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Y = x(x – 3)2 decreases for the values of x given by : ______.

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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`

MCQ
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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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Chapter 6: Application Of Derivatives - Exercise [Page 140]

APPEARS IN

NCERT Exemplar Mathematics Exemplar [English] Class 12
Chapter 6 Application Of Derivatives
Exercise | Q 48 | Page 140

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