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If X = {1, 2, 3}, if n represents any member of X, write the following sets containing all numbers represented by
(i) 4n
(ii) n + 6
(iii) n2
(iv) n – 1
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W e a r e g i v e n t h a t : X = { 1 , 2 , 3 } ( i ) { 4 n | n ∈ X } = { 4 , 8 , 1 2 } ( i i ) { n + 6 | n ∈ X } = { 7 , 8 , 9 } ( i i i ) { n 2 | n ∈ X } = { 1 2 , 1 , 3 2 } ( i v ) { ( n − 1 ) | n ∈ X } = { 0 , 1 , 2 }
Given that N = {1, 2, 3, ., 100}. Then write
(i) The subset of N whose elements are even numbers.
(ii) The subset of N whose element are perfect square numbers.
W e a r e g i v e n t h a t : N = { 1 , 2 , 3 , 4 , 5 , … , 1 0 0 } (i)Required subset whose elements are even = { 2 , 4 , 6 , 8 , … , 1 0 0 } (ii)Required subset whose elements are perfect squares = { 1 , 4 , 9 , 1 6 , 2 5 , 3 6 , … , 1 0 0 }
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
( i ) G i v e n t h a t : A ⊂ U a n d B ⊂ U L e t x ∈ A o r x ∈ B ⇒ x ∈ A ∪ B H e n c e , A ⊂ ( A ∪ B ) ( i i ) I f A ⊂ B T h e n l e t x ∈ A ∪ B ⇒ x ∈ A o r x ∈ B ⇒ A ∪ B ⊂ B … ( i ) B u t B ⊂ A ∪ B … ( i i ) F r o m e q . ( i ) a n d ( i i ) , w e g e t A ∪ B = B L e t y ∈ A ⇒ y ∈ ( A ∪ B ) ⇒ y ∈ B ⇒ y ∈ B ⇔ A ∪ B = B ( i i i ) L e t x ∈ A ∩ B ⇒ x ∈ A a n d x ∈ B ⇒ x ∈ A S o , A ∩ B ⊂ A
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)
G i v e n , L = { 1 , 2 , 3 , 4 } , M = { 3 , 4 , 5 , 6 } a n d N = { 1 , 3 , 5 } ∴ M ∪ N = { 1 , 3 , 4 , 5 , 6 } L − M ∪ N = { 2 } N o w , L − M = { 1 , 2 } , L − N = { 2 , 4 } ∴ ( L − M ) ∩ ( L − N ) = { 2 } H e n c e , L − M ∪ N = ( L − M ) ∩ ( L − N ) .
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}.
( i ) G i v e n t h a t : 3 5 ∈ { 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 3 5 a r e 1 , 5 , 7 , 3 5 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 : 1 2 8 ∈ { 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 1 2 8 a r e 1 , 2 , 4 , 8 , 1 6 , 3 2 , 6 4 a n d 1 2 8 . S u m o f a l l f a c t o r s = 1 + 2 + 4 + 8 + 1 6 + 3 2 + 6 4 + 1 2 8 = 2 5 5 ≠ 2 * 1 2 8 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 ∈ { x | x 4 − 5 x 3 + 2 x 2 − 1 1 2 x + 6 = 0 } ∴ x 4 − 5 x 3 + 2 x 2 − 1 1 2 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 − 1 1 2 ( 3 ) + 6 ⇒ 8 1 − 1 3 5 + 1 8 − 3 3 6 + 6 ⇒ − 3 6 6 ≠ 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 : 4 9 6 ∉ { 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 4 9 6 a r e 1 , 2 , 4 , 8 , 1 6 , 3 1 , 6 2 , 1 2 4 , 2 4 8 a n d 4 9 6 . ∴ 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 + 1 6 + 3 1 + 6 2 + 1 2 4 + 2 4 8 + 4 9 6 = 9 9 2 = 2 * 4 9 6 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 ' .
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.
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, the factors of 2p−1 are 1,2,22,23,…,2p−1 and factors of 2p−1 are 1 and 2p−1 ∴ Y = { 1 , 2 , 2 2 , 2 3 , … , 2 p − 1 , 2 p − 1 } = 2 ( 2 p − 1 ) , 2 2 ( 2 p − 1 ) , … , 2 p − 1 ( 2 p − 1 )
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}
(i)We have, D={t|t3=t,t∈R}∴ t3=t⇒ t3−t=0 ⇒t(t2−1)=0⇒t(t−1)(t+1)=0 ⇒ t=0,1,−1∴ D={−1,0,1}(ii)We have, E={w|w−2w+3=3, w∈R}∴ w−2w+3=3⇒ w−2=3w+9 ⇒w−3w=9+2⇒ −2w=11 ⇒w=−112∴ E={−112}(iii)We have, F={x|x4−5x2+6=0,x∈R}∴ x4−5x2+6=0⇒ x4−3x2−2x2+6=0 ⇒x2(x2−3)−2(x2−3)=0⇒ (x2−3)(x2−2)=0 ⇒x=±3,±2∴ F={−3,−2,2,3}
(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}
( i ) W e h a v e , A = { x : x ∈ R , 2 x + 1 1 = 1 5 } ∴ 2 x + 1 1 = 1 5 ⇒ 2 x = 1 5 − 1 1 ⇒ 2 x = 4 ⇒ x = 2 ∴ A = { 2 } ( i i ) W e h a v e , B = { x | x 2 = x , x ∈ R } ∴ x 2 = x ⇒ x 2 − x = 0 ⇒ x ( x − 1 ) = 0 ⇒ 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, positive factors of a prime number are 1 and the number itself. ∴ C = { 1 , p }
What effect does a catalyst have on the equilibrium position of a reaction?(a) A catalyst favours the formation of products(b) A catalyst favours the formation of reactants(c) A catalyst does not change the equilibrium position of a reaction(d) A catalyst may favour reactants or product formation, depending upon the direction in which the reaction is written.
Answer: (a) A catalyst favours the formation of products
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