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CAT 2018 Slot 2 — QA Question 27

Discriminant and Roots of Quadratic EquationEasy

If a and b are integers such that 2x2 − ax + 2 > 0 and x2 − bx + 8 ≥ 0 for all real numbers x, then the largest possible value of 2a − 6b is

Answer & solution

Correct answer: 36

Answer: 36

Solution

Easy

A quadratic with positive leading coefficient stays positive for all xx exactly when its discriminant is negative (or ≤0\le 0 for “≥0\ge 0”). Apply this to each quadratic to bound the integers aa and bb, then maximize 2a−6b2a-6b by taking aa as large and bb as small as allowed.

1

Bound aa from the first quadratic. 2x2−ax+2>02x^2-ax+2>0 for all xx needs discriminant <0<0.

(−a)2−4⋅2⋅2<0⇒ a2−16<0⇒ −4<a<4(a integer)⇒ amax⁡=3\begin{aligned} &(-a)^2 - 4\cdot 2\cdot 2 < 0\\ &\Rightarrow\ a^2 - 16 < 0\\ &\Rightarrow\ -4 < a < 4 \quad\text{(}a\text{ integer)}\\ &\Rightarrow\ a_{\max} = 3 \end{aligned}
2

Bound bb from the second quadratic. x2−bx+8≥0x^2-bx+8\ge 0 for all xx needs discriminant ≤0\le 0.

(−b)2−4⋅1⋅8≤0⇒ b2−32≤0⇒ −42≤b≤42(≈−5.66≤b≤5.66)⇒ bmin⁡=−5(b integer)\begin{aligned} &(-b)^2 - 4\cdot 1\cdot 8 \le 0\\ &\Rightarrow\ b^2 - 32 \le 0\\ &\Rightarrow\ -4\sqrt2 \le b \le 4\sqrt2 \quad\text{(}\approx -5.66\le b\le 5.66\text{)}\\ &\Rightarrow\ b_{\min} = -5 \quad\text{(}b\text{ integer)} \end{aligned}
3

Maximize 2a−6b2a-6b. Use a=3a=3 (from step 1) and b=−5b=-5 (from step 2); the −6b-6b term is largest when bb is smallest.

2a−6b=2(3)−6(−5)⇒ =6+30=36\begin{aligned} &2a-6b = 2(3) - 6(-5)\\ &\Rightarrow\ = 6 + 30 = 36 \end{aligned}
36\textbf{36}

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CAT 2018 Slot 2 QA Q27 — Discriminant and Roots of Quadratic Equation solution | TheCATExam