Tags Continuity and Differentiability

Continuity and Differentiability

Get insights from 335 questions on Continuity and Differentiability, answered by students, alumni, and experts. You may also ask and answer any question you like about Continuity and Differentiability

Follow Ask Question

335

Questions

Discussions

Active Users

Followers

New answer posted

a year ago

Continuity and Differentiability 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 $y = f (f (x))$

0 Follower 6 Views

alok kumar singh

Contributor-Level 10

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

Sol:

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

0 0

New answer posted

a year ago

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

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

0 Follower 15 Views

alok kumar singh

Contributor-Level 10

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

Sol:

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

0 0

New answer posted

a year ago

Continuity and Differentiability 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 $k$ .

0 Follower 11 Views

alok kumar singh

Contributor-Level 10

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

Sol:

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

0 0

New answer posted

a year ago

Continuity and Differentiability 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 $x = 0$

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

Kindly go through the solution

Sol:

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

0 0

New question posted

a year ago

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

Kindly consider the following

0 Follower 10 Views

New answer posted

a year ago

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

Kindly consider the following

$f (x) = {\begin{matrix} \frac{2^{x + 2} - 16}{4^{x} - 16}, & i f x \neq 2 \\ k, & i f x = 2 \end{matrix}$ at $x = 2$

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} f (x) = \frac{2^{x + 2} - 16}{4^{x} - 16} = \frac{2^{2} . 2^{x} - 16}{{(2^{x})}^{2} - {(4)}^{2}} = \frac{4 (2^{x} - 4)}{(2^{x} - 4) (2^{x} + 4)} \\ = \frac{4}{(2^{x} + 4)} \\ \underset{x \to 2^{-}}{l i m} f (x) = \underset{h \to 0}{l i m} \frac{4}{(2^{2 - h} + 4)} = \frac{4}{2^{2} + 4} = \frac{4}{4 + 4} = \frac{4}{8} = \frac{1}{2} \\ \underset{x \to 2}{l i m} f (x) = k \\ A s t h e f u n c t i o n i s c o n t i n u o u s a t x = 2 \\ ∴ \underset{x \to 2^{-}}{l i m} f (x) = \underset{x \to 2}{l i m} f (x) \\ ∴ k = \frac{1}{2} \\ H e n c e, t h e v a l u e o f k i s \frac{1}{2} . \end{array}$

0 0

New answer posted

a year ago

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

Find the value of $k$ in each of the Exercises 11 to 14 so that the function $f$ is continuous at the indicated point:

$f (x) = {\begin{matrix} 3 x - 8, & i f x \leq 5 \\ 2 k, & i f x > 5 \end{matrix}$ at $x = 5$

0 Follower 6 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} \underset{x \to 5}{l i m} f (x) = 3 x - 8 = \underset{h \to 0}{l i m} 3 (5 - h) - 8 = 15 - 8 = 7 \\ \underset{x \to 5^{+}}{l i m} f (x) = 2 k \\ A s t h e f u n c t i o n i s c o n t i n u o u s a t x = 5 \\ ∴ \underset{x \to 1^{-}}{l i m} f (x) = \underset{x \to 5^{+}}{l i m} f (x) \\ ∴ 7 = 2 k \Rightarrow k = \frac{7}{2} \\ A s \underset{x \to 1^{-}}{l i m} f (x) = \underset{x \to 1^{+}}{l i m} f (x) = \underset{x \to 1}{l i m} f (x) \\ H e n c e, t h e v a l u e o f k i s \frac{7}{2} . \end{array}$

0 0

New answer posted

a year ago

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

Kindly consider the following
$f (x) = | x | + | x - 1 |$ at $x = 1$

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

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

Sol:

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

0 0

New answer posted

a year ago

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

Kindly consider the following

$f (x) = {\begin{matrix} \frac{x^{2}}{2}, i f & 0 \leq x \leq 1 \\ 2 x^{2} - 3 x + \frac{3}{2}, i f & 1 < x \leq 2 \end{matrix}$ at $x = 0$ , $x = 1$

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

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

Sol:

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

0 0

New answer posted

a year ago

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

Kindly consider the following

$f (x) = {\begin{matrix} \frac{e^{\frac{1}{x}}}{1 + e^{\frac{1}{x}}}, & i f x \neq 0 \\ 0, & i f x = 0 \end{matrix}$

0 Follower 5 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} \underset{x \to 0^{-}}{l i m} f (x) = \frac{e^{1 / x}}{1 + e^{1 / x}} \\ = \underset{h \to 0}{l i m} \frac{e^{\frac{1}{0 - h}}}{1 + e^{\frac{1}{0 - h}}} = \underset{h \to 0}{l i m} \frac{e^{- 1 / h}}{1 + e^{- 1 / h}} \\ = \underset{h \to 0}{l i m} \frac{1}{e^{1 / h} (1 - e^{- 1 / h})} = \underset{h \to 0}{l i m} \frac{1}{e^{1 / h} - 1} = \underset{h \to 0}{l i m} \frac{1}{e^{1 / 0} - 1} \\ = \underset{h \to 0}{l i m} \frac{1}{e^{\infty} - 1} = \underset{h \to 0}{l i m} \frac{1}{0 - 1} = - 1 [? e^{\infty} = 0] \\ \underset{x \to 0^{+}}{l i m} f (x) = \frac{e^{1 / x}}{1 + e^{1 / x}} = \underset{h \to 0}{l i m} \frac{e^{\frac{1}{0 + h}}}{1 + e^{\frac{1}{0 + h}}} = \underset{h \to 0}{l i m} \frac{e^{1 / h}}{1 + e^{1 / h}} \\ = \underset{h \to 0}{l i m} \frac{1}{e^{- 1 / h} (1 + e^{1 / h})} = \underset{h \to 0}{l i m} \frac{1}{e^{- 1 / h} + 1} = \underset{h \to 0}{l i m} \frac{1}{e^{- 1 / 0} + 1} \\ = \underset{h \to 0}{l i m} \frac{1}{e^{- \infty} + 1} = \underset{h \to 0}{l i m} \frac{1}{0 + 1} = 1 [? e^{- \infty} = 0] \\ \underset{x \to 0}{l i m} f (x) = 0 \\ A s \underset{x \to 0^{-}}{l i m} f (x) \neq \underset{x \to 0^{+}}{l i m} f (x) \neq \underset{x \to 0}{l i m} f (x) \\ H e n c e, f (x) i s d i s c o n t i n u o u s a t x = 0 . \end{array}$

0 0

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

67k Colleges
1.2k Exams
718k Reviews
1850k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Continuity and Differentiability

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience