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New answer posted•a year ago

Exemplar Solution Class 11th

If $t a n θ + s i n θ = m$ and $t a n θ - s i n θ = n$ , then prove that $m^{2} - n^{2} = 4 s i n θ t a n θ .$ 

[Hint: Use $m + n = 2 t a n θ$ and $m - n = 2 s i n θ$ ., then use $m^{2} - n^{2} = (m + n) (m - n)$ ]

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:

$\begin{array}{l} G i v e n t h a t : t a n θ + s i n θ = m a n d t a n θ - s i n θ = n \\ L . H . S . m^{2} - n^{2} = (m + n) (m - n) \\ = [(t a n θ + s i n θ) + (t a n θ - s i n θ)] . [(t a n θ + s i n θ) - (t a n θ - s i n θ)] \\ = [t a n θ + s i n θ + t a n θ - s i n θ] . [t a n θ + s i n θ - t a n θ + s i n θ] \\ = 2 t a n θ . 2 s i n θ = 4 s i n θ t a n θ R . H . S . \\ L . H . S . = R . H . S . H e n c e p r o v e d . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

If A and B are any two sets, then A – B is equal to ______________.

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} F r o m t h e v e n n d i a g r a m i t i s c l e a r t h a t A - B = A \cap B' \\ H e n c e, t h e f i l l e r i s A \cap B' . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

If A and B are finite sets such that A ⊂ B, then n (A ∪ B) = ______________.

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} Since A \subset B ∴ n (A \cup B) = n (B) \\ H e n c e, t h e f i l l e r i s n (B) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

When A = φ, then number of elements in P(A) is ______________.

6 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} H e r e, A = ? ∴ n (A) = 0 \\ ? n [P (A)] = 2^{n (A)} = 2^{0} = 1 \\ H e n c e, t h e f i l l e r i s 1 . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

If $f$ and  $g$ are real functions defined by  $f (x) = x^{2} + 7$ and $g (x) = 3 x + 5$ , find each of the following

(i) $f (3) + g (- 5)$  

(ii) $f (\frac{1}{2}) \times g (14)$  

(iii) $f (- 2) + g (- 1)$  

(iv) $f (t) - f (- 2)$  

(v) $f (t) - f (- 2) \frac{f (t) - f (5)}{t - 5}, if t \neq 5$

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

$\begin{array}{l} G i v e n t h a t : f (x) = x^{2} + 7 a n d g (x) = 3 x + 5 \\ (i) f (3) + g (- 5) = [{(3)}^{2} + 7] + [3 (- 5) + 5] \\ = (9 + 7) + (- 15 + 5) = 16 - 10 = 6 \\ H e n c e, f (3) + g (- 5) = 6 \\ (i i) f (\frac{1}{2}) \times g (14) = [{(\frac{1}{2})}^{2} + 7] \times [3 \times 14 + 5] \\ = (\frac{1}{4} + 7) \times (42 + 5) = \frac{29}{4} \times 47 = \frac{1363}{4} \\ H e n c e, f (\frac{1}{2}) \times g (14) = \frac{1363}{4} \\ (i i i) f (- 2) + g (- 1) = [{(- 2)}^{2} + 7] + [3 (- 1) + 5] \\ = (4 + 7) + (- 3 + 5) = 11 + 2 = 13 \\ H e n c e, f (- 2) + g (- 1) = 13 \\ (i v) f (t) - f (- 2) = (t^{2} + 7) - [{(- 2)}^{2} + 7] = t^{2} + 7 - 11 \\ = t^{2} - 4 \\ H e n c e, f (t) - f (- 2) = t^{2} - 4 . \\ (v) \frac{f (t) - f (5)}{t - 5}, t \neq 5 = \frac{(t^{2} + 7) - [{(5)}^{2} + 7]}{t - 5} \\ = \frac{t^{2} + 7 - 32}{t - 5} = \frac{t^{2} - 25}{t - 5} = t + 5 \\ H e n c e, \frac{f (t) - f (5)}{t - 5}, t \neq 5 = t + 5 . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

The set {x ∈ R: 1 ≤ x < 2} can be written as ______________.

11 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} T h e s e t {x \in R : 1 \leq x < 2} c a n b e w r i t t e n a s [1, 2) \\ H e n c e, t h e f i l l e r i s [1, 2) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

If X and Y are two sets and X′ denotes the complement of X, then X ∩ (X ∪ Y) ′is equal to

(a) X

(b) Y

(c) φ

(d) X ∩ Y

5 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} L e t x \in X \cap (X \cup Y)' \\ \Rightarrow x \in X \cap (X' \cap Y') \\ \Rightarrow x \in (X \cap X') \cap (X \cap Y') \\ \Rightarrow x \in ? \cap (X \cap Y') [? A \cap A' = ?] \\ \Rightarrow x \in ? \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

Let S = {x | x is a positive multiple of 3 less than 100}

P = {x | x is a prime number less than 20}. Then n(S) + n(P) is

(a) 34

(b) 31

(c) 33

(d) 30

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} G i v e n t h a t : S = {x | x i s a p o s i t i v e m u l t i p l e o f 3 < 100} \\ ∴ S = {3, 6, 9, 12, 15, 18, \dots, 99} \\ \Rightarrow n (S) = 33 \\ T = {x | x i s a p r i m e n u m b e r < 20} \\ ∴ T = {2, 3, 5, 7, 11, 13, 17, 19} \\ \Rightarrow n (T) = 8 \\ S o, n (S) + n (T) = 33 + 8 = 41 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

Is the given relation a function? Give reasons for your answer.

(i) $h = {(4, 6), (3, 9), (- 11, 6), (3, 11)}$  

(ii) $f = {(x, x) ? x is a real number}$  

(iii) $g = {(n, \frac{1}{n}, n) ? n is a positive integer}$  

(iv) $s = {(n, n^{2}) ? n is a positive integer}$  

(v) $t = {(x, 3) ? x is a real number}$

7 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

$\begin{array}{l} (i) G i v e n t h a t : h = {(4, 6), (3, 9), (- 11, 6), (3, 11)} \\ Since i n t h e g i v e n r e l a t i o n 3 h a s t w o i m a g e s 9 a n d 11 . S o, h i s n o t a f u n c t i o n . \\ (i i) f = {(x, x) | x i s a r e a l n u m b e r} . H e r e, w e o b s e r v e t h a t f o r e v e r y e l e m e n t o f d o m a i n h a s a \\ u n i q u e i m a g e . S o, f i s a f u n c t i o n . \\ (i i i) G i v e n t h a t : g = {(n, \frac{1}{n}) | n i s a p o s i t i v e integer} . H e r e, w e o b s e r v e t h a t n i s a p o s i t i v e \\ integer so, f o r e v e r y e l e m e n t o f d o m a i n, t h e r e i s a u n i q u e \frac{1}{n} i m a g e . H e n c e, g i s a f u n c t i o n . \\ (i v) G i v e n t h a t : S = {(n, n^{2}) | n i s a p o s i t i v e integer} . H e r e, w e o b s e r v e t h a t t h e s q u a r e o f a n y \\ integer i s a u n i q u e n u m b e r . S o, e v e r y e l e m e n t i n t h e d o m a i n t h e r e i s a u n i q u e i m a g e . \\ H e n c e, S i s a f u n c t i o n . \\ (v) G i v e n t h a t : t = {(x, 3) | x i s a r e a l n u m b e r} . H e r e, w e o b s e r v e t h a t f o r e v e r y r e a l e l e m e n t \\ i n t h e d o m a i n, t h e r e i s a constant n u m b e r 3 . H e n c e, t i s a constant f u n c t i o n . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

Prove that $s i n 4 A = 4 s i n A c o s^{3} A - 4 c o s A s i n^{3} A .$

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:

$\begin{array}{l} L . H . S . s i n 4 A = s i n (A + 3 A) \\ = s i n A c o s 3 A + c o s A s i n 3 A \\ = s i n A (4 c o s^{3} A - 3 c o s A) + c o s A (3 s i n A - 4 s i n^{3} A) \\ = 4 s i n A c o s^{3} A - 3 s i n A c o s A + 3 s i n A c o s A - 4 c o s A s i n^{3} A \\ = 4 s i n A c o s^{3} A - 4 c o s A s i n^{3} A R . H . S . \\ L . H . S . = R . H . S . H e n c e p r o v e d . \end{array}$

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