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New answer posted

4 months ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

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

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

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

Sol:

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

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New answer posted

4 months ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

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

4 months ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

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

4 months ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

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

4 months ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Fourteen

Identify the quantifiers in the following statements:

i. There exists a triangle that is not equilateral.

ii. For all real numbers $x$ and $y$ , $x y = y x$ .

iii. There exists a real number that is not a rational number.

iv. For every natural number $x$ , $x + 1$ is also a natural number.

v. For all real numbers $x$ with $x > 3$ , $x^{2} > 9$ .

vi. There exists a triangle that is not an isosceles triangle.

vii. For all negative integers $x$ , $x^{3}$ is also a negative integer.

viii. There exists a statement in the above statements that is not true.

ix. There exists an even prime number other

...more

Identify the quantifiers in the following statements:

i. There exists a triangle that is not equilateral.

ii. For all real numbers $x$ and $y$ , $x y = y x$ .

iii. There exists a real number that is not a rational number.

iv. For every natural number $x$ , $x + 1$ is also a natural number.

v. For all real numbers $x$ with $x > 3$ , $x^{2} > 9$ .

vi. There exists a triangle that is not an isosceles triangle.

vii. For all negative integers $x$ , $x^{3}$ is also a negative integer.

viii. There exists a statement in the above statements that is not true.

ix. There exists an even prime number other than 2.

x. There exists a real number $x$ such that $x^{2} + 1 = 0$ .

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} Q u a n t i f i e r m e a n s a p h r a s e l i k e' t h e r e e x i s t s',' f o r a l l' a n d' f o r e v e r y' e t c . \\ (i) T h e r e e x i s t s (i i) F o r a l l \\ (i i i) T h e r e e x i s t s (i v) F o r e v e r y \\ (v) F o r a l l (v i) T h e r e e x i s t s \\ (v i i) F o r a l l (v i i i) T h e r e e x i s t s \\ (i x) T h e r e e x i s t s (x) T h e r e e x i s t s \end{array}$

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New answer posted

4 months ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Fourteen

Write down the converse of the following statements:

i. If a rectangle $R$ is a square, then $R$ is a rhombus.

ii. If today is Monday, then tomorrow is Tuesday.

iii. If you go to Agra, then you must visit the Taj Mahal.

iv. If the sum of squares of two sides of a triangle is equal to the square of the third side of the triangle, then the triangle is right-angled.

v. If all three angles of a triangle are equal, then the triangle is equilateral.

vi. If $x : y = 3 : 2$ , then $2 x = 3 y$ .

vii. If $S$ is a cyclic quadrilateral, then the opposite angles of $S$ are supplementary.

viii. If $x$ is zero, then

...more

Write down the converse of the following statements:

i. If a rectangle $R$ is a square, then $R$ is a rhombus.

ii. If today is Monday, then tomorrow is Tuesday.

iii. If you go to Agra, then you must visit the Taj Mahal.

iv. If the sum of squares of two sides of a triangle is equal to the square of the third side of the triangle, then the triangle is right-angled.

v. If all three angles of a triangle are equal, then the triangle is equilateral.

vi. If $x : y = 3 : 2$ , then $2 x = 3 y$ .

vii. If $S$ is a cyclic quadrilateral, then the opposite angles of $S$ are supplementary.

viii. If $x$ is zero, then $x$ is neither positive nor negative.

ix. If two triangles are similar, then the ratio of their corresponding sides is equal.

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0 Follower 2 Views

Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} (i) I f t h e r e c t a n g l e R i s r h o m b u s, t h e n i t i s a s q u a r e . \\ (i i) I f t o m o r r o w i s T u e s d a y, t h e n t o d a y i s M o n d a y . \\ (i i i) I f y o u m u s t v i s i t t h e T a j M a h a l, t h e n y o u g o t o A g r a . \\ (i v) I f t h e t r i a n g l e i s r i g h t t r i a n g l e, t h e n t h e s u m o f t h e s q u a r e s o f t w o s i d e s o f a t r i a n g l e \\ i s e q u a l t o t h e s q u a r e o f t h e t h i r d s i d e o f t h e t r i a n g l e . \\ (v) I f t h e t r i a n g l e i s e q u i l a t e r a l, t h e n a l l t h r e e a n g l e s o f a t r i a n g l e a r e e q u a l . \\ (v i) I f 2 x = 3 y t h e n x : y = 3 : 2 . \\ (v i i) I f t h e o p p o s i t e a n g l e s o f a q u a d r i l a t e r a l a r e s u p p l e m e n t a r y, t h e n S i s c y c l i c . \\ (v i i i) I f x i s n e i t h e r p o s i t i v e n o r n e g a t i v e t h e n x = 0 . \\ (i x) I f t h e r a t i o o f c o r r e s p o n d i n g s i d e s o f t w o t r i a n g l e s a r e e q u a l, t h e n t r i a n g l e s a r e s i m i l a r . \end{array}$

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New answer posted

4 months ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

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

4 months ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Fourteen

Write down the contrapositive of the following statements:

i. If $x = y$ and $y = 3$ , then $x = 3$ .

ii. If $n$ is a natural number, then $n$ is an integer.

iii. If all three sides of a triangle are equal, then the triangle is equilateral.

iv. If $x$ and $y$ are negative integers, then $x y$ is positive.

v. If a natural number $n$ is divisible by 6, then $n$ is divisible by 2 and 3.

vi. If it snows, then the weather will be cold.

vii. If $x$ is a real number such that $0 < x < 1$ , then $x^{2} < 1$ .

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} W e k n o w t h a t t h e c o n t r a p o s i t i v e o f p \to q i s (\sim q) \to (\sim p) \\ (i) I f x \neq 3, t h e n x \neq y o r y \neq 3 . \\ (i i) I f n i s n o t a n i n t e g e r, t h e n i t i s n o t a n a t u r a l n u m b e r . \\ (i i i) I f t h e t r i a n g l e i s n o t e q u i l a t e r a l, t h e n a l l t h r e e s i d e s o f t h e t r i a n g l e a r e n o t e q u a l . \\ (i v) I f x y i s n o t p o s i t i v e integer, t h e n x o r y i s n o t n e g a t i v e integer. \\ (v) I f n a t u r a l n u m b e r' n' i s n o t d i v i s i b l e b y 2 o r 3, t h e n n i s n o t d i v i s i b l e b y 6 . \\ (v i) T h e w e a t h e r w i l l n o t b e c o l d, i f i t d o e s n o t s n o w . \\ (v i i) I f x^{2} > 1 t h e n x i s n o t r e a l n u m b e r s u c h t h a t 0 < x < 1 . \end{array}$

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New answer posted

4 months ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

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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New answer posted

4 months ago

Maths 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 3 Views

alok kumar singh

Contributor-Level 10

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

Sol:

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

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