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CAT 2018 Slot 2 — QA Question 24

Inequality Maximization / MinimizationEasy

If the sum of squares of two numbers is 97, then which one of the following cannot be their product?

Answer & solution

Correct answer: 64

  • A

    16

  • B

    48

  • C

    –32

  • 64

Solution

Easy

For any two reals with a fixed sum of squares, the product is bounded. Use the AM–GM inequality (equivalently the fact that (a±b)2≥0(a\pm b)^2\ge 0) to find the range of possible products, then spot the option lying outside it.

1

Bound the product. From (a−b)2≥0(a-b)^2\ge 0 and (a+b)2≥0(a+b)^2\ge 0 we get a2+b2≥2aba^2+b^2\ge 2ab and a2+b2≥−2aba^2+b^2\ge -2ab, i.e. a2+b2≥2∣ab∣a^2+b^2\ge 2|ab|.

a2+b2=97⇒ 97≥2∣ab∣(since a2+b2≥2∣ab∣)⇒ ∣ab∣≤48.5\begin{aligned} &a^2+b^2 = 97\\ &\Rightarrow\ 97 \ge 2|ab| \quad\text{(since }a^2+b^2\ge 2|ab|\text{)}\\ &\Rightarrow\ |ab| \le 48.5 \end{aligned}
2

Range of the product. The product abab can be anything from −48.5-48.5 up to 48.548.5.

−48.5≤ab≤48.5\begin{aligned} &-48.5 \le ab \le 48.5 \end{aligned}
3

Test the options. 16, 48, −3216,\ 48,\ -32 all lie within [−48.5, 48.5][-48.5,\,48.5], so each is achievable. But 64>48.564>48.5, so a product of 6464 is impossible.

64>48.5(outside the allowed range)\begin{aligned} &64 > 48.5 \quad\text{(outside the allowed range)} \end{aligned}
ab≠64⇒64ab \ne 64 \quad\Rightarrow\quad \textbf{64}

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CAT 2018 Slot 2 QA Q24 — Inequality Maximization / Minimization solution | TheCATExam