Characters 0/140
The Answer must contain atleast 20 characters.
Characters 0/300
Edit
Get insights from 64 questions on Maths NCERT Exemplar Solutions Class 11th Chapter One, answered by students, alumni, and experts. You may also ask and answer any question you like about Maths NCERT Exemplar Solutions Class 11th Chapter One
New answer posted
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}.
Characters 0/2500
Contributor-Level 10
( 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 }
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else
On Shiksha, get access to
College Comparison
This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.
