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The maximum value of f(x)=x50−x20f(x) = x^{50} - x^{20} in the interval [0,1][0,1], is

Solution

✅ Correct Option: 3

f′(x)=10x19(5x30−2)f'(x)=10x^{19}(5x^{30}-2), zero at x=0x=0 and x30=25x^{30}=\dfrac{2}{5}.

At the interior critical point f<0f<0 (since x50<x20x^{50}<x^{20} for 0<x<10<x<1), so it is a minimum.

Endpoints: f(0)=0f(0)=0, f(1)=0f(1)=0. Hence maximum value =0=0.

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