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Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Five

Using Rolleâ€™s theorem, find the point on the curve $y = x (x - 4)$ , $x \in [0, 4]$ , where the tangent is parallel to the x-axis.

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Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t y = x (x - 4), x \in [0, 4] \\ L e t f (x) = x (x - 4), x \in [0, 4] \\ (i) f (x) b e i n g a n algebraic p o l y n o m i a l i s c o n t i n u o u s f u n c t i o n e v e r y w h e r e . \\ S o, f (x) = x (x - 4) i s c o n t i n u o u s x i n [0, 4] . \\ (i i) f^{'} (x) = 2 x - 4 w h i c h e x i s t s i n (0, 4) \\ S o, f (x) i s d i f f e r e n t i a b l e . \\ (i i i) f (0) = 0 (0 - 4) = 0 \\ f (4) = 0 (4 - 4) = 0 \\ ∴ f (0) = f (4) = 0 \\ A s t h e a b o v e c o n d i t i o n s a r e s a t i s f i e d, t h e n t h e r e m u s t e x i s t a t l e a s t o n e point c \in (0, 4) s u c h t h a t f^{'} (c) = 0 \\ ∴ 2 c - 4 = 0 \Rightarrow c = 2 \in (0, 4) \\ H e n c e, c = 2 i s t h e point i n (0, 4) o n t h e g i v e n c u r v e a t w h i c h t h e tangent i s p a r a l l e l t o t h e x - a x i s . \end{array}$

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Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Five

Find the points on the curve $y = c o s x - 1$ in $[0, 2 π]$ , where the tangent is parallel to the x-axis.

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Asked by Alok Kumar Singh

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Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t y = c o s x - 1 o n [0, 2 π] \\ F i r s t l y, w e h a v e t o f i n d a point c o n t h e g i v e n c u r v e y = c o s x - 1 o n [0, 2 π] s u c h t h a t t h e tangent at \\ c = 0 \in [0, 2 π] i s p a r a l l e l t o x - a x i s i . e ., f^{'} (c) = 0 w h e r e f^{'} (c) i s t h e s l o p e o f t h e tangent . \\ S o, w e h a v e t o v e r i f y t h e R o l l e' s T h e o r e m . \\ (i) y = c o s x - 1, i s t h e c o m b i n a t i o n o f cosine a n d constant f u n c t i o n s . S o, i t i s c o n t i n u o u s o n [0, 2 π] . \\ (i i) \frac{d y}{d x} = - s i n x w h i c h e x i s t s i n (0, 2 π) \\ S o i t i s d i f f e r e n t i a b l e o n (0, 2 π) . \\ (i i i) L e t f (x) = c o s x - 1 \\ f (0) = c o s 0 - 1 \Rightarrow 1 - 1 = 0 \\ f (2 π) = c o s 2 π - 1 \Rightarrow 1 - 1 = 0 \\ ∴ f (0) = f (2 π) = 0 \\ A s t h e a b o v e c o n d i t i o n s a r e s a t i s f i e d, t h e n t h e r e l i e s a point c \in (0, 2 π) s u c h t h a t f^{'} (c) = 0 \\ ∴ - s i n c = 0 \Rightarrow s i n c = 0 \\ ∴ c = n π, n \in I \\ \Rightarrow c = π \in (0, 2 π) \\ H e n c e, c = π i s t h e p o i n t o n t h e c u r v e i n (0, 2 π) a t w h i c h t h e tangent i s p a r a l l e l t o x - a x i s . \end{array}$

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Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Five

Discuss the applicability of Rolleâ€™s theorem on the function given by $f (x) = {\begin{matrix} x^{2} + 1, & 0 \leq x \leq 1 \\ 3 - x, & 1 \leq x \leq 2 \end{matrix}$

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Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} (i) f (x) b e i n g a n algebraic p o l y n o m i a l i s c o n t i n u o u s e v e r y w h e r e . \\ (i i) f (x) m u s t b e d i f f e r e n t i a b l e a t x = 1 \\ L . H . L . = \underset{x \to 1^{-}}{l i m} \frac{f (x) - f (1)}{x - 1} \\ = \underset{x \to 1^{-}}{l i m} \frac{(x^{2} + 1) - (1 + 1)}{x - 1} \\ = \underset{x \to 1^{-}}{l i m} \frac{x^{2} + 1 - 2}{x - 1} = \underset{x \to 1^{-}}{l i m} \frac{x^{2} - 1}{x - 1} \\ = \underset{x \to 1^{-}}{l i m} \frac{(x - 1) (x + 1)}{x - 1} = \underset{x \to 1^{-}}{l i m} (x + 1) = (1 + 1) = 2 \\ R . H . L . = \underset{x \to 1^{+}}{l i m} \frac{f (x) - f (1)}{x - 1} \\ = \underset{x \to 1^{-}}{l i m} \frac{(3 - x) - (1 + 1)}{x - 1} \\ = \underset{x \to 1^{-}}{l i m} \frac{(3 - x) - 2}{x - 1} = \underset{x \to 1^{-}}{l i m} \frac{1 - x}{x - 1} = - 1 \\ ∴ L . H . L . \neq R . H . L . \\ S o, f (x) i s n o t d i f f e r e n t i a b l e a t x = 1 . \\ H e n c e, R o l l e' s T h e o r e m i s n o t a p p l i c a b l e i n [0, 2] . \end{array}$

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Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Five

Kindly consider the following

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Asked by Alok Kumar Singh

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Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

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Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Five

Kindly consider the following <picture><img src="https://images.shiksha.com/mediadata/images/articles/1744268003phpTZR0pj.jpeg" alt="" width="238" height="45"></picture>

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Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Five

Kindly consider the following

$f (x) = x (x + 3) e^{- x / 2}$ in $[- 3, 0]$ .

5 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = x (x + 3) e^{- x / 2} i n [- 3, 0] \\ (i) Algebraic f u n c t i o n s a n d exponential f u n c t i o n s a r e c o n t i n u o u s i n t h e i r d o m a i n s . \\ ∴ f (x) i s c o n t i n u o u s i n [- 3, 0] . \\ (i i) f^{'} (x) = x (x + 3) . \frac{d}{d x} e^{- x / 2} + x . e^{- x / 2} . \frac{d}{d x} (x + 3) + (x + 3) . e^{- x / 2} \frac{d}{d x} . x \\ = x (x + 3) . e^{- x / 2} . (- \frac{1}{2}) + x . e^{- x / 2} . 1 + (x + 3) . e^{- x / 2} . 1 \\ = e^{- x / 2} [\frac{- x (x + 3)}{2} + x + x + 3] \\ = e^{- x / 2} [\frac{- x (x + 3)}{2} + 2 x + 3] = e^{- x / 2} [\frac{- x^{2} - 3 x + 4 x + 6}{2}] \\ = e^{- x / 2} [\frac{- x^{2} + x + 6}{2}] w h i c h e x i s t s i n (- 3, 0) \\ S o f (x) i s d i f f e r e n t i a b l e i n (- 3, 0) . \\ (i i i) f (- 3) = (- 3) (- 3 + 3) e^{- 3 / 2} = 0 \\ f (0) = (0) (0 + 3) e^{- 0 / 2} = 0 \\ ∴ f (- 3) = f (0) = 0 \\ A s t h e a b o v e c o n d i t i o n s a r e s a t i s f i e d, t h e n t h e r e m u s t e x i s t a t l e a s t o n e point \\ c \in (- 3, 0) s u c h t h a t f^{'} (c) = 0 \\ ∴ e^{- c / 2} [\frac{- c^{2} + c + 6}{2}] = 0 \Rightarrow - \frac{e^{- c / 2}}{2} [(c - 3) (c + 2)] = 0 \Rightarrow - e^{- c / 2} \neq 0 \\ ∴ (c - 3) (c + 2) = 0 \Rightarrow c = 3, - 2 \\ c = 3, - 2 \in (- 3, 0) \\ H e n c e, R o l l e' s T h e o r e m i s v e r i f i e d . \end{array}$

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Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Five

Kindly consider the following

$f (x) = l o g (x^{2} + 2) - l o g 3$ in $[- 1, 1]$

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Asked by Alok Kumar Singh

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Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = l o g (x^{2} + 2) - l o g 3 i n [- 1, 1] \\ (i) f (x) = l o g (x^{2} + 2) - l o g 3, b e i n g a logarithm f u n c t i o n s, f (x) i s c o n t i n u o u s i n [- 1, 1] . \\ (i i) f^{'} (x) = \frac{1}{x^{2} + 2} . 2 x - 0 = \frac{2 x}{x^{2} + 2} w h i c h e x i s t s i n (- 1, 1) \\ S o f (x) i s d i f f e r e n t i a b l e i n (- 1, 1) . \\ (i i i) f (- 1) = l o g (1 + 2) - l o g 3 \Rightarrow l o g 3 - l o g 3 = 0 \\ f (1) = l o g (1 + 2) - l o g 3 \Rightarrow l o g 3 - l o g 3 = 0 \\ ∴ f (- 1) = f (1) = 0 \\ A s t h e a b o v e c o n d i t i o n s a r e s a t i s f i e d, t h e n t h e r e m u s t e x i s t a t l e a s t o n e point \\ c \in (- 1, 1) s u c h t h a t f^{'} (c) = 0 \\ ∴ \frac{2 c}{c^{2} + 2} = 0 \Rightarrow 2 c = 0 \Rightarrow c = 0 \in (- 1, 1) \\ H e n c e, R o l l e' s T h e o r e m i s v e r i f i e d . \end{array}$

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Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Five

Kindly consider the following

$f (x) = s i n^{4} x + c o s^{4} x$ in $[0, \frac{π}{2}]$

5 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = s i n^{4} x + c o s^{4} x i n [0, \frac{π}{2}] \\ (i) f (x) = s i n^{4} x + c o s^{4} x, b e i n g sine a n d cosine f u n c t i o n s, f (x) i s c o n t i n u o u s i n [0, \frac{π}{2}] . \\ (i i) f^{'} (x) = 4 s i n^{3} x . c o s x + 4 c o s^{3} x (- s i n x) \\ = 4 s i n^{3} x . c o s x - 4 s i n x c o s^{3} x \\ = 4 s i n x . c o s x (s i n^{2} x - c o s^{2} x) \\ = - 4 s i n x . c o s x (c o s^{2} x - s i n^{2} x) \\ = - 2.2 s i n x . c o s x c o s 2 x [\begin{array}{l} ? c o s 2 x = c o s^{2} x - s i n^{2} x \\ s i n 2 x = 2 s i n x c o s x \end{array}] \\ = - 2 s i n 2 x . c o s 2 x = - s i n 4 x w h i c h e x i s t s i n (0, \frac{π}{2}) \\ S o f (x) i s d i f f e r e n t i a b l e i n (0, \frac{π}{2}) . \\ (i i i) f (0) = s i n^{4} (0) + c o s^{4} (0) = 1 \\ f (\frac{π}{2}) = s i n^{4} (\frac{π}{2}) + c o s^{4} (\frac{π}{2}) = 1 \\ ∴ f (0) = f (\frac{π}{2}) = 1 \\ A s t h e a b o v e c o n d i t i o n s a r e s a t i s f i e d, t h e n t h e r e m u s t e x i s t a t l e a s t o n e point \\ c \in (0, \frac{π}{2}) s u c h t h a t f^{'} (c) = 0 \\ ∴ f^{'} (c) = 0 \Rightarrow - s i n 4 c = 0 \Rightarrow s i n 4 c = 0 \\ \Rightarrow s i n 4 c = s i n 0 \Rightarrow 4 c = n π \\ ∴ c = \frac{n π}{4}, n \in I \\ F o r n = 1, c = \frac{π}{4} \in (0, \frac{π}{2}) \\ H e n c e, R o l l e' s T h e o r e m i s v e r i f i e d . \end{array}$

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Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Five

Verify Rolle's theorem for each of the functions in Exercises 65 to 69.

$f (x) = x {(x - 1)}^{2}$ in $[0, 1]$

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Asked by Alok Kumar Singh

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Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = x {(x - 1)}^{2} i n [0, 1] \\ (i) f (x) = x {(x - 1)}^{2}, b e i n g a n algebraic p o l y n o m i a l, i s c o n t i n u o u s i n [0, 1] . \\ (i i) f^{'} (x) = x . 2 (x - 1) + {(x - 1)}^{2} . 1 \\ = 2 x^{2} - 2 x + x^{2} + 1 - 2 x \\ = 3 x^{2} - 4 x + 1 w h i c h e x i s t s i n (0, 1) \\ (i i i) f (x) = x {(x - 1)}^{2} \\ = 0 {(0 - 1)}^{2} = 0; f (1) = 1 {(1 - 1)}^{2} = 0 \\ \Rightarrow f (0) = f (1) = 0 \\ A s t h e a b o v e c o n d i t i o n s a r e s a t i s f i e d, t h e n t h e r e m u s t e x i s t a t l e a s t o n e point \\ c \in (0, 1) s u c h t h a t f^{'} (c) = 0 \\ ∴ f^{'} (c) = 3 c^{2} - 4 c + 1 = 0 \Rightarrow 3 c^{2} - 3 c - c + 1 = 0 \\ \Rightarrow 3 c (c - 1) - 1 (c - 1) = 0 \Rightarrow (c - 1) (3 c - 1) = 0 \\ c - 1 = 0 \Rightarrow c = 1 \\ 3 c - 1 = 0 \Rightarrow 3 c = 1 ∴ c = \frac{1}{3} \in (0, 1) \\ H e n c e, R o l l e' s T h e o r e m i s v e r i f i e d . \end{array}$

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Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Five

Kindly consider the following

If $y = t a n^{- 1} x$ , find $\frac{d^{2} y}{d x^{2}}$ in terms of $y$ alone.

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Asked by Alok Kumar Singh

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Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t y = t a n^{- 1} x \\ \Rightarrow x = t a n y \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . y \\ \Rightarrow \frac{d x}{d y} = s e c^{2} y \Rightarrow \frac{d y}{d x} = \frac{1}{s e c^{2} y} = co s^{2} y \\ A g a i n d i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x \\ \Rightarrow \frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d x} (co s^{2} y) \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = 2 c o s y (- s i n y) . \frac{d y}{d x} \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = - 2 s i n y c o s y . co s^{2} y \\ ∴ \frac{d^{2} y}{d x^{2}} = - 2 s i n y co s^{3} y \end{array}$

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Using Rolleâ€™s theorem, find the point on the curve $y = x (x - 4)$ , $x \in [0, 4]$ , where the tangent is parallel to the x-axis.

Find the points on the curve $y = c o s x - 1$ in $[0, 2 π]$ , where the tangent is parallel to the x-axis.

Discuss the applicability of Rolleâ€™s theorem on the function given by $f (x) = {\begin{matrix} x^{2} + 1, & 0 \leq x \leq 1 \\ 3 - x, & 1 \leq x \leq 2 \end{matrix}$

Kindly consider the following

<p><strong>Kindly consider the following</strong></p> <p><picture><img src="https://images.shiksha.com/mediadata/images/articles/1744268003phpTZR0pj.jpeg" alt="" width="238" height="45"></picture></p>

Kindly consider the following

$f (x) = x (x + 3) e^{- x / 2}$ in $[- 3, 0]$ .

Kindly consider the following

$f (x) = l o g (x^{2} + 2) - l o g 3$ in $[- 1, 1]$

Kindly consider the following

$f (x) = s i n^{4} x + c o s^{4} x$ in $[0, \frac{π}{2}]$

Verify Rolle's theorem for each of the functions in Exercises 65 to 69.

$f (x) = x {(x - 1)}^{2}$ in $[0, 1]$

Kindly consider the following

If $y = t a n^{- 1} x$ , find $\frac{d^{2} y}{d x^{2}}$ in terms of $y$ alone.