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If minimum value of f(x) = x2 + 2bx + 2c2 is greater than maximum value of g(x) = - x2 - 2cx + b2 then for x is real
  • a)
    ∣c∣ > ∣b∣ √2
  • b)
    ∣c∣ √2 > b
  • c)
    0 < c < √2b
  • d)
    No real value of a
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
If minimum value of f(x) = x2 + 2bx + 2c2 is greater than maximum valu...
We first find the minimum value of f(x) by completing the square:

f(x) = x^2 - 2bx + 2c^2
= (x - b)^2 + 2c^2 - b^2

The minimum value occurs when x = b, and is equal to 2c^2 - b^2.

Similarly, we find the maximum value of g(x) by completing the square:

g(x) = -x^2 - 2cx + b^2
= -(x + c)^2 + b^2 + c^2

The maximum value occurs when x = -c, and is equal to b^2 + c^2.

Thus, we need to have:

2c^2 - b^2 > b^2 + c^2

Simplifying, we get:

c^2 > b^2/3

Since b and c are both real, this means that c cannot be equal to zero. Therefore, the answer is (a) c ≠ 0.
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Community Answer
If minimum value of f(x) = x2 + 2bx + 2c2 is greater than maximum valu...
Since f(x) is curve mouth opening upward it's minimum value will be -D/4a=2c^2-b^2
g(x) is curve mouth opening downward it's maximum value will be -D/4a=b^2+c^2
now take inequality u will get the answer.
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If minimum value of f(x) = x2 + 2bx + 2c2 is greater than maximum value of g(x) = - x2 - 2cx + b2 then for x is reala)∣c∣ > ∣b∣ √2b)∣c∣ √2 > bc)0 < c < √2bd)No real value of aCorrect answer is option 'A'. Can you explain this answer?
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