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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Fourteen

Which of the following is not a statement

(a) Smoking is injurious to health.

(b) 2 + 2 = 4

(c) 2 is the only even prime number.

(d) Come here.

4 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T o g i v e o r d e r c a n n o t b e a s t a t e m e n t . \\ S o,' C o m e h e r e' i s n o t a s t a t e m e n t . \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Fourteen

Which of the following is a statement.

(a) x is a real number.

(b) Switch off the fan.

(c) 6 is a natural number.

(d) Let me go.

2 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} Since s t a t e m e n t i s a s e n t e n c e w h i c h i s e i t h e r t r u e o r f a l s e . \\ S o, 6 i s a n a t u r a l n u m b e r w h i c h i s t r u e . \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Fourteen

Using the contrapositive method, prove that if $n^{2}$  is an even integer, then $n$ is also an even integer.

5 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} L e t p : n^{2} i s a n e v e ninteger . \\ q : n i s a l s o a n e v e n integer . \\ A l s o l e t \sim q i s t r u e i . e ., n i s n o t a n e v e n integer . \\ \Rightarrow n^{2} i s n o t a n e v e n integer . \\ \Rightarrow \sim p i s t r u e . [? S q u a r e o f a n o d d integer i s o d d] \\ H e n c e, \sim q i s t r u e \Rightarrow \sim p i s t r u e . \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Fourteen

Prove by the direct method that for any real numbers $x, y$ , if $x = y$ , then $x^{2} = y^{2}$ .

4 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} L e t p : x = y, x, y \in R \\ O n s q u a r i n g b o t h s i d e s w e h a v e \\ x^{2} = y^{2} : q (s a y) \\ \Rightarrow p = q H e n c e, p r o v e d . \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Fourteen

Prove the following statement by the contradiction method:

 $p$ : The sum of an irrational number and a rational number is irrational.

3 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} L e t p i s f a l s e i . e ., t h e s u m o f a n i r r a t i o n a l n u m b e r a n d a r a t i o n a l n u m b e r i s r a t i o n a l . \\ L e t \sqrt[]{λ} i s i r r a t i o n a l a n d n i s r a t i o n a l n u m b e r \\ \Rightarrow \sqrt[]{λ} + n = r (r a t i o n a l) \\ \Rightarrow \sqrt[]{λ} = r - n \\ W e o b s e r v e t h a t \sqrt[]{λ} i s i r r a t i o n a l w h e r e a s (r - n) i s r a t i o n a l w h i c h i s a c o n t r a d i c t i o n . \\ S o, o u r supposition i s w r o n g . \\ H e n c e, p i s t r u e . \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Fourteen

Check the validity of the following statements:

i. $p$ : 125 is divisible by 5 and 7.

ii. $q$ : 131 is a multiple of 3 or 11.

2 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} (i) G i v e n t h a t : \\ p : 125 i s d i v i s i b l e b y 5 a n d 7 . \\ L e t q : 125 i s d i v i s i b l e b y 5 . \\ a n d r : 125 i s d i v i s i b l e b y 7 . \\ H e r e q i s t r u e a n d r i s f a l s e . \\ \Rightarrow q \land r i s f a l s e . \\ H e n c e, p i s n o t v a l i d . \\ (i i) G i v e n t h a t : \\ q : 131 i s a m u l t i p l e o f 3 o r 11 . \\ L e t p : 131 i s a m u l t i p l e o f 3 . \\ a n d r : 131 i s a m u l t i p l e o f 11 . \\ H e r e p i s n o t t r u e a n d r i s a l s o n o t t r u e i . e ., f a l s e . \\ \Rightarrow p \lor r i s f a l s e . \\ H e n c e, q i s n o t v a l i d . \end{array}$

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New answer posted•a year ago

Exemplar Solution 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.

2 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following
$l i m_{x \to \frac{π}{4}} \frac{s i n x - c o s x}{x - \frac{π}{4}}$

7 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:

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

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following
$\underset{x \to \frac{π}{3}}{l i m} \frac{\sqrt[]{1 - c o s 6 x}}{\sqrt[]{2} (\frac{π}{3} - x)}$

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:

$\begin{array}{l} G i v e n t h a t \underset{x \to \frac{π}{3}}{l i m} \frac{\sqrt[]{1 - c o s 6 x}}{\sqrt[]{2} (\frac{π}{3} - x)} \\ = \underset{x \to \frac{π}{3}}{l i m} \frac{\sqrt[]{2 s i n^{2} 3 x}}{\sqrt[]{2} (\frac{π}{3} - x)} [? 1 - c o s θ = 2 s i n^{2} \frac{θ}{2}] \\ = \underset{x \to \frac{π}{3}}{l i m} \frac{\sqrt[]{2} s i n 3 x}{\sqrt[]{2} (\frac{π}{3} - x)} = \underset{π - 3 x \to 0}{l i m} \frac{3 . s i n (π - 3 x)}{π - 3 x} \\ = 3 [? \underset{x \to 0}{l i m} \frac{s i n x}{x} = 1] \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

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

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:

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

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