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New answer posted•a year ago

Exemplar Solution Class 11th

If A and B are subsets of the universal set U, then show that

(i) A ⊂ A ∪ B

(ii) A ⊂ B ⇔ A ∪ B = B

(iii) (A ∩ B) ⊂ A

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} (i) G i v e n t h a t : A \subset U a n d B \subset U \\ L e t x \in A o r x \in B \\ \Rightarrow x \in A \cup B \\ H e n c e, A \subset (A \cup B) \\ (i i) I f A \subset B \\ T h e n l e t x \in A \cup B \\ \Rightarrow x \in A o r x \in B \\ \Rightarrow A \cup B \subset B \dots (i) \\ B u t B \subset A \cup B \dots (i i) \\ F r o m e q . (i) a n d (i i), w e g e t \\ A \cup B = B \\ L e t y \in A \Rightarrow y \in (A \cup B) \Rightarrow y \in B \\ \Rightarrow y \in B \Leftrightarrow A \cup B = B \\ (i i i) L e t x \in A \cap B \\ \Rightarrow x \in A a n d x \in B \Rightarrow x \in A \\ S o, A \cap B \subset A \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

Given L = {1, 2, 3, 4}, M = {3, 4, 5, 6} and N = {1, 3, 5}

Verify that L – (M ∪ N) = (L – M) ∩ (L – N)

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} G i v e n, L = {1, 2, 3, 4}, M = {3, 4, 5, 6} a n d N = {1, 3, 5} \\ ∴ M \cup N = {1, 3, 4, 5, 6} \\ L - M \cup N = {2} \\ N o w, L - M = {1, 2}, L - N = {2, 4} \\ ∴ (L - M) \cap (L - N) = {2} \\ H e n c e, L - M \cup N = (L - M) \cap (L - N) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

State which of the following statements a true and which are false. Justify your answer.

(i) 35 ∈ {x | x has exactly four positive factors}.

(ii) 128 ∈ {y | the sum of all the positive factors of y is 2y}

(iii) 3 ∉ {x | x4 – 5x3 + 2x2 – 112x + 6 = 0}

(iv) 496 ∉ {y | the sum of all the positive factors of y is 2y}.

6 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} (i) G i v e n t h a t : 35 \in {x | x h a s e x a c t l y f o u r p o s i t i v e f a c t o r} \\ ∴ F a c t o r s o f 35 a r e 1, 5, 7, 35 \\ H e n c e, t h e s t a t e m e n t (i) i s' T r u e' . \\ (i i) G i v e n t h a t : 128 \in {y | t h e s u m o f a l l p o s i t i v e f a c t o r s o f y i s 2 y} \\ ∴ F a c t o r s o f 128 a r e 1, 2, 4, 8, 16, 32, 64 a n d 128 . \\ S u m o f a l l f a c t o r s = 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 \\ = 255 \neq 2 \times 128 \\ H e n c e, t h e s t a t e m e n t (i i) i s' F a l s e' . \\ (i i i) G i v e n t h a t : 3 \in {x | x^{4} - 5 x^{3} + 2 x^{2} - 112 x + 6 = 0} \\ ∴ x^{4} - 5 x^{3} + 2 x^{2} - 112 x + 6 = 0 \\ N o w f o r x = 3, w e h a v e \\ {(3)}^{4} - 5 {(3)}^{3} + 2 {(3)}^{2} - 112 (3) + 6 \\ \Rightarrow 81 - 135 + 18 - 336 + 6 \Rightarrow - 366 \neq 0 \\ H e n c e, t h e s t a t e m e n t (i i i) i s' T r u e' . \\ (i v) G i v e n t h a t : 496 \notin {y | t h e s u m o f a l l p o s i t i v e f a c t o r s o f y i s 2 y} \\ ∴ T h e p o s i t i v e f a c t o r s o f 496 a r e 1, 2, 4, 8, 16, 31, 62, 124, 248 a n d 496 . \\ ∴ T h e s u m o f a l l p o s i t i v e f a c t o r s = 1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248 + 496 \\ = 992 = 2 \times 496 \\ H e n c e, t h e s t a t e m e n t (i v) i s' F a l s e' . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

If Y = {x | x is a positive factor of the number

2p – 1 (2p – 1), where 2p – 1 is a prime number}. Write Y in the roaster form.

12 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} G i v e n, Y = {x | x i s a p o s i t i v e f a c t o r o f t h e n u m b e r 2^{p - 1} (2^{p} - 1), w h e r e 2^{p - 1} i s a p r i m e n u m b e r} . \\ Since, t h e f a c t o r s o f 2^{p - 1} a r e 1, 2, 2^{2}, 2^{3}, \dots, 2^{p - 1} a n d f a c t o r s o f 2^{p} - 1 a r e 1 a n d 2^{p - 1} \\ ∴ Y = {1, 2, 2^{2}, 2^{3}, \dots, 2^{p - 1}, 2^{p} - 1} \\ = 2 (2^{p} - 1), 2^{2} (2^{p} - 1), \dots, 2^{p - 1} (2^{p} - 1) \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

Write the following sets in the roaster form

(i) D = {t | t3 = t, t ∈ R}

(ii) E = {w | =3, w ∈ R}

(iii) F = {x | x4 – 5x2 + 6 = 0, x ∈ R}

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} (i) W e h a v e, D = {t | t^{3} = t, t \in R} \\ ∴ t^{3} = t \\ \Rightarrow t^{3} - t = 0 \Rightarrow t (t^{2} - 1) = 0 \\ \Rightarrow t (t - 1) (t + 1) = 0 \Rightarrow t = 0, 1, - 1 \\ ∴ D = {- 1, 0, 1} \\ (i i) W e h a v e, E = {w | \frac{w - 2}{w + 3} = 3, w \in R} \\ ∴ \frac{w - 2}{w + 3} = 3 \\ \Rightarrow w - 2 = 3 w + 9 \Rightarrow w - 3 w = 9 + 2 \\ \Rightarrow - 2 w = 11 \Rightarrow w = \frac{- 11}{2} \\ ∴ E = {\frac{- 11}{2}} \\ (i i i) W e h a v e, F = {x | x^{4} - 5 x^{2} + 6 = 0, x \in R} \\ ∴ x^{4} - 5 x^{2} + 6 = 0 \\ \Rightarrow x^{4} - 3 x^{2} - 2 x^{2} + 6 = 0 \Rightarrow x^{2} (x^{2} - 3) - 2 (x^{2} - 3) = 0 \\ \Rightarrow (x^{2} - 3) (x^{2} - 2) = 0 \Rightarrow x = \pm \sqrt[]{3}, \pm \sqrt[]{2} \\ ∴ F = {- \sqrt[]{3}, - \sqrt[]{2}, \sqrt[]{2}, \sqrt[]{3}} \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter One

Write the following sets in the roaster form

(i) A = {x: x ∈ R, 2x + 11 = 15}

(ii) B = {x | x2 = x, x ∈ R}

(iii) C = {x | x is a positive factor of a prime number p}

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} (i) W e h a v e, A = {x : x \in R, 2 x + 11 = 15} \\ ∴ 2 x + 11 = 15 \\ \Rightarrow 2 x = 15 - 11 \\ \Rightarrow 2 x = 4 \Rightarrow x = 2 \\ ∴ A = {2} \\ (i i) W e h a v e, B = {x | x^{2} = x, x \in R} \\ ∴ x^{2} = x \\ \Rightarrow x^{2} - x = 0 \\ \Rightarrow x (x - 1) = 0 \\ \Rightarrow x = 0, 1 \\ ∴ B = {0, 1} \\ (i i i) W e h a v e, C = {x | x i s a p o s i t i v e f a c t o r o f p r i m e n u m b e r p .} \\ Since, p o s i t i v e f a c t o r s o f a p r i m e n u m b e r a r e 1 a n d t h e n u m b e r i t s e l f . \\ ∴ C = {1, p} \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

If $A, B$ are square matrices of the same order and $B$ is a skew-symmetric matrix, show that $A^{T} B A$ is skew-symmetric

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

$\begin{array}{l} G i v e n t h a t B i s a s k e w s y m m e t r i c m a t r i x ∴ B^{'} = - B \\ L e t P = A^{'} B A \\ \Rightarrow P^{'} = {(A^{'} B A)}^{'} \\ \Rightarrow = A^{'} B^{'} {(A^{'})}^{'} [{(A^{'} B)}^{'} = B^{'} A^{'}] \\ \Rightarrow = A^{'} (- B) A \\ \Rightarrow = A^{'} B A = - P \\ S o P^{'} = - P \\ H e n c e, A^{'} B A i s a s k e w s y m m e t r i c m a t r i x . \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

If $A$ is a square matrix such that $A^{2} = A$ , show that ${(I + A)}^{3} = 7 A + I$ .

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

$\begin{array}{l} T o s h o w t h a t : {(I + A)}^{3} = 7 A + I \\ L . H . S . {(I + A)}^{3} = I^{3} + A^{3} + 3 I^{2} A + 3 I A^{2} \\ \Rightarrow = I + A^{2} . A + 3 I A + 3 I A^{2} \\ \Rightarrow = I + A . A + 3 I A + 3 I A [? A^{2} = A] \\ \Rightarrow = I + A^{2} + 3 I A + 3 I A \\ \Rightarrow = I + A + 3 I A + 3 I A [? A^{2} = A] \\ \Rightarrow = I + A + 3 A + 3 A \Rightarrow 7 A + I R . H . S . \\ L . H . S . = R . H . S . H e n c e, p r o v e d . \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

If $P (x) = [\begin{matrix} c o s x & s i n x \\ - s i n x & c o s x \end{matrix}]$ , then show that $P (x) P (y) = P (x + y) = P (y) P (x)$ .

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

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

If the matrix $[\begin{matrix} 0 & a & 3 \\ 2 & b & - 1 \\ c & 1 & 0 \end{matrix}]$ is a skew-symmetric matrix, find the values of a, b, and c.

6 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

$\begin{array}{l} L e t A = [\begin{matrix} 0 & a & 3 \\ 2 & b & - 1 \\ c & 1 & 0 \end{matrix}] A^{'} = [\begin{matrix} 0 & 2 & c \\ a & b & 1 \\ 3 & - 1 & 0 \end{matrix}] \\ F o r s k e w s y m m e t r i c m a t r i x, A^{'} = - A \\ [\begin{matrix} 0 & 2 & c \\ a & b & 1 \\ 3 & - 1 & 0 \end{matrix}] = - [\begin{matrix} 0 & a & 3 \\ 2 & b & - 1 \\ c & 1 & 0 \end{matrix}] \\ \Rightarrow [\begin{matrix} 0 & 2 & c \\ a & b & 1 \\ 3 & - 1 & 0 \end{matrix}] = [\begin{matrix} 0 & - a & - 3 \\ - 2 & - b & 1 \\ - c & - 1 & 0 \end{matrix}] \\ E q u a t i n g t h e c o r r e s p o n d i n g e l e m e n t s, w e g e t \\ a = - 2, b = - b \Rightarrow 2 b = 0 \Rightarrow b = 0 a n d c = - 3 \\ H e n c e, a = - 2, b = 0 a n d c = - 3 . \end{array}$

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