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# CAT 2018 Question Paper | Quants Slot 2

###### CAT Previous Year Paper | CAT Inequalities Questions | Question 23

This is a question from the topic: Quadratic Equations. People generally take this topic for granted, however, this question is here to prove you otherwise. If you know the concept of completion of squares, then this is a child's play for you! Try this delightful question which appeared in the CAT 2018 Question Paper, if you aren't able to solve, the detailed video solution is always there to help you. Solve more such questions by clicking here: CAT Question Bank

Question 23 : 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 [TITA]

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##### Method of solving this CAT Question from Inequalities

Given that a and b are integers such that 2x2 − ax + 2 > 0 and x2 − bx + 8 ≥ 0 for all real numbers x
We have to find the largest possible value of 2a – 6b
Using the completion of squares method
2x2 − ax + 2 > 0 can be written as x2 - $$frac{a}{2}$x + 1 > 0 We know that$x – k )2 = x2 – 2kx + k2
Here we have $$frac{a}{2}$ = 2k or k = $\frac{a}{4}$ This $\frac{a^2}{16}$ is added on both sides ⟹ x2 - $\frac{a}{2}$x + $\frac{a^2}{16}$ > $\frac{a^2}{16}$ - 1 ⟹$x - $$frac{a}{4}$)2 > $\frac{a^2}{16}$ - 1 [ this $\frac{a^2}{16}$ - 1 always holds good if it is negative or zero] $\frac{a^2}{16}$ - 1 < 0 a2 < 16 a ∈$−4 , 4)
Since we have to find the largest value as possible and it cannot be 4 so a = 3
Similarly b can be found in the following way
x2 − bx + 8 ≥ 0
x2 − bx + $$frac{b^2}{4}$ ≥ $\frac{b^2}{4}$ - 8 x2 − bx + $\frac{b^2}{4}$ ≥ $\frac{b^2}{4}$ - 8 [$\frac{b^2}{4}$ =$$$frac{b}{2}$)2]$x - $$frac{b}{2}$)2 ≥ $\frac{b^2}{4}$ - $\frac{32}{4}$ b2 ≤ 32 b2 can go from -5 , -4 , -3 , -2 , -1 , 0 , 1 , 2 , 3 , 4 , 5 Since we have to find the largest possible value of 2a – 6b or b have to be very small so the smallest integer b can take -5 ⟹ 2$3) - 6(-5)
⟹ 6 + 30 = 36
The largest possible value of 2a − 6b is 36

The question is "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 [TITA]"

##### Hence, the answer is 36

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