Tags Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Get insights from 92 questions on Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen, answered by students, alumni, and experts. You may also ask and answer any question you like about Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Follow Ask Question

Questions

Discussions

Active Users

Followers

New answer posted

2 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen Exemplar Solution

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:

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

0 0

New answer posted

2 months ago

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

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:

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

0 0

New answer posted

2 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen Exemplar Solution

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:

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

0 0

New answer posted

2 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen Exemplar Solution

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:

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

0 0

New answer posted

2 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen Exemplar Solution

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:

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

0 0

New answer posted

2 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen Exemplar Solution

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:

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

0 0

New answer posted

2 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen Exemplar Solution

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:

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

0 0

New answer posted

2 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen Exemplar Solution

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:

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

0 0

New answer posted

2 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen Exemplar Solution

Kindly consider the following
$l i m_{x \to 0} \frac{∣ s i n x ∣}{x}$ is

(a) 1

(b) –1

(c) Does not exist

(d) None of these

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n, \underset{x \to 0}{l i m} \frac{| s i n x |}{x} \\ L H L = \underset{x \to 0^{-}}{l i m} \frac{- s i n x}{x} = - 1 [? \underset{x \to 0}{l i m} \frac{s i n x}{x} = 1] \\ R H L = \underset{x \to 0^{+}}{l i m} \frac{s i n x}{x} = 1 \\ L H L \neq R H L \\ S o, t h e limit d o e s n o t e x i s t . \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

0 0

New answer posted

2 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen Exemplar Solution

If $f (x) = {\begin{matrix} \frac{s i n [x]}{[x]}, & if [x] \neq 0, \\ 0, & [x] = 0 \end{matrix}$ , where [.] denotes the greatest integer function, then $l i m_{x \to 0} f (x)$ is equal to

(a) 1

(b) 0

(c) –1

(d) None of these

0 Follower 5 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n, f (x) = {\begin{matrix} \frac{s i n [x]}{[x]}, & i f [x] \neq 0, \\ 0, & [x] = 0 \end{matrix} \\ L H L = \underset{x \to 0^{-}}{l i m} \frac{s i n [x]}{[x]} = \underset{h \to 0}{l i m} \frac{s i n [0 - h]}{[0 - h]} = \underset{h \to 0}{l i m} \frac{- s i n [- h]}{[- h]} = - 1 \\ R H L = \underset{x \to 0^{+}}{l i m} \frac{s i n [x]}{[x]} = \underset{h \to 0}{l i m} \frac{s i n [0 + h]}{[0 + h]} = \underset{h \to 0}{l i m} \frac{s i n [h]}{[h]} = - 1 \\ L H L \neq R H L \\ S o, t h e limit d o e s n o t e x i s t . \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

0 0

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

On Shiksha, get access to

65k Colleges
1.2k Exams
688k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience