Tags Maths

Maths

Get insights from 6.5k questions on Maths, answered by students, alumni, and experts. You may also ask and answer any question you like about Maths

Follow Ask Question

6.5k

Questions

Discussions

Active Users

Followers

New answer posted

10 months ago

Maths 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:

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

0 0

New answer posted

10 months ago

Maths 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 8 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

10 months ago

Maths 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:

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

0 0

New answer posted

10 months ago

Maths 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 13 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

10 months ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

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

10 months ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

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 8 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

New answer posted

10 months ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

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

10 months ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

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 3 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

0 0

New answer posted

10 months ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

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

10 months ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

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 3 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

0 0

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

On Shiksha, get access to

66k Colleges
1.2k Exams
691k Reviews
1850k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Maths

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience