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

Show that $f (x) = ∣ x - 5 ∣$ is continuous but not differentiable at $x = 5$ .

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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:

$\begin{array}{l} W e h a v e f (x) = | x - 5 | \\ \Rightarrow f (x) = {\begin{cases} - (x - 5) i f x - 5 < 0 o r x < 5 \\ x - 5 i f x - 5 > 0 o r x > 5 \end{cases} \\ F o r c o n t i n u i t y a t x = 5 \\ L . H . L . \underset{h \to 5^{-}}{l i m} f (x) = - (x - 5) \\ = \underset{h \to 0}{l i m} - (5 - h - 5) = \underset{h \to 0}{l i m} h = 0 \\ R . H . L . \underset{h \to 5^{+}}{l i m} f (x) = x - 5 \\ = \underset{h \to 0}{l i m} (5 + h - 5) = \underset{h \to 0}{l i m} h = 0 \\ L . H . L . = R . H . L . \\ S o, f (x) i s c o n t i n u o u s a t x = 5 . \\ N o w, f o r d i f f e r e n t i a b i l i t y \\ L f' (c) = \underset{h \to 0}{l i m} \frac{f (5 - h) - f (5)}{- h} = \underset{h \to 0}{l i m} \frac{- (5 - h - 5) - (5 - 5)}{- h} \\ = \underset{h \to 0}{l i m} \frac{h}{- h} = \frac{h}{- h} = 1 \\ R f' (5) = \underset{h \to 0}{l i m} \frac{f (5 + h) - f (5)}{h} = \underset{h \to 0}{l i m} \frac{(5 + h - 5) - (5 - 5)}{h} \\ = \underset{h \to 0}{l i m} \frac{h - 0}{h} = \frac{h}{h} = 1 \\ S o, L f' (5) \neq R f' (5) \\ H e n c e, f (x) i s n o t d i f f e r e n t i a b l e a t x = 5 . \end{array}$

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

Examine the differentiability of $f$ , where $f$ is defined by

$f (x) = {\begin{matrix} 1 + x, & i f x \leq 2 \\ 5 - x, & i f x > 2 \end{matrix}$ at $x = 2$

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:

$\begin{array}{l} f (x) i s d i f f e r e n t i a b l e a t x = 2 i f \\ L f' (2) = R f' (2) \\ ∴ L f' (c) = \underset{h \to 0}{l i m} \frac{f (2 - h) - f (2)}{- h} = \underset{h \to 0}{l i m} \frac{(1 + 2 - h) - (1 + 2)}{- h} \\ = \underset{h \to 0}{l i m} \frac{3 - h - 3}{- h} = \frac{- h}{- h} = 1 \\ R f' (c) = \underset{h \to 0}{l i m} \frac{f (2 + h) - f (2)}{h} = \underset{h \to 0}{l i m} \frac{[5 - (2 + h)] - (1 + 2)}{h} \\ = \underset{h \to 0}{l i m} \frac{3 - h - 3}{h} = \frac{- h}{h} = - 1 \\ S o, L f' (c) \neq R f' (c) \\ H e n c e, f (x) i s n o t d i f f e r e n t i a b l e a t x = 2 . \end{array}$

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

Examine the differentiability of $f$ , where $f$ is defined by

$f (x) = {\begin{matrix} x^{2} s i n x \frac{1}{x}, & i f x \neq 0 \\ 0, & i f x = 0 \end{matrix}$ at $x = 0$

7 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:

$\begin{array}{l} G i v e n t h a t f (x) = {\begin{cases} x^{2} s i n \frac{1}{x}, i f x \neq 0 \\ 0, i f x = 0 \end{cases} a t x = 0 . \\ F o r d i f f e r e n t i a b i l i t y w e k n o w t h a t \\ L f' (c) = R f' (c) \\ L f' (c) = \underset{h \to 0}{l i m} \frac{f (0 - h) - f (0)}{- h} = \underset{h \to 0}{l i m} \frac{{(0 - h)}^{2} s i n \frac{1}{(0 - h)} - 0}{- h} = \frac{h^{2} s i n (\frac{1}{- h})}{- h} \\ = \underset{h \to 0}{l i m} h . s i n (\frac{1}{h}) = 0 \times [- 1 \leq s i n (\frac{1}{h}) \leq 1] = 0 \\ R f' (c) = \underset{h \to 0}{l i m} \frac{f (0 + h) - f (0)}{h} = \underset{h \to 0}{l i m} \frac{{(0 + h)}^{2} s i n \frac{1}{(0 + h)} - 0}{h} = \underset{h \to 0}{l i m} \frac{h^{2} s i n (\frac{1}{h})}{h} \\ = \underset{h \to 0}{l i m} h . s i n (\frac{1}{h}) = 0 \times [- 1 \leq s i n (\frac{1}{h}) \leq 1] = 0 \\ S o, L f' (c) = R f' (c) = 0 \\ H e n c e, f (x) i s d i f f e r e n t i a b l e a t x = 0 . \end{array}$

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

Examine the differentiability of $f$ , where $f$ is defined by

$f (x) = {\begin{matrix} x [x], & 0 \leq x < 2 \\ (x - 1) x, & 2 \leq x < 3 \end{matrix}$ at $x = 2$ .

19 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:

$\begin{array}{l} W e k n o w t h a t a f u n c t i o n f i s d i f f e r e n t i a b l e a t a point' a' i n i t s d o m a i n i f \\ L f' (c) = R f' (c) \\ w h e r e L f' (c) = \underset{h \to 0}{l i m} \frac{f (a - h) - f (a)}{- h} a n d R f' (c) = \underset{h \to 0}{l i m} \frac{f (a + h) - f (a)}{h} \\ H e r e, f (x) = {\begin{cases} x [x], i f 0 \leq x \leq 2 \\ (x - 1) x, i f 2 \leq x \leq 2 \end{cases} a t x = 2 . \\ L f' (c) = \underset{h \to 0}{l i m} \frac{f (2 - h) - f (2)}{- h} = \underset{h \to 0}{l i m} \frac{(2 - h) [2 - h] - (2 - 1) 2}{- h} \\ = \underset{h \to 0}{l i m} \frac{(2 - h) . 1 - 2}{- h} [? [2 - h] = 1] \\ = \underset{h \to 0}{l i m} \frac{2 - h - 2}{- h} = 1 \\ R f' (c) = \underset{h \to 0}{l i m} \frac{f (2 + h) - f (2)}{h} = \underset{h \to 0}{l i m} \frac{(2 + h - 1) (2 + h) - (2 - 1) . 2}{h} \\ = \underset{h \to 0}{l i m} \frac{(1 + h) (2 + h) - 2}{h} = \underset{h \to 0}{l i m} \frac{2 + h + 2 h + h^{2} - 2}{h} \\ = \underset{h \to 0}{l i m} \frac{3 h + h^{2}}{h} \underset{h \to 0}{= l i m} \frac{h (3 + h)}{h} = 3 \\ L f' (c) \neq R f' (c) \\ H e n c e, f (x) i s n o t d i f f e r e n t i a b l e a t x = 2 . \end{array}$

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

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

4 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:

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

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

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

8 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:

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

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Ncert Solutions Maths class 12th Maths 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 $y = f (f (x))$

6 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:

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

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

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

15 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:

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

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

Prove that the function $f$ defined by $f (x) = {\begin{matrix} \frac{x}{| x | + 2 x^{2}}, & x \neq 0 \\ k, & x = 0 \end{matrix}$ remains discontinuous at $x = 0$ , regardless of the choice of $k$ .

11 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:

$\begin{array}{l} \underset{x \to 0^{-}}{l i m} f (x) = \frac{x}{| x | + 2 x^{2}} = \underset{h \to 0}{l i m} \frac{0 - h}{| 0 - h | + 2 {(0 - h)}^{2}} \\ = \underset{h \to 0}{l i m} \frac{- h}{h + 2 h^{2}} = \underset{h \to 0}{l i m} \frac{- h}{h (1 + 2 h)} \\ = \underset{h \to 0}{l i m} \frac{- 1}{1 + 2 h} = \frac{- 1}{1 + 2 (0)} = - 1 \\ \underset{x \to 0^{+}}{l i m} f (x) = \frac{x}{| x | + 2 x^{2}} = \underset{h \to 0}{l i m} \frac{0 + h}{| 0 + h | + 2 {(0 + h)}^{2}} \\ = \underset{h \to 0}{l i m} \frac{h}{h + 2 h^{2}} = \underset{h \to 0}{l i m} \frac{h}{h (1 + 2 h)} = \frac{1}{1 + 0} = 1 \\ A s \underset{x \to 0^{-}}{l i m} f (x) \neq \underset{x \to 0^{+}}{l i m} f (x) \\ H e n c e, f (x) i s d i s c o n t i n u o u s a t x = 0 r e g a r d l e s s t h e c h o i c e o f k . \end{array}$

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

Kindly consider the following

$f (x) = {\begin{matrix} \frac{1 - c o s k x}{x s i n x}, & i f x \neq 0 \\ \frac{1}{2}, & i f x = 0 \end{matrix}$ at $x = 0$

4 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Kindly go through the solution

Sol:

$\begin{array}{l} \underset{x \to 0^{-}}{l i m} f (x) = \frac{1 - c o s k x}{x s i n x} = \underset{h \to 0}{l i m} \frac{1 - c o s k (0 - h)}{(0 - h) s i n (0 - h)} \\ = \underset{h \to 0}{l i m} \frac{1 - c o s (- k h)}{(- h) s i n (- h)} = \underset{h \to 0}{l i m} \frac{1 - c o s k h}{h s i n h} [\begin{array}{l} ? s i n (- θ) = - s i n θ \\ c o s (- θ) = c o s θ \end{array}] \\ = \underset{h \to 0}{l i m} \frac{2 s i n^{2} \frac{k h}{2}}{h s i n h} = \underset{\begin{array}{l} h \to 0 \\ k h \to 0 \end{array}}{l i m} \frac{2 s i n \frac{k h}{2}}{\frac{k h}{2}} \times \frac{k h}{2} \times \frac{s i n \frac{k h}{2}}{\frac{k h}{2}} \times \frac{k h}{2} . \frac{1}{h . \frac{s i n h}{h} . h} \\ = 2.1 . \frac{k h}{2} . 1 . \frac{k h}{2} . \frac{1}{h^{2}} . 1 [\begin{array}{l} \underset{h \to 0}{l i m} \frac{s i n h}{h} = 1 a n d \\ \underset{k h \to 0}{l i m} \frac{s i n k h}{k h} = 1 \end{array}] \\ = \frac{k^{2}}{2} \\ \underset{x \to 0}{l i m} f (x) = \frac{1}{2} \\ ∴ \underset{x \to 0^{-}}{l i m} f (x) = \underset{x \to 0}{l i m} f (x) \\ ∴ \frac{k^{2}}{2} = \frac{1}{2} \Rightarrow k^{2} = 1 \Rightarrow k = \pm 1 \\ H e n c e, t h e v a l u e o f k i s \pm 1 . \end{array}$

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Ncert Solutions Maths class 12th

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Show that $f (x) = ∣ x - 5 ∣$ is continuous but not differentiable at $x = 5$ .

Examine the differentiability of $f$ , where $f$ is defined by

$f (x) = {\begin{matrix} 1 + x, & i f x \leq 2 \\ 5 - x, & i f x > 2 \end{matrix}$ at $x = 2$

Examine the differentiability of $f$ , where $f$ is defined by

$f (x) = {\begin{matrix} x^{2} s i n x \frac{1}{x}, & i f x \neq 0 \\ 0, & i f x = 0 \end{matrix}$ at $x = 0$

Examine the differentiability of $f$ , where $f$ is defined by

$f (x) = {\begin{matrix} x [x], & 0 \leq x < 2 \\ (x - 1) x, & 2 \leq x < 3 \end{matrix}$ at $x = 2$ .

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

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

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

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

Prove that the function $f$ defined by $f (x) = {\begin{matrix} \frac{x}{| x | + 2 x^{2}}, & x \neq 0 \\ k, & x = 0 \end{matrix}$ remains discontinuous at $x = 0$ , regardless of the choice of $k$ .

Kindly consider the following

$f (x) = {\begin{matrix} \frac{1 - c o s k x}{x s i n x}, & i f x \neq 0 \\ \frac{1}{2}, & i f x = 0 \end{matrix}$ at $x = 0$