80000
Define \(f(x)=\left\{\begin{array}{cc}x^{2}+b x+c, x\lt 1 \\ x, x \geq 1\end{array}\right.\). If \(f(x)\) is differentiable at \(x=1\), then \((b-c)=\)
1 -2
2 0
3 1
4 2
Explanation:
(A) : Given, \(f(x)=\left\{\begin{array}{cc}x^{2}+b x+c, x\lt 1 \\ x, x \geq 1\end{array}\right.\) \(\mathrm{f}(\mathrm{x})\) is differentiable at \(\mathrm{x}=1\) \(f^{\prime}(x)=\left\{\begin{array}{cc}2 x+b, x\lt 1 \\ 1, x \geq 1\end{array}\right.\) Now, \(\mathrm{f}(\mathrm{x})\) is differentiable at \(\mathrm{x}=1\) \(\mathrm{f}^{\prime}\left(1^{-}\right)=\mathrm{f}^{\prime \prime}\left(1^{+}\right)\) \(2+b=1 \Rightarrow b=-1\) \(f(x)\) is continuous at \(x=1\) \(\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=f(1)\) \(\lim _{x \rightarrow 1^{-}} x^{2}+b x+c=\lim _{x \rightarrow 1^{+}} x=1\) \(1+\mathrm{b}+\mathrm{c}=1\) \(1-1+\mathrm{c}=1 \Rightarrow \mathrm{c}=1\) Hence, \(\mathrm{b}-\mathrm{c}=-1-1 \Rightarrow \mathrm{b}-\mathrm{c}=-2\)
AP EAMCET-2016
Limits, Continuity and Differentiability
80001
If \(f(x)=\frac{x}{2}-1\), then on the interval \([0, \pi]\)
1 \(\tan [f(\mathrm{x})]\) and \(\frac{1}{f(\mathrm{x})}\) are both continuous
2 \(\tan [f(\mathrm{x})]\) and \(\frac{1}{f(\mathrm{x})}\) are both discontinuous
3 \(\tan [f(\mathrm{x})]\) is continuous but \(\frac{1}{f(\mathrm{x})}\) is not continuous
4 \(\tan [f(x)]\) is not continuous but \(\frac{1}{f(x)}\) is continuous
Explanation:
(C) : We have, \(f(x)=\frac{x}{2}-1 \Rightarrow \frac{x-2}{2}\) \(\frac{1}{f(x)}=\frac{2}{x-2}\) \(\tan \{f(x)\}=\tan \left[\frac{x-2}{2}\right]\) Here, \(\frac{1}{\mathrm{f}(\mathrm{x})}\) is not continuous of \(\mathrm{x}=2 \in[0, \pi]\) Also, \(\tan \theta\) is discontinuous at \(\theta=\frac{-\pi}{2}\) and \(\frac{\pi}{2}\) \(\frac{x-2}{2}=\frac{\pi}{2}\) \(x=2+\pi \in[0, \pi]\) And, \(\quad \frac{x-2}{2}=\frac{-\pi}{2}\) \(\mathrm{x}=2-\pi \in[0, \pi]\) So, \(\tan \mathrm{f}(\mathrm{x})\) is continuous when \(\mathrm{x} \in[0, \pi]\).
AMU-2016
Limits, Continuity and Differentiability
80002
The function \(f(x)=\frac{\ln (1+a x)-\ln (1-b x)}{x}\) is not defined at \(x=0\). The value which should be assigned to \(f\) at \(x=0\), so that it is continuous at \(\mathbf{x}=\mathbf{0}\), is;
80003
Let \(f(\mathrm{x})\) be a function such that \(f(\mathrm{x}+\mathrm{y})=f(\mathrm{x})+\) \(f(y)\) and \(f(x)=\sin x g(x)\) for \(x, y \in R\). If \(g(x)\) is a continuous function such that \(g(0)=C\) then \(f^{\prime}(\mathbf{x})=\)
80000
Define \(f(x)=\left\{\begin{array}{cc}x^{2}+b x+c, x\lt 1 \\ x, x \geq 1\end{array}\right.\). If \(f(x)\) is differentiable at \(x=1\), then \((b-c)=\)
1 -2
2 0
3 1
4 2
Explanation:
(A) : Given, \(f(x)=\left\{\begin{array}{cc}x^{2}+b x+c, x\lt 1 \\ x, x \geq 1\end{array}\right.\) \(\mathrm{f}(\mathrm{x})\) is differentiable at \(\mathrm{x}=1\) \(f^{\prime}(x)=\left\{\begin{array}{cc}2 x+b, x\lt 1 \\ 1, x \geq 1\end{array}\right.\) Now, \(\mathrm{f}(\mathrm{x})\) is differentiable at \(\mathrm{x}=1\) \(\mathrm{f}^{\prime}\left(1^{-}\right)=\mathrm{f}^{\prime \prime}\left(1^{+}\right)\) \(2+b=1 \Rightarrow b=-1\) \(f(x)\) is continuous at \(x=1\) \(\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=f(1)\) \(\lim _{x \rightarrow 1^{-}} x^{2}+b x+c=\lim _{x \rightarrow 1^{+}} x=1\) \(1+\mathrm{b}+\mathrm{c}=1\) \(1-1+\mathrm{c}=1 \Rightarrow \mathrm{c}=1\) Hence, \(\mathrm{b}-\mathrm{c}=-1-1 \Rightarrow \mathrm{b}-\mathrm{c}=-2\)
AP EAMCET-2016
Limits, Continuity and Differentiability
80001
If \(f(x)=\frac{x}{2}-1\), then on the interval \([0, \pi]\)
1 \(\tan [f(\mathrm{x})]\) and \(\frac{1}{f(\mathrm{x})}\) are both continuous
2 \(\tan [f(\mathrm{x})]\) and \(\frac{1}{f(\mathrm{x})}\) are both discontinuous
3 \(\tan [f(\mathrm{x})]\) is continuous but \(\frac{1}{f(\mathrm{x})}\) is not continuous
4 \(\tan [f(x)]\) is not continuous but \(\frac{1}{f(x)}\) is continuous
Explanation:
(C) : We have, \(f(x)=\frac{x}{2}-1 \Rightarrow \frac{x-2}{2}\) \(\frac{1}{f(x)}=\frac{2}{x-2}\) \(\tan \{f(x)\}=\tan \left[\frac{x-2}{2}\right]\) Here, \(\frac{1}{\mathrm{f}(\mathrm{x})}\) is not continuous of \(\mathrm{x}=2 \in[0, \pi]\) Also, \(\tan \theta\) is discontinuous at \(\theta=\frac{-\pi}{2}\) and \(\frac{\pi}{2}\) \(\frac{x-2}{2}=\frac{\pi}{2}\) \(x=2+\pi \in[0, \pi]\) And, \(\quad \frac{x-2}{2}=\frac{-\pi}{2}\) \(\mathrm{x}=2-\pi \in[0, \pi]\) So, \(\tan \mathrm{f}(\mathrm{x})\) is continuous when \(\mathrm{x} \in[0, \pi]\).
AMU-2016
Limits, Continuity and Differentiability
80002
The function \(f(x)=\frac{\ln (1+a x)-\ln (1-b x)}{x}\) is not defined at \(x=0\). The value which should be assigned to \(f\) at \(x=0\), so that it is continuous at \(\mathbf{x}=\mathbf{0}\), is;
80003
Let \(f(\mathrm{x})\) be a function such that \(f(\mathrm{x}+\mathrm{y})=f(\mathrm{x})+\) \(f(y)\) and \(f(x)=\sin x g(x)\) for \(x, y \in R\). If \(g(x)\) is a continuous function such that \(g(0)=C\) then \(f^{\prime}(\mathbf{x})=\)
80000
Define \(f(x)=\left\{\begin{array}{cc}x^{2}+b x+c, x\lt 1 \\ x, x \geq 1\end{array}\right.\). If \(f(x)\) is differentiable at \(x=1\), then \((b-c)=\)
1 -2
2 0
3 1
4 2
Explanation:
(A) : Given, \(f(x)=\left\{\begin{array}{cc}x^{2}+b x+c, x\lt 1 \\ x, x \geq 1\end{array}\right.\) \(\mathrm{f}(\mathrm{x})\) is differentiable at \(\mathrm{x}=1\) \(f^{\prime}(x)=\left\{\begin{array}{cc}2 x+b, x\lt 1 \\ 1, x \geq 1\end{array}\right.\) Now, \(\mathrm{f}(\mathrm{x})\) is differentiable at \(\mathrm{x}=1\) \(\mathrm{f}^{\prime}\left(1^{-}\right)=\mathrm{f}^{\prime \prime}\left(1^{+}\right)\) \(2+b=1 \Rightarrow b=-1\) \(f(x)\) is continuous at \(x=1\) \(\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=f(1)\) \(\lim _{x \rightarrow 1^{-}} x^{2}+b x+c=\lim _{x \rightarrow 1^{+}} x=1\) \(1+\mathrm{b}+\mathrm{c}=1\) \(1-1+\mathrm{c}=1 \Rightarrow \mathrm{c}=1\) Hence, \(\mathrm{b}-\mathrm{c}=-1-1 \Rightarrow \mathrm{b}-\mathrm{c}=-2\)
AP EAMCET-2016
Limits, Continuity and Differentiability
80001
If \(f(x)=\frac{x}{2}-1\), then on the interval \([0, \pi]\)
1 \(\tan [f(\mathrm{x})]\) and \(\frac{1}{f(\mathrm{x})}\) are both continuous
2 \(\tan [f(\mathrm{x})]\) and \(\frac{1}{f(\mathrm{x})}\) are both discontinuous
3 \(\tan [f(\mathrm{x})]\) is continuous but \(\frac{1}{f(\mathrm{x})}\) is not continuous
4 \(\tan [f(x)]\) is not continuous but \(\frac{1}{f(x)}\) is continuous
Explanation:
(C) : We have, \(f(x)=\frac{x}{2}-1 \Rightarrow \frac{x-2}{2}\) \(\frac{1}{f(x)}=\frac{2}{x-2}\) \(\tan \{f(x)\}=\tan \left[\frac{x-2}{2}\right]\) Here, \(\frac{1}{\mathrm{f}(\mathrm{x})}\) is not continuous of \(\mathrm{x}=2 \in[0, \pi]\) Also, \(\tan \theta\) is discontinuous at \(\theta=\frac{-\pi}{2}\) and \(\frac{\pi}{2}\) \(\frac{x-2}{2}=\frac{\pi}{2}\) \(x=2+\pi \in[0, \pi]\) And, \(\quad \frac{x-2}{2}=\frac{-\pi}{2}\) \(\mathrm{x}=2-\pi \in[0, \pi]\) So, \(\tan \mathrm{f}(\mathrm{x})\) is continuous when \(\mathrm{x} \in[0, \pi]\).
AMU-2016
Limits, Continuity and Differentiability
80002
The function \(f(x)=\frac{\ln (1+a x)-\ln (1-b x)}{x}\) is not defined at \(x=0\). The value which should be assigned to \(f\) at \(x=0\), so that it is continuous at \(\mathbf{x}=\mathbf{0}\), is;
80003
Let \(f(\mathrm{x})\) be a function such that \(f(\mathrm{x}+\mathrm{y})=f(\mathrm{x})+\) \(f(y)\) and \(f(x)=\sin x g(x)\) for \(x, y \in R\). If \(g(x)\) is a continuous function such that \(g(0)=C\) then \(f^{\prime}(\mathbf{x})=\)
80000
Define \(f(x)=\left\{\begin{array}{cc}x^{2}+b x+c, x\lt 1 \\ x, x \geq 1\end{array}\right.\). If \(f(x)\) is differentiable at \(x=1\), then \((b-c)=\)
1 -2
2 0
3 1
4 2
Explanation:
(A) : Given, \(f(x)=\left\{\begin{array}{cc}x^{2}+b x+c, x\lt 1 \\ x, x \geq 1\end{array}\right.\) \(\mathrm{f}(\mathrm{x})\) is differentiable at \(\mathrm{x}=1\) \(f^{\prime}(x)=\left\{\begin{array}{cc}2 x+b, x\lt 1 \\ 1, x \geq 1\end{array}\right.\) Now, \(\mathrm{f}(\mathrm{x})\) is differentiable at \(\mathrm{x}=1\) \(\mathrm{f}^{\prime}\left(1^{-}\right)=\mathrm{f}^{\prime \prime}\left(1^{+}\right)\) \(2+b=1 \Rightarrow b=-1\) \(f(x)\) is continuous at \(x=1\) \(\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=f(1)\) \(\lim _{x \rightarrow 1^{-}} x^{2}+b x+c=\lim _{x \rightarrow 1^{+}} x=1\) \(1+\mathrm{b}+\mathrm{c}=1\) \(1-1+\mathrm{c}=1 \Rightarrow \mathrm{c}=1\) Hence, \(\mathrm{b}-\mathrm{c}=-1-1 \Rightarrow \mathrm{b}-\mathrm{c}=-2\)
AP EAMCET-2016
Limits, Continuity and Differentiability
80001
If \(f(x)=\frac{x}{2}-1\), then on the interval \([0, \pi]\)
1 \(\tan [f(\mathrm{x})]\) and \(\frac{1}{f(\mathrm{x})}\) are both continuous
2 \(\tan [f(\mathrm{x})]\) and \(\frac{1}{f(\mathrm{x})}\) are both discontinuous
3 \(\tan [f(\mathrm{x})]\) is continuous but \(\frac{1}{f(\mathrm{x})}\) is not continuous
4 \(\tan [f(x)]\) is not continuous but \(\frac{1}{f(x)}\) is continuous
Explanation:
(C) : We have, \(f(x)=\frac{x}{2}-1 \Rightarrow \frac{x-2}{2}\) \(\frac{1}{f(x)}=\frac{2}{x-2}\) \(\tan \{f(x)\}=\tan \left[\frac{x-2}{2}\right]\) Here, \(\frac{1}{\mathrm{f}(\mathrm{x})}\) is not continuous of \(\mathrm{x}=2 \in[0, \pi]\) Also, \(\tan \theta\) is discontinuous at \(\theta=\frac{-\pi}{2}\) and \(\frac{\pi}{2}\) \(\frac{x-2}{2}=\frac{\pi}{2}\) \(x=2+\pi \in[0, \pi]\) And, \(\quad \frac{x-2}{2}=\frac{-\pi}{2}\) \(\mathrm{x}=2-\pi \in[0, \pi]\) So, \(\tan \mathrm{f}(\mathrm{x})\) is continuous when \(\mathrm{x} \in[0, \pi]\).
AMU-2016
Limits, Continuity and Differentiability
80002
The function \(f(x)=\frac{\ln (1+a x)-\ln (1-b x)}{x}\) is not defined at \(x=0\). The value which should be assigned to \(f\) at \(x=0\), so that it is continuous at \(\mathbf{x}=\mathbf{0}\), is;
80003
Let \(f(\mathrm{x})\) be a function such that \(f(\mathrm{x}+\mathrm{y})=f(\mathrm{x})+\) \(f(y)\) and \(f(x)=\sin x g(x)\) for \(x, y \in R\). If \(g(x)\) is a continuous function such that \(g(0)=C\) then \(f^{\prime}(\mathbf{x})=\)
