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CBSE Class 12 Math 2022 Term I Solved Paper

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SECTION - B In this Section attempt any 16 questions out of the Questions 21-40. Each question is of one mark .
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Question : 21 of 50
Marks: +1, -0
The function f(x)=2x3−15x2+36x+6f(x)=2x^3-15x^2+36x+6 is increasing in the interval
Solution:     👈: Video Solution
Explanation: Given, f(x)=2x3−15x2+36x+6f(x)=2x^3-15x^2+36x+6
∴f′(x)=6x2−30x+36\therefore f'(x)=6x^2-30x+36
It f′(x)≥0f'(x) \ge 0, then f(x)f(x) is increasing.
So, 6x2−30x+36≥06x^2-30x+36 \ge 0
 or, x2−5x+6≥0\text{ or, } x^2-5x+6 \ge 0
 or, (x−3)(x−2)≥0\text{ or, } (x-3)(x-2) \ge 0
∴x∈(−∞,2]∪[3,∞)\therefore x \in (-\infty, 2] \cup [3, \infty)
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