Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
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:
MathematicsApplication of DerivativesJEE MainJEE Main 2019 (12 Apr Shift 1)
Options:
  • A 4, 33
  • B 5, 36
  • C 3, 33
  • D 4, 32
Solution:
2625 Upvotes Verified Answer
The correct answer is: 4, 33
fx=xkx-x2  f'x=3kx-4x22kx-x2
As per the given condition f'x0 for x0, 3
3kx-4x20 for x0, 3
3k-4x 0 for x0, 3
k4x3 for x0, 3
k4 . 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)

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.