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Continuity and Differentiability Maths NCERT Exemplar Solutions Class 12th Chapter Five

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alok kumar singh

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Continuity and Differentiability Maths NCERT Exemplar Solutions Class 12th Chapter Five

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$f (x) = s i n x - s i n 2 x$ in $[0, π]$ .

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alok kumar singh

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Continuity and Differentiability Maths NCERT Exemplar Solutions Class 12th Chapter Five

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$f (x) = x^{3} - 2 x^{2} - x + 3$ in $[0, 1]$ .

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alok kumar singh

Contributor-Level 10

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

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

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Continuity and Differentiability Maths NCERT Exemplar Solutions Class 12th Chapter Five

Verify the Mean Value Theorem for each of the functions given in Exercises 73 to 76.

$f (x) = \frac{1}{4 x - 1}$ in $[1, 4]$ .

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alok kumar singh

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Continuity and Differentiability 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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alok kumar singh

Contributor-Level 10

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

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

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Continuity and Differentiability 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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alok kumar singh

Contributor-Level 10

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

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

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New answer posted

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Continuity and Differentiability 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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alok kumar singh

Contributor-Level 10

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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New answer posted

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Continuity and Differentiability Maths NCERT Exemplar Solutions Class 12th Chapter Five

Kindly consider the following

0 Follower 5 Views

alok kumar singh

Contributor-Level 10

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

Sol:

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New question posted

a year ago

Continuity and Differentiability Maths NCERT Exemplar Solutions Class 12th Chapter Five

Kindly consider the following

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New answer posted

a year ago

Continuity and Differentiability Maths NCERT Exemplar Solutions Class 12th Chapter Five

Kindly consider the following

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

0 Follower 5 Views

alok kumar singh

Contributor-Level 10

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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