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4 months ago

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

Prove by the direct method that for any integer $' n'$ , $n^{3} - n$ is always even.
Hint: Two cases (i) $n$ is even, (ii) $n$ is odd.

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} W e h a v e t w o c a s e s : \\ C a s e I : I f n i s e v e n \\ L e t n = 2 k w h e r e k \in N \\ ∴ n^{3} - n = {(2 k)}^{3} - (2 k) \\ = 2 k (4 k^{2} - 1) = 2 m, \\ w h e r e m = k (4 k^{2} - 1) \\ T h e r e f o r e, (n^{3} - n) i s e v e n . \\ C a s e I I : I f n i s o d d \\ L e t n = (2 k + 1), k \in N \\ ∴ n^{3} - n = {(2 k + 1)}^{3} - (2 k + 1) \\ = (2 k + 1) [{(2 k + 1)}^{2} - 1] \\ = (2 k + 1) [4 k^{2} + 4 k + 1 - 1] \\ = (2 k + 1) (4 k^{2} + 4 k) = 4 k (2 k + 1) (k + 1) \\ = 2.2 k (2 k + 1) (k + 1) = 2 λ w h e r e λ = 2 k (2 k + 1) (k + 1) \\ T h e r e f o r e, (n^{3} - n) i s e v e n . \\ H e n c e, (n^{3} - n) i s a l w a y s e v e n . \end{array}$

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New answer posted

4 months ago

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

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

4 months ago

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

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

4 months ago

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

Kindly consider the following

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

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

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

Sol:

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

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New answer posted

4 months ago

Class 11th 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

Class 11th 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

Class 11th 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

Class 11th 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

Class 11th 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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0 Follower 2 Views

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

Class 11th 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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