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← IIT-JEE 2012 · Paper 1
MathematicsSingle correct+3 / -1Question 43 of 60

IIT-JEE 2012 · Paper 1, Mathematics question 43

The function f:[0,3]→[1,29]f:[0,3]\to[1,29], defined by f(x)=2x3−15x2+36x+1f(x)=2x^3-15x^2+36x+1, is

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Answer and worked solution

Official answer: B

Monotonicity. f′(x)=6x2−30x+36=6(x−2)(x−3)f'(x)=6x^2-30x+36=6(x-2)(x-3): increasing on [0,2][0,2], decreasing on [2,3][2,3].

Values. f(0)=1f(0)=1, f(2)=29f(2)=29, f(3)=28f(3)=28. The range is [1,29][1,29], so ff is onto.

Not one-one. Every value in [28,29)[28,29) is taken once on (0,2)(0,2) and once on (2,3](2,3].

(B)