JEE Main 2015MathematicsApplication of DerivativesMediumMCQ

JEE Main 2015 — Application of Derivatives Question with Solution

JEE Main 2015 (11 Apr Online)

Question

Let k and K be the minimum and the maximum values of the function fx=1+x0.61+x0.6 in 0, 1, respectively, then the ordered pair (k, K) is equal to:

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Show full solutionCorrect option: A
Correct answer
A2-0.4,1

Step-by-step explanation

fx= 1+x0.61+x0.6

f′x= 0.61+x-0.4 1+x0.6-0.6 x0.41+x0.61+x0.62

=0.6 1+x-0.4 1+x0.6-x-0.4 1+x21+x0.62=0.6x0.4-11+x0.41+x0.62

Clearly it is always negative in [0, 1]

minimum value would be at x=1.

(1+1)0.61+10.6= 20.62=2-0.4=k

& maximum values would be at x=0

(1+0)0.61+00.6=11=K

Hence 2-0.4,1

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About this question

This is a previous-year question from JEE Main 2015, covering the Application of Derivatives chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.