CUET · MATHS · PYQ PAPER 2025
If \(f(x)=2 x^3-15 x^2+36 x+1, x \in[1,5]\), then the absolute minimum value of \(f(x)\) is:
- A 22
- B 28
- C 24
- D 29
Answer & Solution
Correct Answer
(C) 24
Step-by-step Solution
Detailed explanation
\(f'(x) = 6x^2 - 30x + 36\) \(6x^2 - 30x + 36 = 0 \Rightarrow x^2 - 5x + 6 = 0 \Rightarrow (x-2)(x-3) = 0\) Critical points: \(x=2, x=3\). Both are in \([1,5]\). \(f(1) = 2(1)^3 - 15(1)^2 + 36(1) + 1 = 24\) \(f(2) = 2(2)^3 - 15(2)^2 + 36(2) + 1 = 16 - 60 + 72 + 1 = 29\)…
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