QBManager

Application of Derivatives

85746 If $f (x) = \log (\sin x), x \in [\frac{π}{6}, \frac{5 π}{6}]$ , then value of ' $c$ ' by applying L.M.V.T. is

1

\frac{2 π}{3}

2

\frac{π}{2}

3

\frac{π}{4}

4

\frac{3 π}{4}

Explanation:

(B) : Given,
$f (x) = \log (\sin x) x \in [\frac{π}{6}, \frac{5 π}{6}]$
By Lagrange's mean value theorem,
$f^{'} (c) = \frac{f (b) - f (a)}{b - a} = \frac{f (\frac{5 π}{6}) - f (\frac{π}{6})}{\frac{5 π}{6} - \frac{π}{6}}$
f $(b) = \log (\sin \frac{π}{6}) = \log (\frac{1}{2})$
$f (a) = \log (\sin \frac{5 π}{6}) = \log (\frac{1}{2})$
Now, $f^{'} (x) = \frac{1}{\sin x} \cdot \cos x = \cot x$
$∴ f^{'} (c) = \cot c$ $∴ f^{'} (c) = \frac{\log \frac{1}{2} - \log \frac{1}{2}}{\frac{5 π}{6} - \frac{π}{6}}$
$\cot c$
So, $c = \frac{π}{2}$

MHT CET-2020

Application of Derivatives

85747 IF the L.M.V.T. holds for the function $f (x) = x + \frac{1}{x}, x \in [1, 3]$ , then $c =$

1 2

2 3

3

\sqrt{3}

4 -3

Explanation:

(C) : Given,
$f (x) = x + \frac{1}{x} x \in [1, 3]$
By Lagrange's mean value theorem,
$f^{'} (c) = \frac{f (b) - f (a)}{b - a}$
$f (3) = 3 + \frac{1}{3} = \frac{10}{3}$
$f (1) = 1 + \frac{1}{1} = 2$
Now, $f^{'} (x) = 1 - \frac{1}{x^{2}}$
$f^{'} (c) = 1 - \frac{1}{c^{2}} \Rightarrow f^{'} (c) = \frac{f (b) - f (a)}{b - a}$
$1 - \frac{1}{c^{2}} = \frac{\frac{10}{3} - 2}{3 - 1} = \frac{2}{3} \Rightarrow 1 - \frac{2}{3} = \frac{1}{c^{2}}$
$c = \pm \sqrt{3} (∵ x \in [1, 3])$
$∴ c = \sqrt{3}$

Manipal-2014

Application of Derivatives

85748 If Rolle's theorem holds for the function $f (x) = \cos x + \sin x + 7, x \in [0, 2 π]$ and
$0 < c < 2 π$ such that $f^{'} (c) = 0$ , then the number of possible value of $c$ is

1 1

2 2

3 0

4 3

Explanation:

(B) : Given,
$f (x) = \cos x + \sin x + 7 x \in [0, 2 π]$
$f (0) = \cos 0 + \sin 0 + 7 = 1 + 0 + 7 = 8$
$f (2 π) = \sin 2 π + \cos 2 π + 7$
$= 8$
$∴ f^{'} (c) = \frac{f (b) - f (a)}{b - a} = \frac{8 - 8}{2 π - 0} = 0$
Rolles's theorem holds for the function
$f (x) \Rightarrow f^{'} (c) = 0$
So,
$f^{'} (c) = \cos c - \sin c$
$0 = \cos c - \sin c$
$\cos c = \sin c$
$\tan c = 1$
$c = n π + \frac{π}{4} \Rightarrow∴ c = \frac{π}{4}, \frac{5 π}{4}, \frac{9 π}{4}$
But, $0 < c < 2 π \Rightarrow$ Hence, $c = \frac{π}{4}, \frac{5 π}{4}$

MHT CET-2019

Application of Derivatives

85749 If Rolle's theorem for $f (x) = e^{x} (\sin x - \cos x)$ is verified on $[\frac{π}{4}, \frac{5 π}{4}]$ , then the value of $c$ is

1

\frac{π}{3}

2

\frac{π}{2}

3

\frac{3 π}{4}

4

π

Explanation:

(D) : Given,
We know that,
$f^{'} (c) = \frac{f (b) - f (a)}{b - a}$
$f (\frac{π}{4}) = e^{π / 4} (\sin \frac{π}{4} - \cos \frac{π}{4})$
$= e^{π / 4} (\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) = 0$
$f (\frac{5 π}{4}) = e^{5 π / 4} (\sin \frac{5 π}{4} - \cos \frac{5 π}{4})$
$= e^{5 π / 4} (- \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}})$
$= 0$
$f (\frac{5 π}{4}) = 0$
$f^{'} (x) = e^{x} (\cos x + \sin x) + (\sin x - \cos x) e^{x}$
$f^{'} (c) = 2 e^{c} \sin c$
By the Rolle's theorem,
$2 e^{c} \sin c = 0$
$\sin c = 0$
$c = π$
$∵ π \in [π / 4, 5 π / 4]$

[JCECE-2009

Application of Derivatives

85750 Let $f (x) = {\frac{x^{p}}{(\sin x)^{q}}$ , if $0 < x \leq \frac{π}{2}$
(p, q $\in R)$ . Then, Lagrange's mean value theorem applicable to $f (x)$ in closed interval [0, $x]$

1 For all

p, q

2 Only when

p > q

3 Only when

P < q

4 For no value of

p, q

Explanation:

(B) : Given,
$f (x) = {\frac{x^{p}}{(\sin x)^{q}}, if 0 < x \leq \frac{π}{2}$
Since, Lagrange's mean value theorem is applicable on $f (x)$ .
$∴ lim_{x \to 0} \frac{x^{p}}{(\sin x)^{q}} = f (0) \Rightarrow lim_{x \to 0} \frac{x^{p}}{(\sin x)^{q}} = 0$
$lim_{x \to 0} \frac{x^{p} \cdot x^{q} \cdot x^{- q}}{(\sin x)^{q}} \Rightarrow lim_{x \to 0} \frac{x^{q} \cdot x^{p - q}}{(\sin x)^{q}}$
$lim_{x \to 0} \frac{x^{q}}{(\sin x)^{q}} \times lim_{x \to 0} x^{p - q}$
$lim_{x \to 0} {(\frac{x}{(\sin x)})}^{q} \times lim_{x \to 0} x^{p - q}$
So, $lim_{x \to 0} f (x) = 0$ only when $p > q$
Hence, L.M.V Theorem is applicable to $f (x)$ in $[0, x]$ only when $p > q$

WB JEE-2017

Application of Derivatives

85746 If $f (x) = \log (\sin x), x \in [\frac{π}{6}, \frac{5 π}{6}]$ , then value of ' $c$ ' by applying L.M.V.T. is

1

\frac{2 π}{3}

2

\frac{π}{2}

3

\frac{π}{4}

4

\frac{3 π}{4}

Explanation:

(B) : Given,
$f (x) = \log (\sin x) x \in [\frac{π}{6}, \frac{5 π}{6}]$
By Lagrange's mean value theorem,
$f^{'} (c) = \frac{f (b) - f (a)}{b - a} = \frac{f (\frac{5 π}{6}) - f (\frac{π}{6})}{\frac{5 π}{6} - \frac{π}{6}}$
f $(b) = \log (\sin \frac{π}{6}) = \log (\frac{1}{2})$
$f (a) = \log (\sin \frac{5 π}{6}) = \log (\frac{1}{2})$
Now, $f^{'} (x) = \frac{1}{\sin x} \cdot \cos x = \cot x$
$∴ f^{'} (c) = \cot c$ $∴ f^{'} (c) = \frac{\log \frac{1}{2} - \log \frac{1}{2}}{\frac{5 π}{6} - \frac{π}{6}}$
$\cot c$
So, $c = \frac{π}{2}$

MHT CET-2020

Application of Derivatives

85747 IF the L.M.V.T. holds for the function $f (x) = x + \frac{1}{x}, x \in [1, 3]$ , then $c =$

1 2

2 3

3

\sqrt{3}

4 -3

Explanation:

(C) : Given,
$f (x) = x + \frac{1}{x} x \in [1, 3]$
By Lagrange's mean value theorem,
$f^{'} (c) = \frac{f (b) - f (a)}{b - a}$
$f (3) = 3 + \frac{1}{3} = \frac{10}{3}$
$f (1) = 1 + \frac{1}{1} = 2$
Now, $f^{'} (x) = 1 - \frac{1}{x^{2}}$
$f^{'} (c) = 1 - \frac{1}{c^{2}} \Rightarrow f^{'} (c) = \frac{f (b) - f (a)}{b - a}$
$1 - \frac{1}{c^{2}} = \frac{\frac{10}{3} - 2}{3 - 1} = \frac{2}{3} \Rightarrow 1 - \frac{2}{3} = \frac{1}{c^{2}}$
$c = \pm \sqrt{3} (∵ x \in [1, 3])$
$∴ c = \sqrt{3}$

Manipal-2014

Application of Derivatives

85748 If Rolle's theorem holds for the function $f (x) = \cos x + \sin x + 7, x \in [0, 2 π]$ and
$0 < c < 2 π$ such that $f^{'} (c) = 0$ , then the number of possible value of $c$ is

1 1

2 2

3 0

4 3

Explanation:

(B) : Given,
$f (x) = \cos x + \sin x + 7 x \in [0, 2 π]$
$f (0) = \cos 0 + \sin 0 + 7 = 1 + 0 + 7 = 8$
$f (2 π) = \sin 2 π + \cos 2 π + 7$
$= 8$
$∴ f^{'} (c) = \frac{f (b) - f (a)}{b - a} = \frac{8 - 8}{2 π - 0} = 0$
Rolles's theorem holds for the function
$f (x) \Rightarrow f^{'} (c) = 0$
So,
$f^{'} (c) = \cos c - \sin c$
$0 = \cos c - \sin c$
$\cos c = \sin c$
$\tan c = 1$
$c = n π + \frac{π}{4} \Rightarrow∴ c = \frac{π}{4}, \frac{5 π}{4}, \frac{9 π}{4}$
But, $0 < c < 2 π \Rightarrow$ Hence, $c = \frac{π}{4}, \frac{5 π}{4}$

MHT CET-2019

Application of Derivatives

85749 If Rolle's theorem for $f (x) = e^{x} (\sin x - \cos x)$ is verified on $[\frac{π}{4}, \frac{5 π}{4}]$ , then the value of $c$ is

1

\frac{π}{3}

2

\frac{π}{2}

3

\frac{3 π}{4}

4

π

Explanation:

(D) : Given,
We know that,
$f^{'} (c) = \frac{f (b) - f (a)}{b - a}$
$f (\frac{π}{4}) = e^{π / 4} (\sin \frac{π}{4} - \cos \frac{π}{4})$
$= e^{π / 4} (\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) = 0$
$f (\frac{5 π}{4}) = e^{5 π / 4} (\sin \frac{5 π}{4} - \cos \frac{5 π}{4})$
$= e^{5 π / 4} (- \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}})$
$= 0$
$f (\frac{5 π}{4}) = 0$
$f^{'} (x) = e^{x} (\cos x + \sin x) + (\sin x - \cos x) e^{x}$
$f^{'} (c) = 2 e^{c} \sin c$
By the Rolle's theorem,
$2 e^{c} \sin c = 0$
$\sin c = 0$
$c = π$
$∵ π \in [π / 4, 5 π / 4]$

[JCECE-2009

Application of Derivatives

85750 Let $f (x) = {\frac{x^{p}}{(\sin x)^{q}}$ , if $0 < x \leq \frac{π}{2}$
(p, q $\in R)$ . Then, Lagrange's mean value theorem applicable to $f (x)$ in closed interval [0, $x]$

1 For all

p, q

2 Only when

p > q

3 Only when

P < q

4 For no value of

p, q

Explanation:

(B) : Given,
$f (x) = {\frac{x^{p}}{(\sin x)^{q}}, if 0 < x \leq \frac{π}{2}$
Since, Lagrange's mean value theorem is applicable on $f (x)$ .
$∴ lim_{x \to 0} \frac{x^{p}}{(\sin x)^{q}} = f (0) \Rightarrow lim_{x \to 0} \frac{x^{p}}{(\sin x)^{q}} = 0$
$lim_{x \to 0} \frac{x^{p} \cdot x^{q} \cdot x^{- q}}{(\sin x)^{q}} \Rightarrow lim_{x \to 0} \frac{x^{q} \cdot x^{p - q}}{(\sin x)^{q}}$
$lim_{x \to 0} \frac{x^{q}}{(\sin x)^{q}} \times lim_{x \to 0} x^{p - q}$
$lim_{x \to 0} {(\frac{x}{(\sin x)})}^{q} \times lim_{x \to 0} x^{p - q}$
So, $lim_{x \to 0} f (x) = 0$ only when $p > q$
Hence, L.M.V Theorem is applicable to $f (x)$ in $[0, x]$ only when $p > q$

WB JEE-2017

Application of Derivatives

85746 If $f (x) = \log (\sin x), x \in [\frac{π}{6}, \frac{5 π}{6}]$ , then value of ' $c$ ' by applying L.M.V.T. is

1

\frac{2 π}{3}

2

\frac{π}{2}

3

\frac{π}{4}

4

\frac{3 π}{4}

Explanation:

(B) : Given,
$f (x) = \log (\sin x) x \in [\frac{π}{6}, \frac{5 π}{6}]$
By Lagrange's mean value theorem,
$f^{'} (c) = \frac{f (b) - f (a)}{b - a} = \frac{f (\frac{5 π}{6}) - f (\frac{π}{6})}{\frac{5 π}{6} - \frac{π}{6}}$
f $(b) = \log (\sin \frac{π}{6}) = \log (\frac{1}{2})$
$f (a) = \log (\sin \frac{5 π}{6}) = \log (\frac{1}{2})$
Now, $f^{'} (x) = \frac{1}{\sin x} \cdot \cos x = \cot x$
$∴ f^{'} (c) = \cot c$ $∴ f^{'} (c) = \frac{\log \frac{1}{2} - \log \frac{1}{2}}{\frac{5 π}{6} - \frac{π}{6}}$
$\cot c$
So, $c = \frac{π}{2}$

MHT CET-2020

Application of Derivatives

85747 IF the L.M.V.T. holds for the function $f (x) = x + \frac{1}{x}, x \in [1, 3]$ , then $c =$

1 2

2 3

3

\sqrt{3}

4 -3

Explanation:

(C) : Given,
$f (x) = x + \frac{1}{x} x \in [1, 3]$
By Lagrange's mean value theorem,
$f^{'} (c) = \frac{f (b) - f (a)}{b - a}$
$f (3) = 3 + \frac{1}{3} = \frac{10}{3}$
$f (1) = 1 + \frac{1}{1} = 2$
Now, $f^{'} (x) = 1 - \frac{1}{x^{2}}$
$f^{'} (c) = 1 - \frac{1}{c^{2}} \Rightarrow f^{'} (c) = \frac{f (b) - f (a)}{b - a}$
$1 - \frac{1}{c^{2}} = \frac{\frac{10}{3} - 2}{3 - 1} = \frac{2}{3} \Rightarrow 1 - \frac{2}{3} = \frac{1}{c^{2}}$
$c = \pm \sqrt{3} (∵ x \in [1, 3])$
$∴ c = \sqrt{3}$

Manipal-2014

Application of Derivatives

85748 If Rolle's theorem holds for the function $f (x) = \cos x + \sin x + 7, x \in [0, 2 π]$ and
$0 < c < 2 π$ such that $f^{'} (c) = 0$ , then the number of possible value of $c$ is

1 1

2 2

3 0

4 3

Explanation:

(B) : Given,
$f (x) = \cos x + \sin x + 7 x \in [0, 2 π]$
$f (0) = \cos 0 + \sin 0 + 7 = 1 + 0 + 7 = 8$
$f (2 π) = \sin 2 π + \cos 2 π + 7$
$= 8$
$∴ f^{'} (c) = \frac{f (b) - f (a)}{b - a} = \frac{8 - 8}{2 π - 0} = 0$
Rolles's theorem holds for the function
$f (x) \Rightarrow f^{'} (c) = 0$
So,
$f^{'} (c) = \cos c - \sin c$
$0 = \cos c - \sin c$
$\cos c = \sin c$
$\tan c = 1$
$c = n π + \frac{π}{4} \Rightarrow∴ c = \frac{π}{4}, \frac{5 π}{4}, \frac{9 π}{4}$
But, $0 < c < 2 π \Rightarrow$ Hence, $c = \frac{π}{4}, \frac{5 π}{4}$

MHT CET-2019

Application of Derivatives

85749 If Rolle's theorem for $f (x) = e^{x} (\sin x - \cos x)$ is verified on $[\frac{π}{4}, \frac{5 π}{4}]$ , then the value of $c$ is

1

\frac{π}{3}

2

\frac{π}{2}

3

\frac{3 π}{4}

4

π

Explanation:

(D) : Given,
We know that,
$f^{'} (c) = \frac{f (b) - f (a)}{b - a}$
$f (\frac{π}{4}) = e^{π / 4} (\sin \frac{π}{4} - \cos \frac{π}{4})$
$= e^{π / 4} (\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) = 0$
$f (\frac{5 π}{4}) = e^{5 π / 4} (\sin \frac{5 π}{4} - \cos \frac{5 π}{4})$
$= e^{5 π / 4} (- \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}})$
$= 0$
$f (\frac{5 π}{4}) = 0$
$f^{'} (x) = e^{x} (\cos x + \sin x) + (\sin x - \cos x) e^{x}$
$f^{'} (c) = 2 e^{c} \sin c$
By the Rolle's theorem,
$2 e^{c} \sin c = 0$
$\sin c = 0$
$c = π$
$∵ π \in [π / 4, 5 π / 4]$

[JCECE-2009

Application of Derivatives

85750 Let $f (x) = {\frac{x^{p}}{(\sin x)^{q}}$ , if $0 < x \leq \frac{π}{2}$
(p, q $\in R)$ . Then, Lagrange's mean value theorem applicable to $f (x)$ in closed interval [0, $x]$

1 For all

p, q

2 Only when

p > q

3 Only when

P < q

4 For no value of

p, q

Explanation:

(B) : Given,
$f (x) = {\frac{x^{p}}{(\sin x)^{q}}, if 0 < x \leq \frac{π}{2}$
Since, Lagrange's mean value theorem is applicable on $f (x)$ .
$∴ lim_{x \to 0} \frac{x^{p}}{(\sin x)^{q}} = f (0) \Rightarrow lim_{x \to 0} \frac{x^{p}}{(\sin x)^{q}} = 0$
$lim_{x \to 0} \frac{x^{p} \cdot x^{q} \cdot x^{- q}}{(\sin x)^{q}} \Rightarrow lim_{x \to 0} \frac{x^{q} \cdot x^{p - q}}{(\sin x)^{q}}$
$lim_{x \to 0} \frac{x^{q}}{(\sin x)^{q}} \times lim_{x \to 0} x^{p - q}$
$lim_{x \to 0} {(\frac{x}{(\sin x)})}^{q} \times lim_{x \to 0} x^{p - q}$
So, $lim_{x \to 0} f (x) = 0$ only when $p > q$
Hence, L.M.V Theorem is applicable to $f (x)$ in $[0, x]$ only when $p > q$

WB JEE-2017

Application of Derivatives

85746 If $f (x) = \log (\sin x), x \in [\frac{π}{6}, \frac{5 π}{6}]$ , then value of ' $c$ ' by applying L.M.V.T. is

1

\frac{2 π}{3}

2

\frac{π}{2}

3

\frac{π}{4}

4

\frac{3 π}{4}

Explanation:

(B) : Given,
$f (x) = \log (\sin x) x \in [\frac{π}{6}, \frac{5 π}{6}]$
By Lagrange's mean value theorem,
$f^{'} (c) = \frac{f (b) - f (a)}{b - a} = \frac{f (\frac{5 π}{6}) - f (\frac{π}{6})}{\frac{5 π}{6} - \frac{π}{6}}$
f $(b) = \log (\sin \frac{π}{6}) = \log (\frac{1}{2})$
$f (a) = \log (\sin \frac{5 π}{6}) = \log (\frac{1}{2})$
Now, $f^{'} (x) = \frac{1}{\sin x} \cdot \cos x = \cot x$
$∴ f^{'} (c) = \cot c$ $∴ f^{'} (c) = \frac{\log \frac{1}{2} - \log \frac{1}{2}}{\frac{5 π}{6} - \frac{π}{6}}$
$\cot c$
So, $c = \frac{π}{2}$

MHT CET-2020

Application of Derivatives

85747 IF the L.M.V.T. holds for the function $f (x) = x + \frac{1}{x}, x \in [1, 3]$ , then $c =$

1 2

2 3

3

\sqrt{3}

4 -3

Explanation:

(C) : Given,
$f (x) = x + \frac{1}{x} x \in [1, 3]$
By Lagrange's mean value theorem,
$f^{'} (c) = \frac{f (b) - f (a)}{b - a}$
$f (3) = 3 + \frac{1}{3} = \frac{10}{3}$
$f (1) = 1 + \frac{1}{1} = 2$
Now, $f^{'} (x) = 1 - \frac{1}{x^{2}}$
$f^{'} (c) = 1 - \frac{1}{c^{2}} \Rightarrow f^{'} (c) = \frac{f (b) - f (a)}{b - a}$
$1 - \frac{1}{c^{2}} = \frac{\frac{10}{3} - 2}{3 - 1} = \frac{2}{3} \Rightarrow 1 - \frac{2}{3} = \frac{1}{c^{2}}$
$c = \pm \sqrt{3} (∵ x \in [1, 3])$
$∴ c = \sqrt{3}$

Manipal-2014

Application of Derivatives

85748 If Rolle's theorem holds for the function $f (x) = \cos x + \sin x + 7, x \in [0, 2 π]$ and
$0 < c < 2 π$ such that $f^{'} (c) = 0$ , then the number of possible value of $c$ is

1 1

2 2

3 0

4 3

Explanation:

(B) : Given,
$f (x) = \cos x + \sin x + 7 x \in [0, 2 π]$
$f (0) = \cos 0 + \sin 0 + 7 = 1 + 0 + 7 = 8$
$f (2 π) = \sin 2 π + \cos 2 π + 7$
$= 8$
$∴ f^{'} (c) = \frac{f (b) - f (a)}{b - a} = \frac{8 - 8}{2 π - 0} = 0$
Rolles's theorem holds for the function
$f (x) \Rightarrow f^{'} (c) = 0$
So,
$f^{'} (c) = \cos c - \sin c$
$0 = \cos c - \sin c$
$\cos c = \sin c$
$\tan c = 1$
$c = n π + \frac{π}{4} \Rightarrow∴ c = \frac{π}{4}, \frac{5 π}{4}, \frac{9 π}{4}$
But, $0 < c < 2 π \Rightarrow$ Hence, $c = \frac{π}{4}, \frac{5 π}{4}$

MHT CET-2019

Application of Derivatives

85749 If Rolle's theorem for $f (x) = e^{x} (\sin x - \cos x)$ is verified on $[\frac{π}{4}, \frac{5 π}{4}]$ , then the value of $c$ is

1

\frac{π}{3}

2

\frac{π}{2}

3

\frac{3 π}{4}

4

π

Explanation:

(D) : Given,
We know that,
$f^{'} (c) = \frac{f (b) - f (a)}{b - a}$
$f (\frac{π}{4}) = e^{π / 4} (\sin \frac{π}{4} - \cos \frac{π}{4})$
$= e^{π / 4} (\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) = 0$
$f (\frac{5 π}{4}) = e^{5 π / 4} (\sin \frac{5 π}{4} - \cos \frac{5 π}{4})$
$= e^{5 π / 4} (- \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}})$
$= 0$
$f (\frac{5 π}{4}) = 0$
$f^{'} (x) = e^{x} (\cos x + \sin x) + (\sin x - \cos x) e^{x}$
$f^{'} (c) = 2 e^{c} \sin c$
By the Rolle's theorem,
$2 e^{c} \sin c = 0$
$\sin c = 0$
$c = π$
$∵ π \in [π / 4, 5 π / 4]$

[JCECE-2009

Application of Derivatives

85750 Let $f (x) = {\frac{x^{p}}{(\sin x)^{q}}$ , if $0 < x \leq \frac{π}{2}$
(p, q $\in R)$ . Then, Lagrange's mean value theorem applicable to $f (x)$ in closed interval [0, $x]$

1 For all

p, q

2 Only when

p > q

3 Only when

P < q

4 For no value of

p, q

Explanation:

(B) : Given,
$f (x) = {\frac{x^{p}}{(\sin x)^{q}}, if 0 < x \leq \frac{π}{2}$
Since, Lagrange's mean value theorem is applicable on $f (x)$ .
$∴ lim_{x \to 0} \frac{x^{p}}{(\sin x)^{q}} = f (0) \Rightarrow lim_{x \to 0} \frac{x^{p}}{(\sin x)^{q}} = 0$
$lim_{x \to 0} \frac{x^{p} \cdot x^{q} \cdot x^{- q}}{(\sin x)^{q}} \Rightarrow lim_{x \to 0} \frac{x^{q} \cdot x^{p - q}}{(\sin x)^{q}}$
$lim_{x \to 0} \frac{x^{q}}{(\sin x)^{q}} \times lim_{x \to 0} x^{p - q}$
$lim_{x \to 0} {(\frac{x}{(\sin x)})}^{q} \times lim_{x \to 0} x^{p - q}$
So, $lim_{x \to 0} f (x) = 0$ only when $p > q$
Hence, L.M.V Theorem is applicable to $f (x)$ in $[0, x]$ only when $p > q$

WB JEE-2017

Application of Derivatives

85746 If $f (x) = \log (\sin x), x \in [\frac{π}{6}, \frac{5 π}{6}]$ , then value of ' $c$ ' by applying L.M.V.T. is

1

\frac{2 π}{3}

2

\frac{π}{2}

3

\frac{π}{4}

4

\frac{3 π}{4}

Explanation:

(B) : Given,
$f (x) = \log (\sin x) x \in [\frac{π}{6}, \frac{5 π}{6}]$
By Lagrange's mean value theorem,
$f^{'} (c) = \frac{f (b) - f (a)}{b - a} = \frac{f (\frac{5 π}{6}) - f (\frac{π}{6})}{\frac{5 π}{6} - \frac{π}{6}}$
f $(b) = \log (\sin \frac{π}{6}) = \log (\frac{1}{2})$
$f (a) = \log (\sin \frac{5 π}{6}) = \log (\frac{1}{2})$
Now, $f^{'} (x) = \frac{1}{\sin x} \cdot \cos x = \cot x$
$∴ f^{'} (c) = \cot c$ $∴ f^{'} (c) = \frac{\log \frac{1}{2} - \log \frac{1}{2}}{\frac{5 π}{6} - \frac{π}{6}}$
$\cot c$
So, $c = \frac{π}{2}$

MHT CET-2020

Application of Derivatives

85747 IF the L.M.V.T. holds for the function $f (x) = x + \frac{1}{x}, x \in [1, 3]$ , then $c =$

1 2

2 3

3

\sqrt{3}

4 -3

Explanation:

(C) : Given,
$f (x) = x + \frac{1}{x} x \in [1, 3]$
By Lagrange's mean value theorem,
$f^{'} (c) = \frac{f (b) - f (a)}{b - a}$
$f (3) = 3 + \frac{1}{3} = \frac{10}{3}$
$f (1) = 1 + \frac{1}{1} = 2$
Now, $f^{'} (x) = 1 - \frac{1}{x^{2}}$
$f^{'} (c) = 1 - \frac{1}{c^{2}} \Rightarrow f^{'} (c) = \frac{f (b) - f (a)}{b - a}$
$1 - \frac{1}{c^{2}} = \frac{\frac{10}{3} - 2}{3 - 1} = \frac{2}{3} \Rightarrow 1 - \frac{2}{3} = \frac{1}{c^{2}}$
$c = \pm \sqrt{3} (∵ x \in [1, 3])$
$∴ c = \sqrt{3}$

Manipal-2014

Application of Derivatives

85748 If Rolle's theorem holds for the function $f (x) = \cos x + \sin x + 7, x \in [0, 2 π]$ and
$0 < c < 2 π$ such that $f^{'} (c) = 0$ , then the number of possible value of $c$ is

1 1

2 2

3 0

4 3

Explanation:

(B) : Given,
$f (x) = \cos x + \sin x + 7 x \in [0, 2 π]$
$f (0) = \cos 0 + \sin 0 + 7 = 1 + 0 + 7 = 8$
$f (2 π) = \sin 2 π + \cos 2 π + 7$
$= 8$
$∴ f^{'} (c) = \frac{f (b) - f (a)}{b - a} = \frac{8 - 8}{2 π - 0} = 0$
Rolles's theorem holds for the function
$f (x) \Rightarrow f^{'} (c) = 0$
So,
$f^{'} (c) = \cos c - \sin c$
$0 = \cos c - \sin c$
$\cos c = \sin c$
$\tan c = 1$
$c = n π + \frac{π}{4} \Rightarrow∴ c = \frac{π}{4}, \frac{5 π}{4}, \frac{9 π}{4}$
But, $0 < c < 2 π \Rightarrow$ Hence, $c = \frac{π}{4}, \frac{5 π}{4}$

MHT CET-2019

Application of Derivatives

85749 If Rolle's theorem for $f (x) = e^{x} (\sin x - \cos x)$ is verified on $[\frac{π}{4}, \frac{5 π}{4}]$ , then the value of $c$ is

1

\frac{π}{3}

2

\frac{π}{2}

3

\frac{3 π}{4}

4

π

Explanation:

(D) : Given,
We know that,
$f^{'} (c) = \frac{f (b) - f (a)}{b - a}$
$f (\frac{π}{4}) = e^{π / 4} (\sin \frac{π}{4} - \cos \frac{π}{4})$
$= e^{π / 4} (\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) = 0$
$f (\frac{5 π}{4}) = e^{5 π / 4} (\sin \frac{5 π}{4} - \cos \frac{5 π}{4})$
$= e^{5 π / 4} (- \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}})$
$= 0$
$f (\frac{5 π}{4}) = 0$
$f^{'} (x) = e^{x} (\cos x + \sin x) + (\sin x - \cos x) e^{x}$
$f^{'} (c) = 2 e^{c} \sin c$
By the Rolle's theorem,
$2 e^{c} \sin c = 0$
$\sin c = 0$
$c = π$
$∵ π \in [π / 4, 5 π / 4]$

[JCECE-2009

Application of Derivatives

85750 Let $f (x) = {\frac{x^{p}}{(\sin x)^{q}}$ , if $0 < x \leq \frac{π}{2}$
(p, q $\in R)$ . Then, Lagrange's mean value theorem applicable to $f (x)$ in closed interval [0, $x]$

1 For all

p, q

2 Only when

p > q

3 Only when

P < q

4 For no value of

p, q

Explanation:

(B) : Given,
$f (x) = {\frac{x^{p}}{(\sin x)^{q}}, if 0 < x \leq \frac{π}{2}$
Since, Lagrange's mean value theorem is applicable on $f (x)$ .
$∴ lim_{x \to 0} \frac{x^{p}}{(\sin x)^{q}} = f (0) \Rightarrow lim_{x \to 0} \frac{x^{p}}{(\sin x)^{q}} = 0$
$lim_{x \to 0} \frac{x^{p} \cdot x^{q} \cdot x^{- q}}{(\sin x)^{q}} \Rightarrow lim_{x \to 0} \frac{x^{q} \cdot x^{p - q}}{(\sin x)^{q}}$
$lim_{x \to 0} \frac{x^{q}}{(\sin x)^{q}} \times lim_{x \to 0} x^{p - q}$
$lim_{x \to 0} {(\frac{x}{(\sin x)})}^{q} \times lim_{x \to 0} x^{p - q}$
So, $lim_{x \to 0} f (x) = 0$ only when $p > q$
Hence, L.M.V Theorem is applicable to $f (x)$ in $[0, x]$ only when $p > q$

WB JEE-2017

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: