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Question

# Let x,y,z,t be real numbers x2+y2=9,z2+t2=4, and xt−yz=6. Then the greatest value of P=xz is

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Solution

## The correct option is B 3Let x,y,z,t,θ,ϕ∈R with x2+y2=9;z2+t2=4;xt−yz=6The following substitutions are valid;x=3cosθ,y=3sinθ,z=2cosϕ,t=2sinϕxt−yz=6⇒(3cosθ)(2sinϕ)−(3sinθ)(2cosϕ)=6⇒6sinϕcosθ−6cosϕsinθ=6⇒sin(ϕ−θ)=1⇒ϕ−θ=2nπ+π2 or (2m+1)π−π2m,n∈zxz=(3cosθ)(2cosϕ) =6cosθcosϕ =6cosθcos(2nπ+π2+θ) =−6cosθsinθ =−3sin2θ(or)xz=6cosθcos((2m+1)π−π2+θ) =−6cosθsinθ =−3sin2θ(xz)max=3 and (xz)min=−3

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