { is continuous at } x=0\) \(\therefore \lim _{\mathrm{x} \rightarrow 0} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow 0} \mathrm{f}(\mathrm{x})=\mathrm{f}(0)\) \(\lim _{x \rightarrow 0} \frac{e^{3 x}-1}{4 x}=\frac{k}{4}\) \(\lim _{\mathrm{h} \rightarrow 0} \frac{3 \mathrm{e}^{3 \mathrm{~h}}}{4}=\frac{\mathrm{k}}{4}\left(\right.\) Using \(L-\) Hospital rule in \(\frac{0}{0}\) form \()\) So, \(\mathrm{k}=3\)
COMEDK-2012
Limits, Continuity and Differentiability
80202
If \(\mathrm{f}(\mathrm{x})=\left\{\begin{array}{c}{[\mathrm{x}]+[-\mathrm{x}], \mathrm{x} \neq 2} \\ \mathrm{~K}, \mathrm{x}=2\end{array}\right.\) then \(f(\mathrm{x})\) is continuous at \(x=2\), provided \(K\) is equal to
1 2
2 1
3 -1
4 0
Explanation:
(C) : Given, \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=2\) \(\lim _{x \rightarrow 2^{+}} f(x)=\left[2^{+}\right]+\left[-2^{+}\right]=2-3=-1\) \(\begin{aligned} \lim _{x \rightarrow 2^{-}} f(x) =\left[2^{-}\right]+\left[-2^{-}\right] \\ =1-2 \\ =-1\end{aligned}\) For \(f(x)\) to be continuous \(\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{-}} f(x)=f(x)=k\) \(\therefore \quad k=-1\)
COMEDK-2014
Limits, Continuity and Differentiability
80203
If \(f(x)=\left\{\begin{array}{cr}x \sin 1 / x, & x \neq 0 \\ k \quad, & x=0\end{array}\right.\) is continuous at \(x=0\) , then the value of \(k\) will be
1 1
2 -1
3 0
4 None of these
Explanation:
(C) : Since \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=0\), so \(f(0)=\lim _{x \rightarrow 0} f(x)\) \(\mathrm{k} =\lim _{\mathrm{x} \rightarrow 0}(\mathrm{x} \sin 1 / \mathrm{x})\) \(=0 \times(\mathrm{a} \text { finite quantity })\) \(=0\)
80200
If \(f(x)\) and \(g(x)\) are two functions with \(g(x)=x-\frac{1}{x}\) and \(f o g(x)=x^{3}-\frac{1}{x^{3}}\), then \(f^{\prime}(x)=\)
1 \(3 x^{2}+3\)
2 \(x^{2}-\frac{1}{x^{2}}\)
3 \(1+\frac{1}{\mathrm{x}^{2}}\)
4 \(3 x^{2}+\frac{3}{x^{4}}\)
Explanation:
(A) : Given, \(g(x)=x-\frac{1}{x}\) And, \(f \circ g(x)=x^{3}-\frac{1}{x^{3}}\) Using \((a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)\) Then, \(\left(x-\frac{1}{x}\right)^{3}=x^{3}-\frac{1}{x^{3}}-3 x \cdot \frac{1}{x}\left(x-\frac{1}{x}\right)\) \(\mathrm{x}^{3}-\frac{1}{\mathrm{x}^{3}}=\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)^{3}+3\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)\) We have, \(f o g(x)=x^{3}-\frac{1}{x^{3}}\) \(f o g(x)=\left(x-\frac{1}{x}\right)^{3}+3\left(x-\frac{1}{x}\right)\) As \(g(x)=x-\frac{1}{x}\), this yields \(f\left(x-\frac{1}{x}\right)=\left(x-\frac{1}{x}\right)^{3}+3\left(x-\frac{1}{x}\right)\) Putting, \(x-\frac{1}{x}=\mathrm{t}\) \(f(t)=t^{3}+3 t\) Thus, \(f(x)=x^{3}+3 x\) \(f^{\prime}(x)=3 x^{2}+3\)
Karnataka CET-2006
Limits, Continuity and Differentiability
80201
If \(f(x)=\left\{\begin{array}{l}\frac{e^{3 x}-1}{4 x} \text { for } x \neq 0 \\ \frac{k+x}{4} \text { for } x=0\end{array}\right.\) is continuous at \(x=0\),
1 5
2 3
3 2
4 0
Explanation:
(B)\((x)=\left\{\begin{array}{ll}\frac{e^{3 x}-1}{4 x}, & \text { for } x \neq 0 \\ \frac{k+x}{4}, & \text { for } x=0\end{array}\right\}\) \(f(x) \text { is continuous at } x=0\) \(\therefore \lim _{\mathrm{x} \rightarrow 0} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow 0} \mathrm{f}(\mathrm{x})=\mathrm{f}(0)\) \(\lim _{x \rightarrow 0} \frac{e^{3 x}-1}{4 x}=\frac{k}{4}\) \(\lim _{\mathrm{h} \rightarrow 0} \frac{3 \mathrm{e}^{3 \mathrm{~h}}}{4}=\frac{\mathrm{k}}{4}\left(\right.\) Using \(L-\) Hospital rule in \(\frac{0}{0}\) form \()\) So, \(\mathrm{k}=3\)
COMEDK-2012
Limits, Continuity and Differentiability
80202
If \(\mathrm{f}(\mathrm{x})=\left\{\begin{array}{c}{[\mathrm{x}]+[-\mathrm{x}], \mathrm{x} \neq 2} \\ \mathrm{~K}, \mathrm{x}=2\end{array}\right.\) then \(f(\mathrm{x})\) is continuous at \(x=2\), provided \(K\) is equal to
1 2
2 1
3 -1
4 0
Explanation:
(C) : Given, \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=2\) \(\lim _{x \rightarrow 2^{+}} f(x)=\left[2^{+}\right]+\left[-2^{+}\right]=2-3=-1\) \(\begin{aligned} \lim _{x \rightarrow 2^{-}} f(x) =\left[2^{-}\right]+\left[-2^{-}\right] \\ =1-2 \\ =-1\end{aligned}\) For \(f(x)\) to be continuous \(\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{-}} f(x)=f(x)=k\) \(\therefore \quad k=-1\)
COMEDK-2014
Limits, Continuity and Differentiability
80203
If \(f(x)=\left\{\begin{array}{cr}x \sin 1 / x, & x \neq 0 \\ k \quad, & x=0\end{array}\right.\) is continuous at \(x=0\) , then the value of \(k\) will be
1 1
2 -1
3 0
4 None of these
Explanation:
(C) : Since \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=0\), so \(f(0)=\lim _{x \rightarrow 0} f(x)\) \(\mathrm{k} =\lim _{\mathrm{x} \rightarrow 0}(\mathrm{x} \sin 1 / \mathrm{x})\) \(=0 \times(\mathrm{a} \text { finite quantity })\) \(=0\)
80200
If \(f(x)\) and \(g(x)\) are two functions with \(g(x)=x-\frac{1}{x}\) and \(f o g(x)=x^{3}-\frac{1}{x^{3}}\), then \(f^{\prime}(x)=\)
1 \(3 x^{2}+3\)
2 \(x^{2}-\frac{1}{x^{2}}\)
3 \(1+\frac{1}{\mathrm{x}^{2}}\)
4 \(3 x^{2}+\frac{3}{x^{4}}\)
Explanation:
(A) : Given, \(g(x)=x-\frac{1}{x}\) And, \(f \circ g(x)=x^{3}-\frac{1}{x^{3}}\) Using \((a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)\) Then, \(\left(x-\frac{1}{x}\right)^{3}=x^{3}-\frac{1}{x^{3}}-3 x \cdot \frac{1}{x}\left(x-\frac{1}{x}\right)\) \(\mathrm{x}^{3}-\frac{1}{\mathrm{x}^{3}}=\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)^{3}+3\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)\) We have, \(f o g(x)=x^{3}-\frac{1}{x^{3}}\) \(f o g(x)=\left(x-\frac{1}{x}\right)^{3}+3\left(x-\frac{1}{x}\right)\) As \(g(x)=x-\frac{1}{x}\), this yields \(f\left(x-\frac{1}{x}\right)=\left(x-\frac{1}{x}\right)^{3}+3\left(x-\frac{1}{x}\right)\) Putting, \(x-\frac{1}{x}=\mathrm{t}\) \(f(t)=t^{3}+3 t\) Thus, \(f(x)=x^{3}+3 x\) \(f^{\prime}(x)=3 x^{2}+3\)
Karnataka CET-2006
Limits, Continuity and Differentiability
80201
If \(f(x)=\left\{\begin{array}{l}\frac{e^{3 x}-1}{4 x} \text { for } x \neq 0 \\ \frac{k+x}{4} \text { for } x=0\end{array}\right.\) is continuous at \(x=0\),
1 5
2 3
3 2
4 0
Explanation:
(B)\((x)=\left\{\begin{array}{ll}\frac{e^{3 x}-1}{4 x}, & \text { for } x \neq 0 \\ \frac{k+x}{4}, & \text { for } x=0\end{array}\right\}\) \(f(x) \text { is continuous at } x=0\) \(\therefore \lim _{\mathrm{x} \rightarrow 0} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow 0} \mathrm{f}(\mathrm{x})=\mathrm{f}(0)\) \(\lim _{x \rightarrow 0} \frac{e^{3 x}-1}{4 x}=\frac{k}{4}\) \(\lim _{\mathrm{h} \rightarrow 0} \frac{3 \mathrm{e}^{3 \mathrm{~h}}}{4}=\frac{\mathrm{k}}{4}\left(\right.\) Using \(L-\) Hospital rule in \(\frac{0}{0}\) form \()\) So, \(\mathrm{k}=3\)
COMEDK-2012
Limits, Continuity and Differentiability
80202
If \(\mathrm{f}(\mathrm{x})=\left\{\begin{array}{c}{[\mathrm{x}]+[-\mathrm{x}], \mathrm{x} \neq 2} \\ \mathrm{~K}, \mathrm{x}=2\end{array}\right.\) then \(f(\mathrm{x})\) is continuous at \(x=2\), provided \(K\) is equal to
1 2
2 1
3 -1
4 0
Explanation:
(C) : Given, \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=2\) \(\lim _{x \rightarrow 2^{+}} f(x)=\left[2^{+}\right]+\left[-2^{+}\right]=2-3=-1\) \(\begin{aligned} \lim _{x \rightarrow 2^{-}} f(x) =\left[2^{-}\right]+\left[-2^{-}\right] \\ =1-2 \\ =-1\end{aligned}\) For \(f(x)\) to be continuous \(\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{-}} f(x)=f(x)=k\) \(\therefore \quad k=-1\)
COMEDK-2014
Limits, Continuity and Differentiability
80203
If \(f(x)=\left\{\begin{array}{cr}x \sin 1 / x, & x \neq 0 \\ k \quad, & x=0\end{array}\right.\) is continuous at \(x=0\) , then the value of \(k\) will be
1 1
2 -1
3 0
4 None of these
Explanation:
(C) : Since \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=0\), so \(f(0)=\lim _{x \rightarrow 0} f(x)\) \(\mathrm{k} =\lim _{\mathrm{x} \rightarrow 0}(\mathrm{x} \sin 1 / \mathrm{x})\) \(=0 \times(\mathrm{a} \text { finite quantity })\) \(=0\)
80200
If \(f(x)\) and \(g(x)\) are two functions with \(g(x)=x-\frac{1}{x}\) and \(f o g(x)=x^{3}-\frac{1}{x^{3}}\), then \(f^{\prime}(x)=\)
1 \(3 x^{2}+3\)
2 \(x^{2}-\frac{1}{x^{2}}\)
3 \(1+\frac{1}{\mathrm{x}^{2}}\)
4 \(3 x^{2}+\frac{3}{x^{4}}\)
Explanation:
(A) : Given, \(g(x)=x-\frac{1}{x}\) And, \(f \circ g(x)=x^{3}-\frac{1}{x^{3}}\) Using \((a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)\) Then, \(\left(x-\frac{1}{x}\right)^{3}=x^{3}-\frac{1}{x^{3}}-3 x \cdot \frac{1}{x}\left(x-\frac{1}{x}\right)\) \(\mathrm{x}^{3}-\frac{1}{\mathrm{x}^{3}}=\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)^{3}+3\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)\) We have, \(f o g(x)=x^{3}-\frac{1}{x^{3}}\) \(f o g(x)=\left(x-\frac{1}{x}\right)^{3}+3\left(x-\frac{1}{x}\right)\) As \(g(x)=x-\frac{1}{x}\), this yields \(f\left(x-\frac{1}{x}\right)=\left(x-\frac{1}{x}\right)^{3}+3\left(x-\frac{1}{x}\right)\) Putting, \(x-\frac{1}{x}=\mathrm{t}\) \(f(t)=t^{3}+3 t\) Thus, \(f(x)=x^{3}+3 x\) \(f^{\prime}(x)=3 x^{2}+3\)
Karnataka CET-2006
Limits, Continuity and Differentiability
80201
If \(f(x)=\left\{\begin{array}{l}\frac{e^{3 x}-1}{4 x} \text { for } x \neq 0 \\ \frac{k+x}{4} \text { for } x=0\end{array}\right.\) is continuous at \(x=0\),
1 5
2 3
3 2
4 0
Explanation:
(B)\((x)=\left\{\begin{array}{ll}\frac{e^{3 x}-1}{4 x}, & \text { for } x \neq 0 \\ \frac{k+x}{4}, & \text { for } x=0\end{array}\right\}\) \(f(x) \text { is continuous at } x=0\) \(\therefore \lim _{\mathrm{x} \rightarrow 0} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow 0} \mathrm{f}(\mathrm{x})=\mathrm{f}(0)\) \(\lim _{x \rightarrow 0} \frac{e^{3 x}-1}{4 x}=\frac{k}{4}\) \(\lim _{\mathrm{h} \rightarrow 0} \frac{3 \mathrm{e}^{3 \mathrm{~h}}}{4}=\frac{\mathrm{k}}{4}\left(\right.\) Using \(L-\) Hospital rule in \(\frac{0}{0}\) form \()\) So, \(\mathrm{k}=3\)
COMEDK-2012
Limits, Continuity and Differentiability
80202
If \(\mathrm{f}(\mathrm{x})=\left\{\begin{array}{c}{[\mathrm{x}]+[-\mathrm{x}], \mathrm{x} \neq 2} \\ \mathrm{~K}, \mathrm{x}=2\end{array}\right.\) then \(f(\mathrm{x})\) is continuous at \(x=2\), provided \(K\) is equal to
1 2
2 1
3 -1
4 0
Explanation:
(C) : Given, \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=2\) \(\lim _{x \rightarrow 2^{+}} f(x)=\left[2^{+}\right]+\left[-2^{+}\right]=2-3=-1\) \(\begin{aligned} \lim _{x \rightarrow 2^{-}} f(x) =\left[2^{-}\right]+\left[-2^{-}\right] \\ =1-2 \\ =-1\end{aligned}\) For \(f(x)\) to be continuous \(\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{-}} f(x)=f(x)=k\) \(\therefore \quad k=-1\)
COMEDK-2014
Limits, Continuity and Differentiability
80203
If \(f(x)=\left\{\begin{array}{cr}x \sin 1 / x, & x \neq 0 \\ k \quad, & x=0\end{array}\right.\) is continuous at \(x=0\) , then the value of \(k\) will be
1 1
2 -1
3 0
4 None of these
Explanation:
(C) : Since \(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=0\), so \(f(0)=\lim _{x \rightarrow 0} f(x)\) \(\mathrm{k} =\lim _{\mathrm{x} \rightarrow 0}(\mathrm{x} \sin 1 / \mathrm{x})\) \(=0 \times(\mathrm{a} \text { finite quantity })\) \(=0\)
