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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

$l i m_{x \to - 3} \frac{x^{3} + 27}{x^{5} + 243}$

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to - 3}{l i m} \frac{x^{3} + 27}{x^{5} + 243} \\ D i v i d i n g t h e n u m e r a t o r a n d denominator, b y x - 3 w e g e t \\ = \underset{x \to - 3}{l i m} \frac{\frac{x^{3} + {(3)}^{3}}{x - 3}}{\frac{x^{5} + {(3)}^{5}}{x - 3}} = \frac{\underset{x \to - 3}{l i m} (\frac{x^{3} - {(- 3)}^{3}}{x - 3})}{\underset{x \to - 3}{l i m} (\frac{x^{5} - {(- 3)}^{5}}{x - 3})} [? \underset{x \to a}{l i m} \frac{f (x)}{g (x)} = \frac{\underset{x \to a}{l i m} f (x)}{\underset{x \to a}{l i m} g (x)}] \\ = \frac{3 {(- 3)}^{3 - 1}}{5 {(- 3)}^{5 - 1}} [? \underset{x \to a}{l i m} \frac{x^{n} - a^{n}}{x - a} = n . a^{n - 1}] \\ = \frac{3 * {(- 3)}^{2}}{5 * {(- 3)}^{4}} = \frac{1}{5 * 3} = \frac{1}{15} \end{array}$

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

$l i m_{x \to 0} \frac{\sqrt[]{1 + x^{3}} - \sqrt[]{1 - x^{3}}}{x^{2}}$

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{\sqrt[]{1 + x^{3}} - \sqrt[]{1 - x^{3}}}{x^{2}} \\ R a t i o n a l i z i n g t h e denominator, w e g e t \\ = \underset{x \to 0}{l i m} \frac{\sqrt[]{1 + x^{3}} - \sqrt[]{1 - x^{3}}}{x^{2}} * \frac{\sqrt[]{1 + x^{3}} + \sqrt[]{1 - x^{3}}}{\sqrt[]{1 + x^{3}} + \sqrt[]{1 - x^{3}}} \\ = \underset{x \to 0}{l i m} \frac{(1 + x^{3}) - (1 - x^{3})}{x^{2} [\sqrt[]{1 + x^{3}} + \sqrt[]{1 - x^{3}}]} = \underset{x \to 0}{l i m} \frac{1 + x^{3} - 1 + x^{3}}{x^{2} [\sqrt[]{1 + x^{3}} + \sqrt[]{1 - x^{3}}]} \\ = \underset{x \to 0}{l i m} \frac{2 x^{3}}{x^{2} [\sqrt[]{1 + x^{3}} + \sqrt[]{1 - x^{3}}]} = \underset{x \to 0}{l i m} \frac{2 x}{\sqrt[]{1 + x^{3}} + \sqrt[]{1 - x^{3}}} = 0 \end{array}$

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Fourteen

Form the biconditional statement $p \leftrightarrow q$ , where:

$i. p$ : The unit digit of an integer is zero.

It is divisible by 5.

$ii. p$ : A natural number $n$ is odd.

$q$ : Natural number $n$ is not divisible by 2.

$iii. p$ : A triangle is an equilateral triangle.

$q$ : All three sides of a triangle are equal.

0 Follower 10 Views

Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} (i) p \leftrightarrow q : U n i t d i g i t o f a n integer i s z e r o i f a n d o n l y i f i t i s d i v i s i b l e b y 5 . \\ (i i) p \leftrightarrow q : A n a t u r a l n u m b e r i s o d d i f a n d o n l y i f i t i s n o t d i v i s i b l e b y 2 . \\ (i i i) p \leftrightarrow q : A t r i a n g l e i s a n e q u i l a t e r a l t r i a n g l e i f a n d o n l y i f a l l t h r e e s i d e s o f t r i a n g l e \\ a r e e q u a l . \end{array}$

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

$l i m_{x \to 3} \frac{x^{7} - 2 x^{5} - 1}{x^{3} + 3 x^{2} + 2}$

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 1}{l i m} \frac{x^{7} - 2 x^{5} - 1}{x^{3} + 3 x^{2} + 2} [\frac{0}{0} f o r m] \\ = \underset{x \to 1}{l i m} \frac{x^{7} - x^{5} - x^{5} - 1}{x^{3} - x^{2} - 2 x^{2} + 2} = \underset{x \to 1}{l i m} \frac{x^{5} (x^{2} - 1) - 1 (x^{5} - 1)}{x^{2} (x - 1) - 2 (x^{2} - 1)} \\ D i v i d i n g t h e n u m e r a t o r a n d denominator b y (x - 1) w e g e t \\ = \underset{x \to 1}{l i m} \frac{x^{5} (\frac{x^{2} - 1}{x - 1}) - 1 (\frac{x^{5} - 1}{x - 1})}{x^{2} (\frac{x - 1}{x - 1}) - 2 (\frac{x^{2} - 1}{x - 1})} = \frac{\underset{x \to 1}{l i m} x^{5} (x + 1) - \underset{x \to 1}{l i m} (\frac{x^{5} - {(1)}^{5}}{x - 1})}{\underset{x \to 1}{l i m} x^{2} - 1 - 2 \underset{x \to 1}{l i m} (x + 1)} \\ = \frac{1 (2) - 5 . {(1)}^{5 - 1}}{1 - 2 (2)} = \frac{2 - 5}{1 - 4} = \frac{- 3}{- 3} = 1 \end{array}$

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

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

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to \sqrt[]{2}}{l i m} \frac{x^{4} - 4}{x^{2} + 3 \sqrt[]{2} x - 8} \\ = \underset{x \to \sqrt[]{2}}{l i m} \frac{(x^{2} - 2) (x^{2} + 2)}{x^{2} + 4 \sqrt[]{2} x - \sqrt[]{2} x - 8} \\ = \underset{x \to \sqrt[]{2}}{l i m} \frac{(x + \sqrt[]{2}) (x - \sqrt[]{2}) (x^{2} + 2)}{x (x + 4 \sqrt[]{2}) - \sqrt[]{2} (x + 4 \sqrt[]{2})} \\ = \underset{x \to \sqrt[]{2}}{l i m} \frac{(x + \sqrt[]{2}) (x - \sqrt[]{2}) (x^{2} + 2)}{(x + 4 \sqrt[]{2}) (x - \sqrt[]{2})} = \underset{x \to \sqrt[]{2}}{l i m} \frac{(x + \sqrt[]{2}) (x^{2} + 2)}{(x + 4 \sqrt[]{2})} \\ T a k i n g, w e h a v e \\ = \frac{(\sqrt[]{2} + \sqrt[]{2}) (2 + 2)}{\sqrt[]{2} + 4 \sqrt[]{2}} = \frac{2 \sqrt[]{2} * 4}{5 \sqrt[]{2}} = \frac{8}{5} . \end{array}$

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

$l i m_{x \to 2} \frac{x^{2} - 4}{\sqrt[]{3 x - 2} - \sqrt[]{x + 2}}$

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 2}{l i m} \frac{x^{2} - 4}{\sqrt[]{3 x - 2} - \sqrt[]{x + 2}} \\ R a t i o n a l i z i n g t h e denominator w e g e t \\ = \underset{x \to 2}{l i m} \frac{(x - 2) (x + 2)}{\sqrt[]{3 x - 2} - \sqrt[]{x + 2}} * \frac{\sqrt[]{3 x - 2} + \sqrt[]{x + 2}}{\sqrt[]{3 x - 2} + \sqrt[]{x + 2}} \\ = \underset{x \to 2}{l i m} \frac{(x - 2) (x + 2) (\sqrt[]{3 x - 2} + \sqrt[]{x + 2})}{3 x - 2 - x - 2} \\ = \underset{x \to 2}{l i m} \frac{(x - 2) (x + 2) (\sqrt[]{3 x - 2} + \sqrt[]{x + 2})}{2 x - 4} \\ = \underset{x \to 2}{l i m} \frac{(x - 2) (x + 2) (\sqrt[]{3 x - 2} + \sqrt[]{x + 2})}{2 (x - 2)} = \underset{x \to 2}{l i m} \frac{(x + 2) (\sqrt[]{3 x - 2} + \sqrt[]{x + 2})}{2} \\ T a k i n g limits, w e h a v e \\ = \frac{(2 + 2) (\sqrt[]{6 - 2} + \sqrt[]{2 + 2})}{2} = \frac{4 (2 + 2)}{2} = 8 \end{array}$

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

$l i m_{x \to 1} \frac{x^{4} - \sqrt[]{x}}{\sqrt[]{x} - 1}$

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 1}{l i m} \frac{x^{4} - \sqrt[]{x}}{\sqrt[]{x} - 1} \\ = \underset{x \to 1}{l i m} \frac{\sqrt[]{x} [{(x)}^{7 / 2} - 1]}{\sqrt[]{x} - 1} \\ D i v i d i n g t h e n u m e r a t o r a n d denominator b y x - 1 \\ = \underset{x \to 1}{l i m} \frac{\frac{\sqrt[]{x} [{(x)}^{7 / 2} - {(1)}^{7 / 2}]}{x - 1}}{\frac{{(x)}^{1 / 2} - {(1)}^{1 / 2}}{x - 1}} \\ = \underset{x \to 1}{l i m} \frac{\frac{{(x)}^{7 / 2} - {(1)}^{7 / 2}}{x - 1}}{\frac{{(x)}^{1 / 2} - {(1)}^{1 / 2}}{x - 1}} * \underset{x \to 1}{l i m} \sqrt[]{x} [? \underset{x \to a}{l i m} f (x) . g (x) = \underset{x \to a}{l i m} f (x) . \underset{x \to a}{l i m} g (x)] \\ = \frac{\frac{7}{2} {(1)}^{7 / 2 - 1}}{\frac{1}{2} {(1)}^{1 / 2 - 1}} * \sqrt[]{1} = \frac{7 / 2}{1 / 2} = 7 . \end{array}$

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Fourteen

Rewrite each of the following statements in the form of conditional statements:

i. The square of an odd number is odd.

ii. You will get a sweet dish after the dinner.

iii. You will fail if you do not study.

iv. The unit digit of an integer is 0 or 5 if it is divisible by 5.

v. The square of a prime number is not prime.

vi. 2b = a + c if a, b, and c are in A.P.

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} (i) I f p, t h e n q \Rightarrow I f t h e n u m b e r i s o d d, t h e n i t s s q u a r e i s o d d n u m b e r . \\ (i i) q i f p \Rightarrow I f t a k e t h e d i n n e r, t h e n y o u w i l l g e t s w e e t d i s h . \\ (i i i) p o n l y i f q \Rightarrow I f y o u d o n o t s t u d y, t h e n y o u w i l l f a i l . \\ (i v) p i s s u f f i c i e n t f o r q \Rightarrow I f a n integer i s d i v i s i b l e b y 5, t h e n i t s u n i t d i g i t s a r e 0 a n d 5 . \\ (v) q i s n e c e s s a r y f o r p \Rightarrow I f a n y n u m b e r i s p r i m e, t h e n i t s s q u a r e i s n o t p r i m e . \\ (v i) q i m p l i e s p \Rightarrow I f a, b, c a r e i n A . P t h e n 2 b = a + c . \end{array}$

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

$\underset{x \to a}{l i m} \frac{{(2 + x)}^{\frac{5}{2}} - {(a + 2)}^{\frac{5}{2}}}{(x - a)}$

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

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

Sol:

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

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

$l i m_{x \to 1} \frac{{(1 + x)}^{6} - 1}{{(1 + x)}^{2} - 1}$

0 Follower 6 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{{(1 + x)}^{6} - 1}{{(1 + x)}^{2} - 1} \\ D i v i d i n g t h e n u m e r a t o r a n d denominator b y x, w e g e t \\ = \underset{x \to 0}{l i m} \frac{\frac{{(1 + x)}^{6} - 1}{x}}{\frac{{(1 + x)}^{2} - 1}{x}} \\ P u t 1 + x = y \Rightarrow x = y - 1 \\ = \underset{y - 1 \to 0}{l i m} \frac{\frac{y^{6} - {(1)}^{6}}{y - 1}}{\frac{y^{2} - {(1)}^{2}}{y - 1}} = \frac{\underset{y \to 1}{l i m} \frac{y^{6} - {(1)}^{6}}{y - 1}}{\underset{y \to 1}{l i m} \frac{y^{2} - {(1)}^{2}}{y - 1}} [\underset{x \to a}{l i m} \frac{f (x)}{g (x)} = \frac{\underset{x \to a}{l i m} f (x)}{\underset{x \to a}{l i m} g (x)}] \\ = \frac{6 . {(1)}^{6 - 1}}{2 . {(1)}^{2 - 1}} = \frac{6}{2} = 3 [\underset{x \to a}{l i m} \frac{x^{n} - a^{n}}{x - a} = n . a^{n - 1}] \end{array}$

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