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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
Correct answer is '36'. Can you explain this answer?
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If a and b are integers such that 2x2- ax + 2 > 0 and x2 - bx + 8 &...
Let f(x) = 2x2 - ax + 2. We can see that f(x) is a quadratic function.
For, f(x) > 0, Discriminant (D) < />
=> (-a)2 – 4*2*2< />
=> (a-4)(a+4) < />
=> a ε (-4,4)
Therefore, integer values that 'a' can take = {-3, -2, -1,0,1,2,3}
Let g(x) = x2- bx + 8. We can see that g(x) is also a quadratic function.
For, g(x)≥0, Discriminant (D) ≤ 0
=> (-b)2– 4*8* 1 < />
=>(b - √32 ) (b + √32 ) < />
=> b ε(-√32 ,√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.
amax = 3, bmin = -5
Therefore, the largest possible value of 2a - 6b – 2*3 – 6*(-5) = 36.
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If a and b are integers such that 2x2- ax + 2 > 0 and x2 - bx + 8 &...
It seems like the equation you provided is incomplete. Could you please provide the full equation so that I can assist you further?
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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 isCorrect answer is '36'. Can you explain this answer?
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