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Physics NCERT Exemplar Solutions Class 12th Chapter Five

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4 months ago

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

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 $x = 0$ , $x = 1$

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alok kumar singh

Contributor-Level 10

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

Sol:

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

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4 months ago

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

Kindly consider the following

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

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alok kumar singh

Contributor-Level 10

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

Sol:

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

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4 months ago

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

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 $x = 0$ , $x = a$

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alok kumar singh

Contributor-Level 10

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

Sol:

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

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New answer posted

4 months ago

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

Kindly consider the following

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

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alok kumar singh

Contributor-Level 10

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

Sol:

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

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New answer posted

4 months ago

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

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

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alok kumar singh

Contributor-Level 10

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

Sol:

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

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New answer posted

4 months ago

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

Kindly consider the following

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

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alok kumar singh

Contributor-Level 10

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

Sol:

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

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New answer posted

4 months ago

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

$f (x) = {\begin{matrix} \frac{1 - c o s 2 x}{x^{2}}, & i f x \neq 0 \\ 5, & i f x = 0 \end{matrix}$ at $x = 2$ , $x = 0$

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} \underset{x \to 0^{-}}{l i m} f (x) = \frac{1 - c o s 2 x}{x^{2}} = \underset{h \to 0}{l i m} \frac{1 - c o s 2 (0 - h)}{{(0 - h)}^{2}} \\ = \underset{h \to 0}{l i m} \frac{1 - c o s (- 2 h)}{h^{2}} = \underset{h \to 0}{l i m} \frac{1 - c o s (2 h)}{h^{2}} \\ = \underset{h \to 0}{l i m} \frac{2 s i n^{2} h}{h^{2}} [? 1 - c o s θ = 2 s i n^{2} \frac{θ}{2}] \\ = \underset{h \to 0}{l i m} \frac{2 s i n h}{h} . \frac{s i n h}{h} = 2.1.1 = 2 [\underset{x \to 0}{l i m} \frac{s i n x}{x} = 1] \\ \underset{x \to 0^{+}}{l i m} f (x) = \frac{1 - c o s 2 x}{x^{2}} \\ = \underset{h \to 0}{l i m} \frac{1 - c o s 2 (0 + h)}{{(0 + h)}^{2}} = \underset{h \to 0}{l i m} \frac{1 - c o s (2 h)}{h^{2}} \\ = \underset{h \to 0}{l i m} \frac{2 s i n^{2} h}{h^{2}} = \underset{h \to 0}{l i m} \frac{2 s i n h}{h} . \frac{s i n h}{h} = 2.1.1 = 2 \\ \underset{x \to 0}{l i m} f (x) = 5 \\ A s \underset{x \to 0^{-}}{l i m} f (x) = \underset{x \to 0^{+}}{l i m} f (x) \neq \underset{x \to 0}{l i m} f (x) \\ H e n c e, f (x) i s d i s c o n t i n u o u s a t x = 0 . \end{array}$

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New answer posted

4 months ago

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

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

4 months ago

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

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:

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