Exemplar Solution

Get insights from 1.3k questions on Exemplar Solution, answered by students, alumni, and experts. You may also ask and answer any question you like about Exemplar Solution

1.3k
Questions
0
Discussions
0
Active Users
0
Followers

Have a question about Exemplar Solution? Ask our experts

Ask Your Question

Keep it short & simple. Type complete word. Avoid abusive language.

New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

$(3 x + 5) (1 + t a n x)$

4 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:

$\begin{array}{l} L e t y = (3 x + 5) (1 + t a n x) \\ \frac{d y}{d x} = \frac{d}{d x} (3 x + 5) (1 + t a n x) \\ = (3 x + 5) \frac{d}{d x} (1 + t a n x) + (1 + t a n x) \frac{d}{d x} (3 x + 5) [\begin{array}{l} ? \frac{d}{d x} [f (x) . g (x)] \\ = f (x) . g' (x) + g (x) . f' (x) \end{array}] \\ = (3 x + 5) s e c^{2} x + (1 + t a n x) (3) \\ = 3 x s e c^{2} x + 5 s e c^{2} x + 3 + 3 t a n x \end{array}$

Upvote

New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following
${(x + \frac{1}{x})}^{3}$

4 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:

$\begin{array}{l} L e t y = {(x + \frac{1}{x})}^{3} \\ \frac{d y}{d x} = \frac{d}{d x} {(x + \frac{1}{x})}^{3} \\ = \frac{d}{d x} (x^{3} + \frac{1}{x^{3}} + 3 x + \frac{3}{x}) \\ = \frac{d}{d x} (x^{3} + x^{- 3} + 3 x + 3 . x^{- 1}) = 3 x^{2} - 3 x^{- 4} + 3 - 3 . x^{- 2} \\ = 3 x^{2} - \frac{3}{x^{4}} + 3 - \frac{3}{x^{2}} \end{array}$

Upvote

New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Differentiate each of the functions w.r.t. $x$ in Exercises 29 to 42.

$\frac{x^{4} + x^{3} + x^{2} + x + 1}{x}$

14 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:

$\begin{array}{l} L e t y = (\frac{x^{4} + x^{3} + x^{2} + x + 1}{x}) \\ ∴ \frac{d y}{d x} = \frac{d}{d x} (\frac{x^{4} + x^{3} + x^{2} + x + 1}{x}) \\ = \frac{d}{d x} (x^{3} + x^{2} + x + \frac{1}{x}) = 3 x^{2} + 2 x + 1 - \frac{1}{x^{2}} \end{array}$

Upvote

New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

If $l i m_{x \to 1} \frac{x^{4} - 1}{x - 1} = l i m_{x \to k} \frac{x^{3} - k^{3}}{x^{2} - k^{2}}$ , then find the value of $k$ .

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:

$\begin{array}{l} G i v e n t h a t \underset{x \to 1}{l i m} \frac{x^{4} - 1}{x - 1} = \underset{x \to k}{l i m} \frac{x^{3} - k^{3}}{x^{2} - k^{2}} \\ \Rightarrow 4 {(1)}^{4 - 1} = \underset{x \to k}{l i m} \frac{(x - k) (x^{2} + k^{2} + k x)}{(x - k) (x + k)} \\ \Rightarrow 4 = \underset{x \to k}{l i m} \frac{x^{2} + k^{2} + k x}{x + k} \Rightarrow 4 = \frac{k^{2} + k^{2} + k^{2}}{2 k} \\ \Rightarrow 4 = \frac{3 k^{2}}{2 k} \Rightarrow 4 = \frac{3}{2} k \Rightarrow k = \frac{8}{3} . \end{array}$

Upvote

New question posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

<p><strong>If <math><mrow><mi>l</mi><mi>i</mi><msub><mrow><mi>m</mi></mrow><mrow><mi>x</mi><mo>→</mo><mn>1</mn></mrow></msub><mfrac><mrow><msup><mrow><mi>x</mi></mrow><mrow><mn>4</mn></mrow></msup><mo>−</mo><mn>1</mn></mrow><mrow><mi>x</mi><mo>−</mo><mn>1</mn></mrow></mfrac><mo>=</mo><mi>l</mi><mi>i</mi><msub><mrow><mi>m</mi></mrow><mrow><mi>x</mi><mo>→</mo><mi>k</mi></mrow></msub><mfrac><mrow><msup><mrow><mi>x</mi></mrow><mrow><mn>3</mn></mrow></msup><mo>−</mo><msup><mrow><mi>k</mi></mrow><mrow><mn>3</mn></mrow></msup></mrow><mrow><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>−</mo><msup><mrow><mi>k</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac></mrow></math> , then find the value of <math><mrow><mi>k</mi></mrow></math> .</strong></p>

4 Views|Posted a year ago

Asked by Alok Kumar Singh

New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following
$l i m_{x \to 0} \frac{s i n x - 2 s i n 3 x + s i n 5 x}{x}$

4 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:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{s i n x - 2 s i n 3 x + s i n 5 x}{x} \\ = \underset{x \to 0}{l i m} \frac{s i n x}{x} - \frac{2 s i n 3 x}{x} + \frac{s i n 5 x}{x} = \underset{x \to 0}{l i m} \frac{s i n x}{x} - \underset{3 x \to 0}{l i m} 2 (\frac{s i n 3 x}{3 x}) \times 3 + \underset{5 x \to 0}{l i m} (\frac{s i n 5 x}{5 x}) \times 5 \\ = 1 - 6 + 5 = 0 . \end{array}$

Upvote

New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

$l i m_{x \to 0} \frac{\sqrt[]{2} - \sqrt[]{1 + c o s x}}{s i n^{2} x}$

4 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:

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

Upvote

New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

$l i m_{x \to \frac{π}{6}} \frac{c o t^{3} x - 3}{c o s e c x - 2}$

4 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:

$\begin{array}{l} G i v e n t h a t \underset{x \to \frac{π}{6}}{l i m} \frac{c o t^{2} x - 3}{c o s e c x - 2} \\ = \underset{x \to \frac{π}{6}}{l i m} \frac{c o s e c^{2} x - 1 - 3}{c o s e c x - 2} = \underset{x \to \frac{π}{6}}{l i m} \frac{c o s e c^{2} x - 4}{c o s e c x - 2} \\ = \underset{x \to \frac{π}{6}}{l i m} \frac{(c o s e c x - 2) (c o s e c x + 2)}{(c o s e c x - 2)} = \underset{x \to \frac{π}{6}}{l i m} (c o s e c x + 2) \\ T a k i n g, w e h a v e \\ = c o s e c \frac{π}{6} + 2 = 2 + 2 = 4 \end{array}$

Upvote

New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following
$l i m_{x \to a} \frac{s i n x - s i n a}{\sqrt[]{x} - \sqrt[]{a}}$

3 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:

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

Upvote

New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following
$\underset{x \to 0}{l i m} \frac{s i n 2 x + 3 x}{2 x + t a n 3 x}$

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:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{s i n 2 x + 3 x}{2 x + t a n 3 x} \\ = \underset{x \to 0}{l i m} \frac{(\frac{s i n 2 x + 3 x}{2 x} \times 2 x)}{(\frac{2 x + t a n 3 x}{3 x} \times 3 x)} = \underset{x \to 0}{l i m} \frac{(\frac{s i n 2 x}{2 x} + \frac{3 x}{2 x}) \times 2 x}{(\frac{2 x}{3 x} + \frac{t a n 3 x}{3 x}) \times 3 x} \\ = \frac{(\underset{2 x \to 0}{l i m} \frac{s i n 2 x}{2 x} + \frac{3}{2})}{[\frac{2}{3} + \underset{3 x \to 0}{l i m} \frac{t a n 3 x}{3 x}]} \times \frac{2}{3} [? \underset{x \to 0}{l i m} \frac{s i n x}{x} = 1] \\ = (\frac{1 + \frac{3}{2}}{\frac{2}{3} + 1}) \times \frac{2}{3} [? \underset{x \to 0}{l i m} \frac{t a n x}{x} = 1] \\ = \frac{5 / 2}{5 / 3} \times \frac{2}{3} = \frac{3}{2} \times \frac{2}{3} = 1 \end{array}$

Upvote

Exemplar Solution

Questions

Discussions

Active Users

Followers