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Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

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2 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen Exemplar Solution

Kindly consider the following
$l i m_{x \to \frac{π}{6}} \frac{\sqrt[]{3} s i n x - c o s x}{x - \frac{π}{6}}$

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to \frac{π}{6}}{l i m} \frac{\sqrt[]{3} s i n x - c o s x}{x - \frac{π}{6}} \\ = \underset{x \to \frac{π}{6}}{l i m} \frac{2 [\frac{\sqrt[]{3}}{2} s i n x - \frac{1}{2} c o s x]}{x - \frac{π}{6}} = \underset{x \to \frac{π}{6}}{l i m} \frac{2 [c o s \frac{π}{6} s i n x - s i n \frac{π}{6} c o s x]}{x - \frac{π}{6}} \\ = \underset{x \to \frac{π}{6}}{l i m} \frac{2 s i n (x - \frac{π}{6})}{x - \frac{π}{6}} [? s i n (A - B) = s i n A c o s B - c o s A s i n B] \\ = 2 \underset{x \to \frac{π}{6}}{l i m} \frac{s i n (x - \frac{π}{6})}{x - \frac{π}{6}} = 2.1 = 2 [? \underset{x \to 0}{l i m} \frac{s i n x}{x} = 1] \end{array}$

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2 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen Exemplar Solution

Kindly consider the following
$l i m_{x \to \frac{π}{4}} \frac{s i n x - c o s x}{x - \frac{π}{4}}$

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alok kumar singh

Contributor-Level 10

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

Sol:

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

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New answer posted

2 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen Exemplar Solution

Kindly consider the following
$\underset{x \to \frac{π}{3}}{l i m} \frac{\sqrt[]{1 - c o s 6 x}}{\sqrt[]{2} (\frac{π}{3} - x)}$

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to \frac{π}{3}}{l i m} \frac{\sqrt[]{1 - c o s 6 x}}{\sqrt[]{2} (\frac{π}{3} - x)} \\ = \underset{x \to \frac{π}{3}}{l i m} \frac{\sqrt[]{2 s i n^{2} 3 x}}{\sqrt[]{2} (\frac{π}{3} - x)} [? 1 - c o s θ = 2 s i n^{2} \frac{θ}{2}] \\ = \underset{x \to \frac{π}{3}}{l i m} \frac{\sqrt[]{2} s i n 3 x}{\sqrt[]{2} (\frac{π}{3} - x)} = \underset{π - 3 x \to 0}{l i m} \frac{3 . s i n (π - 3 x)}{π - 3 x} \\ = 3 [? \underset{x \to 0}{l i m} \frac{s i n x}{x} = 1] \end{array}$

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New answer posted

2 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen Exemplar Solution

Kindly consider the following

$l i m_{x \to 0} \frac{1 - c o s m x}{1 - c o s n x}$

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alok kumar singh

Contributor-Level 10

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

Sol:

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

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New answer posted

2 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen Exemplar Solution

Kindly consider the following

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

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{2 s i n x - s i n 2 x}{x^{3}} \\ = \underset{x \to 0}{l i m} \frac{2 s i n x - 2 s i n x c o s x}{x^{3}} = \underset{x \to 0}{l i m} \frac{2 s i n x (1 - c o s x)}{x^{3}} \\ = \underset{x \to 0}{l i m} \frac{2 s i n x}{x} (\frac{1 - c o s x}{x^{2}}) = \underset{x \to 0}{l i m} 2 (\frac{s i n x}{x}) (\frac{2 s i n^{2} x / 2}{x^{2}}) \\ = \underset{x \to 0}{l i m} 2 (\frac{s i n x}{x}) (2 \frac{s i n^{2} x / 2}{\frac{x^{2}}{4}} * \frac{1}{4}) = \underset{x \to 0}{l i m} 2 (\frac{s i n x}{x}) 2 {(\frac{s i n x / 2}{\frac{x}{2}})}^{2} * \frac{1}{4} \\ = \underset{x \to 0}{l i m} \frac{4}{4} (\frac{s i n x}{x}) \underset{\frac{x}{2} \to 0}{l i m} {(\frac{s i n x / 2}{\frac{x}{2}})}^{2} = 1.1 . {(1)}^{2} = 1 [? \underset{x \to 0}{l i m} \frac{s i n x}{x} = 1] \end{array}$

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New answer posted

2 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen Exemplar Solution

Kindly consider the following
$l i m_{x \to 0} \frac{1 - c o s 2 x}{x^{2}}$

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

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

Sol:

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

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New answer posted

2 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen Exemplar Solution

Kindly consider the following

$l i m_{x \to 0} \frac{s i n^{2} 2 x}{s i n^{2} 4 x}$

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{s i n^{2} 2 x}{s i n^{2} 4 x} \\ = \underset{x \to 0}{l i m} \frac{s i n^{2} 2 x}{s i n^{2} 2 (2 x)} \\ = \underset{x \to 0}{l i m} \frac{s i n^{2} 2 x}{4 s i n^{2} 2 x . c o s^{2} 2 x} [s i n 2 x = 2 s i n x c o s x] \\ = \underset{x \to 0}{l i m} \frac{1}{4 c o s^{2} 2 x} \\ T a k i n g limit, w e h a v e \\ = \frac{1}{4 . c o s^{2} 0} = \frac{1}{4} \end{array}$

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New answer posted

2 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen Exemplar Solution

Kindly consider the following

$l i m_{x \to a} \frac{s i n 3 x}{s i n 7 x}$

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{s i n 3 x}{s i n 7 x} \\ = \underset{x \to 0}{l i m} \frac{\frac{s i n 3 x}{3 x} * 3 x}{\frac{s i n 7 x}{7 x} * 7 x} = \frac{\underset{3 x \to 0}{l i m} (\frac{s i n 3 x}{3 x})}{\underset{3 x \to 0}{l i m} (\frac{s i n 7 x}{7 x})} * \frac{3}{7} \\ = \frac{1}{1} * \frac{3}{7} = \frac{3}{7} [? \underset{x \to 0}{l i m} \frac{s i n x}{x} = 1] \end{array}$

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New answer posted

2 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen Exemplar Solution

Find $n$ , if $l i m_{x \to 2} \frac{x^{n} - 2^{n}}{x - 2} = 80$ , $n \inN$
.

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 2}{l i m} \frac{x^{n} - 2^{n}}{x - 2} = 80 \\ n . {(2)}^{n - 1} = 80 [? \underset{x \to a}{l i m} \frac{x^{n} - a^{n}}{x - a} = n . a^{n - 1}] \\ \Rightarrow n * 2^{n - 1} = 5 * {(2)}^{5 - 1} \\ ∴ n = 5 \end{array}$

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New answer posted

2 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen Exemplar Solution

Kindly consider the following

$l i m_{x \to \frac{1}{2}} \frac{8 x - 3}{2 x - 1} - \frac{4 x^{2} + 1}{4 x^{2} - 1}$

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to \frac{1}{2}}{l i m} (\frac{8 x - 3}{2 x - 1} - \frac{4 x^{2} + 1}{4 x^{2} - 1}) \\ = \underset{x \to \frac{1}{2}}{l i m} [\frac{(8 x - 3) (2 x + 1) - (4 x^{2} + 1)}{(4 x^{2} - 1)}] \\ = \underset{x \to \frac{1}{2}}{l i m} [\frac{16 x^{2} - 6 x + 8 x - 3 - 4 x^{2} - 1}{4 x^{2} - 1}] \\ = \underset{x \to \frac{1}{2}}{l i m} [\frac{12 x^{2} + 2 x - 4}{4 x^{2} - 1}] = \underset{x \to \frac{1}{2}}{l i m} \frac{2 (6 x^{2} + x - 2)}{4 x^{2} - 1} \\ = \underset{x \to \frac{1}{2}}{l i m} \frac{2 (6 x^{2} + 4 x - 3 x - 2)}{(2 x + 1) (2 x - 1)} = \underset{x \to \frac{1}{2}}{l i m} \frac{2 [2 x (3 x + 2) - 1 (3 x + 2)]}{(2 x + 1) (2 x - 1)} \\ = \underset{x \to \frac{1}{2}}{l i m} \frac{2 (3 x + 2) (2 x - 1)}{(2 x + 1) (2 x - 1)} = \underset{x \to \frac{1}{2}}{l i m} \frac{2 (3 x + 2)}{(2 x + 1)} \\ T a k i n g limit, w e h a v e \\ = \frac{2 (3 * \frac{1}{2} + 2)}{2 * \frac{1}{2} + 1} = \frac{2 (\frac{7}{2})}{2} = \frac{7}{2} \end{array}$

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