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

Kindly consider the following
$\underset{x \to 1}{l i m} \frac{(\sqrt[]{x} - 1) (2 x - 3)}{2 x^{2} + x - 3}$ is

(a) $\frac{1}{10}$
(b) $- \frac{1}{10}$
(c) 1
(d) None of these

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 1}{l i m} \frac{(\sqrt[]{x} - 1) (2 x - 3)}{2 x^{2} + x - 3} \\ = \underset{x \to 1}{l i m} \frac{(\sqrt[]{x} - 1) (2 x - 3)}{2 x^{2} + 3 x - 2 x - 3} = \underset{x \to 1}{l i m} \frac{(\sqrt[]{x} - 1) (2 x - 3)}{x (2 x + 3) - 1 (2 x + 3)} \\ = \underset{x \to 1}{l i m} \frac{(\sqrt[]{x} - 1) (2 x - 3)}{(2 x + 3) (x - 1)} = \underset{x \to 1}{l i m} \frac{(\sqrt[]{x} - 1) (\sqrt[]{x} + 1) (2 x - 3)}{(2 x + 3) (x - 1) (\sqrt[]{x} + 1)} \\ = \underset{x \to 1}{l i m} \frac{(x - 1) (2 x - 3)}{(x - 1) (\sqrt[]{x} + 1) (2 x + 3)} = \underset{x \to 1}{l i m} \frac{2 x - 3}{(\sqrt[]{x} + 1) (2 x + 3)} \\ T a k i n g limits w e h a v e \\ = \frac{2 (1) - 3}{(\sqrt[]{1} + 1) (2 * 1 + 3)} = \frac{- 1}{2 * 5} = \frac{- 1}{10} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

0 0

New answer posted

2 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen Exemplar Solution

Kindly consider the following

$\underset{x \to \frac{π}{4}}{l i m} \frac{s e c^{2} x - 2}{t a n x - 1}$ is

(a) 3
(b) 1
(c) 0
(d) √2

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

0 0

New answer posted

2 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen Exemplar Solution

Kindly consider the following

$\underset{x \to 0}{l i m} \frac{s i n x}{\sqrt[]{x + 1} - \sqrt[]{1 - x}}$ is

(a) 2
(b) 0
(c) 1
(d) -1

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{s i n x}{\sqrt[]{x + 1} - \sqrt[]{1 - x}} \\ = \underset{x \to 0}{l i m} \frac{s i n x}{\sqrt[]{x + 1} - \sqrt[]{1 - x}} * \frac{\sqrt[]{x + 1} + \sqrt[]{1 - x}}{\sqrt[]{x + 1} + \sqrt[]{1 - x}} \\ = \underset{x \to 0}{l i m} \frac{s i n x [\sqrt[]{x + 1} + \sqrt[]{1 - x}]}{x + 1 - 1 + x} = \underset{x \to 0}{l i m} \frac{s i n x [\sqrt[]{x + 1} + \sqrt[]{1 - x}]}{2 x} \\ = \frac{1}{2} . \underset{x \to 0}{l i m} \frac{s i n x}{x} [\sqrt[]{x + 1} + \sqrt[]{1 - x}] \\ T a k i n g limit, w e g e t \\ = \frac{1}{2} * 1 * [\sqrt[]{0 + 1} + \sqrt[]{1 - 0}] = \frac{1}{2} * 1 * 2 = 1 \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

0 0

New answer posted

2 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen Exemplar Solution

Kindly consider the following
$l i m_{x \to 0} \frac{c o s e c x - c o t x}{x}$ is

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

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

0 0

New answer posted

2 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen Exemplar Solution

Kindly consider the following

$l i m_{θ \to 0} \frac{1 - c o s 4 θ}{1 - c o s 6 θ}$ is

(a) $\frac{4}{9}$
(b) $\frac{1}{2}$
(c) $- \frac{1}{2}$
(d) -1

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{θ \to 0}{l i m} \frac{1 - c o s 4 θ}{1 - c o s 6 θ} \\ = \underset{θ \to 0}{l i m} \frac{2 s i n^{2} 2 θ}{2 s i n^{2} 3 θ} [? 1 - c o s θ = 2 s i n^{2} \frac{θ}{2}] \\ = \underset{θ \to 0}{l i m} \frac{s i n^{2} 2 θ}{s i n^{2} 3 θ} = \underset{θ \to 0}{l i m} {[\frac{s i n 2 θ}{s i n 3 θ}]}^{2} = \underset{\begin{array}{l} θ \to 0 \\ 2 θ \to 0 \\ 3 θ \to 0 \end{array}}{l i m} {[\frac{\frac{s i n 2 θ}{2 θ} * 2 θ}{\frac{s i n 3 θ}{3 θ} * 3 θ}]}^{2} \\ = {[\frac{2 θ}{3 θ}]}^{2} = {(\frac{2}{3})}^{2} = \frac{4}{9} \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

0 0

New answer posted

2 months ago

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

Kindly consider the following
$l i m_{x \to 1} \frac{x^{m} - 1}{x^{n} - 1}$ is

(a) 1
(b) $\frac{m}{n}$
(c) $\frac{m}{n} - 1$
(d) $\frac{m^{2}}{n^{2}}$

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

0 0

New answer posted

2 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen Exemplar Solution

$l i m_{x \to 0} \frac{x^{2} c o s x}{1 - c o s x}$ is

(a) 2

(b) $\frac{3}{2}$
(c) $- \frac{3}{2}$
(d) 1

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{x^{2} c o s x}{1 - c o s x} \\ = \underset{x \to 0}{l i m} \frac{x^{2} c o s x}{2 s i n^{2} \frac{x}{2}} = - 1 [? 1 - c o s x = 2 s i n^{2} \frac{x}{2}] \\ = \underset{x \to 0}{l i m} \frac{\frac{x^{2}}{4} * 4 c o s x}{2 s i n^{2} \frac{x}{2}} = \underset{\begin{array}{l} x \to 0 \\ \frac{x}{2} \to 0 \end{array}}{l i m} \frac{{(\frac{x}{2})}^{2} * 2 c o s x}{s i n^{2} \frac{x}{2}} \\ = \underset{\frac{x}{2} \to 0}{l i m} {(\frac{\frac{x}{2}}{s i n \frac{x}{2}})}^{2} * 2 c o s x = 2 c o s 0 = 2 * 1 = 2 [? \underset{x \to 0}{l i m} \frac{x}{s i n x} = 1] \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

0 0

New answer posted

2 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen Exemplar Solution

Choose the correct answer from the given four options in each of the Exercises 54 to 76:

$l i m_{x \to π} \frac{s i n x}{x - π}$ is

(a) 1

(b) 2

(c) -1

(d) 2

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to π}{l i m} \frac{s i n x}{x - π} \\ = \underset{x \to π}{l i m} \frac{s i n (π - x)}{- (π - x)} = - 1 [? \underset{x \to 0}{l i m} \frac{s i n x}{x} = 1 a n d π - x \to 0 \Rightarrow x \to π] \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

0 0

New answer posted

2 months ago

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

Let $f (x) = {\begin{matrix} x^{} + 2 & if x \leq - 1, \\ c x^{2} & if x > - 1 \end{matrix}$ , find ' $c$ ' if $l i m_{x \to - 1} f (x)$ exists

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n f (x) = {\begin{matrix} x + 2 & i f x \leq - 1, \\ c x^{2} & i f x > - 1 \end{matrix} \\ L H L f (x) = \underset{x \to - 1^{-}}{l i m} (x + 2) = \underset{h \to 0}{l i m} (- 1 - h + 2) \\ = \underset{h \to 0}{l i m} (1 - h) = 1 \\ R H L f (x) = \underset{x \to - 1^{+}}{l i m} c x^{2} = \underset{h \to 0}{l i m} c {(- 1 + h)}^{2} = c \\ Since t h e limits e x i t . \\ ∴ L H L = R H L \\ ∴ c = 1 \end{array}$

0 0

New answer posted

2 months ago

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

Let $f (x) = {\begin{matrix} \frac{k c o s x}{π - 2 x} & when x \neq \frac{π}{2}, \\ 3 & when x = \frac{π}{2} \end{matrix}$ , and if $l i m_{x \to \frac{π}{2}} f (x) = f (\frac{π}{2})$ , find the value of $k$

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n f (x) = {\begin{matrix} \frac{k c o s x}{π - 2 x} & w h e n x \neq \frac{π}{2}, \\ 3 & w h e n x = \frac{π}{2} \end{matrix} \\ L H L f (x) = \underset{x \to {\frac{π}{2}}^{-}}{l i m} \frac{k c o s x}{π - 2 x} = \underset{h \to 0}{l i m} \frac{k c o s (\frac{π}{2} - h)}{π - 2 (\frac{π}{2} - h)} \\ = \underset{h \to 0}{l i m} \frac{k s i n h}{π - π + 2 h} = \underset{h \to 0}{l i m} \frac{k s i n h}{2 h} = \frac{k}{2} . 1 = \frac{k}{2} [? \underset{x \to 0}{l i m} \frac{s i n x}{x} = 1] \\ R H L f (x) = \underset{x \to {\frac{π}{2}}^{+}}{l i m} \frac{k c o s x}{π - 2 x} = \underset{h \to 0}{l i m} \frac{k c o s (\frac{π}{2} + h)}{π - 2 (\frac{π}{2} + h)} \\ = \underset{h \to 0}{l i m} \frac{- k s i n h}{π - π - 2 h} = \underset{h \to 0}{l i m} \frac{- k s i n h}{- 2 h} = \frac{k}{2} [? \underset{x \to 0}{l i m} \frac{s i n x}{x} = 1] \\ W e a r e g i v e n t h a t \underset{x \to \frac{π}{2}}{l i m} f (x) = 3 \\ S o, \frac{k}{2} = 3 \Rightarrow k = 6 \end{array}$

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