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Question

Let f(x)=b3+b2b22b2+5b+6x2 ;0x<1 3x4 ;1x3
where bR.

Then which of the following option is INCORRECT?

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

The correct option is D f(x) has minimum value at x=1 for b(3,2)
f(x)={2x;0x<1 3;1x3

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 limx1f(x)f(1)
limx1x2+b3+b2b22b2+5b+61

1+b3+b2b22b2+5b+61

(b2)(b2+1)(b+2)(b+3)0
(b2)(b+2)(b+3)0 as b2+1>0
b(3,2)[2,)

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