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CBSE Class 12 Math 2024 All Sets Solved Paper

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Question : 5 of 20
Marks: +1, -0
The function f(x)=x3−3x2+12x−18f(x) = x^3 - 3x^2 + 12x - 18 is
Solution:  
f(x)=x3−3x2+12x−18f(x) = x^3 - 3x^2 + 12 x - 18
f′(x)=3x2−6x+12=3(x2−2x+4)f'(x) = 3x^2 - 6x +12 = 3(x^2 - 2x +4)
Case I: f is strictly increasing (f'(x) > 0)
3(x2−2x+4)>0→x2−2x+4>0→x2−2x+1+3>03(x^2 - 2x + 4) > 0 \rightarrow x^2 - 2x + 4 > 0 \rightarrow x^2 - 2x + 1 + 3 > 0
(x−1)2+3>0(x-1)^2 + 3 > 0
(x−1)2>−3→(x-1)^2 > -3 \rightarrow Always True because positive real number is always greater than negative real number.
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