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Question Number 60514 by tanmay last updated on 21/May/19

i found some interesting basic question hence sharing... 1)if A∈[1,4] A^2 ∈ ? ←find interval 2)if A ∈ [−1,4] A^2 ∈ ? 3) y=(1/(A )) and A∈ [1,4] y∈ ? 4)y=(1/(∣A∣)) A∈[−1,4] y∈ ?

ifoundsomeinterestingbasicquestionhencesharing...1)ifA∈[1,4]A2∈?←findinterval2)ifA∈[−1,4]A2∈?3)y=1AandA∈[1,4]y∈?4)y=1∣A∣A∈[−1,4]y∈?

Commented by kaivan.ahmadi last updated on 22/May/19

1. A^2 ∈[1,16] 2. A^2 ∈[0,16] 3. 1≤A≤4⇒1≥(1/A)≥(1/4)⇒(1/4)≤y≤1 4. if A≠0 −1≤A<0⇒0<∣A∣≤1⇒(1/(∣A∣))≥1 0<A≤4⇒(1/(∣A∣))≥(1/4) ⇒y≥(1/4)

1.A2∈[1,16]2.A2∈[0,16]3.1⩽A⩽4⇒1⩾1A⩾14⇒14⩽y⩽14.ifA≠0−1⩽A<0⇒0<∣A∣⩽1⇒1∣A∣⩾10<A⩽4⇒1∣A∣⩾14⇒y⩾14

Commented by tanmay last updated on 22/May/19

1)let A=x and y=x^2 A∈[1,4] so A^2 ∈[1,16] 2)A∈[−1,4] from graph y=x^2 minimum value of x^2 =0 so A^2 ∈[0,16]

1)letA=xandy=x2A∈[1,4]soA2∈[1,16]2)A∈[−1,4]fromgraphy=x2minimumvalueofx2=0soA2∈[0,16]

Commented by tanmay last updated on 22/May/19

Commented by tanmay last updated on 22/May/19

3)let x=A and y=(1/A) A∈[1,4] (1/A)∈[0.25,1]

3)letx=Aandy=1AA∈[1,4]1A∈[0.25,1]

Commented by tanmay last updated on 22/May/19

thank you sir

thankyousir

Answered by tanmay last updated on 22/May/19

4) A=x y=(1/(∣x∣)) when x=−1 y=(1/(∣−1∣))=1 when x=4 y=(1/4) A∈[−1,4] (1/([A]))∈ [1,∞) ∩ [(1/4),∞)

4)A=xy=1∣x∣whenx=−1y=1∣−1∣=1whenx=4y=14A∈[−1,4]1[A]∈[1,∞)∩[14,∞)

Commented by tanmay last updated on 22/May/19

Commented by MJS last updated on 22/May/19

[1, ∞)∩[(1/4), ∞) = [1, ∞) so your conclusion is wrong. the path is ok. (1/(∣A∣))∈[(1/4), ∞) is the right answer

[1,∞)∩[14,∞)=[1,∞)soyourconclusioniswrong.thepathisok.1∣A∣∈[14,∞)istherightanswer

Commented by tanmay last updated on 22/May/19

thank you sir...

thankyousir...

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