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JEE Advanced Mathematics Application of Derivatives 2021 JEE Advanced 2021 (Paper 2)

JEE Advanced Mathematics Question (2021) — Solution

Question

Let f_ 1 :(0, ) R and f_ 2 :(0, ) R be defined by f_ 1 (x)= _ 0 ^ x _ j=1 ^ 21 (t-j)^ j d t, x > 0 and f_ 2 (x)=98(x-1)^ 50 -600(x-1)^ 49 +2450, x > 0 where, for any positive integer n and real numbers a_ 1 , a_ 2 , , a_ n , _ i=1 ^ n a_ i denotes the product of a_ 1 , a_ 2 , , a_ n . Let m_ i and n_ i , respectively, denote the number of points of local minima and the number of points of local maxima of function f_ i , i=1,2, in the interval (0, ). The value of  6 m 2 + 4 n 2 + 8 m 2 n 2  is ____.

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

f 2 x = 98 x - 1 50 - 600 x - 1 49 + 2450 ,    x > 0 f 2 ' x = 98 × 50 x − 1 49 − 600 × 49 x − 1 48 = 4900 x − 1 48 x − 1 − 6 = 4900 x − 1 48 x − 7 So extrema is at  x = 7  only, which is minima m 2 = 1 , n 2 = 0 Hence  6 m 2 + 4 n 2 + 8 m 2 n 2 = 6

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