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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∈ ?

$${i}\:{found}\:{some}\:{interesting}\:{basic}\:{question} \\ $$$${hence}\:{sharing}... \\ $$$$\left.\mathrm{1}\right){if}\:{A}\in\left[\mathrm{1},\mathrm{4}\right]\:\:{A}^{\mathrm{2}} \:\in\:\:?\:\leftarrow{find}\:{interval}\: \\ $$$$\left.\mathrm{2}\right){if}\:{A}\:\in\:\left[−\mathrm{1},\mathrm{4}\right]\:\:{A}^{\mathrm{2}} \:\in\:? \\ $$$$\left.\mathrm{3}\right)\:{y}=\frac{\mathrm{1}}{{A}\:\:}\:\:{and}\:{A}\in\:\:\:\:\left[\mathrm{1},\mathrm{4}\right]\:\:{y}\in\:? \\ $$$$\left.\mathrm{4}\right){y}=\frac{\mathrm{1}}{\mid{A}\mid}\:\:{A}\in\left[−\mathrm{1},\mathrm{4}\right]\:\:\:{y}\in\:? \\ $$

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)

$$\mathrm{1}.\:\:{A}^{\mathrm{2}} \in\left[\mathrm{1},\mathrm{16}\right] \\ $$$$\mathrm{2}.\:\:{A}^{\mathrm{2}} \in\left[\mathrm{0},\mathrm{16}\right] \\ $$$$\mathrm{3}.\:\:\mathrm{1}\leqslant{A}\leqslant\mathrm{4}\Rightarrow\mathrm{1}\geqslant\frac{\mathrm{1}}{{A}}\geqslant\frac{\mathrm{1}}{\mathrm{4}}\Rightarrow\frac{\mathrm{1}}{\mathrm{4}}\leqslant{y}\leqslant\mathrm{1} \\ $$$$\mathrm{4}.\:\:{if}\:{A}\neq\mathrm{0} \\ $$$$−\mathrm{1}\leqslant{A}<\mathrm{0}\Rightarrow\mathrm{0}<\mid{A}\mid\leqslant\mathrm{1}\Rightarrow\frac{\mathrm{1}}{\mid{A}\mid}\geqslant\mathrm{1} \\ $$$$\mathrm{0}<{A}\leqslant\mathrm{4}\Rightarrow\frac{\mathrm{1}}{\mid{A}\mid}\geqslant\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow{y}\geqslant\frac{\mathrm{1}}{\mathrm{4}} \\ $$

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]

$$\left.\mathrm{1}\right){let}\:{A}={x}\:\:{and}\:\:{y}={x}^{\mathrm{2}} \\ $$$${A}\in\left[\mathrm{1},\mathrm{4}\right]\:\:\:{so}\:{A}^{\mathrm{2}} \in\left[\mathrm{1},\mathrm{16}\right] \\ $$$$\left.\mathrm{2}\right){A}\in\left[−\mathrm{1},\mathrm{4}\right]\:\:{from}\:{graph}\:{y}={x}^{\mathrm{2}} \: \\ $$$${minimum}\:{value}\:{of}\:{x}^{\mathrm{2}} =\mathrm{0} \\ $$$${so}\:{A}^{\mathrm{2}} \in\left[\mathrm{0},\mathrm{16}\right] \\ $$

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]

$$\left.\mathrm{3}\right){let}\:{x}={A}\:\:{and}\:{y}=\frac{\mathrm{1}}{{A}} \\ $$$${A}\in\left[\mathrm{1},\mathrm{4}\right]\:\:\frac{\mathrm{1}}{{A}}\in\left[\mathrm{0}.\mathrm{25},\mathrm{1}\right] \\ $$$$ \\ $$

Commented by tanmay last updated on 22/May/19

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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),∞)

$$\left.\mathrm{4}\right)\:{A}={x}\:\:\:{y}=\frac{\mathrm{1}}{\mid{x}\mid} \\ $$$${when}\:{x}=−\mathrm{1}\:\:{y}=\frac{\mathrm{1}}{\mid−\mathrm{1}\mid}=\mathrm{1} \\ $$$${when}\:{x}=\mathrm{4}\:\:{y}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${A}\in\left[−\mathrm{1},\mathrm{4}\right] \\ $$$$\frac{\mathrm{1}}{\left[{A}\right]}\in\:\left[\mathrm{1},\infty\right)\:\cap\:\left[\frac{\mathrm{1}}{\mathrm{4}},\infty\right) \\ $$

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

$$\left[\mathrm{1},\:\infty\right)\cap\left[\frac{\mathrm{1}}{\mathrm{4}},\:\infty\right)\:=\:\left[\mathrm{1},\:\infty\right) \\ $$$$\mathrm{so}\:\mathrm{your}\:\mathrm{conclusion}\:\mathrm{is}\:\mathrm{wrong}.\:\mathrm{the}\:\mathrm{path}\:\mathrm{is}\:\mathrm{ok}. \\ $$$$\frac{\mathrm{1}}{\mid{A}\mid}\in\left[\frac{\mathrm{1}}{\mathrm{4}},\:\infty\right)\:\mathrm{is}\:\mathrm{the}\:\mathrm{right}\:\mathrm{answer} \\ $$

Commented by tanmay last updated on 22/May/19

thank you sir...

$${thank}\:{you}\:{sir}... \\ $$

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