If \( x \) and \( y \) are real numbers such that \( x > 1 \) and \( y < -1 \), then which of the following inequalities must be true?

A. \(\frac{x}{y} > 1\)

B. \(|x|^2 > |y|\)

C. \(\frac{x}{3} - 5 > \frac{y}{3} - 5\)

D. \(x^2 + 1 > y^2 + 1\)

E. \(x^{-2} > y^{-2}\)

1\ \wedge\ y < -1\\\\A.\ \frac{x}{y} > 1-FALSE;example:x=2;\ y=-2\ then\\L=\frac{2}{-2}=-1;R=1;\ L < R\\\\B.\ |x|^2 > |y|-FALSE;example:x=2;\ y=-6\ then\\L=|2|^2=2^2=4;R=|-6|=6;\ L < R\\\\C.\ \frac{x}{3}-5 > \frac{y}{3}-5-TRUE;\frac{x}{3}-5 > \frac{y}{3}-5\to\frac{x}{3} > \frac{y}{3}\to x > y"> x > 1 ∧ y < − 1 A . y x ​ > 1 − F A L SE ; e x am pl e : x = 2 ; y = − 2 t h e n L = − 2 2 ​ = − 1 ; R = 1 ; L < R B . ∣ x ∣ 2 > ∣ y ∣ − F A L SE ; e x am pl e : x = 2 ; y = − 6 t h e n L = ∣2 ∣ 2 = 2 2 = 4 ; R = ∣ − 6∣ = 6 ; L < R C . 3 x ​ − 5 > 3 y ​ − 5 − TR U E ; 3 x ​ − 5 > 3 y ​ − 5 → 3 x ​ > 3 y ​ → x > y
y^2+1-FALSE;example:x=2;\ y=-3\ then\\L=2^2+1=4+1=5;R=(-3)^2+1=9+1=10;L < R\\\\E.\ x^{-2} > y^{-2}-FALSE;example:x=3;\ y=-2\ then\\L=3^{-2}=\frac{1}{9};R=(-2)^{-2}=\frac{1}{4};\ L < R"> D . x 2 + 1 > y 2 + 1 − F A L SE ; e x am pl e : x = 2 ; y = − 3 t h e n L = 2 2 + 1 = 4 + 1 = 5 ; R = ( − 3 ) 2 + 1 = 9 + 1 = 10 ; L < R E . x − 2 > y − 2 − F A L SE ; e x am pl e : x = 3 ; y = − 2 t h e n L = 3 − 2 = 9 1 ​ ; R = ( − 2 ) − 2 = 4 1 ​ ; L < R

Answered by Anonymous | 2024-06-24

The correct answer is option C: \frac{y}{3} - 5"> 3 x ​ − 5 > 3 y ​ − 5 , because it simplifies to y"> x > y , which is always true under the given conditions. All other options are false based on specific counterexamples.
;

Answered by Anonymous | 2024-10-09