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# CAT 2018 Question Paper | Quants Slot 2

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

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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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