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Continuity and Differentiability 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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alok kumar singh

Contributor-Level 10

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

Sol:

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

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Continuity and Differentiability 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 $[0, \frac{π}{2}]$

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alok kumar singh

Contributor-Level 10

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

Sol:

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

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New answer posted

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Continuity and Differentiability 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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alok kumar singh

Contributor-Level 10

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

Sol:

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

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New answer posted

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Continuity and Differentiability 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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alok kumar singh

Contributor-Level 10

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

Sol:

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

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New answer posted

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

Kindly consider the following

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alok kumar singh

Contributor-Level 10

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

Sol:

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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

If $x s i n (a + y) + s i n a c o s (a + y) = 0$ , prove that

$\frac{d y}{d x} = \frac{2 s i n^{2} (a + y)}{s i n a}$

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alok kumar singh

Contributor-Level 10

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

Sol:

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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

If $y = {(c o s x)}^{{(c o s x)}^{(c o s x) \dots \infty}}$ , show that $\frac{d y}{d x} = \frac{y^{2} t a n x}{y l o g c o s x - 1}$

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t y = {(c o s x)}^{{(c o s x)}^{(c o s x) \dots \infty}} \\ \Rightarrow y = {(c o s x)}^{y} [? y = {(c o s x)}^{{(c o s x)}^{(c o s x) \dots \infty}}] \\ T a k i n g l o g o n b o t h t h e s i d e s, l o g y = y . l o g (c o s x) \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x \\ \frac{1}{y} . \frac{d y}{d x} = y . \frac{d}{d x} l o g (c o s x) + l o g (c o s x) . \frac{d y}{d x} \\ \Rightarrow \frac{1}{y} . \frac{d y}{d x} = y . \frac{1}{c o s x} . \frac{d}{d x} (c o s x) + l o g (c o s x) . \frac{d y}{d x} \\ \Rightarrow \frac{1}{y} . \frac{d y}{d x} = y . \frac{1}{c o s x} . (- s i n x) + l o g (c o s x) . \frac{d y}{d x} \\ \Rightarrow \frac{1}{y} . \frac{d y}{d x} - l o g (c o s x) . \frac{d y}{d x} = - y t a n x \\ \Rightarrow [\frac{1}{y} - l o g (c o s x)] \frac{d y}{d x} = - y t a n x \\ \Rightarrow \frac{d y}{d x} = \frac{- y t a n x}{\frac{1}{y} - l o g (c o s x)} = \frac{y^{2} t a n x}{y l o g c o s x - 1} \\ H e n c e, \frac{d y}{d x} = \frac{y^{2} t a n x}{y l o g c o s x - 1} \\ H e n c e, p r o v e d . \end{array}$

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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

If $x^{y} = e^{y - x}$ , prove that $\frac{d y}{d x} = \frac{{(1 + l o g y)}^{2}}{l o g y}$

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t y^{x} = e^{y - x} \\ T a k i n g l o g o n b o t h t h e s i d e s, \\ l o g y^{x} = l o g e^{y - x} \\ \Rightarrow x l o g y = (y - x) l o g e \Rightarrow x l o g y = y - x [? l o g e = 1] \\ \Rightarrow x l o g y + x = y \\ \Rightarrow x (l o g y + 1) = y \\ \Rightarrow x = \frac{y}{l o g y + 1} \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . y \\ \frac{d x}{d y} = \frac{d}{d y} (\frac{y}{l o g y + 1}) \\ \Rightarrow = \frac{(l o g y + 1) . 1 - y . \frac{d}{d y} (l o g y + 1)}{{(l o g y + 1)}^{2}} \\ \Rightarrow = \frac{(l o g y + 1) - y . \frac{1}{y}}{{(l o g y + 1)}^{2}} = \frac{l o g y + 1 - 1}{{(l o g y + 1)}^{2}} = \frac{l o g y}{{(l o g y + 1)}^{2}} \\ W e k n o w t h a t \\ H e n c e, \frac{d y}{d x} = \frac{1}{\frac{d x}{d y}} = \frac{1}{\frac{l o g y}{{(l o g y + 1)}^{2}}} = \frac{{(l o g y + 1)}^{2}}{l o g y} \\ H e n c e, p r o v e d . \end{array}$

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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

If $x = e^{\frac{x}{y}}$ , prove that $\frac{d y}{d x} = \frac{x - y}{x l o g x}$

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t x = e^{\frac{x}{y}} \\ T a k i n g l o g o n b o t h t h e s i d e s, \\ l o g x = l o g e^{\frac{x}{y}} \\ \Rightarrow l o g x = \frac{x}{y} l o g e \Rightarrow l o g x = \frac{x}{y} [? l o g e = 1] \dots (1) \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x \\ \frac{d}{d x} l o g x = \frac{d}{d x} (\frac{x}{y}) \\ \Rightarrow \frac{1}{x} = \frac{y . 1 - x . \frac{d y}{d x}}{y^{2}} \Rightarrow y^{2} = x (y - x . \frac{d y}{d x}) \\ \Rightarrow y^{2} = x y - x^{2} . \frac{d y}{d x} \Rightarrow - x^{2} . \frac{d y}{d x} = y^{2} - x y \\ \Rightarrow \frac{d y}{d x} = \frac{y^{2} - x y}{- x^{2}} = \frac{- y^{2} + x y}{x^{2}} = \frac{y (x - y)}{x^{2}} \\ \Rightarrow \frac{d y}{d x} = \frac{y}{x} . \frac{(x - y)}{x} = \frac{1}{l o g x} . (\frac{x - y}{x}) [? l o g x = \frac{x}{y} f r o m e q n (1)] \\ H e n c e, \frac{d y}{d x} = \frac{x - y}{x l o g x} . H e n c e, p r o v e d . \end{array}$

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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

If $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$ , then show that $\frac{d y}{d x} \cdot \frac{d x}{d y} = 1$ .

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0 \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x \\ \frac{d}{d x} (a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c) = \frac{d}{d x} (0) \\ \Rightarrow a . 2 x + 2 h (x . \frac{d y}{d x} + y . 1) + b . 2 y . \frac{d y}{d x} + 2 g . 1 + 2 f . \frac{d y}{d x} + 0 = 0 \\ \Rightarrow 2 a x + 2 h x . \frac{d y}{d x} + 2 h y + 2 b y . \frac{d y}{d x} + 2 g + 2 f . \frac{d y}{d x} = 0 \\ \Rightarrow 2 h x . \frac{d y}{d x} + 2 b y . \frac{d y}{d x} + 2 f . \frac{d y}{d x} = - 2 a x - 2 h y - 2 g \\ \Rightarrow (2 h x + 2 b y + 2 f) . \frac{d y}{d x} = - 2 (a x + h y + g) \\ \Rightarrow \frac{d y}{d x} = \frac{- 2 (a x + h y + g)}{2 (h x + b y + f)} = \frac{- (a x + h y + g)}{(h x + b y + f)} \\ \Rightarrow \frac{d y}{d x} = \frac{- (a x + h y + g)}{(h x + b y + f)} \\ N o w, d i f f e r e n t i a t i n g b o t h s i d e s w . r . t . y \\ \frac{d}{d y} (a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c) = \frac{d}{d y} (0) \\ \Rightarrow 2 a x . \frac{d x}{d y} + 2 h (y . \frac{d x}{d y} + x . 1) + 2 b y + 2 g . \frac{d x}{d y} + 2 f . 1 + 0 = 0 \\ \Rightarrow 2 a x . \frac{d x}{d y} + 2 h y . \frac{d x}{d y} + 2 h x + 2 b y + 2 g . \frac{d x}{d y} + 2 f = 0 \\ \Rightarrow 2 a x \frac{d x}{d y} + 2 h y . \frac{d x}{d y} + 2 g . \frac{d x}{d y} = - 2 h x - 2 b y - 2 f \\ \Rightarrow (2 a x + 2 h y + 2 g) . \frac{d x}{d y} = - 2 (h x + b y + f) \\ \Rightarrow \frac{d x}{d y} = \frac{- 2 (h x + b y + f)}{2 (a x + h y + g)} = \frac{- (h x + b y + f)}{(a x + h y + g)} \\ \Rightarrow \frac{d x}{d y} = \frac{- (h x + b y + f)}{(a x + h y + g)} \\ ∴ \frac{d y}{d x} . \frac{d x}{d y} = [\frac{- (a x + h y + g)}{(h x + b y + f)}] . [\frac{- (h x + b y + f)}{(a x + h y + g)}] = 1 \\ H e n c e, \frac{d y}{d x} . \frac{d x}{d y} = 1 . H e n c e, p r o v e d . \end{array}$

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