Ncert Solutions Maths class 12th

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New question posted•a year ago

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

Kindly consider the following <picture><img src="https://images.shiksha.com/mediadata/images/articles/1744265928phpipMdMh.jpeg" alt="" width="271" height="78"></picture>

10 Views|Posted a year ago

Asked by Alok Kumar Singh

New answer posted•a year ago

Ncert Solutions Maths class 12th 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 $x = 2$

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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New answer posted•a year ago

Ncert Solutions Maths class 12th 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 $x = 5$

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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New answer posted•a year ago

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

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

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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New answer posted•a year ago

Ncert Solutions Maths class 12th 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 $x = 0$ , $x = 1$

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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New answer posted•a year ago

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

Kindly consider the following

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

7 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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New answer posted•a year ago

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

Kindly consider the following

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

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

Upvote

New answer posted•a year ago

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

Kindly consider the following

$f (x) = {\begin{matrix} | x | c o s \frac{1}{x}, & i f x \neq 0 \\ 0, & i f x = 0 \end{matrix}$

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

Upvote

New answer posted•a year ago

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

Kindly consider the following
$f (x) = {\begin{matrix} \frac{| x - 4 |}{2 (x - 4)}, & x \neq 4 \\ 0, & x = 4 \end{matrix}$ at $x = 2$ , $x = 4$

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} \underset{x \to 4^{-}}{l i m} f (x) = \frac{| x - 4 |}{2 (x - 4)} [\begin{array}{l} f o r x < 4, | x - 4 | = - (x - 4) \\ f o r x > 4, | x - 4 | = (x - 4) \end{array}] \\ = \underset{h \to 0}{l i m} \frac{- [4 - h - 4]}{2 [4 - h - 4]} = \underset{h \to 0}{l i m} \frac{h}{- 2 h} = - \frac{1}{2} \\ \underset{x \to 4^{+}}{l i m} f (x) = \frac{| x - 4 |}{2 (x - 4)} = \underset{h \to 0}{l i m} \frac{[4 + h - 4]}{2 [4 + h - 4]} = \underset{h \to 0}{l i m} \frac{h}{2 h} = \frac{1}{2} \\ \underset{x \to 4}{l i m} f (x) = 0 \\ ∴ \underset{x \to 4^{-}}{l i m} f (x) \neq \underset{x \to 4^{+}}{l i m} f (x) \neq \underset{x \to 4}{l i m} f (x) \\ H e n c e, f (x) i s d i s c o n t i n u o u s a t x = 4 . \end{array}$

Upvote

New answer posted•a year ago

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

Kindly consider the following
$f (x) = {\begin{matrix} \frac{2 x^{2} - 3 x + 2}{x - 2}, & x \neq 2 \\ 5, & x = 2 \end{matrix}$

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} f (x) = \frac{2 x^{2} - 3 x - 2}{x - 2} \\ = \frac{2 x^{2} - 4 x + x - 2}{x - 2} = \frac{2 x (x - 2) + 1 (x - 2)}{x - 2} \\ = \frac{(2 x + 1) (x - 2)}{x - 2} = 2 x + 1 \\ \underset{x \to 2^{-}}{l i m} f (x) = 2 x + 1 = \underset{h \to 0}{l i m} 2 (2 - h) + 1 = 4 + 1 = 5 \\ \underset{x \to 2^{+}}{l i m} f (x) = 2 x + 1 = \underset{h \to 0}{l i m} 2 (2 + h) + 1 = 4 + 1 = 5 \\ \underset{x \to 2}{l i m} f (x) = 5 \\ A s \underset{x \to 2^{-}}{l i m} f (x) = \underset{x \to 2^{+}}{l i m} f (x) = \underset{x \to 2}{l i m} f (x) = 5 \\ H e n c e, f (x) i s c o n t i n u o u s a t x = 2 . \end{array}$

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