79973
The values of \(p\) and \(q\) so that \(f(x)=\left\{\begin{array}{cc}x^2+3 x+p, & x \leq 1 \\ q x+2, & x>1\end{array}\right.\) is differentiable at \(x=1\) are respectively
79974
The value of \(\mathrm{k}\), so that the function \(f\) defined \(f(x)=\left\{\begin{array}{cc}\frac{1-\cos 4 x}{8 x^{2}}, x \neq 0 \\ k x=0\end{array}\right.\) becomes continuous at \(x=0\), is
79975
If \(\left\{\begin{array}{cl}x e^{-\left(\frac{1}{|x|}+\frac{1}{x}\right),} \text { if } x \neq 0 \text {, then } \\ 0 \text { if } x=0\end{array}\right.\) which of the following is correct?
1 \(f(x)\) is continuous and \(f^{\prime}(0)\) does not exist
2 \(f(x)\) is not continuous
3 \(f(x)\) is continuous and \(f^{\prime}(0)\) also exist
4 None of the above
Explanation:
(A) : Given, \(f(x)=x e^{-\left(\frac{1}{|x|}+\frac{1}{x}\right)}\) \(f(0)=0\) Let, \(\quad \mathrm{x}=\mathrm{h}, \mathrm{h} \rightarrow 0\) \(f(h)=h e^{-\left(\frac{1}{h}+\frac{1}{h}\right)}\) \(f(h)=h e^{-\frac{2}{h}}\) \(\mathrm{f}(0)=0\) And, \(\quad \mathrm{f}(-\mathrm{h})=-h \mathrm{e}^{-\left(-\frac{1}{\mathrm{~h}}+\frac{1}{\mathrm{~h}}\right)}\) \(f(-h)=-h e^{0}\) \(\mathrm{f}(-0)=0\) The function is continuous at \(\mathrm{x}=0\) \(f^{\prime}\left(0^{+}\right)=\lim _{h \rightarrow 0} \frac{h e^{-2 / h}}{h}=\lim _{h \rightarrow 0} e^{-2 / h}=0\) \(f^{\prime}\left(0^{-}\right)=\lim _{h \rightarrow 0} \frac{-h e^{0}}{-h}=e^{0}=1\) \(\because \quad \mathrm{f}^{\prime}\left(0^{+}\right) \neq \mathrm{f}^{\prime}\left(0^{-}\right)\) So, \(\mathrm{f}^{\prime}(\mathrm{x})\) does not exists
Limits, Continuity and Differentiability
79976
If \(f(x)=\cos ^{-1}\left\{\frac{1-\left(\log _{e} x\right)^{2}}{1+\left(\log _{e} x\right)^{2}}\right\}\), then \(f^{\prime}(e)\)
79973
The values of \(p\) and \(q\) so that \(f(x)=\left\{\begin{array}{cc}x^2+3 x+p, & x \leq 1 \\ q x+2, & x>1\end{array}\right.\) is differentiable at \(x=1\) are respectively
79974
The value of \(\mathrm{k}\), so that the function \(f\) defined \(f(x)=\left\{\begin{array}{cc}\frac{1-\cos 4 x}{8 x^{2}}, x \neq 0 \\ k x=0\end{array}\right.\) becomes continuous at \(x=0\), is
79975
If \(\left\{\begin{array}{cl}x e^{-\left(\frac{1}{|x|}+\frac{1}{x}\right),} \text { if } x \neq 0 \text {, then } \\ 0 \text { if } x=0\end{array}\right.\) which of the following is correct?
1 \(f(x)\) is continuous and \(f^{\prime}(0)\) does not exist
2 \(f(x)\) is not continuous
3 \(f(x)\) is continuous and \(f^{\prime}(0)\) also exist
4 None of the above
Explanation:
(A) : Given, \(f(x)=x e^{-\left(\frac{1}{|x|}+\frac{1}{x}\right)}\) \(f(0)=0\) Let, \(\quad \mathrm{x}=\mathrm{h}, \mathrm{h} \rightarrow 0\) \(f(h)=h e^{-\left(\frac{1}{h}+\frac{1}{h}\right)}\) \(f(h)=h e^{-\frac{2}{h}}\) \(\mathrm{f}(0)=0\) And, \(\quad \mathrm{f}(-\mathrm{h})=-h \mathrm{e}^{-\left(-\frac{1}{\mathrm{~h}}+\frac{1}{\mathrm{~h}}\right)}\) \(f(-h)=-h e^{0}\) \(\mathrm{f}(-0)=0\) The function is continuous at \(\mathrm{x}=0\) \(f^{\prime}\left(0^{+}\right)=\lim _{h \rightarrow 0} \frac{h e^{-2 / h}}{h}=\lim _{h \rightarrow 0} e^{-2 / h}=0\) \(f^{\prime}\left(0^{-}\right)=\lim _{h \rightarrow 0} \frac{-h e^{0}}{-h}=e^{0}=1\) \(\because \quad \mathrm{f}^{\prime}\left(0^{+}\right) \neq \mathrm{f}^{\prime}\left(0^{-}\right)\) So, \(\mathrm{f}^{\prime}(\mathrm{x})\) does not exists
Limits, Continuity and Differentiability
79976
If \(f(x)=\cos ^{-1}\left\{\frac{1-\left(\log _{e} x\right)^{2}}{1+\left(\log _{e} x\right)^{2}}\right\}\), then \(f^{\prime}(e)\)
79973
The values of \(p\) and \(q\) so that \(f(x)=\left\{\begin{array}{cc}x^2+3 x+p, & x \leq 1 \\ q x+2, & x>1\end{array}\right.\) is differentiable at \(x=1\) are respectively
79974
The value of \(\mathrm{k}\), so that the function \(f\) defined \(f(x)=\left\{\begin{array}{cc}\frac{1-\cos 4 x}{8 x^{2}}, x \neq 0 \\ k x=0\end{array}\right.\) becomes continuous at \(x=0\), is
79975
If \(\left\{\begin{array}{cl}x e^{-\left(\frac{1}{|x|}+\frac{1}{x}\right),} \text { if } x \neq 0 \text {, then } \\ 0 \text { if } x=0\end{array}\right.\) which of the following is correct?
1 \(f(x)\) is continuous and \(f^{\prime}(0)\) does not exist
2 \(f(x)\) is not continuous
3 \(f(x)\) is continuous and \(f^{\prime}(0)\) also exist
4 None of the above
Explanation:
(A) : Given, \(f(x)=x e^{-\left(\frac{1}{|x|}+\frac{1}{x}\right)}\) \(f(0)=0\) Let, \(\quad \mathrm{x}=\mathrm{h}, \mathrm{h} \rightarrow 0\) \(f(h)=h e^{-\left(\frac{1}{h}+\frac{1}{h}\right)}\) \(f(h)=h e^{-\frac{2}{h}}\) \(\mathrm{f}(0)=0\) And, \(\quad \mathrm{f}(-\mathrm{h})=-h \mathrm{e}^{-\left(-\frac{1}{\mathrm{~h}}+\frac{1}{\mathrm{~h}}\right)}\) \(f(-h)=-h e^{0}\) \(\mathrm{f}(-0)=0\) The function is continuous at \(\mathrm{x}=0\) \(f^{\prime}\left(0^{+}\right)=\lim _{h \rightarrow 0} \frac{h e^{-2 / h}}{h}=\lim _{h \rightarrow 0} e^{-2 / h}=0\) \(f^{\prime}\left(0^{-}\right)=\lim _{h \rightarrow 0} \frac{-h e^{0}}{-h}=e^{0}=1\) \(\because \quad \mathrm{f}^{\prime}\left(0^{+}\right) \neq \mathrm{f}^{\prime}\left(0^{-}\right)\) So, \(\mathrm{f}^{\prime}(\mathrm{x})\) does not exists
Limits, Continuity and Differentiability
79976
If \(f(x)=\cos ^{-1}\left\{\frac{1-\left(\log _{e} x\right)^{2}}{1+\left(\log _{e} x\right)^{2}}\right\}\), then \(f^{\prime}(e)\)
79973
The values of \(p\) and \(q\) so that \(f(x)=\left\{\begin{array}{cc}x^2+3 x+p, & x \leq 1 \\ q x+2, & x>1\end{array}\right.\) is differentiable at \(x=1\) are respectively
79974
The value of \(\mathrm{k}\), so that the function \(f\) defined \(f(x)=\left\{\begin{array}{cc}\frac{1-\cos 4 x}{8 x^{2}}, x \neq 0 \\ k x=0\end{array}\right.\) becomes continuous at \(x=0\), is
79975
If \(\left\{\begin{array}{cl}x e^{-\left(\frac{1}{|x|}+\frac{1}{x}\right),} \text { if } x \neq 0 \text {, then } \\ 0 \text { if } x=0\end{array}\right.\) which of the following is correct?
1 \(f(x)\) is continuous and \(f^{\prime}(0)\) does not exist
2 \(f(x)\) is not continuous
3 \(f(x)\) is continuous and \(f^{\prime}(0)\) also exist
4 None of the above
Explanation:
(A) : Given, \(f(x)=x e^{-\left(\frac{1}{|x|}+\frac{1}{x}\right)}\) \(f(0)=0\) Let, \(\quad \mathrm{x}=\mathrm{h}, \mathrm{h} \rightarrow 0\) \(f(h)=h e^{-\left(\frac{1}{h}+\frac{1}{h}\right)}\) \(f(h)=h e^{-\frac{2}{h}}\) \(\mathrm{f}(0)=0\) And, \(\quad \mathrm{f}(-\mathrm{h})=-h \mathrm{e}^{-\left(-\frac{1}{\mathrm{~h}}+\frac{1}{\mathrm{~h}}\right)}\) \(f(-h)=-h e^{0}\) \(\mathrm{f}(-0)=0\) The function is continuous at \(\mathrm{x}=0\) \(f^{\prime}\left(0^{+}\right)=\lim _{h \rightarrow 0} \frac{h e^{-2 / h}}{h}=\lim _{h \rightarrow 0} e^{-2 / h}=0\) \(f^{\prime}\left(0^{-}\right)=\lim _{h \rightarrow 0} \frac{-h e^{0}}{-h}=e^{0}=1\) \(\because \quad \mathrm{f}^{\prime}\left(0^{+}\right) \neq \mathrm{f}^{\prime}\left(0^{-}\right)\) So, \(\mathrm{f}^{\prime}(\mathrm{x})\) does not exists
Limits, Continuity and Differentiability
79976
If \(f(x)=\cos ^{-1}\left\{\frac{1-\left(\log _{e} x\right)^{2}}{1+\left(\log _{e} x\right)^{2}}\right\}\), then \(f^{\prime}(e)\)
79973
The values of \(p\) and \(q\) so that \(f(x)=\left\{\begin{array}{cc}x^2+3 x+p, & x \leq 1 \\ q x+2, & x>1\end{array}\right.\) is differentiable at \(x=1\) are respectively
79974
The value of \(\mathrm{k}\), so that the function \(f\) defined \(f(x)=\left\{\begin{array}{cc}\frac{1-\cos 4 x}{8 x^{2}}, x \neq 0 \\ k x=0\end{array}\right.\) becomes continuous at \(x=0\), is
79975
If \(\left\{\begin{array}{cl}x e^{-\left(\frac{1}{|x|}+\frac{1}{x}\right),} \text { if } x \neq 0 \text {, then } \\ 0 \text { if } x=0\end{array}\right.\) which of the following is correct?
1 \(f(x)\) is continuous and \(f^{\prime}(0)\) does not exist
2 \(f(x)\) is not continuous
3 \(f(x)\) is continuous and \(f^{\prime}(0)\) also exist
4 None of the above
Explanation:
(A) : Given, \(f(x)=x e^{-\left(\frac{1}{|x|}+\frac{1}{x}\right)}\) \(f(0)=0\) Let, \(\quad \mathrm{x}=\mathrm{h}, \mathrm{h} \rightarrow 0\) \(f(h)=h e^{-\left(\frac{1}{h}+\frac{1}{h}\right)}\) \(f(h)=h e^{-\frac{2}{h}}\) \(\mathrm{f}(0)=0\) And, \(\quad \mathrm{f}(-\mathrm{h})=-h \mathrm{e}^{-\left(-\frac{1}{\mathrm{~h}}+\frac{1}{\mathrm{~h}}\right)}\) \(f(-h)=-h e^{0}\) \(\mathrm{f}(-0)=0\) The function is continuous at \(\mathrm{x}=0\) \(f^{\prime}\left(0^{+}\right)=\lim _{h \rightarrow 0} \frac{h e^{-2 / h}}{h}=\lim _{h \rightarrow 0} e^{-2 / h}=0\) \(f^{\prime}\left(0^{-}\right)=\lim _{h \rightarrow 0} \frac{-h e^{0}}{-h}=e^{0}=1\) \(\because \quad \mathrm{f}^{\prime}\left(0^{+}\right) \neq \mathrm{f}^{\prime}\left(0^{-}\right)\) So, \(\mathrm{f}^{\prime}(\mathrm{x})\) does not exists
Limits, Continuity and Differentiability
79976
If \(f(x)=\cos ^{-1}\left\{\frac{1-\left(\log _{e} x\right)^{2}}{1+\left(\log _{e} x\right)^{2}}\right\}\), then \(f^{\prime}(e)\)
