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CBSE Class 12 Math 2025 All Sets Solved Paper

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Question : 12 of 20
Marks: +1, -0
The absolute maximum value of function f(x)=x3−3x+2f(x)=x^{3}-3x+2 in [0, 2] is
Solution:  
f(x)=x3−3x+2,x∈[0,2]f(x)=x^{3}-3x+2, x\in[0,2]
To check extremum, differentiate,
f′(x)=3x2−3f'(x)=3x^{2}-3
⇒f′(x)=0⇒3(x2−1)=x=±1\Rightarrow f'(x)=0 \Rightarrow 3(x^{2}-1)=x=\pm 1
⇒\Rightarrow Critical points are -1, 1 and end points 0, 2
⇒f(0)=2\Rightarrow f(0)=2
f(1) = 0
f(2)=23−3(2)+2=8−6+2=4f(2)=2^{3}-3(2)+2=8-6+2=4
absolute maximum value is 4
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