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
Correct Answer: 36
Let f(x) = $$2x^2−ax+2$$. 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 $$\epsilon$$ (-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) $$\leq$$ 0
$$\Rightarrow$$ $$(-b)^2-4*8*1<0$$
$$\Rightarrow$$ $$(b-\sqrt{32})(b+\sqrt{32})<0$$
$$\Rightarrow$$ b $$\epsilon$$ (-$$\sqrt{32}$$, $$\sqrt{32}$$)
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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