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CAT 2019 Slot 2 — QA Question 3

Number TheoryEasy

Let a, b, x, y be real numbers such that a2 + b2 = 25 , x2 + y2 = 169 and ax + by = 65. If k = ay - bx, then

Answer & solution

Correct answer: k = 0

  • k = 0

  • B

    k > 513

  • C

    k = 513

  • D

    0 < k ≤ 513

Solution

Easy

Notice (a2+b2)(x2+y2)=25×169=652=(ax+by)2(a^2+b^2)(x^2+y^2)=25\times169=65^2=(ax+by)^2. This is the equality case of the Lagrange / Cauchy–Schwarz identity, which forces the cross term ay−bxay-bx to vanish.

1

Multiply the two sum-of-squares. Use a2+b2=25a^2+b^2=25 and x2+y2=169x^2+y^2=169.

(a2+b2)(x2+y2)=25×169=4225=652\begin{aligned} &(a^{2}+b^{2})(x^{2}+y^{2})=25\times 169=4225=65^{2} \end{aligned}
2

Apply the Lagrange identity. The product of two sums of squares splits exactly as

(a2+b2)(x2+y2)=(ax+by)2+(ay−bx)2\begin{aligned} &(a^{2}+b^{2})(x^{2}+y^{2})=(ax+by)^{2}+(ay-bx)^{2} \end{aligned}
3

Substitute the known values. The left side equals 65265^2 (step 1) and ax+by=65ax+by=65 is given, so the first square already accounts for everything.

652=652+(ay−bx)2⇒ (ay−bx)2=0⇒ k=ay−bx=0\begin{aligned} &65^{2}=65^{2}+(ay-bx)^{2}\\ &\Rightarrow\ (ay-bx)^{2}=0\\ &\Rightarrow\ k=ay-bx=0 \end{aligned}
k=0k=0

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CAT 2019 Slot 2 QA Q3: Let a, b, x, y be real numbers such that a 2 + b 2 = 25 , x 2 + y 2 = 169 and ax + by = 65. If k = ay - bx, th — Solution | TheCATExam