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Question
y = x(x – 3)2 decreases for the values of x given by : ______.
Options
1 < x < 3
x < 0
x > 0
`0 < x < 3/2`
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Solution
y = x(x – 3)2 decreases for the values of x given by : 1 < x < 3.
Explanation:
Here y = x(x – 3)2
`"dy"/"dx" = x * 2(x - 3) + (x - 3)^2 * 1`
⇒ `"dy"/"dx" = 2x(x - 3) + (x - 3)^2`
For increasing and decreasing `"dy"/"dx"` = 0
∴ 2x(x – 3) + (x – 3)2 = 0
⇒ (x – 3)(2x + x – 3) = 0
⇒ (x – 3)(3x – 3) = 0
⇒ 3(x – 3)(x – 1) = 0
∴ x = 1, 3
∴ Possible intervals are `(– oo, 1), (1, 3), (3, oo)`
`"dy"/"dx"` = (x – 3)(x – 1)
For `(– oo, 1)` = (–) (–) = (+) increasing
For (1, 3) = (–) (+) = (–) decreasing
For `(3, oo)` = (+) (+) = (+) increasing
So the function decreases in (1, 3) or 1 < x < 3
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