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Class 11th

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New answer posted

a year ago

Class 11th Exemplar Solution

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

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

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

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

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

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

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

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New answer posted

a year ago

Class 11th Exemplar Solution

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

0 Follower 11 Views

Vishal Baghel

Contributor-Level 10

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

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New answer posted

a year ago

Class 11th Exemplar Solution

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

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

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

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New answer posted

a year ago

Class 11th Exemplar Solution

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

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

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

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New answer posted

a year ago

Class 11th Exemplar Solution

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

0 Follower 7 Views

Payal Gupta

Contributor-Level 10

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

Class 11th Exemplar Solution

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

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

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

Sol:

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

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New answer posted

a year ago

Class 11th Exemplar Solution

If A = {1, 3, 5, 7, 9, 11, 13, 15, 17} B = {2, 4, ., 18} and N the set of natural numbers is the universal set, then A′ ∪(A ∪ B) ∩ B′) is

(a) φ

(b) N

(c) A

(d) B

0 Follower 12 Views

Vishal Baghel

Contributor-Level 10

$\begin{array}{l} G i v e n t h a t : A = {1, 3, 5, 7, 9, 11, 13, 15, 17} \\ B = {2, 4, \dots, 18} \\ U = N = {1, 2, 3, 4, 5, \dots} \\ A' \cup (A \cup B) \cap B' = A' \cup [(A \cap B') \cup (B \cap B')] \\ = A' \cup (A \cap B') \cup ? [? B \cap B' = ?] \\ = A' \cup (A \cap B') \\ = (A' \cup A) \cap (A' \cup B') \\ = N \cap (A' \cup B') [? A' \cup A = N] \\ = A' \cup B' = (A \cap B)' = (?)' = N [? A \cap B = ?] \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

If A and B are two sets, then A ∩ (A ∪ B) equals

(a) A

(b) B

(c) φ

(d) A ∩ B

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

$\begin{array}{l} G i v e n t h a t : A \cap (A \cup B) \\ L e t x \in A \cap (A \cup B) \\ \Rightarrow x \in A a n d x \in (A \cup B) \\ \Rightarrow x \in A a n d (x \in A o r x \in B) \\ \Rightarrow (x \in A a n d x \in A) o r (x \in A a n d x \in B) \\ \Rightarrow x \in A o r x \in (A \cap B) \Rightarrow x \in A \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

If $R_{3} = {(x, x^{2}) ? x is a real number}$ is a relation. Then find the domain and range of $R_{3}$ .

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} G i v e n t h a t : R_{3} = {(x, | x) i s a r e a l n u m b e r} \\ H e n c e, D o m a i n o f R_{3} = R \\ a n d R a n g e o f R_{3} i s (0, \infty) [? | x | = R_{+}] \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

If sets A and B are defined as

A = {(x, y) | y = 1/x, 0 ≠ x ∈ R} B = {(x, y) | y = – x, x ∈ R}, then

(a) A ∩ B = A

(b) A ∩ B = B

(c) A ∩ B = φ

(d) A ∪ B = A

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

$\begin{array}{l} G i v e n t h a t : A = {(x, y) | y = \frac{1}{x}, 0 \neq x \in R} \\ B = {(x, y) | y = - x, x \in R} \\ H e r e, i t i s c l e a r t h a t y = \frac{1}{x} a n d y = - x \\ ? \frac{1}{x} \neq - x \\ ∴ A \cap B = ? \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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