JEE Main 2026 Jan 22 Shift 1, Mathematics Q5: Maxima & Minima

JEE Main2026Jan 22, Shift 1Mathematics

Let $f (x) = x^{2025} - x^{2000}, x \in [0, 1]$ and the minimum value of the function $f (x)$ in the interval $[0, 1]$ be $(80)^{80} (n)^{- 81}$ . Then n is equal to

A
−81
B
−40
C
−41
D
−80

Solution

$f^{'} (x) = 2025 x^{2024} - 2000 x^{1999} = x^{1999} (2025 x^{25} - 2000)$ .

Critical point on $(0, 1)$ : $x^{25} = \frac{2000}{2025} = \frac{80}{81}$ , so $x_{0} = (\frac{80}{81})^{1/25}$ .

$f (0) = 0$ , $f (1) = 0$ . So the minimum (negative value) is at $x_{0}$ .

$f (x_{0}) = x_{0}^{2000} (x_{0}^{25} - 1) = (\frac{80}{81})^{80} (\frac{80}{81} - 1) = (\frac{80}{81})^{80} (- \frac{1}{81}) = \frac{8 0 ^{80} \cdot ( - 1 )}{8 1 ^{81}} = 8 0^{80} \cdot (- 81)^{- 81}$ .

Comparing with $(80)^{80} (n)^{- 81}$ : $n = - 81$ .

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JEE Main 2026 Jan 22 Shift 1, Mathematics Q5: Maxima & Minima

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