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If g(x) = max(y2 − xy) (where 0 ≤ y ≤ 1), then the minimum value of g(x) (for real x) is:
- a)1/4
- b)0
- c)3 + √8
- d)1/2
Correct answer is option 'B'. Can you explain this answer?
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Ifg(x) = max(y2 − xy) (where0 ≤ y ≤ 1), then the minimum v...
Explanation:
1. Finding the Maximum Value:
- To find the maximum value of g(x), we need to maximize the expression y^2 - xy within the given range, 0 ≤ y ≤ 1.
- For a fixed x, the maximum value of y^2 - xy occurs at y = x/2, which lies within the given range.
2. Substitute the Value:
- Substituting y = x/2 in the expression y^2 - xy, we get:
g(x) = (x/2)^2 - x(x/2)
= x^2/4 - x^2/2
= x^2(1/4 - 1/2)
= x^2(-1/4)
= -x^2/4
3. Finding the Minimum Value:
- The minimum value of g(x) will be the negative of the maximum value, as the function is in the form of max(y^2 - xy).
- Therefore, the minimum value of g(x) is -(-x^2/4) = x^2/4.
4. Answer:
- Hence, the minimum value of g(x) is 1/4, which is option 'a'.
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Ifg(x) = max(y2 − xy) (where0 ≤ y ≤ 1), then the minimum v...
Given, g(x) = max(y2 − xy)
Let f(y) = y2 − xy
For maximum of f(y),
f′(y) = 2y − x = 0
⇒ y = x2.
So, maximum can be achieved at either of the points y = 0, x/2 or 1.
Hence, the minimum value of g(x) is 0.
Hence, the minimum value of g(x) is 0.
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