85751
For the function \(f(x)=e^{\cos x}\), Rolle's Theorem is
1 applicable when \(\frac{\pi}{2} \leq x \leq \frac{3 \pi}{2}\)
2 applicable when \(0 \leq x \leq \frac{\pi}{2}\)
3 applicable when \(0 \leq \mathrm{x} \leq \pi\)
4 applicable when \(\frac{\pi}{4} \leq x \leq \frac{\pi}{2}\)
Explanation:
(A) : Given, \(f(x)=e^{\cos x}\) \(\mathrm{f}(0)=\mathrm{e}^{\cos 0}=\mathrm{e}^{1}=e\) \(\mathrm{f}\left(\frac{\pi}{2}\right)=\mathrm{e}^{\cos \frac{\pi}{2}}=\mathrm{e}^{0}=1\) \(\mathrm{f}\left(\frac{3 \pi}{2}\right)=\mathrm{e}^{\cos \frac{3 \pi}{2}}=\mathrm{e}^{0}=1\) \(\therefore \mathrm{f}\left(\frac{\pi}{2}\right)=\mathrm{f}\left(\frac{3 \pi}{2}\right)\) Hence, Roll's theorem is applicable when \(\frac{\pi}{2} \leq x \leq \frac{3 \pi}{2}\).
WB JEE-2011
Application of Derivatives
85752
The value of \(x\) in the interval \([4,9]\) at which the function \(f(x)=\sqrt{x}\) satisfies the mean value theorem is
1 \(\frac{13}{4}\)
2 \(\frac{17}{4}\)
3 \(\frac{21}{4}\)
4 \(\frac{25}{4}\)
Explanation:
(D) : Given function (i) \(f(x)=\sqrt{x}\) is continuous in [4,9] (ii) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2 \sqrt{\mathrm{x}}}\) Thus \(\mathrm{f}(\mathrm{x})\) is differentiable in \((4,9)\) (iii) \(\mathrm{f}(4) \neq \mathrm{f}(9)\). All the three conditions of LMV theorem satisfied then there exist at least one \(c \in(4,9)\) such that. \(f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\) \(\frac{1}{2 \sqrt{c}}=\frac{1}{5} \Rightarrow 2 \sqrt{c}=5 \Rightarrow \sqrt{c}=\frac{5}{2}\) Squaring both side we get \(\mathrm{c}=\frac{25}{4}\)
VITEEE-2019
Application of Derivatives
85753
A value of \(c\) for which conclusion of Mean Value Theorem holds for the function \(f(x)=\) \(\log _{e} x\) on the interval \([1,3]\) is
1 \(\log _{3} \mathrm{e}\)
2 \(\log _{\mathrm{e}} 3\)
3 \(2 \log _{3} \mathrm{e}\)
4 \(\frac{1}{2} \log _{3} \mathrm{e}\)
Explanation:
(C) : Using Lagrange's Mean Value Theorem Let \(f(x)\) be a function defined on \([a, b]\) Then, \(\mathrm{f}^{\prime}(\mathrm{c})=\frac{\mathrm{f}(\mathrm{b})-\mathrm{f}(\mathrm{a})}{\mathrm{b}-\mathrm{a}} \tag{i}\) \(\mathrm{c} \in[\mathrm{a}, \mathrm{b}]\) Given, \(f(x)=\log _{e}{ }^{x}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{x}}\) From equation (i), we get - \(\frac{1}{c}=\frac{f(3)-f(1)}{3-1}\) \(\frac{1}{\mathrm{c}}=\frac{\log _{\mathrm{e}}{ }^{3}-\log _{\mathrm{e}}{ }^{1}}{2} \quad\left(\because \log \mathrm{a}-\log \mathrm{b}=\log \frac{\mathrm{a}}{\mathrm{b}}\right)\) \(\frac{1}{\mathrm{c}}=\frac{\log _{\mathrm{e}} 3}{2}\) \(\mathrm{c}=\frac{2}{\log _{\mathrm{e}} 3} \quad\left(\because \log _{\mathrm{b}}^{\mathrm{a}}=\frac{1}{\log _{\mathrm{a}}^{\mathrm{b}}}\right)\) \(\mathrm{c}=2 \log _{3} \mathrm{e}\)
Kerala CEE-2019
Application of Derivatives
85754
Find c of Lagrange's means value theorem for the function \(f(x)=3 x^2+5 x+7\) in the interval \([1,3]\).
1 \(\frac{7}{3}\)
2 2
3 \(\frac{3}{2}\)
4 \(\frac{4}{3}\)
Explanation:
(B): Given, \(f(x)=3 x^2+5 x+7\) For interval \([1,3]\) \(f(1)=3 \times 1^2+5 \times 1+7=15\) And \(\quad f(3)=3 \times 3^2+5 \times 3+7=49\) Now, \(\quad f^{\prime}(x)=6 x+5\) \(f^{\prime}(c)=6 c+5\) From Lagrange's mean value theorem \(f^{\prime}(c)=\frac{f(3)-f(1)}{3-1}\) \(6 c+5=\frac{49-15}{3-1}=17\) \(6 c=12, \quad c=2\)
85751
For the function \(f(x)=e^{\cos x}\), Rolle's Theorem is
1 applicable when \(\frac{\pi}{2} \leq x \leq \frac{3 \pi}{2}\)
2 applicable when \(0 \leq x \leq \frac{\pi}{2}\)
3 applicable when \(0 \leq \mathrm{x} \leq \pi\)
4 applicable when \(\frac{\pi}{4} \leq x \leq \frac{\pi}{2}\)
Explanation:
(A) : Given, \(f(x)=e^{\cos x}\) \(\mathrm{f}(0)=\mathrm{e}^{\cos 0}=\mathrm{e}^{1}=e\) \(\mathrm{f}\left(\frac{\pi}{2}\right)=\mathrm{e}^{\cos \frac{\pi}{2}}=\mathrm{e}^{0}=1\) \(\mathrm{f}\left(\frac{3 \pi}{2}\right)=\mathrm{e}^{\cos \frac{3 \pi}{2}}=\mathrm{e}^{0}=1\) \(\therefore \mathrm{f}\left(\frac{\pi}{2}\right)=\mathrm{f}\left(\frac{3 \pi}{2}\right)\) Hence, Roll's theorem is applicable when \(\frac{\pi}{2} \leq x \leq \frac{3 \pi}{2}\).
WB JEE-2011
Application of Derivatives
85752
The value of \(x\) in the interval \([4,9]\) at which the function \(f(x)=\sqrt{x}\) satisfies the mean value theorem is
1 \(\frac{13}{4}\)
2 \(\frac{17}{4}\)
3 \(\frac{21}{4}\)
4 \(\frac{25}{4}\)
Explanation:
(D) : Given function (i) \(f(x)=\sqrt{x}\) is continuous in [4,9] (ii) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2 \sqrt{\mathrm{x}}}\) Thus \(\mathrm{f}(\mathrm{x})\) is differentiable in \((4,9)\) (iii) \(\mathrm{f}(4) \neq \mathrm{f}(9)\). All the three conditions of LMV theorem satisfied then there exist at least one \(c \in(4,9)\) such that. \(f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\) \(\frac{1}{2 \sqrt{c}}=\frac{1}{5} \Rightarrow 2 \sqrt{c}=5 \Rightarrow \sqrt{c}=\frac{5}{2}\) Squaring both side we get \(\mathrm{c}=\frac{25}{4}\)
VITEEE-2019
Application of Derivatives
85753
A value of \(c\) for which conclusion of Mean Value Theorem holds for the function \(f(x)=\) \(\log _{e} x\) on the interval \([1,3]\) is
1 \(\log _{3} \mathrm{e}\)
2 \(\log _{\mathrm{e}} 3\)
3 \(2 \log _{3} \mathrm{e}\)
4 \(\frac{1}{2} \log _{3} \mathrm{e}\)
Explanation:
(C) : Using Lagrange's Mean Value Theorem Let \(f(x)\) be a function defined on \([a, b]\) Then, \(\mathrm{f}^{\prime}(\mathrm{c})=\frac{\mathrm{f}(\mathrm{b})-\mathrm{f}(\mathrm{a})}{\mathrm{b}-\mathrm{a}} \tag{i}\) \(\mathrm{c} \in[\mathrm{a}, \mathrm{b}]\) Given, \(f(x)=\log _{e}{ }^{x}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{x}}\) From equation (i), we get - \(\frac{1}{c}=\frac{f(3)-f(1)}{3-1}\) \(\frac{1}{\mathrm{c}}=\frac{\log _{\mathrm{e}}{ }^{3}-\log _{\mathrm{e}}{ }^{1}}{2} \quad\left(\because \log \mathrm{a}-\log \mathrm{b}=\log \frac{\mathrm{a}}{\mathrm{b}}\right)\) \(\frac{1}{\mathrm{c}}=\frac{\log _{\mathrm{e}} 3}{2}\) \(\mathrm{c}=\frac{2}{\log _{\mathrm{e}} 3} \quad\left(\because \log _{\mathrm{b}}^{\mathrm{a}}=\frac{1}{\log _{\mathrm{a}}^{\mathrm{b}}}\right)\) \(\mathrm{c}=2 \log _{3} \mathrm{e}\)
Kerala CEE-2019
Application of Derivatives
85754
Find c of Lagrange's means value theorem for the function \(f(x)=3 x^2+5 x+7\) in the interval \([1,3]\).
1 \(\frac{7}{3}\)
2 2
3 \(\frac{3}{2}\)
4 \(\frac{4}{3}\)
Explanation:
(B): Given, \(f(x)=3 x^2+5 x+7\) For interval \([1,3]\) \(f(1)=3 \times 1^2+5 \times 1+7=15\) And \(\quad f(3)=3 \times 3^2+5 \times 3+7=49\) Now, \(\quad f^{\prime}(x)=6 x+5\) \(f^{\prime}(c)=6 c+5\) From Lagrange's mean value theorem \(f^{\prime}(c)=\frac{f(3)-f(1)}{3-1}\) \(6 c+5=\frac{49-15}{3-1}=17\) \(6 c=12, \quad c=2\)
85751
For the function \(f(x)=e^{\cos x}\), Rolle's Theorem is
1 applicable when \(\frac{\pi}{2} \leq x \leq \frac{3 \pi}{2}\)
2 applicable when \(0 \leq x \leq \frac{\pi}{2}\)
3 applicable when \(0 \leq \mathrm{x} \leq \pi\)
4 applicable when \(\frac{\pi}{4} \leq x \leq \frac{\pi}{2}\)
Explanation:
(A) : Given, \(f(x)=e^{\cos x}\) \(\mathrm{f}(0)=\mathrm{e}^{\cos 0}=\mathrm{e}^{1}=e\) \(\mathrm{f}\left(\frac{\pi}{2}\right)=\mathrm{e}^{\cos \frac{\pi}{2}}=\mathrm{e}^{0}=1\) \(\mathrm{f}\left(\frac{3 \pi}{2}\right)=\mathrm{e}^{\cos \frac{3 \pi}{2}}=\mathrm{e}^{0}=1\) \(\therefore \mathrm{f}\left(\frac{\pi}{2}\right)=\mathrm{f}\left(\frac{3 \pi}{2}\right)\) Hence, Roll's theorem is applicable when \(\frac{\pi}{2} \leq x \leq \frac{3 \pi}{2}\).
WB JEE-2011
Application of Derivatives
85752
The value of \(x\) in the interval \([4,9]\) at which the function \(f(x)=\sqrt{x}\) satisfies the mean value theorem is
1 \(\frac{13}{4}\)
2 \(\frac{17}{4}\)
3 \(\frac{21}{4}\)
4 \(\frac{25}{4}\)
Explanation:
(D) : Given function (i) \(f(x)=\sqrt{x}\) is continuous in [4,9] (ii) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2 \sqrt{\mathrm{x}}}\) Thus \(\mathrm{f}(\mathrm{x})\) is differentiable in \((4,9)\) (iii) \(\mathrm{f}(4) \neq \mathrm{f}(9)\). All the three conditions of LMV theorem satisfied then there exist at least one \(c \in(4,9)\) such that. \(f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\) \(\frac{1}{2 \sqrt{c}}=\frac{1}{5} \Rightarrow 2 \sqrt{c}=5 \Rightarrow \sqrt{c}=\frac{5}{2}\) Squaring both side we get \(\mathrm{c}=\frac{25}{4}\)
VITEEE-2019
Application of Derivatives
85753
A value of \(c\) for which conclusion of Mean Value Theorem holds for the function \(f(x)=\) \(\log _{e} x\) on the interval \([1,3]\) is
1 \(\log _{3} \mathrm{e}\)
2 \(\log _{\mathrm{e}} 3\)
3 \(2 \log _{3} \mathrm{e}\)
4 \(\frac{1}{2} \log _{3} \mathrm{e}\)
Explanation:
(C) : Using Lagrange's Mean Value Theorem Let \(f(x)\) be a function defined on \([a, b]\) Then, \(\mathrm{f}^{\prime}(\mathrm{c})=\frac{\mathrm{f}(\mathrm{b})-\mathrm{f}(\mathrm{a})}{\mathrm{b}-\mathrm{a}} \tag{i}\) \(\mathrm{c} \in[\mathrm{a}, \mathrm{b}]\) Given, \(f(x)=\log _{e}{ }^{x}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{x}}\) From equation (i), we get - \(\frac{1}{c}=\frac{f(3)-f(1)}{3-1}\) \(\frac{1}{\mathrm{c}}=\frac{\log _{\mathrm{e}}{ }^{3}-\log _{\mathrm{e}}{ }^{1}}{2} \quad\left(\because \log \mathrm{a}-\log \mathrm{b}=\log \frac{\mathrm{a}}{\mathrm{b}}\right)\) \(\frac{1}{\mathrm{c}}=\frac{\log _{\mathrm{e}} 3}{2}\) \(\mathrm{c}=\frac{2}{\log _{\mathrm{e}} 3} \quad\left(\because \log _{\mathrm{b}}^{\mathrm{a}}=\frac{1}{\log _{\mathrm{a}}^{\mathrm{b}}}\right)\) \(\mathrm{c}=2 \log _{3} \mathrm{e}\)
Kerala CEE-2019
Application of Derivatives
85754
Find c of Lagrange's means value theorem for the function \(f(x)=3 x^2+5 x+7\) in the interval \([1,3]\).
1 \(\frac{7}{3}\)
2 2
3 \(\frac{3}{2}\)
4 \(\frac{4}{3}\)
Explanation:
(B): Given, \(f(x)=3 x^2+5 x+7\) For interval \([1,3]\) \(f(1)=3 \times 1^2+5 \times 1+7=15\) And \(\quad f(3)=3 \times 3^2+5 \times 3+7=49\) Now, \(\quad f^{\prime}(x)=6 x+5\) \(f^{\prime}(c)=6 c+5\) From Lagrange's mean value theorem \(f^{\prime}(c)=\frac{f(3)-f(1)}{3-1}\) \(6 c+5=\frac{49-15}{3-1}=17\) \(6 c=12, \quad c=2\)
85751
For the function \(f(x)=e^{\cos x}\), Rolle's Theorem is
1 applicable when \(\frac{\pi}{2} \leq x \leq \frac{3 \pi}{2}\)
2 applicable when \(0 \leq x \leq \frac{\pi}{2}\)
3 applicable when \(0 \leq \mathrm{x} \leq \pi\)
4 applicable when \(\frac{\pi}{4} \leq x \leq \frac{\pi}{2}\)
Explanation:
(A) : Given, \(f(x)=e^{\cos x}\) \(\mathrm{f}(0)=\mathrm{e}^{\cos 0}=\mathrm{e}^{1}=e\) \(\mathrm{f}\left(\frac{\pi}{2}\right)=\mathrm{e}^{\cos \frac{\pi}{2}}=\mathrm{e}^{0}=1\) \(\mathrm{f}\left(\frac{3 \pi}{2}\right)=\mathrm{e}^{\cos \frac{3 \pi}{2}}=\mathrm{e}^{0}=1\) \(\therefore \mathrm{f}\left(\frac{\pi}{2}\right)=\mathrm{f}\left(\frac{3 \pi}{2}\right)\) Hence, Roll's theorem is applicable when \(\frac{\pi}{2} \leq x \leq \frac{3 \pi}{2}\).
WB JEE-2011
Application of Derivatives
85752
The value of \(x\) in the interval \([4,9]\) at which the function \(f(x)=\sqrt{x}\) satisfies the mean value theorem is
1 \(\frac{13}{4}\)
2 \(\frac{17}{4}\)
3 \(\frac{21}{4}\)
4 \(\frac{25}{4}\)
Explanation:
(D) : Given function (i) \(f(x)=\sqrt{x}\) is continuous in [4,9] (ii) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2 \sqrt{\mathrm{x}}}\) Thus \(\mathrm{f}(\mathrm{x})\) is differentiable in \((4,9)\) (iii) \(\mathrm{f}(4) \neq \mathrm{f}(9)\). All the three conditions of LMV theorem satisfied then there exist at least one \(c \in(4,9)\) such that. \(f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\) \(\frac{1}{2 \sqrt{c}}=\frac{1}{5} \Rightarrow 2 \sqrt{c}=5 \Rightarrow \sqrt{c}=\frac{5}{2}\) Squaring both side we get \(\mathrm{c}=\frac{25}{4}\)
VITEEE-2019
Application of Derivatives
85753
A value of \(c\) for which conclusion of Mean Value Theorem holds for the function \(f(x)=\) \(\log _{e} x\) on the interval \([1,3]\) is
1 \(\log _{3} \mathrm{e}\)
2 \(\log _{\mathrm{e}} 3\)
3 \(2 \log _{3} \mathrm{e}\)
4 \(\frac{1}{2} \log _{3} \mathrm{e}\)
Explanation:
(C) : Using Lagrange's Mean Value Theorem Let \(f(x)\) be a function defined on \([a, b]\) Then, \(\mathrm{f}^{\prime}(\mathrm{c})=\frac{\mathrm{f}(\mathrm{b})-\mathrm{f}(\mathrm{a})}{\mathrm{b}-\mathrm{a}} \tag{i}\) \(\mathrm{c} \in[\mathrm{a}, \mathrm{b}]\) Given, \(f(x)=\log _{e}{ }^{x}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{x}}\) From equation (i), we get - \(\frac{1}{c}=\frac{f(3)-f(1)}{3-1}\) \(\frac{1}{\mathrm{c}}=\frac{\log _{\mathrm{e}}{ }^{3}-\log _{\mathrm{e}}{ }^{1}}{2} \quad\left(\because \log \mathrm{a}-\log \mathrm{b}=\log \frac{\mathrm{a}}{\mathrm{b}}\right)\) \(\frac{1}{\mathrm{c}}=\frac{\log _{\mathrm{e}} 3}{2}\) \(\mathrm{c}=\frac{2}{\log _{\mathrm{e}} 3} \quad\left(\because \log _{\mathrm{b}}^{\mathrm{a}}=\frac{1}{\log _{\mathrm{a}}^{\mathrm{b}}}\right)\) \(\mathrm{c}=2 \log _{3} \mathrm{e}\)
Kerala CEE-2019
Application of Derivatives
85754
Find c of Lagrange's means value theorem for the function \(f(x)=3 x^2+5 x+7\) in the interval \([1,3]\).
1 \(\frac{7}{3}\)
2 2
3 \(\frac{3}{2}\)
4 \(\frac{4}{3}\)
Explanation:
(B): Given, \(f(x)=3 x^2+5 x+7\) For interval \([1,3]\) \(f(1)=3 \times 1^2+5 \times 1+7=15\) And \(\quad f(3)=3 \times 3^2+5 \times 3+7=49\) Now, \(\quad f^{\prime}(x)=6 x+5\) \(f^{\prime}(c)=6 c+5\) From Lagrange's mean value theorem \(f^{\prime}(c)=\frac{f(3)-f(1)}{3-1}\) \(6 c+5=\frac{49-15}{3-1}=17\) \(6 c=12, \quad c=2\)
