80020
\(\lim _{x \rightarrow 0}(\sin x)^{2 \tan x}\) is equal to
1 2
2 1
3 0
4 does not exist
Explanation:
(B) : Let, \(y=\lim _{x \rightarrow 0}(\sin x)^{2 \tan x}\) \(\log y=\lim _{x \rightarrow 0} \log (\sin x)^{2 t}\) \(\log y=2 \lim _{x \rightarrow 0} \tan x \log \sin x\) \(\log y=2 \lim _{x \rightarrow 0} \frac{\log \sin x}{\cot x} \quad\left[\frac{0}{0} \text { form }\right]\) On using L- Hospital rule, we get - \(\log y=2 \lim _{x \rightarrow 0} \frac{\frac{1}{\sin x} \cdot \cos x}{-\operatorname{cosec}^{2} x}\) \(\log y=2 \lim _{x \rightarrow 0}(-\sin x \cos x)\) \(\log y=2 \times 0=0\) \(\therefore \quad \log \mathrm{y}_{0}=0\) \(\mathrm{y}=\mathrm{e}^{0}=1\) \(\left\{-2 \sin x, \quad\right.\) if \(x \leq-\frac{\pi}{2}\)
WB JEE-2017
Limits, Continuity and Differentiability
80021
Let \(f(x)=\left\{A \sin x+B\right.\), if \(-\frac{\pi}{2}\lt x\lt \frac{\pi}{2}\). Then, \(\cos x, \quad \text { if } x \geq \frac{\pi}{2}\)
1 \(f\) is discontinuous for all \(\mathrm{A}\) and \(\mathrm{B}\)
2 \(f\) is continuous for all \(A=-1\) and \(B=1\)
3 \(f\) is continuous for all \(A=1\) and \(B=-1\)
4 \(f\) is continuous for all real values of \(A, B\)
Explanation:
(B) : We have, \(f(x)=\left\{\begin{array}{cc}-2 \sin x, \text { if } x \leq-\frac{\pi}{2} \\ A \sin x+B, \text { if }-\frac{\pi}{2}\lt x\lt \frac{\pi}{2} \\ \cos x, \text { if } x \geq \frac{\pi}{2}\end{array}\right.\) At, \(\quad \mathrm{x}=-\frac{\pi}{2}\) \(\mathrm{RHL}=-\mathrm{A}+\mathrm{B}\) For continuity, \(\mathrm{LHL}=\mathrm{RHL}=\mathrm{f}\left(-\frac{\pi}{2}\right)\) \(=-A+B=2 \tag{i}\) At, \(\quad \mathrm{x}=\frac{\pi}{2}\) \(\mathrm{LHL}=\mathrm{A}+\mathrm{B}\) \(\mathrm{RHL}=\mathrm{O}\) For continuity, \(\mathrm{LHL}=\mathrm{RHL}=\mathrm{f}\left(\frac{\pi}{2}\right)\) \(A+B=0 \tag{ii}\) On solving equation (i) and (ii), we get - \(\mathrm{A}=-1 \text { and } \mathrm{B}=1\)
WB JEE-2018
Limits, Continuity and Differentiability
80022
Let \(f:[\mathrm{a}, \mathrm{b}] \rightarrow \mathrm{R}\) be such that \(f\) is differentiable in \((a, b), f\) is continuous at \(x=a\) and \(x=b\) and moreover \(f(\mathrm{a})=0=f(\mathrm{~b})\). Then
1 There exists at least one point \(\mathrm{c}\) in \((\mathrm{a}, \mathrm{b})\) such that \(f^{\prime}(c)=f(c)\)
2 \(f^{\prime}(x)=f(x)\) does not hold at any point in (a, b
3 at every point of (a, b), \(f^{\prime}(x)>f(x)\)
4 at every point of (a,b), \(f^{\prime}(x)\lt f(x)\)
Explanation:
(A) : Let \(\mathrm{g}(\mathrm{x})=\mathrm{e}^{-\mathrm{x}} \mathrm{f}(\mathrm{x})\) Such that \(g(a)=0, \quad g(b)=0\) And \(\mathrm{g}(\mathrm{x})\) is continuous and differentiable Then, For at least one value of \(c \in(a, b)\). such that \(g^{\prime}\) (c) \(=0\) New, \(\quad g^{\prime}(x)=e^{-x} f^{\prime}(x)+\left(-e^{-x}\right) f(x)\) \(g^{\prime}(c)=e^{-c} f^{\prime}(c)+\left(-e^{-c}\right) f(c)=0\) \(e^{-c^{-}} f^{\prime}(c)=e^{-c^{\prime}} f^{\prime}(c)\) \(f^{\prime}(c)=f(c)\)
Limits, Continuity and Differentiability
80023
Let \(f ;[\mathrm{a}, \mathrm{b}] \rightarrow \mathrm{R}\) be differentiable on \([\mathrm{a}, \mathrm{b}]\) and \(k \in R\). Let \(f(a)=0=f(b)\). Also let \(\mathrm{J}(\mathrm{x})=\boldsymbol{f}^{\prime}(\mathrm{x})+\mathbf{k f}(\mathrm{x})\). Then
1 \(J(x)>0\) for all \(x \in[a, b]\)
2 \(\mathrm{J}(\mathrm{x})\lt 0\) for all \(\mathrm{x} \in[\mathrm{a}, \mathrm{b}]\)
3 \(J(x)=0\) has at least one root in (a,b)
4 \(\mathrm{J}(\mathrm{x})>0\) through \([\mathrm{a}, \mathrm{b}]\)
Explanation:
(C) : We have, \(\mathrm{f}:[\mathrm{a}, \mathrm{b}] \rightarrow \mathrm{R}\) be differentiable on \([\mathrm{a}, \mathrm{b}]\) and \(\mathrm{k} \in \mathrm{R}\), also \(\mathrm{f}(\mathrm{a})=0=\mathrm{f}(\mathrm{b})\) And \(\mathrm{J}(\mathrm{x})=\mathrm{f}^{\prime}(\mathrm{x})+\mathrm{kf}(\mathrm{x})\) Let \(g(x)=\operatorname{kxf}(\mathrm{x})\) which is continuous in \([\mathrm{a}, \mathrm{b}]\) and differentiable in \((a, b)\) such that \(g(a)=0=g(b)\) Then, for every \(c \in(a, b), g^{\prime}(c)=0\) [by Rolle's theorem] Now, \(\quad g^{\prime}(x)=k f(x)+\operatorname{kxf}^{\prime}(x)\) \(\mathrm{g}^{\prime}(\mathrm{c})=\mathrm{kf}(\mathrm{c})+\mathrm{kcf}^{\prime}(\mathrm{c})\) \(\mathrm{kf}(\mathrm{c})+\mathrm{kcf}^{\prime}(\mathrm{c})=0\) \(f(x)=0\) for every \(x=c \in(a, b)\) \(\therefore \mathrm{J}(\mathrm{x})=0\) has at least one root in (a,b)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Limits, Continuity and Differentiability
80020
\(\lim _{x \rightarrow 0}(\sin x)^{2 \tan x}\) is equal to
1 2
2 1
3 0
4 does not exist
Explanation:
(B) : Let, \(y=\lim _{x \rightarrow 0}(\sin x)^{2 \tan x}\) \(\log y=\lim _{x \rightarrow 0} \log (\sin x)^{2 t}\) \(\log y=2 \lim _{x \rightarrow 0} \tan x \log \sin x\) \(\log y=2 \lim _{x \rightarrow 0} \frac{\log \sin x}{\cot x} \quad\left[\frac{0}{0} \text { form }\right]\) On using L- Hospital rule, we get - \(\log y=2 \lim _{x \rightarrow 0} \frac{\frac{1}{\sin x} \cdot \cos x}{-\operatorname{cosec}^{2} x}\) \(\log y=2 \lim _{x \rightarrow 0}(-\sin x \cos x)\) \(\log y=2 \times 0=0\) \(\therefore \quad \log \mathrm{y}_{0}=0\) \(\mathrm{y}=\mathrm{e}^{0}=1\) \(\left\{-2 \sin x, \quad\right.\) if \(x \leq-\frac{\pi}{2}\)
WB JEE-2017
Limits, Continuity and Differentiability
80021
Let \(f(x)=\left\{A \sin x+B\right.\), if \(-\frac{\pi}{2}\lt x\lt \frac{\pi}{2}\). Then, \(\cos x, \quad \text { if } x \geq \frac{\pi}{2}\)
1 \(f\) is discontinuous for all \(\mathrm{A}\) and \(\mathrm{B}\)
2 \(f\) is continuous for all \(A=-1\) and \(B=1\)
3 \(f\) is continuous for all \(A=1\) and \(B=-1\)
4 \(f\) is continuous for all real values of \(A, B\)
Explanation:
(B) : We have, \(f(x)=\left\{\begin{array}{cc}-2 \sin x, \text { if } x \leq-\frac{\pi}{2} \\ A \sin x+B, \text { if }-\frac{\pi}{2}\lt x\lt \frac{\pi}{2} \\ \cos x, \text { if } x \geq \frac{\pi}{2}\end{array}\right.\) At, \(\quad \mathrm{x}=-\frac{\pi}{2}\) \(\mathrm{RHL}=-\mathrm{A}+\mathrm{B}\) For continuity, \(\mathrm{LHL}=\mathrm{RHL}=\mathrm{f}\left(-\frac{\pi}{2}\right)\) \(=-A+B=2 \tag{i}\) At, \(\quad \mathrm{x}=\frac{\pi}{2}\) \(\mathrm{LHL}=\mathrm{A}+\mathrm{B}\) \(\mathrm{RHL}=\mathrm{O}\) For continuity, \(\mathrm{LHL}=\mathrm{RHL}=\mathrm{f}\left(\frac{\pi}{2}\right)\) \(A+B=0 \tag{ii}\) On solving equation (i) and (ii), we get - \(\mathrm{A}=-1 \text { and } \mathrm{B}=1\)
WB JEE-2018
Limits, Continuity and Differentiability
80022
Let \(f:[\mathrm{a}, \mathrm{b}] \rightarrow \mathrm{R}\) be such that \(f\) is differentiable in \((a, b), f\) is continuous at \(x=a\) and \(x=b\) and moreover \(f(\mathrm{a})=0=f(\mathrm{~b})\). Then
1 There exists at least one point \(\mathrm{c}\) in \((\mathrm{a}, \mathrm{b})\) such that \(f^{\prime}(c)=f(c)\)
2 \(f^{\prime}(x)=f(x)\) does not hold at any point in (a, b
3 at every point of (a, b), \(f^{\prime}(x)>f(x)\)
4 at every point of (a,b), \(f^{\prime}(x)\lt f(x)\)
Explanation:
(A) : Let \(\mathrm{g}(\mathrm{x})=\mathrm{e}^{-\mathrm{x}} \mathrm{f}(\mathrm{x})\) Such that \(g(a)=0, \quad g(b)=0\) And \(\mathrm{g}(\mathrm{x})\) is continuous and differentiable Then, For at least one value of \(c \in(a, b)\). such that \(g^{\prime}\) (c) \(=0\) New, \(\quad g^{\prime}(x)=e^{-x} f^{\prime}(x)+\left(-e^{-x}\right) f(x)\) \(g^{\prime}(c)=e^{-c} f^{\prime}(c)+\left(-e^{-c}\right) f(c)=0\) \(e^{-c^{-}} f^{\prime}(c)=e^{-c^{\prime}} f^{\prime}(c)\) \(f^{\prime}(c)=f(c)\)
Limits, Continuity and Differentiability
80023
Let \(f ;[\mathrm{a}, \mathrm{b}] \rightarrow \mathrm{R}\) be differentiable on \([\mathrm{a}, \mathrm{b}]\) and \(k \in R\). Let \(f(a)=0=f(b)\). Also let \(\mathrm{J}(\mathrm{x})=\boldsymbol{f}^{\prime}(\mathrm{x})+\mathbf{k f}(\mathrm{x})\). Then
1 \(J(x)>0\) for all \(x \in[a, b]\)
2 \(\mathrm{J}(\mathrm{x})\lt 0\) for all \(\mathrm{x} \in[\mathrm{a}, \mathrm{b}]\)
3 \(J(x)=0\) has at least one root in (a,b)
4 \(\mathrm{J}(\mathrm{x})>0\) through \([\mathrm{a}, \mathrm{b}]\)
Explanation:
(C) : We have, \(\mathrm{f}:[\mathrm{a}, \mathrm{b}] \rightarrow \mathrm{R}\) be differentiable on \([\mathrm{a}, \mathrm{b}]\) and \(\mathrm{k} \in \mathrm{R}\), also \(\mathrm{f}(\mathrm{a})=0=\mathrm{f}(\mathrm{b})\) And \(\mathrm{J}(\mathrm{x})=\mathrm{f}^{\prime}(\mathrm{x})+\mathrm{kf}(\mathrm{x})\) Let \(g(x)=\operatorname{kxf}(\mathrm{x})\) which is continuous in \([\mathrm{a}, \mathrm{b}]\) and differentiable in \((a, b)\) such that \(g(a)=0=g(b)\) Then, for every \(c \in(a, b), g^{\prime}(c)=0\) [by Rolle's theorem] Now, \(\quad g^{\prime}(x)=k f(x)+\operatorname{kxf}^{\prime}(x)\) \(\mathrm{g}^{\prime}(\mathrm{c})=\mathrm{kf}(\mathrm{c})+\mathrm{kcf}^{\prime}(\mathrm{c})\) \(\mathrm{kf}(\mathrm{c})+\mathrm{kcf}^{\prime}(\mathrm{c})=0\) \(f(x)=0\) for every \(x=c \in(a, b)\) \(\therefore \mathrm{J}(\mathrm{x})=0\) has at least one root in (a,b)
80020
\(\lim _{x \rightarrow 0}(\sin x)^{2 \tan x}\) is equal to
1 2
2 1
3 0
4 does not exist
Explanation:
(B) : Let, \(y=\lim _{x \rightarrow 0}(\sin x)^{2 \tan x}\) \(\log y=\lim _{x \rightarrow 0} \log (\sin x)^{2 t}\) \(\log y=2 \lim _{x \rightarrow 0} \tan x \log \sin x\) \(\log y=2 \lim _{x \rightarrow 0} \frac{\log \sin x}{\cot x} \quad\left[\frac{0}{0} \text { form }\right]\) On using L- Hospital rule, we get - \(\log y=2 \lim _{x \rightarrow 0} \frac{\frac{1}{\sin x} \cdot \cos x}{-\operatorname{cosec}^{2} x}\) \(\log y=2 \lim _{x \rightarrow 0}(-\sin x \cos x)\) \(\log y=2 \times 0=0\) \(\therefore \quad \log \mathrm{y}_{0}=0\) \(\mathrm{y}=\mathrm{e}^{0}=1\) \(\left\{-2 \sin x, \quad\right.\) if \(x \leq-\frac{\pi}{2}\)
WB JEE-2017
Limits, Continuity and Differentiability
80021
Let \(f(x)=\left\{A \sin x+B\right.\), if \(-\frac{\pi}{2}\lt x\lt \frac{\pi}{2}\). Then, \(\cos x, \quad \text { if } x \geq \frac{\pi}{2}\)
1 \(f\) is discontinuous for all \(\mathrm{A}\) and \(\mathrm{B}\)
2 \(f\) is continuous for all \(A=-1\) and \(B=1\)
3 \(f\) is continuous for all \(A=1\) and \(B=-1\)
4 \(f\) is continuous for all real values of \(A, B\)
Explanation:
(B) : We have, \(f(x)=\left\{\begin{array}{cc}-2 \sin x, \text { if } x \leq-\frac{\pi}{2} \\ A \sin x+B, \text { if }-\frac{\pi}{2}\lt x\lt \frac{\pi}{2} \\ \cos x, \text { if } x \geq \frac{\pi}{2}\end{array}\right.\) At, \(\quad \mathrm{x}=-\frac{\pi}{2}\) \(\mathrm{RHL}=-\mathrm{A}+\mathrm{B}\) For continuity, \(\mathrm{LHL}=\mathrm{RHL}=\mathrm{f}\left(-\frac{\pi}{2}\right)\) \(=-A+B=2 \tag{i}\) At, \(\quad \mathrm{x}=\frac{\pi}{2}\) \(\mathrm{LHL}=\mathrm{A}+\mathrm{B}\) \(\mathrm{RHL}=\mathrm{O}\) For continuity, \(\mathrm{LHL}=\mathrm{RHL}=\mathrm{f}\left(\frac{\pi}{2}\right)\) \(A+B=0 \tag{ii}\) On solving equation (i) and (ii), we get - \(\mathrm{A}=-1 \text { and } \mathrm{B}=1\)
WB JEE-2018
Limits, Continuity and Differentiability
80022
Let \(f:[\mathrm{a}, \mathrm{b}] \rightarrow \mathrm{R}\) be such that \(f\) is differentiable in \((a, b), f\) is continuous at \(x=a\) and \(x=b\) and moreover \(f(\mathrm{a})=0=f(\mathrm{~b})\). Then
1 There exists at least one point \(\mathrm{c}\) in \((\mathrm{a}, \mathrm{b})\) such that \(f^{\prime}(c)=f(c)\)
2 \(f^{\prime}(x)=f(x)\) does not hold at any point in (a, b
3 at every point of (a, b), \(f^{\prime}(x)>f(x)\)
4 at every point of (a,b), \(f^{\prime}(x)\lt f(x)\)
Explanation:
(A) : Let \(\mathrm{g}(\mathrm{x})=\mathrm{e}^{-\mathrm{x}} \mathrm{f}(\mathrm{x})\) Such that \(g(a)=0, \quad g(b)=0\) And \(\mathrm{g}(\mathrm{x})\) is continuous and differentiable Then, For at least one value of \(c \in(a, b)\). such that \(g^{\prime}\) (c) \(=0\) New, \(\quad g^{\prime}(x)=e^{-x} f^{\prime}(x)+\left(-e^{-x}\right) f(x)\) \(g^{\prime}(c)=e^{-c} f^{\prime}(c)+\left(-e^{-c}\right) f(c)=0\) \(e^{-c^{-}} f^{\prime}(c)=e^{-c^{\prime}} f^{\prime}(c)\) \(f^{\prime}(c)=f(c)\)
Limits, Continuity and Differentiability
80023
Let \(f ;[\mathrm{a}, \mathrm{b}] \rightarrow \mathrm{R}\) be differentiable on \([\mathrm{a}, \mathrm{b}]\) and \(k \in R\). Let \(f(a)=0=f(b)\). Also let \(\mathrm{J}(\mathrm{x})=\boldsymbol{f}^{\prime}(\mathrm{x})+\mathbf{k f}(\mathrm{x})\). Then
1 \(J(x)>0\) for all \(x \in[a, b]\)
2 \(\mathrm{J}(\mathrm{x})\lt 0\) for all \(\mathrm{x} \in[\mathrm{a}, \mathrm{b}]\)
3 \(J(x)=0\) has at least one root in (a,b)
4 \(\mathrm{J}(\mathrm{x})>0\) through \([\mathrm{a}, \mathrm{b}]\)
Explanation:
(C) : We have, \(\mathrm{f}:[\mathrm{a}, \mathrm{b}] \rightarrow \mathrm{R}\) be differentiable on \([\mathrm{a}, \mathrm{b}]\) and \(\mathrm{k} \in \mathrm{R}\), also \(\mathrm{f}(\mathrm{a})=0=\mathrm{f}(\mathrm{b})\) And \(\mathrm{J}(\mathrm{x})=\mathrm{f}^{\prime}(\mathrm{x})+\mathrm{kf}(\mathrm{x})\) Let \(g(x)=\operatorname{kxf}(\mathrm{x})\) which is continuous in \([\mathrm{a}, \mathrm{b}]\) and differentiable in \((a, b)\) such that \(g(a)=0=g(b)\) Then, for every \(c \in(a, b), g^{\prime}(c)=0\) [by Rolle's theorem] Now, \(\quad g^{\prime}(x)=k f(x)+\operatorname{kxf}^{\prime}(x)\) \(\mathrm{g}^{\prime}(\mathrm{c})=\mathrm{kf}(\mathrm{c})+\mathrm{kcf}^{\prime}(\mathrm{c})\) \(\mathrm{kf}(\mathrm{c})+\mathrm{kcf}^{\prime}(\mathrm{c})=0\) \(f(x)=0\) for every \(x=c \in(a, b)\) \(\therefore \mathrm{J}(\mathrm{x})=0\) has at least one root in (a,b)
80020
\(\lim _{x \rightarrow 0}(\sin x)^{2 \tan x}\) is equal to
1 2
2 1
3 0
4 does not exist
Explanation:
(B) : Let, \(y=\lim _{x \rightarrow 0}(\sin x)^{2 \tan x}\) \(\log y=\lim _{x \rightarrow 0} \log (\sin x)^{2 t}\) \(\log y=2 \lim _{x \rightarrow 0} \tan x \log \sin x\) \(\log y=2 \lim _{x \rightarrow 0} \frac{\log \sin x}{\cot x} \quad\left[\frac{0}{0} \text { form }\right]\) On using L- Hospital rule, we get - \(\log y=2 \lim _{x \rightarrow 0} \frac{\frac{1}{\sin x} \cdot \cos x}{-\operatorname{cosec}^{2} x}\) \(\log y=2 \lim _{x \rightarrow 0}(-\sin x \cos x)\) \(\log y=2 \times 0=0\) \(\therefore \quad \log \mathrm{y}_{0}=0\) \(\mathrm{y}=\mathrm{e}^{0}=1\) \(\left\{-2 \sin x, \quad\right.\) if \(x \leq-\frac{\pi}{2}\)
WB JEE-2017
Limits, Continuity and Differentiability
80021
Let \(f(x)=\left\{A \sin x+B\right.\), if \(-\frac{\pi}{2}\lt x\lt \frac{\pi}{2}\). Then, \(\cos x, \quad \text { if } x \geq \frac{\pi}{2}\)
1 \(f\) is discontinuous for all \(\mathrm{A}\) and \(\mathrm{B}\)
2 \(f\) is continuous for all \(A=-1\) and \(B=1\)
3 \(f\) is continuous for all \(A=1\) and \(B=-1\)
4 \(f\) is continuous for all real values of \(A, B\)
Explanation:
(B) : We have, \(f(x)=\left\{\begin{array}{cc}-2 \sin x, \text { if } x \leq-\frac{\pi}{2} \\ A \sin x+B, \text { if }-\frac{\pi}{2}\lt x\lt \frac{\pi}{2} \\ \cos x, \text { if } x \geq \frac{\pi}{2}\end{array}\right.\) At, \(\quad \mathrm{x}=-\frac{\pi}{2}\) \(\mathrm{RHL}=-\mathrm{A}+\mathrm{B}\) For continuity, \(\mathrm{LHL}=\mathrm{RHL}=\mathrm{f}\left(-\frac{\pi}{2}\right)\) \(=-A+B=2 \tag{i}\) At, \(\quad \mathrm{x}=\frac{\pi}{2}\) \(\mathrm{LHL}=\mathrm{A}+\mathrm{B}\) \(\mathrm{RHL}=\mathrm{O}\) For continuity, \(\mathrm{LHL}=\mathrm{RHL}=\mathrm{f}\left(\frac{\pi}{2}\right)\) \(A+B=0 \tag{ii}\) On solving equation (i) and (ii), we get - \(\mathrm{A}=-1 \text { and } \mathrm{B}=1\)
WB JEE-2018
Limits, Continuity and Differentiability
80022
Let \(f:[\mathrm{a}, \mathrm{b}] \rightarrow \mathrm{R}\) be such that \(f\) is differentiable in \((a, b), f\) is continuous at \(x=a\) and \(x=b\) and moreover \(f(\mathrm{a})=0=f(\mathrm{~b})\). Then
1 There exists at least one point \(\mathrm{c}\) in \((\mathrm{a}, \mathrm{b})\) such that \(f^{\prime}(c)=f(c)\)
2 \(f^{\prime}(x)=f(x)\) does not hold at any point in (a, b
3 at every point of (a, b), \(f^{\prime}(x)>f(x)\)
4 at every point of (a,b), \(f^{\prime}(x)\lt f(x)\)
Explanation:
(A) : Let \(\mathrm{g}(\mathrm{x})=\mathrm{e}^{-\mathrm{x}} \mathrm{f}(\mathrm{x})\) Such that \(g(a)=0, \quad g(b)=0\) And \(\mathrm{g}(\mathrm{x})\) is continuous and differentiable Then, For at least one value of \(c \in(a, b)\). such that \(g^{\prime}\) (c) \(=0\) New, \(\quad g^{\prime}(x)=e^{-x} f^{\prime}(x)+\left(-e^{-x}\right) f(x)\) \(g^{\prime}(c)=e^{-c} f^{\prime}(c)+\left(-e^{-c}\right) f(c)=0\) \(e^{-c^{-}} f^{\prime}(c)=e^{-c^{\prime}} f^{\prime}(c)\) \(f^{\prime}(c)=f(c)\)
Limits, Continuity and Differentiability
80023
Let \(f ;[\mathrm{a}, \mathrm{b}] \rightarrow \mathrm{R}\) be differentiable on \([\mathrm{a}, \mathrm{b}]\) and \(k \in R\). Let \(f(a)=0=f(b)\). Also let \(\mathrm{J}(\mathrm{x})=\boldsymbol{f}^{\prime}(\mathrm{x})+\mathbf{k f}(\mathrm{x})\). Then
1 \(J(x)>0\) for all \(x \in[a, b]\)
2 \(\mathrm{J}(\mathrm{x})\lt 0\) for all \(\mathrm{x} \in[\mathrm{a}, \mathrm{b}]\)
3 \(J(x)=0\) has at least one root in (a,b)
4 \(\mathrm{J}(\mathrm{x})>0\) through \([\mathrm{a}, \mathrm{b}]\)
Explanation:
(C) : We have, \(\mathrm{f}:[\mathrm{a}, \mathrm{b}] \rightarrow \mathrm{R}\) be differentiable on \([\mathrm{a}, \mathrm{b}]\) and \(\mathrm{k} \in \mathrm{R}\), also \(\mathrm{f}(\mathrm{a})=0=\mathrm{f}(\mathrm{b})\) And \(\mathrm{J}(\mathrm{x})=\mathrm{f}^{\prime}(\mathrm{x})+\mathrm{kf}(\mathrm{x})\) Let \(g(x)=\operatorname{kxf}(\mathrm{x})\) which is continuous in \([\mathrm{a}, \mathrm{b}]\) and differentiable in \((a, b)\) such that \(g(a)=0=g(b)\) Then, for every \(c \in(a, b), g^{\prime}(c)=0\) [by Rolle's theorem] Now, \(\quad g^{\prime}(x)=k f(x)+\operatorname{kxf}^{\prime}(x)\) \(\mathrm{g}^{\prime}(\mathrm{c})=\mathrm{kf}(\mathrm{c})+\mathrm{kcf}^{\prime}(\mathrm{c})\) \(\mathrm{kf}(\mathrm{c})+\mathrm{kcf}^{\prime}(\mathrm{c})=0\) \(f(x)=0\) for every \(x=c \in(a, b)\) \(\therefore \mathrm{J}(\mathrm{x})=0\) has at least one root in (a,b)