85746
If \(f(x)=\log (\sin x), x \in\left[\frac{\pi}{6}, \frac{5 \pi}{6}\right]\), then value of ' \(c\) ' by applying L.M.V.T. is
1 \(\frac{2 \pi}{3}\)
2 \(\frac{\pi}{2}\)
3 \(\frac{\pi}{4}\)
4 \(\frac{3 \pi}{4}\)
Explanation:
(B) : Given, \(f(x)=\log (\sin x) x \in\left[\frac{\pi}{6}, \frac{5 \pi}{6}\right]\) By Lagrange's mean value theorem, \(f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}=\frac{f\left(\frac{5 \pi}{6}\right)-f\left(\frac{\pi}{6}\right)}{\frac{5 \pi}{6}-\frac{\pi}{6}}\) f \((\mathrm{b})=\log \left(\sin \frac{\pi}{6}\right)=\log \left(\frac{1}{2}\right)\) \(f(a)=\log \left(\sin \frac{5 \pi}{6}\right)=\log \left(\frac{1}{2}\right)\) Now, \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\sin \mathrm{x}} \cdot \cos \mathrm{x}=\cot \mathrm{x}\) \(\therefore \quad \mathrm{f}^{\prime}(\mathrm{c})=\cot \mathrm{c}\) \(\therefore \mathrm{f}^{\prime}(\mathrm{c})=\frac{\log \frac{1}{2}-\log \frac{1}{2}}{\frac{5 \pi}{6}-\frac{\pi}{6}}\) \(\cot \mathrm{c}\) So, \(\quad \mathrm{c}=\frac{\pi}{2}\)
MHT CET-2020
Application of Derivatives
85747
IF the L.M.V.T. holds for the function \(\mathrm{f}(\mathrm{x})=\mathrm{x}+\frac{1}{\mathrm{x}}, \mathrm{x} \in[1,3]\), then \(\mathrm{c}=\)
1 2
2 3
3 \(\sqrt{3}\)
4 -3
Explanation:
(C) : Given, \(f(x)=x+\frac{1}{x} \quad x \in[1,3]\) By Lagrange's mean value theorem, \(f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\) \(f(3)=3+\frac{1}{3}=\frac{10}{3}\) \(\mathrm{f}(1)=1+\frac{1}{1}=2\) Now, \(\quad f^{\prime}(x)=1-\frac{1}{x^{2}}\) \(\mathrm{f}^{\prime}(\mathrm{c})=1-\frac{1}{\mathrm{c}^{2}} \Rightarrow \mathrm{f}^{\prime}(\mathrm{c})=\frac{\mathrm{f}(\mathrm{b})-\mathrm{f}(\mathrm{a})}{\mathrm{b}-\mathrm{a}}\) \(1-\frac{1}{\mathrm{c}^{2}}=\frac{\frac{10}{3}-2}{3-1}=\frac{2}{3} \Rightarrow 1-\frac{2}{3}=\frac{1}{\mathrm{c}^{2}}\) \(\mathrm{c}= \pm \sqrt{3} \quad(\because \mathrm{x} \in[1,3])\) \(\therefore \quad \mathrm{c}=\sqrt{3}\)
Manipal-2014
Application of Derivatives
85748
If Rolle's theorem holds for the function \(f(x)=\cos x+\sin x+7, x \in[0,2 \pi]\) and \(0\lt \mathbf{c}\lt 2 \pi\) such that \(\mathbf{f}^{\prime}(\mathrm{c})=\mathbf{0}\), then the number of possible value of \(c\) is
85750
Let \(f(x)=\left\{\frac{x^{p}}{(\sin x)^{q}}\right.\), if \(0\lt x \leq \frac{\pi}{2}\) (p, q \(\in \mathbf{R})\). Then, Lagrange's mean value theorem applicable to \(f(x)\) in closed interval [0, \(\mathbf{x ]}\)
1 For all \(p, q\)
2 Only when \(\mathrm{p}>\mathrm{q}\)
3 Only when \(\mathrm{P}\lt \mathrm{q}\)
4 For no value of \(p, q\)
Explanation:
(B) : Given, \(f(x)=\left\{\frac{x^{p}}{(\sin x)^{q}}, \text { if } 0\lt x \leq \frac{\pi}{2}\right.\) Since, Lagrange's mean value theorem is applicable on \(\mathrm{f}(\mathrm{x})\). \(\therefore \quad \lim _{x \rightarrow 0} \frac{x^{p}}{(\sin x)^{q}}=f(0) \Rightarrow \lim _{x \rightarrow 0} \frac{x^{p}}{(\sin x)^{q}}=0\) \(\lim _{x \rightarrow 0} \frac{x^{p} \cdot x^{q} \cdot x^{-q}}{(\sin x)^{q}} \Rightarrow \lim _{x \rightarrow 0} \frac{x^{q} \cdot x^{p-q}}{(\sin x)^{q}}\) \(\lim _{x \rightarrow 0} \frac{x^{q}}{(\sin x)^{q}} \times \lim _{x \rightarrow 0} x^{p-q}\) \(\lim _{x \rightarrow 0}\left(\frac{x}{(\sin x)}\right)^{q} \times \lim _{x \rightarrow 0} x^{p-q}\) So, \(\quad \lim _{\mathrm{x} \rightarrow 0} \mathrm{f}(\mathrm{x})=0\) only when \(\mathrm{p}>\mathrm{q}\) Hence, L.M.V Theorem is applicable to \(\mathrm{f}(\mathrm{x})\) in \([0, \mathrm{x}]\) only when \(\mathrm{p}>\mathrm{q}\)
85746
If \(f(x)=\log (\sin x), x \in\left[\frac{\pi}{6}, \frac{5 \pi}{6}\right]\), then value of ' \(c\) ' by applying L.M.V.T. is
1 \(\frac{2 \pi}{3}\)
2 \(\frac{\pi}{2}\)
3 \(\frac{\pi}{4}\)
4 \(\frac{3 \pi}{4}\)
Explanation:
(B) : Given, \(f(x)=\log (\sin x) x \in\left[\frac{\pi}{6}, \frac{5 \pi}{6}\right]\) By Lagrange's mean value theorem, \(f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}=\frac{f\left(\frac{5 \pi}{6}\right)-f\left(\frac{\pi}{6}\right)}{\frac{5 \pi}{6}-\frac{\pi}{6}}\) f \((\mathrm{b})=\log \left(\sin \frac{\pi}{6}\right)=\log \left(\frac{1}{2}\right)\) \(f(a)=\log \left(\sin \frac{5 \pi}{6}\right)=\log \left(\frac{1}{2}\right)\) Now, \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\sin \mathrm{x}} \cdot \cos \mathrm{x}=\cot \mathrm{x}\) \(\therefore \quad \mathrm{f}^{\prime}(\mathrm{c})=\cot \mathrm{c}\) \(\therefore \mathrm{f}^{\prime}(\mathrm{c})=\frac{\log \frac{1}{2}-\log \frac{1}{2}}{\frac{5 \pi}{6}-\frac{\pi}{6}}\) \(\cot \mathrm{c}\) So, \(\quad \mathrm{c}=\frac{\pi}{2}\)
MHT CET-2020
Application of Derivatives
85747
IF the L.M.V.T. holds for the function \(\mathrm{f}(\mathrm{x})=\mathrm{x}+\frac{1}{\mathrm{x}}, \mathrm{x} \in[1,3]\), then \(\mathrm{c}=\)
1 2
2 3
3 \(\sqrt{3}\)
4 -3
Explanation:
(C) : Given, \(f(x)=x+\frac{1}{x} \quad x \in[1,3]\) By Lagrange's mean value theorem, \(f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\) \(f(3)=3+\frac{1}{3}=\frac{10}{3}\) \(\mathrm{f}(1)=1+\frac{1}{1}=2\) Now, \(\quad f^{\prime}(x)=1-\frac{1}{x^{2}}\) \(\mathrm{f}^{\prime}(\mathrm{c})=1-\frac{1}{\mathrm{c}^{2}} \Rightarrow \mathrm{f}^{\prime}(\mathrm{c})=\frac{\mathrm{f}(\mathrm{b})-\mathrm{f}(\mathrm{a})}{\mathrm{b}-\mathrm{a}}\) \(1-\frac{1}{\mathrm{c}^{2}}=\frac{\frac{10}{3}-2}{3-1}=\frac{2}{3} \Rightarrow 1-\frac{2}{3}=\frac{1}{\mathrm{c}^{2}}\) \(\mathrm{c}= \pm \sqrt{3} \quad(\because \mathrm{x} \in[1,3])\) \(\therefore \quad \mathrm{c}=\sqrt{3}\)
Manipal-2014
Application of Derivatives
85748
If Rolle's theorem holds for the function \(f(x)=\cos x+\sin x+7, x \in[0,2 \pi]\) and \(0\lt \mathbf{c}\lt 2 \pi\) such that \(\mathbf{f}^{\prime}(\mathrm{c})=\mathbf{0}\), then the number of possible value of \(c\) is
85750
Let \(f(x)=\left\{\frac{x^{p}}{(\sin x)^{q}}\right.\), if \(0\lt x \leq \frac{\pi}{2}\) (p, q \(\in \mathbf{R})\). Then, Lagrange's mean value theorem applicable to \(f(x)\) in closed interval [0, \(\mathbf{x ]}\)
1 For all \(p, q\)
2 Only when \(\mathrm{p}>\mathrm{q}\)
3 Only when \(\mathrm{P}\lt \mathrm{q}\)
4 For no value of \(p, q\)
Explanation:
(B) : Given, \(f(x)=\left\{\frac{x^{p}}{(\sin x)^{q}}, \text { if } 0\lt x \leq \frac{\pi}{2}\right.\) Since, Lagrange's mean value theorem is applicable on \(\mathrm{f}(\mathrm{x})\). \(\therefore \quad \lim _{x \rightarrow 0} \frac{x^{p}}{(\sin x)^{q}}=f(0) \Rightarrow \lim _{x \rightarrow 0} \frac{x^{p}}{(\sin x)^{q}}=0\) \(\lim _{x \rightarrow 0} \frac{x^{p} \cdot x^{q} \cdot x^{-q}}{(\sin x)^{q}} \Rightarrow \lim _{x \rightarrow 0} \frac{x^{q} \cdot x^{p-q}}{(\sin x)^{q}}\) \(\lim _{x \rightarrow 0} \frac{x^{q}}{(\sin x)^{q}} \times \lim _{x \rightarrow 0} x^{p-q}\) \(\lim _{x \rightarrow 0}\left(\frac{x}{(\sin x)}\right)^{q} \times \lim _{x \rightarrow 0} x^{p-q}\) So, \(\quad \lim _{\mathrm{x} \rightarrow 0} \mathrm{f}(\mathrm{x})=0\) only when \(\mathrm{p}>\mathrm{q}\) Hence, L.M.V Theorem is applicable to \(\mathrm{f}(\mathrm{x})\) in \([0, \mathrm{x}]\) only when \(\mathrm{p}>\mathrm{q}\)
85746
If \(f(x)=\log (\sin x), x \in\left[\frac{\pi}{6}, \frac{5 \pi}{6}\right]\), then value of ' \(c\) ' by applying L.M.V.T. is
1 \(\frac{2 \pi}{3}\)
2 \(\frac{\pi}{2}\)
3 \(\frac{\pi}{4}\)
4 \(\frac{3 \pi}{4}\)
Explanation:
(B) : Given, \(f(x)=\log (\sin x) x \in\left[\frac{\pi}{6}, \frac{5 \pi}{6}\right]\) By Lagrange's mean value theorem, \(f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}=\frac{f\left(\frac{5 \pi}{6}\right)-f\left(\frac{\pi}{6}\right)}{\frac{5 \pi}{6}-\frac{\pi}{6}}\) f \((\mathrm{b})=\log \left(\sin \frac{\pi}{6}\right)=\log \left(\frac{1}{2}\right)\) \(f(a)=\log \left(\sin \frac{5 \pi}{6}\right)=\log \left(\frac{1}{2}\right)\) Now, \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\sin \mathrm{x}} \cdot \cos \mathrm{x}=\cot \mathrm{x}\) \(\therefore \quad \mathrm{f}^{\prime}(\mathrm{c})=\cot \mathrm{c}\) \(\therefore \mathrm{f}^{\prime}(\mathrm{c})=\frac{\log \frac{1}{2}-\log \frac{1}{2}}{\frac{5 \pi}{6}-\frac{\pi}{6}}\) \(\cot \mathrm{c}\) So, \(\quad \mathrm{c}=\frac{\pi}{2}\)
MHT CET-2020
Application of Derivatives
85747
IF the L.M.V.T. holds for the function \(\mathrm{f}(\mathrm{x})=\mathrm{x}+\frac{1}{\mathrm{x}}, \mathrm{x} \in[1,3]\), then \(\mathrm{c}=\)
1 2
2 3
3 \(\sqrt{3}\)
4 -3
Explanation:
(C) : Given, \(f(x)=x+\frac{1}{x} \quad x \in[1,3]\) By Lagrange's mean value theorem, \(f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\) \(f(3)=3+\frac{1}{3}=\frac{10}{3}\) \(\mathrm{f}(1)=1+\frac{1}{1}=2\) Now, \(\quad f^{\prime}(x)=1-\frac{1}{x^{2}}\) \(\mathrm{f}^{\prime}(\mathrm{c})=1-\frac{1}{\mathrm{c}^{2}} \Rightarrow \mathrm{f}^{\prime}(\mathrm{c})=\frac{\mathrm{f}(\mathrm{b})-\mathrm{f}(\mathrm{a})}{\mathrm{b}-\mathrm{a}}\) \(1-\frac{1}{\mathrm{c}^{2}}=\frac{\frac{10}{3}-2}{3-1}=\frac{2}{3} \Rightarrow 1-\frac{2}{3}=\frac{1}{\mathrm{c}^{2}}\) \(\mathrm{c}= \pm \sqrt{3} \quad(\because \mathrm{x} \in[1,3])\) \(\therefore \quad \mathrm{c}=\sqrt{3}\)
Manipal-2014
Application of Derivatives
85748
If Rolle's theorem holds for the function \(f(x)=\cos x+\sin x+7, x \in[0,2 \pi]\) and \(0\lt \mathbf{c}\lt 2 \pi\) such that \(\mathbf{f}^{\prime}(\mathrm{c})=\mathbf{0}\), then the number of possible value of \(c\) is
85750
Let \(f(x)=\left\{\frac{x^{p}}{(\sin x)^{q}}\right.\), if \(0\lt x \leq \frac{\pi}{2}\) (p, q \(\in \mathbf{R})\). Then, Lagrange's mean value theorem applicable to \(f(x)\) in closed interval [0, \(\mathbf{x ]}\)
1 For all \(p, q\)
2 Only when \(\mathrm{p}>\mathrm{q}\)
3 Only when \(\mathrm{P}\lt \mathrm{q}\)
4 For no value of \(p, q\)
Explanation:
(B) : Given, \(f(x)=\left\{\frac{x^{p}}{(\sin x)^{q}}, \text { if } 0\lt x \leq \frac{\pi}{2}\right.\) Since, Lagrange's mean value theorem is applicable on \(\mathrm{f}(\mathrm{x})\). \(\therefore \quad \lim _{x \rightarrow 0} \frac{x^{p}}{(\sin x)^{q}}=f(0) \Rightarrow \lim _{x \rightarrow 0} \frac{x^{p}}{(\sin x)^{q}}=0\) \(\lim _{x \rightarrow 0} \frac{x^{p} \cdot x^{q} \cdot x^{-q}}{(\sin x)^{q}} \Rightarrow \lim _{x \rightarrow 0} \frac{x^{q} \cdot x^{p-q}}{(\sin x)^{q}}\) \(\lim _{x \rightarrow 0} \frac{x^{q}}{(\sin x)^{q}} \times \lim _{x \rightarrow 0} x^{p-q}\) \(\lim _{x \rightarrow 0}\left(\frac{x}{(\sin x)}\right)^{q} \times \lim _{x \rightarrow 0} x^{p-q}\) So, \(\quad \lim _{\mathrm{x} \rightarrow 0} \mathrm{f}(\mathrm{x})=0\) only when \(\mathrm{p}>\mathrm{q}\) Hence, L.M.V Theorem is applicable to \(\mathrm{f}(\mathrm{x})\) in \([0, \mathrm{x}]\) only when \(\mathrm{p}>\mathrm{q}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Application of Derivatives
85746
If \(f(x)=\log (\sin x), x \in\left[\frac{\pi}{6}, \frac{5 \pi}{6}\right]\), then value of ' \(c\) ' by applying L.M.V.T. is
1 \(\frac{2 \pi}{3}\)
2 \(\frac{\pi}{2}\)
3 \(\frac{\pi}{4}\)
4 \(\frac{3 \pi}{4}\)
Explanation:
(B) : Given, \(f(x)=\log (\sin x) x \in\left[\frac{\pi}{6}, \frac{5 \pi}{6}\right]\) By Lagrange's mean value theorem, \(f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}=\frac{f\left(\frac{5 \pi}{6}\right)-f\left(\frac{\pi}{6}\right)}{\frac{5 \pi}{6}-\frac{\pi}{6}}\) f \((\mathrm{b})=\log \left(\sin \frac{\pi}{6}\right)=\log \left(\frac{1}{2}\right)\) \(f(a)=\log \left(\sin \frac{5 \pi}{6}\right)=\log \left(\frac{1}{2}\right)\) Now, \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\sin \mathrm{x}} \cdot \cos \mathrm{x}=\cot \mathrm{x}\) \(\therefore \quad \mathrm{f}^{\prime}(\mathrm{c})=\cot \mathrm{c}\) \(\therefore \mathrm{f}^{\prime}(\mathrm{c})=\frac{\log \frac{1}{2}-\log \frac{1}{2}}{\frac{5 \pi}{6}-\frac{\pi}{6}}\) \(\cot \mathrm{c}\) So, \(\quad \mathrm{c}=\frac{\pi}{2}\)
MHT CET-2020
Application of Derivatives
85747
IF the L.M.V.T. holds for the function \(\mathrm{f}(\mathrm{x})=\mathrm{x}+\frac{1}{\mathrm{x}}, \mathrm{x} \in[1,3]\), then \(\mathrm{c}=\)
1 2
2 3
3 \(\sqrt{3}\)
4 -3
Explanation:
(C) : Given, \(f(x)=x+\frac{1}{x} \quad x \in[1,3]\) By Lagrange's mean value theorem, \(f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\) \(f(3)=3+\frac{1}{3}=\frac{10}{3}\) \(\mathrm{f}(1)=1+\frac{1}{1}=2\) Now, \(\quad f^{\prime}(x)=1-\frac{1}{x^{2}}\) \(\mathrm{f}^{\prime}(\mathrm{c})=1-\frac{1}{\mathrm{c}^{2}} \Rightarrow \mathrm{f}^{\prime}(\mathrm{c})=\frac{\mathrm{f}(\mathrm{b})-\mathrm{f}(\mathrm{a})}{\mathrm{b}-\mathrm{a}}\) \(1-\frac{1}{\mathrm{c}^{2}}=\frac{\frac{10}{3}-2}{3-1}=\frac{2}{3} \Rightarrow 1-\frac{2}{3}=\frac{1}{\mathrm{c}^{2}}\) \(\mathrm{c}= \pm \sqrt{3} \quad(\because \mathrm{x} \in[1,3])\) \(\therefore \quad \mathrm{c}=\sqrt{3}\)
Manipal-2014
Application of Derivatives
85748
If Rolle's theorem holds for the function \(f(x)=\cos x+\sin x+7, x \in[0,2 \pi]\) and \(0\lt \mathbf{c}\lt 2 \pi\) such that \(\mathbf{f}^{\prime}(\mathrm{c})=\mathbf{0}\), then the number of possible value of \(c\) is
85750
Let \(f(x)=\left\{\frac{x^{p}}{(\sin x)^{q}}\right.\), if \(0\lt x \leq \frac{\pi}{2}\) (p, q \(\in \mathbf{R})\). Then, Lagrange's mean value theorem applicable to \(f(x)\) in closed interval [0, \(\mathbf{x ]}\)
1 For all \(p, q\)
2 Only when \(\mathrm{p}>\mathrm{q}\)
3 Only when \(\mathrm{P}\lt \mathrm{q}\)
4 For no value of \(p, q\)
Explanation:
(B) : Given, \(f(x)=\left\{\frac{x^{p}}{(\sin x)^{q}}, \text { if } 0\lt x \leq \frac{\pi}{2}\right.\) Since, Lagrange's mean value theorem is applicable on \(\mathrm{f}(\mathrm{x})\). \(\therefore \quad \lim _{x \rightarrow 0} \frac{x^{p}}{(\sin x)^{q}}=f(0) \Rightarrow \lim _{x \rightarrow 0} \frac{x^{p}}{(\sin x)^{q}}=0\) \(\lim _{x \rightarrow 0} \frac{x^{p} \cdot x^{q} \cdot x^{-q}}{(\sin x)^{q}} \Rightarrow \lim _{x \rightarrow 0} \frac{x^{q} \cdot x^{p-q}}{(\sin x)^{q}}\) \(\lim _{x \rightarrow 0} \frac{x^{q}}{(\sin x)^{q}} \times \lim _{x \rightarrow 0} x^{p-q}\) \(\lim _{x \rightarrow 0}\left(\frac{x}{(\sin x)}\right)^{q} \times \lim _{x \rightarrow 0} x^{p-q}\) So, \(\quad \lim _{\mathrm{x} \rightarrow 0} \mathrm{f}(\mathrm{x})=0\) only when \(\mathrm{p}>\mathrm{q}\) Hence, L.M.V Theorem is applicable to \(\mathrm{f}(\mathrm{x})\) in \([0, \mathrm{x}]\) only when \(\mathrm{p}>\mathrm{q}\)
85746
If \(f(x)=\log (\sin x), x \in\left[\frac{\pi}{6}, \frac{5 \pi}{6}\right]\), then value of ' \(c\) ' by applying L.M.V.T. is
1 \(\frac{2 \pi}{3}\)
2 \(\frac{\pi}{2}\)
3 \(\frac{\pi}{4}\)
4 \(\frac{3 \pi}{4}\)
Explanation:
(B) : Given, \(f(x)=\log (\sin x) x \in\left[\frac{\pi}{6}, \frac{5 \pi}{6}\right]\) By Lagrange's mean value theorem, \(f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}=\frac{f\left(\frac{5 \pi}{6}\right)-f\left(\frac{\pi}{6}\right)}{\frac{5 \pi}{6}-\frac{\pi}{6}}\) f \((\mathrm{b})=\log \left(\sin \frac{\pi}{6}\right)=\log \left(\frac{1}{2}\right)\) \(f(a)=\log \left(\sin \frac{5 \pi}{6}\right)=\log \left(\frac{1}{2}\right)\) Now, \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\sin \mathrm{x}} \cdot \cos \mathrm{x}=\cot \mathrm{x}\) \(\therefore \quad \mathrm{f}^{\prime}(\mathrm{c})=\cot \mathrm{c}\) \(\therefore \mathrm{f}^{\prime}(\mathrm{c})=\frac{\log \frac{1}{2}-\log \frac{1}{2}}{\frac{5 \pi}{6}-\frac{\pi}{6}}\) \(\cot \mathrm{c}\) So, \(\quad \mathrm{c}=\frac{\pi}{2}\)
MHT CET-2020
Application of Derivatives
85747
IF the L.M.V.T. holds for the function \(\mathrm{f}(\mathrm{x})=\mathrm{x}+\frac{1}{\mathrm{x}}, \mathrm{x} \in[1,3]\), then \(\mathrm{c}=\)
1 2
2 3
3 \(\sqrt{3}\)
4 -3
Explanation:
(C) : Given, \(f(x)=x+\frac{1}{x} \quad x \in[1,3]\) By Lagrange's mean value theorem, \(f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\) \(f(3)=3+\frac{1}{3}=\frac{10}{3}\) \(\mathrm{f}(1)=1+\frac{1}{1}=2\) Now, \(\quad f^{\prime}(x)=1-\frac{1}{x^{2}}\) \(\mathrm{f}^{\prime}(\mathrm{c})=1-\frac{1}{\mathrm{c}^{2}} \Rightarrow \mathrm{f}^{\prime}(\mathrm{c})=\frac{\mathrm{f}(\mathrm{b})-\mathrm{f}(\mathrm{a})}{\mathrm{b}-\mathrm{a}}\) \(1-\frac{1}{\mathrm{c}^{2}}=\frac{\frac{10}{3}-2}{3-1}=\frac{2}{3} \Rightarrow 1-\frac{2}{3}=\frac{1}{\mathrm{c}^{2}}\) \(\mathrm{c}= \pm \sqrt{3} \quad(\because \mathrm{x} \in[1,3])\) \(\therefore \quad \mathrm{c}=\sqrt{3}\)
Manipal-2014
Application of Derivatives
85748
If Rolle's theorem holds for the function \(f(x)=\cos x+\sin x+7, x \in[0,2 \pi]\) and \(0\lt \mathbf{c}\lt 2 \pi\) such that \(\mathbf{f}^{\prime}(\mathrm{c})=\mathbf{0}\), then the number of possible value of \(c\) is
85750
Let \(f(x)=\left\{\frac{x^{p}}{(\sin x)^{q}}\right.\), if \(0\lt x \leq \frac{\pi}{2}\) (p, q \(\in \mathbf{R})\). Then, Lagrange's mean value theorem applicable to \(f(x)\) in closed interval [0, \(\mathbf{x ]}\)
1 For all \(p, q\)
2 Only when \(\mathrm{p}>\mathrm{q}\)
3 Only when \(\mathrm{P}\lt \mathrm{q}\)
4 For no value of \(p, q\)
Explanation:
(B) : Given, \(f(x)=\left\{\frac{x^{p}}{(\sin x)^{q}}, \text { if } 0\lt x \leq \frac{\pi}{2}\right.\) Since, Lagrange's mean value theorem is applicable on \(\mathrm{f}(\mathrm{x})\). \(\therefore \quad \lim _{x \rightarrow 0} \frac{x^{p}}{(\sin x)^{q}}=f(0) \Rightarrow \lim _{x \rightarrow 0} \frac{x^{p}}{(\sin x)^{q}}=0\) \(\lim _{x \rightarrow 0} \frac{x^{p} \cdot x^{q} \cdot x^{-q}}{(\sin x)^{q}} \Rightarrow \lim _{x \rightarrow 0} \frac{x^{q} \cdot x^{p-q}}{(\sin x)^{q}}\) \(\lim _{x \rightarrow 0} \frac{x^{q}}{(\sin x)^{q}} \times \lim _{x \rightarrow 0} x^{p-q}\) \(\lim _{x \rightarrow 0}\left(\frac{x}{(\sin x)}\right)^{q} \times \lim _{x \rightarrow 0} x^{p-q}\) So, \(\quad \lim _{\mathrm{x} \rightarrow 0} \mathrm{f}(\mathrm{x})=0\) only when \(\mathrm{p}>\mathrm{q}\) Hence, L.M.V Theorem is applicable to \(\mathrm{f}(\mathrm{x})\) in \([0, \mathrm{x}]\) only when \(\mathrm{p}>\mathrm{q}\)