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New answer posted

2 months ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

If $f (x) = \frac{x^{n} - a^{n}}{x - a}$ for some constant $a$ , then $f^{'} (a)$ is

(a) 1

(b) 0

(c) Does not exist

(d) $\frac{1}{2}$

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = \frac{x^{n} - a^{n}}{x - a} \\ f' (x) = \frac{(x - a) (n . x^{n - 1}) - (x^{n} - a^{n}) . 1}{{(x - a)}^{2}} \\ ∴ f' (a) = \frac{(a - a) (n . a^{n - 1}) - (a^{n} - a^{n}) . 1}{{(a - a)}^{2}} \\ S o, f' (a) = \frac{0}{0} = d o e s n o t e x i s t \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted

2 months ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

If $f (x) = 1 + x + \frac{x^{2}}{2} + \dots + \frac{x^{100}}{100}$ , then $f^{'} (1)$ is equal to

(a) $\frac{1}{100}$
(b) 100
(c) Does not exist
(d) 0

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = 1 + x + \frac{x^{2}}{2} + \dots + \frac{x^{100}}{100} \\ f' (x) = 1 + \frac{2 x}{2} + \dots + \frac{100 x^{99}}{100} \\ ∴ f' (1) = 1 + 1 + 1 + \dots + 1 (100 t i m e s) = 100 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted

2 months ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

If $y = \frac{s i n (x + 9)}{c o s x}$ , then $\frac{d y}{d x}$ at $x = 0$ is

(a) cos9

(b) sin9

(c) 0

(d) 1

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t y = \frac{s i n (x + 9)}{c o s x} \\ ∴ \frac{d y}{d x} = \frac{c o s x . c o s (x + 9) - s i n (x + 9) (- s i n x)}{c o s^{2} x} \\ = \frac{c o s x c o s (x + 9) + s i n x s i n (x + 9)}{c o s^{2} x} \\ = \frac{c o s (x + 9 - x)}{c o s^{2} x} = \frac{c o s 9}{c o s^{2} x} \\ ∴ {(\frac{d y}{d x})}_{a t x = 0} = \frac{c o s 9}{c o s^{2} 0} = \frac{c o s 9}{{(1)}^{2}} = c o s 9 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted

2 months ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

If $y = \frac{s i n x + c o s x}{s i n x - c o s x}$ , then $\frac{d y}{d x}$ at $x = 0$ is

(a) –2

(b) 0

(c) $\frac{1}{2}$

(d) Does not exist

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

2 months ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

If $y = \frac{1 + \frac{1}{x^{2}}}{1 - \frac{1}{x^{2}}}$ , then $\frac{d y}{d x}$ is

(a) $\frac{- 4 x}{(x^{2} - 1)^{2}}$
(b) $\frac{- 4 x}{x^{2} - 1}$
(c) $\frac{1 - x^{2}}{4 x}$
(d) $\frac{4}{x^{2} - 1}$

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t y = \frac{1 + \frac{1}{x^{2}}}{1 - \frac{1}{x^{2}}} \\ \Rightarrow y = \frac{x^{2} + 1}{x^{2} - 1} \\ ∴ \frac{d y}{d x} = \frac{(x^{2} - 1) . 2 x - (x^{2} + 1) . 2 x}{{(x^{2} - 1)}^{2}} \\ = \frac{2 x (x^{2} - 1 - x^{2} - 1)}{{(x^{2} - 1)}^{2}} = \frac{2 x (- 2)}{{(x^{2} - 1)}^{2}} = \frac{- 4 x}{{(x^{2} - 1)}^{2}} \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted

2 months ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

If $f (x) = \frac{x - 4}{2 \sqrt x}$ , then $f^{'} (1)$ is

(a) $\frac{5}{4}$

(b) $\frac{4}{5}$

(c) 1

(d) 0

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = \frac{x - 4}{2 \sqrt[]{x}} \\ ∴ f' (x) = \frac{1}{2} [\frac{\sqrt[]{x} . 1 - (x - 4) . \frac{1}{2 \sqrt[]{x}}}{x}] \\ = \frac{1}{2} [\frac{2 x - x + 4}{2 \sqrt[]{x} . x}] = \frac{1}{2} [\frac{x + 4}{2 {(x)}^{3 / 2}}] \\ ∴ f' (x) a t x = 1 = \frac{1}{2} [\frac{1 + 4}{2 * 1}] = \frac{5}{4} \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted

2 months ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

If $y = \sqrt[]{x} + \frac{1}{\sqrt x}$ , then $\frac{d y}{d x}$ at $x = 1$ is

(a) 1

(b) $\frac{1}{2}$

(c) $\frac{1}{\sqrt 2}$

(d) 0

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

2 months ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Let $f (x) = x - [x]$ ; $x \in ℝ$

, then $f^{'} (\frac{1}{2})$ i

(a) $\frac{3}{2}$

(b) 1

(c) 0

(d) –1

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n, f (x) = x - [x] \\ W e h a v e t o f i r s t c h e c k d i f f e r e n t i a b i l i t y o f f (x) a t x = \frac{1}{2} \\ ∴ L f' (\frac{1}{2}) = L H D = \underset{h \to 0}{l i m} \frac{f [\frac{1}{2} - h] - f [\frac{1}{2}]}{- h} \\ = \underset{h \to 0}{l i m} \frac{(\frac{1}{2} - h) - [\frac{1}{2} - h] - \frac{1}{2} + [\frac{1}{2}]}{- h} = \underset{h \to 0}{l i m} \frac{\frac{1}{2} - h - 0 - \frac{1}{2} + 0}{- h} = \frac{- h}{- h} = 1 \\ L f' (\frac{1}{2}) = R H D = \underset{h \to 0}{l i m} \frac{f (\frac{1}{2} + h) - f (\frac{1}{2})}{h} \\ = \underset{h \to 0}{l i m} \frac{(\frac{1}{2} + h) - [\frac{1}{2} + h] - \frac{1}{2} + [\frac{1}{2}]}{h} = \underset{h \to 0}{l i m} \frac{\frac{1}{2} + h - 1 - \frac{1}{2} + 1}{h} = \frac{h}{h} = 1 \\ L H D = R H D \\ ∴ f' (\frac{1}{2}) = 1 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted

2 months ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

$l i m_{x \to 0} \frac{t a n 2 x - x}{3 x - s i n x}$ is

(a) 2

(b) $\frac{1}{2}$

(c) $- \frac{1}{2}$

(d) $\frac{1}{4}$

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n, \underset{x \to 0}{l i m} \frac{t a n 2 x - x}{3 x - s i n x} \\ = \underset{x \to 0}{l i m} \frac{x [\frac{t a n 2 x}{x} - 1]}{x [3 - \frac{s i n x}{x}]} = \underset{\begin{array}{l} x \to 0 \\ ∴ 2 x \to 0 \end{array}}{l i m} \frac{\frac{t a n 2 x}{2 x} * 2 - 1}{3 - \frac{s i n x}{x}} \\ = \frac{1.2 - 1}{3 - 1} = \frac{2 - 1}{2} = \frac{1}{2} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted

2 months ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Let $f (x) = {\begin{matrix} x^{2} - 1, & 0 < x < 2, \\ 2 x + 3, & 2 \leq x < 3 \end{matrix}$ , the quadratic equation whose roots are $l i m_{x \to 2^{-}} f (x)$ and $l i m_{x \to 2^{+}} f (x)$ is

(a) $x^{2} - 6 x + 9 = 0$
(b) $x^{2} - 7 x + 8 = 0$
(c) $x^{2} - 14 x + 49 = 0$
(d) $x^{2} - 10 x + 21 = 0$

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n, f (x) = {\begin{matrix} x^{2} - 1, & 0 < x < 2 \\ 2 x + 3, & 2 \leq x < 3 \end{matrix} \\ \underset{x \to 2^{-}}{l i m} f (x) = \underset{x \to 2^{-}}{l i m} (x^{2} - 1) = \underset{h \to 0}{l i m} [{(2 - h)}^{2} - 1] = \underset{h \to 0}{l i m} (4 + h^{2} - 4 h - 1) \\ = \underset{h \to 0}{l i m} (h^{2} - 4 h + 3) = 3 \\ a n d \underset{x \to 2^{+}}{l i m} f (x) = \underset{x \to 2^{+}}{l i m} (2 x + 3) = \underset{h \to 0}{l i m} [2 (2 + h) + 3] = 7 \\ T h e r e f o r e, t h e q u a d r a t i c e q u a t i o n w h o s e r o o t s a r e 3 a n d 7 i s \\ x^{2} - (3 + 7) x + 3 * 7 = 0 i . e ., x^{2} - 10 x + 21 = 0 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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