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New answer posted•a year ago

Exemplar Solution 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

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted•a year ago

Exemplar Solution 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

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted•a year ago

Exemplar Solution 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

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted•a year ago

Exemplar Solution 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

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

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

8 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

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) $\frac{m^{2}}{n^{2}}$

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

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

(a) 2

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

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

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

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

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

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

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 $k$

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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