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= (1-2)x' D X where D = 2(c-yd), X = 2λ(yx + y), (1-2)=C+Y where C = 2(d- yc), Y = 2λ(x + YY), where x' = dx ds' = dy ds x²+Y2-2yXY - 4c(1-2)x + 4d(1-2)Y + C² + D² + 2xCD = (1 − y²)². - C² + 2xCD + D² = (1-y²)² and C = −λR. Deduce that if y² > 1, then there are always two real solutions of these equations, and if y² < 1, then there are two real solutions if AR<√√12 and none if AR > √1 – y².

= (1-2)x' D X where D = 2(c-yd), X = 2λ(yx + y), (1-2)=C+Y where C = 2(d- yc), Y = 2λ(x + YY), where x' = dx ds' = dy ds x²+Y2-2yXY - 4c(1-2)x + 4d(1-2)Y + C² + D² + 2xCD = (1 − y²)². - C² + 2xCD + D² = (1-y²)² and C = −λR. Deduce that if y² > 1, then there are always two real solutions of these equations, and if y² < 1, then there are two real solutions if AR<√√12 and none if AR > √1 – y².

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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= (1-2)x' D X where D = 2(c-yd), X = 2λ(yx + y), (1-2)=C+Y where C = 2(d- yc), Y = 2λ(x + YY), where x' = dx ds' = dy ds x²+Y2-2yXY - 4c(1-2)x + 4d(1-2)Y + C² + D² + 2xCD = (1 − y²)². - C² + 2xCD + D² = (1-y²)² and C = −λR. Deduce that if y² > 1, then there are always two real solutions of these equations, and if y² < 1, then there are two real solutions if AR<√√12 and none if AR > √1 – y².
Transcribed Image Text:= (1-2)x' D X where D = 2(c-yd), X = 2λ(yx + y), (1-2)=C+Y where C = 2(d- yc), Y = 2λ(x + YY), where x' = dx ds' = dy ds x²+Y2-2yXY - 4c(1-2)x + 4d(1-2)Y + C² + D² + 2xCD = (1 − y²)². - C² + 2xCD + D² = (1-y²)² and C = −λR. Deduce that if y² > 1, then there are always two real solutions of these equations, and if y² < 1, then there are two real solutions if AR<√√12 and none if AR > √1 – y².
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