If m is the minimum value of k for which the function f x = x k x - x 2   is increasing in the interval…

If m is the minimum value of k for which the function fx=xkx-x2  is increasing in the interval [0, 3] and M is the maximum value of f in [0, 3] when k=m, then the ordered pair (m, M) is equal to:
  1. 4, 33
  2. 5, 36
  3. 3, 33
  4. 4, 32

Solution

fx=xkx-x2 ⇒ f'x=3kx-4x22kx-x2
As per the given condition f'x≥0 for x∈0, 3
⇒3kx-4x2≥0 for x∈0, 3
⇒3k-4x ≥0 for x∈0, 3
⇒k≥4x3 for x∈0, 3
⇒k≥4 . So minimum value of k is m=4.
Now fx=x4x-x2
Since given function in increasing hence maximum value will occur at x=3
⇒f(3)=34×3-32=33, M=33
Hence (m, M) = (4, 33)

Asked in: JEE Main 2019 (12 Apr Shift 1)

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