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Question

# Let f(x)=⎧⎨⎩−x2+b3+b−2b2−2b2+5b+6;0≤x<13x−4;1≤x≤3 where b∈R. Which of the following option(s) is/are correct?

A
f(x) is strictly increasing on (0,1)
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B
f(x) has minimum value at x=1 for b(3,2)[2,)
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C
f(x) is not differentiable at x=1
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D
f(x) is strictly decreasing on (1,3)
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Solution

## The correct options are B f(x) has minimum value at x=1 for b∈(−3,−2)∪[2,∞) C f(x) is not differentiable at x=1f′(x)={−2x;0≤x<13;1≤x≤3 f is not differentiable at x=1. As f′(x)<0 for x∈(0,1) and f′(x)>0 for x∈(1,3), ⇒f(x) is strictly decreasing on (0,1) and strictly increasing on (1,3). ⇒f(x)≥f(1) for x∈[1,3] For f(x) to have the smallest value at x=1 for x∈[0,3], we must have limx→1−f(x)≥f(1) ⇒limx→1−−x2+b3+b−2b2−2b2+5b+6≥−1 ⇒−1+b3+b−2b2−2b2+5b+6≥−1 ⇒(b−2)(b2+1)(b+2)(b+3)≥0 ⇒(b−2)(b+2)(b+3)≥0 as b2+1>0 ⇒b∈(−3,−2)∪[2,∞)

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