Limits and Derivatives

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New answer posted•a year ago

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

Kindly consider the following

$\frac{3 x + 4}{5 x^{2} - 7 x + 9}$

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:

$\begin{array}{l} L e t y = \frac{3 x + 4}{5 x^{2} - 7 x + 9} \\ \frac{d y}{d x} = \frac{d}{d x} (\frac{3 x + 4}{5 x^{2} - 7 x + 9}) \\ = \frac{(5 x^{2} - 7 x + 9) \frac{d}{d x} (3 x + 4) - (3 x + 4) \frac{d}{d x} (5 x^{2} - 7 x + 9)}{{(5 x^{2} - 7 x + 9)}^{2}} [Using q u o t i e n t r u l e] \\ = \frac{(5 x^{2} - 7 x + 9) (3) - (3 x + 4) (10 x - 7)}{{(5 x^{2} - 7 x + 9)}^{2}} \\ = \frac{15 x^{2} - 21 x + 27 - 30 x^{2} + 21 x - 40 x + 28}{{(5 x^{2} - 7 x + 9)}^{2}} \\ = \frac{- 15 x^{2} - 40 x + 55}{{(5 x^{2} - 7 x + 9)}^{2}} \\ = \frac{55 - 40 x - 15 x^{2}}{{(5 x^{2} - 7 x + 9)}^{2}} \end{array}$

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New answer posted•a year ago

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

Kindly consider the following
$(s e c x - 1) (s e c x + 1)$

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} L e t y = (s e c x - 1) (s e c x + 1) \\ = (s e c^{2} x - 1) [? a^{2} - b^{2} = (a - b) (a + b)] \\ = t a n^{2} x [? t a n^{2} x = s e c^{2} x - 1] \\ \frac{d y}{d x} = \frac{d}{d x} t a n^{2} x \\ = 2 t a n x \frac{d}{d x} t a n x = 2 t a n x . s e c^{2} x \end{array}$

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New answer posted•a year ago

Limits and Derivatives 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:

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

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New answer posted•a year ago

Limits and Derivatives 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:

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

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New answer posted•a year ago

Limits and Derivatives 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:

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

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New answer posted•a year ago

Limits and Derivatives 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:

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

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New question posted•a year ago

Limits and Derivatives 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

Limits and Derivatives 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:

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

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New answer posted•a year ago

Limits and Derivatives 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:

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

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New answer posted•a year ago

Limits and Derivatives 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:

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

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