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

Find which of the functions in Exercises 2 to 10 is continuous or discontinuous at the indicated points:

$f (x) = {\begin{matrix} 3 x + 5, & if x \geq 2 \\ x^{2}, & if x < 2 \end{matrix}$

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} \underset{x \to 2^{+}}{l i m} f (x) = 3 x + 5 \\ = \underset{h \to 0}{l i m} 3 (2 + h) + 5 = 11 \\ \underset{x \to 2}{l i m} f (x) = 3 x + 5 = 3 (2) + 5 = 11 \\ \underset{x \to 2^{-}}{l i m} f (x) = x^{2} = \underset{h \to 0}{l i m} {(2 - h)}^{2} \\ = \underset{h \to 0}{l i m} {(2)}^{2} + h^{2} - 4 h = {(2)}^{2} = 4 \\ Since \underset{x \to 2^{+}}{l i m} f (x) = \underset{x \to 2}{l i m} f (x) \neq \underset{x \to 2^{-}}{l i m} f (x) \\ H e n c e, f (x) i s d i s c o n t i n u o u s a t x = 2 . \end{array}$

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New answer posted

a year ago

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

Examine the continuity of the function
$f (x) = x^{3} + 2 x^{2} - 1$ at $x = 1$

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} W e k n o w t h a t y = f (x) w i l l b e c o n t i n u o u s a t x = a i f \\ \underset{x \to a^{-}}{l i m} f (x) = \underset{x \to a}{l i m} f (x) = \underset{x \to a^{+}}{l i m} f (x) \\ G i v e n f (x) = x^{3} + 2 x^{2} - 1 \\ \underset{x \to 1^{-}}{l i m} f (x) = \underset{h \to 0}{l i m} {(1 + h)}^{3} + 2 {(1 + h)}^{2} - 1 = 1 + 2 - 1 = 2 \\ \underset{x \to 1}{l i m} f (x) = {(1)}^{3} + 2 {(1)}^{2} - 1 = 1 + 2 - 1 = 2 \\ \underset{x \to 1^{+}}{l i m} f (x) = \underset{h \to 0}{l i m} {(1 + h)}^{3} + 2 {(1 + h)}^{2} - 1 = 1 + 2 - 1 = 2 \\ \underset{x \to 1^{-}}{l i m} f (x) = \underset{x \to 1}{l i m} f (x) = \underset{x \to 1^{+}}{l i m} f (x) = 2 . \\ H e n c e, f (x) i s c o n t i n u o u s a t x = 1 . \end{array}$

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New answer posted

a year ago

Class 12th 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 Long Answer Type Questions as classified in NCERT Exemplar

Sol:

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New answer posted

a year ago

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

If $x = s i n t$ and $y = s i n p t$ , prove that $(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} + p^{2} y = 0$

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t x = s i n t a n d y = s i n p t \\ 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 . t \\ \Rightarrow \frac{d x}{d t} = c o s t a n d \frac{d y}{d t} = c o s p t . p = p . c o s p t \\ \Rightarrow \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{p . c o s p t}{c o s t} \\ \Rightarrow \frac{d y}{d x} = \frac{p . c o s p t}{c o s t} \\ 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}) = p . \frac{d}{d x} (\frac{c o s p t}{c o s t}) \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = p . [\frac{c o s t . \frac{d}{d x} (c o s p t) - c o s p t . \frac{d}{d x} (c o s t)}{c o s^{2} t}] \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = p . [\frac{c o s t . (- s i n p t) . p \frac{d t}{d x} - c o s p t . (- s i n t) . \frac{d t}{d x}}{c o s^{2} t}] \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = p . [\frac{- p c o s t . s i n p t + c o s p t . s i n t}{c o s^{2} t}] \frac{d t}{d x} \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = p . [\frac{- p c o s t . s i n p t + c o s p t . s i n t}{c o s^{2} t}] \frac{1}{c o s t} \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = p . (\frac{- p c o s t . s i n p t + c o s p t . s i n t}{c o s^{3} t}) \\ N o w w e h a v e t o p r o v e t h a t \\ (1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} + p^{2} y = 0 \\ L . H . S . = (1 - x^{2}) [p . (\frac{- p c o s t . s i n p t + c o s p t . s i n t}{c o s^{3} t})] - x (\frac{p . c o s p t}{c o s t}) + p^{2} y \\ \Rightarrow = (1 - s i n^{2} t) [p . (\frac{- p c o s t . s i n p t + c o s p t . s i n t}{c o s^{3} t})] - \frac{p . s i n t c o s p t}{c o s t} + p^{2} . s i n p t \\ \Rightarrow = c o s^{2} t [\frac{- p^{2} c o s t . s i n p t + p c o s p t . s i n t}{c o s^{3} t}] - \frac{p . s i n t c o s p t}{c o s t} + p^{2} . s i n p t \\ \Rightarrow = \frac{- p^{2} c o s t . s i n p t + p c o s p t . s i n t}{c o s t} - \frac{p . s i n t c o s p t}{c o s t} + p^{2} . s i n p t \\ \Rightarrow = \frac{- p^{2} c o s t . s i n p t + p c o s p t . s i n t - p . s i n t c o s p t + p^{2} . s i n p t c o s t}{c o s t} \\ \Rightarrow = \frac{0}{c o s t} = 0 = R . H . S . \\ H e n c e, p r o v e d . \end{array}$

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New answer posted

a year ago

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

If $x^{m} y^{n} = {(x + y)}^{m + n}$ , prove that

(i) $\frac{d y}{d x} = \frac{y}{x}$

(ii) $\frac{d^{2} y}{d x^{2}} = 0$

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alok kumar singh

Contributor-Level 10

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

Sol:

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

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

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New answer posted

a year ago

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

Find the values of $p$ and $q$ so that $f (x) = {\begin{matrix} x^{2} + 3 x + p, & x \leq 1 \\ q x + 2, & x > 1 \end{matrix}$ is differentiable at $x = 1$ .

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = {\begin{matrix} x^{2} + 3 x + p, & x \leq 1 \\ q x + 2, & x > 1 \end{matrix} a t x = 1 . \\ L . H . L . f' (c) = \underset{x \to 1^{-}}{l i m} \frac{f (x) - f (c)}{x - c} \\ f' (1) = \underset{x \to 1^{-}}{l i m} \frac{f (x) - f (1)}{x - 1} \\ = \underset{x \to 1^{-}}{l i m} \frac{(x^{2} + 3 x + p) - (1 + 3 + p)}{x - 1} \\ = \underset{h \to 0}{l i m} \frac{[{(1 - h)}^{2} + 3 (1 - h) + p] - (1 + 3 + p)}{1 - h - 1} \\ = \underset{h \to 0}{l i m} \frac{[1 + h^{2} - 2 h + 3 - 3 h + p] - (4 + p)}{- h} \\ = \underset{h \to 0}{l i m} \frac{[h^{2} - 5 h + 4 + p] - [4 + p]}{- h} \\ = \underset{h \to 0}{l i m} \frac{h^{2} - 5 h + 4 + p - 4 - p}{- h} \\ = \underset{h \to 0}{l i m} \frac{h^{2} - 5 h}{- h} = \underset{h \to 0}{l i m} \frac{h [h - 5]}{- h} = 5 \\ R . H . L . f' (1) = \underset{x \to 1^{+}}{l i m} \frac{f (x) - f (1)}{x - 1} \\ = \underset{x \to 1^{+}}{l i m} \frac{(q x + 2) - (1 + 3 + p)}{x - 1} \\ = \underset{h \to 0}{l i m} \frac{[q (1 + h) + 2] - [4 + p]}{1 + h - 1} \\ = \underset{h \to 0}{l i m} \frac{q + q h + 2 - 4 - p}{h} = \underset{h \to 0}{l i m} \frac{q + q h - 2 - p}{h} \\ F o r e x i s t i n g t h e limit \\ q - 2 - p = 0 \Rightarrow q - p = 0 \dots (1) \\ \underset{h \to 0}{\Rightarrow l i m} \frac{q h - 0}{h} = q \\ I f L . H . L . f' (1) = R . H . L . f' (1) t h e n q = 5 . \\ N o w p \end{array}$

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New answer posted

a year ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Five

Show that the function $f (x) = | s i n x + c o s x |$ is continuous at $x = π$ .

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = | s i n x + c o s x | a t x = π \\ P u t g (x) = s i n x + c o s x a n d h (x) = | x | \\ ∴ h [g (x)] = h (s i n x + c o s x) = | s i n x + c o s x | \\ N o w, g (x) = s i n x + c o s x i s a c o n t i n u o u s f u n c t i o n since s i n x a n d c o s x a r e t w o c o n t i n u o u s \\ f u n c t i o n s a t x = π . \\ W e k n o w t h a t e v e r y modulus f u n c t i o n i s a c o n t i n u o u s f u n c t i o n e v e r y w h e r e . \\ H e n c e, f (x) = | s i n x + c o s x | i s a c o n t i n u o u s f u n c t i o n a t x = π . \end{array}$

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New answer posted

a year ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Five

Find all points of discontinuity of the function $f (t) = \frac{1}{t^{2} + t - 2}$ where $t = \frac{1}{x - 1}$

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} W e h a v e f (x) = \frac{1}{t^{2} + t - 2} \\ \Rightarrow f (t) = \frac{1}{{(\frac{1}{x - 1})}^{2} + \frac{1}{x - 1} - 2} [P u t t i n g t = \frac{1}{x - 1}] \\ = \frac{1}{\frac{1 + x - 1 - 2 {(x - 1)}^{2}}{{(x - 1)}^{2}}} = \frac{{(x - 1)}^{2}}{x - 2 x^{2} - 2 + 4 x} \\ = \frac{{(x - 1)}^{2}}{- 2 x^{2} + 5 x - 2} = \frac{{(x - 1)}^{2}}{- (2 x^{2} - 5 x + 2)} \\ = \frac{{(x - 1)}^{2}}{- [2 x^{2} - 4 x - x + 2]} = \frac{{(x - 1)}^{2}}{- [2 x (x - 2) - 1 (x - 2)]} \\ = \frac{{(x - 1)}^{2}}{- (x - 2) (2 x - 1)} = \frac{{(x - 1)}^{2}}{(2 - x) (2 x - 1)} \\ S o, i f f (t) i s d i s c o n t i n u o u s, t h e n 2 - x = 0 ∴ x = 2 \\ a n d 2 x - 1 = 0 ∴ x = \frac{1}{2} \\ H e n c e, t h e r e q u i r e d o f d i s c o n t i n u i t y a r e 2 a n d \frac{1}{2} . \end{array}$

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New answer posted

a year ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Five

Given the function $f (x) = \frac{1}{x + 2}$ . Find the points of discontinuity of the composite function $y = f (f (x))$

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} f (x) = \frac{1}{x + 2} \\ f [f (x)] = \frac{1}{f (x) + 2} = \frac{1}{\frac{1}{x + 2} + 2} = \frac{1}{\frac{1 + 2 x + 4}{x + 2}} = \frac{x + 2}{2 x + 5} \\ ∴ f [f (x)] = \frac{x + 2}{2 x + 5} \\ T h i s f u n c t i o n w i l l n o t b e d e f i n e d a n d c o n t i n u o u s w h e r e 2 x + 5 = 0 \\ \Rightarrow x = \frac{- 5}{2} . \\ H e n c e, x = \frac{- 5}{2} i s t h e point o f d i s c o n t i n u i t y . \end{array}$

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New answer posted

a year ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Five

Find the values of $a$ and $b$ such that the function $f$ defined by

0 Follower 14 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} \underset{x \to 4^{-}}{l i m} f (x) = \frac{x - 4}{| x - 4 |} + a = \underset{h \to 0}{l i m} \frac{4 - h - 4}{| 4 - h - 4 |} + a \\ = \underset{h \to 0}{l i m} \frac{- h}{h} + a = - 1 + a \\ \underset{x \to 4}{l i m} f (x) = a + b \\ \underset{x \to 4^{+}}{l i m} f (x) = \frac{x - 4}{| x - 4 |} + b = \underset{h \to 0}{l i m} \frac{4 + h - 4}{| 4 + h - 4 |} + b \\ = \underset{h \to 0}{l i m} \frac{h}{h} + b = 1 + b \\ A s t h e f u n c t i o n i s c o n t i n u o u s a t x = 4 . \\ ∴ \underset{x \to 4^{-}}{l i m} f (x) = \underset{x \to 4}{l i m} f (x) = \underset{x \to 4^{+}}{l i m} f (x) \\ - 1 + a = a + b = 1 + b \\ ∴ - 1 + a = a + b \Rightarrow b = - 1 \\ 1 + b = a + b \Rightarrow a = 1 \\ H e n c e, t h e v a l u e o f a = 1 a n d b = - 1 . \end{array}$

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