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Question 23 : If a and b are integers such that 2x^{2} − ax + 2 ＞ 0 and x^{2} − bx + 8 ≥ 0 for all real numbers x, then the largest possible value of 2a − 6b is [TITA]

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Given that a and b are integers such that 2x^{2} − ax + 2 > 0 and x^{2} − 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

2x^{2} − ax + 2 > 0 can be written as x^{2} - \\frac{a}{2})x + 1 > 0

We know that (x – k )^{2} = x^{2} – 2kx + k^{2}

Here we have \\frac{a}{2}) = 2k or k = \\frac{a}{4})

This \\frac{a^2}{16}) is added on both sides

⟹ x^{2} - \\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

a^{2} < 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

x^{2} − bx + 8 ≥ 0

x^{2} − bx + \\frac{b^2}{4}) ≥ \\frac{b^2}{4}) - 8

x^{2} − 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})

b^{2} ≤ 32

b^{2} 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 2x ^{2} − ax + 2 ＞ 0 and x^{2} − bx + 8 ≥ 0 for all real numbers x, then the largest possible value of 2a − 6b is [TITA]"**

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