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Mathematics · 2025 · 5 marks
CBSE 2025 · Region 1 · Set 1 · Q33
Find the absolute maximum and absolute minimum of function f (x)=2 x^3-15 x^2+36 x+1 on [1,5].
Find the absolute maximum and absolute minimum of function $\displaystyle \mathrm{f}(x)=2 x^{3}-15 x^{2}+36 x+1$ on $\displaystyle [1,5]$.
Marking-scheme solution
\[\begin{aligned}
& \mathrm{f}(x)=2 x^{3}-15 x^{2}+36 x+1 \\
& \Rightarrow \mathrm{f}^{\prime}(x)=6\left(x^{2}-5 x+6\right)=6(x-2)(x-3) \\
& \mathrm{f}^{\prime}(x)=0 \Rightarrow x=2,3 \in[1,5] \\
& \text { Now } \mathrm{f}(1)=24, \mathrm{f}(2)=29, \mathrm{f}(3)=28, \mathrm{f}(5)=56
\end{aligned}
\]
Hence, the absolute maximum value is $\displaystyle 56$ and the absolute minimum value is $\displaystyle 24$ .
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