Applications of Derivatives – Maxima and Minima – JEE Main 22 Jan 2026 Shift 1

Solution:

To find the minimum value of $f(x) = x^{2025} – x^{2000}$ on $[0, 1]$, we first find its critical points by taking the derivative.
$$ f'(x) = 2025 x^{2024} – 2000 x^{1999} $$

Set $f'(x) = 0$:
$$ x^{1999} (2025 x^{25} – 2000) = 0 $$

This gives $x = 0$ or $2025 x^{25} = 2000$.
$$ x^{25} = \frac{2000}{2025} = \frac{80}{81} $$
Let this critical point be $\alpha$. So, $\alpha^{25} = \frac{80}{81}$.
Since $0 < \frac{80}{81} < 1$, the point $\alpha$ lies in the interval $(0, 1)$.
Now we evaluate $f(x)$ at the endpoints and the critical point.
$f(0) = 0$
$f(1) = 1 – 1 = 0$

At $x = \alpha$:
$$ f(\alpha) = \alpha^{2025} – \alpha^{2000} $$
$$ f(\alpha) = (\alpha^{25})^{81} – (\alpha^{25})^{80} $$

Substitute $\alpha^{25} = \frac{80}{81}$:
$$ f(\alpha) = \left( \frac{80}{81} \right)^{81} – \left( \frac{80}{81} \right)^{80} $$

Factor out $\left( \frac{80}{81} \right)^{80}$:
$$ f(\alpha) = \left( \frac{80}{81} \right)^{80} \left[ \frac{80}{81} – 1 \right] $$
$$ f(\alpha) = \left( \frac{80}{81} \right)^{80} \left[ \frac{80 – 81}{81} \right] $$
$$ f(\alpha) = \frac{80^{80}}{81^{80}} \cdot \left( -\frac{1}{81} \right) $$
$$ f(\alpha) = – \frac{80^{80}}{81^{81}} $$

We can rewrite this in the form $(80)^{80} (n)^{-81}$:
$$ f(\alpha) = 80^{80} \cdot (-1) \cdot (81)^{-81} $$
$$ f(\alpha) = 80^{80} \cdot (-(81)^{-81}) $$
Since $(-81)^{-81} = -(81)^{-81}$ (because the exponent is odd), we can write:
$$ f(\alpha) = 80^{80} \cdot (-81)^{-81} $$

Comparing this with the given form $(80)^{80}(n)^{-81}$, we get:
$$ n = -81 $$

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