# CAT 2018 Quant Questions

Question:
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

Let f(x) = $2{x^2} - ax + 2$. We can see that f(x) is a quadratic function.

For, f(x) > 0, Discriminant (D) < 0

$\Rightarrow$ ${( - a)^2} - 4*2*2 < 0$

$\Rightarrow$ (a-4)(a+4)<0

$\Rightarrow$ a  (-4, 4)

Therefore, integer values that 'a' can take = {-3, -2, -1, 0, 1, 2, 3}

Let g(x) = ${x^2} - bx + 8$. We can see that g(x) is also a quadratic function.

For, g(x)≥0, Discriminant (D) $\le$ 0

$\Rightarrow$ ${( - b)^2} - 4*8*1 < 0$

$\Rightarrow$ $(b - \sqrt {32} )(b + \sqrt {32} ) < 0$

$\Rightarrow$ b  (-$\sqrt {32}$, $\sqrt {32}$)

Therefore, integer values that 'b' can take = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}

We have to find out the largest possible value of $2a−6b$. The largest possible value will occur when 'a' is maximum and 'b' is minimum.

a$_{max}$ = 3, b$_{min}$ = -5

Therefore, the largest possible value of $2a−6b$ = 2*3 - 6*(-5) = 36.

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