79965
If \(y=\frac{1}{t^{2}-t-6}\) and \(t=\frac{1}{x-2}\), then the value of \(x\) which make the function \(y\) discontinuous are
1 \(2, \frac{2}{3}, \frac{7}{3}\)
2 \(2, \frac{3}{2}, \frac{7}{3}\)
3 \(2, \frac{3}{2}, \frac{3}{7}\)
4 None of the above
Explanation:
(B) : Given, \(y=\frac{1}{t^{2}-t-6} \text { and } t=\frac{1}{x-2}\) \(\because \mathrm{t}^{2}-\mathrm{t}-6=0\) \(\mathrm{t}^{2}-3 \mathrm{t}+2 \mathrm{t}-6=0\) \(\mathrm{t}(\mathrm{t}-3)+2(\mathrm{t}-3)=0\) \((\mathrm{t}-3)(\mathrm{t}+2)=0\) \(\mathrm{t}=3,-2\) When, \(\mathrm{t}=-2\) \(\frac{1}{x-2}=-2 \Rightarrow x-2=-\frac{1}{2}\) \(x=2-\frac{1}{2} \Rightarrow x=3 / 2\) When, \(\mathrm{t}=3\) \(\frac{1}{x-2}=3 \Rightarrow x-2=\frac{1}{3}\) \(x=2+\frac{1}{3} \Rightarrow x=7 / 3\) When, \(\mathrm{x}-2=0 \Rightarrow \mathrm{x}=2\) The value of \(x\) which make the function \(y\) discontinuous are \(\mathrm{x}=2,3 / 2\) and \(7 / 3\).
[JCECE-2012]
Limits, Continuity and Differentiability
79966
A function is defined as follows \(f(x)=\left\{\begin{array}{cc}x^{m} \sin \left(\frac{1}{x}\right), x \neq 0 \\ 0, x=0\end{array}\right\}\) what condition should be imposed on \(\mathrm{m}\), so that \(f(x)\) may be continuous at \(\mathrm{x}=0\) ?
1 \(\mathrm{m}>0\)
2 \(\mathrm{m}\lt 0\)
3 \(\mathrm{m}=0\)
4 any value of \(m\)
Explanation:
(A) : Given, \(f(x)=x^{m} \sin \left(\frac{1}{x}\right)\) \(f(0)=0\) For \(f(x)\) is continuous at \(x=0\) \(\lim _{x \rightarrow 0} f(x)=f(0)\) \(\lim _{x \rightarrow 0} x^{m} \sin \frac{1}{x}=0\) Which is possible only when \(\mathrm{m}>0\) Hence, \(f(x)\) is continuous when \(m>0\)
[JCECE-2011]
Limits, Continuity and Differentiability
79968
Let \(f(x)=\left\{\begin{array}{ll}\frac{\sin \pi x}{5 x}, & x \neq 0 \\ k, & x=0\end{array}\right.\) if \(f(x)\) is continuous at \(x=0\), then \(k\) is equal to
79965
If \(y=\frac{1}{t^{2}-t-6}\) and \(t=\frac{1}{x-2}\), then the value of \(x\) which make the function \(y\) discontinuous are
1 \(2, \frac{2}{3}, \frac{7}{3}\)
2 \(2, \frac{3}{2}, \frac{7}{3}\)
3 \(2, \frac{3}{2}, \frac{3}{7}\)
4 None of the above
Explanation:
(B) : Given, \(y=\frac{1}{t^{2}-t-6} \text { and } t=\frac{1}{x-2}\) \(\because \mathrm{t}^{2}-\mathrm{t}-6=0\) \(\mathrm{t}^{2}-3 \mathrm{t}+2 \mathrm{t}-6=0\) \(\mathrm{t}(\mathrm{t}-3)+2(\mathrm{t}-3)=0\) \((\mathrm{t}-3)(\mathrm{t}+2)=0\) \(\mathrm{t}=3,-2\) When, \(\mathrm{t}=-2\) \(\frac{1}{x-2}=-2 \Rightarrow x-2=-\frac{1}{2}\) \(x=2-\frac{1}{2} \Rightarrow x=3 / 2\) When, \(\mathrm{t}=3\) \(\frac{1}{x-2}=3 \Rightarrow x-2=\frac{1}{3}\) \(x=2+\frac{1}{3} \Rightarrow x=7 / 3\) When, \(\mathrm{x}-2=0 \Rightarrow \mathrm{x}=2\) The value of \(x\) which make the function \(y\) discontinuous are \(\mathrm{x}=2,3 / 2\) and \(7 / 3\).
[JCECE-2012]
Limits, Continuity and Differentiability
79966
A function is defined as follows \(f(x)=\left\{\begin{array}{cc}x^{m} \sin \left(\frac{1}{x}\right), x \neq 0 \\ 0, x=0\end{array}\right\}\) what condition should be imposed on \(\mathrm{m}\), so that \(f(x)\) may be continuous at \(\mathrm{x}=0\) ?
1 \(\mathrm{m}>0\)
2 \(\mathrm{m}\lt 0\)
3 \(\mathrm{m}=0\)
4 any value of \(m\)
Explanation:
(A) : Given, \(f(x)=x^{m} \sin \left(\frac{1}{x}\right)\) \(f(0)=0\) For \(f(x)\) is continuous at \(x=0\) \(\lim _{x \rightarrow 0} f(x)=f(0)\) \(\lim _{x \rightarrow 0} x^{m} \sin \frac{1}{x}=0\) Which is possible only when \(\mathrm{m}>0\) Hence, \(f(x)\) is continuous when \(m>0\)
[JCECE-2011]
Limits, Continuity and Differentiability
79968
Let \(f(x)=\left\{\begin{array}{ll}\frac{\sin \pi x}{5 x}, & x \neq 0 \\ k, & x=0\end{array}\right.\) if \(f(x)\) is continuous at \(x=0\), then \(k\) is equal to
79965
If \(y=\frac{1}{t^{2}-t-6}\) and \(t=\frac{1}{x-2}\), then the value of \(x\) which make the function \(y\) discontinuous are
1 \(2, \frac{2}{3}, \frac{7}{3}\)
2 \(2, \frac{3}{2}, \frac{7}{3}\)
3 \(2, \frac{3}{2}, \frac{3}{7}\)
4 None of the above
Explanation:
(B) : Given, \(y=\frac{1}{t^{2}-t-6} \text { and } t=\frac{1}{x-2}\) \(\because \mathrm{t}^{2}-\mathrm{t}-6=0\) \(\mathrm{t}^{2}-3 \mathrm{t}+2 \mathrm{t}-6=0\) \(\mathrm{t}(\mathrm{t}-3)+2(\mathrm{t}-3)=0\) \((\mathrm{t}-3)(\mathrm{t}+2)=0\) \(\mathrm{t}=3,-2\) When, \(\mathrm{t}=-2\) \(\frac{1}{x-2}=-2 \Rightarrow x-2=-\frac{1}{2}\) \(x=2-\frac{1}{2} \Rightarrow x=3 / 2\) When, \(\mathrm{t}=3\) \(\frac{1}{x-2}=3 \Rightarrow x-2=\frac{1}{3}\) \(x=2+\frac{1}{3} \Rightarrow x=7 / 3\) When, \(\mathrm{x}-2=0 \Rightarrow \mathrm{x}=2\) The value of \(x\) which make the function \(y\) discontinuous are \(\mathrm{x}=2,3 / 2\) and \(7 / 3\).
[JCECE-2012]
Limits, Continuity and Differentiability
79966
A function is defined as follows \(f(x)=\left\{\begin{array}{cc}x^{m} \sin \left(\frac{1}{x}\right), x \neq 0 \\ 0, x=0\end{array}\right\}\) what condition should be imposed on \(\mathrm{m}\), so that \(f(x)\) may be continuous at \(\mathrm{x}=0\) ?
1 \(\mathrm{m}>0\)
2 \(\mathrm{m}\lt 0\)
3 \(\mathrm{m}=0\)
4 any value of \(m\)
Explanation:
(A) : Given, \(f(x)=x^{m} \sin \left(\frac{1}{x}\right)\) \(f(0)=0\) For \(f(x)\) is continuous at \(x=0\) \(\lim _{x \rightarrow 0} f(x)=f(0)\) \(\lim _{x \rightarrow 0} x^{m} \sin \frac{1}{x}=0\) Which is possible only when \(\mathrm{m}>0\) Hence, \(f(x)\) is continuous when \(m>0\)
[JCECE-2011]
Limits, Continuity and Differentiability
79968
Let \(f(x)=\left\{\begin{array}{ll}\frac{\sin \pi x}{5 x}, & x \neq 0 \\ k, & x=0\end{array}\right.\) if \(f(x)\) is continuous at \(x=0\), then \(k\) is equal to
79965
If \(y=\frac{1}{t^{2}-t-6}\) and \(t=\frac{1}{x-2}\), then the value of \(x\) which make the function \(y\) discontinuous are
1 \(2, \frac{2}{3}, \frac{7}{3}\)
2 \(2, \frac{3}{2}, \frac{7}{3}\)
3 \(2, \frac{3}{2}, \frac{3}{7}\)
4 None of the above
Explanation:
(B) : Given, \(y=\frac{1}{t^{2}-t-6} \text { and } t=\frac{1}{x-2}\) \(\because \mathrm{t}^{2}-\mathrm{t}-6=0\) \(\mathrm{t}^{2}-3 \mathrm{t}+2 \mathrm{t}-6=0\) \(\mathrm{t}(\mathrm{t}-3)+2(\mathrm{t}-3)=0\) \((\mathrm{t}-3)(\mathrm{t}+2)=0\) \(\mathrm{t}=3,-2\) When, \(\mathrm{t}=-2\) \(\frac{1}{x-2}=-2 \Rightarrow x-2=-\frac{1}{2}\) \(x=2-\frac{1}{2} \Rightarrow x=3 / 2\) When, \(\mathrm{t}=3\) \(\frac{1}{x-2}=3 \Rightarrow x-2=\frac{1}{3}\) \(x=2+\frac{1}{3} \Rightarrow x=7 / 3\) When, \(\mathrm{x}-2=0 \Rightarrow \mathrm{x}=2\) The value of \(x\) which make the function \(y\) discontinuous are \(\mathrm{x}=2,3 / 2\) and \(7 / 3\).
[JCECE-2012]
Limits, Continuity and Differentiability
79966
A function is defined as follows \(f(x)=\left\{\begin{array}{cc}x^{m} \sin \left(\frac{1}{x}\right), x \neq 0 \\ 0, x=0\end{array}\right\}\) what condition should be imposed on \(\mathrm{m}\), so that \(f(x)\) may be continuous at \(\mathrm{x}=0\) ?
1 \(\mathrm{m}>0\)
2 \(\mathrm{m}\lt 0\)
3 \(\mathrm{m}=0\)
4 any value of \(m\)
Explanation:
(A) : Given, \(f(x)=x^{m} \sin \left(\frac{1}{x}\right)\) \(f(0)=0\) For \(f(x)\) is continuous at \(x=0\) \(\lim _{x \rightarrow 0} f(x)=f(0)\) \(\lim _{x \rightarrow 0} x^{m} \sin \frac{1}{x}=0\) Which is possible only when \(\mathrm{m}>0\) Hence, \(f(x)\) is continuous when \(m>0\)
[JCECE-2011]
Limits, Continuity and Differentiability
79968
Let \(f(x)=\left\{\begin{array}{ll}\frac{\sin \pi x}{5 x}, & x \neq 0 \\ k, & x=0\end{array}\right.\) if \(f(x)\) is continuous at \(x=0\), then \(k\) is equal to