JEE Mains · Maths · STD 12 - 6. Application of derivatives
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 \(-80\)
Answer & Solution
Correct Answer
(A) \(-81\)
Step-by-step Solution
Detailed explanation
\(f(x)=x^{2025}-x^{2000}\) \(f ^{\prime}( x )=0 \Rightarrow x =\left(\frac{2000}{2025}\right)^{1 / 25}=\alpha(\)say\()\) \(\therefore f(0)=0, f(1)=0, f(\alpha)=\left(\frac{80}{81}\right)^{80} \cdot \frac{-1}{81}=80^{80} \cdot(-81)^{-81}\)
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