Let ⨍ (x)=-x3+b3-b2+b-1b2+3b+2,0≤x<1 2x-31≤x≤3If ⨍ (x) attains minimum at x=1, then b∈

LHL ≥ ⨍ (1) -1+b3-b2+b-1b2+3b+2≥-1 b2(b-1)+(b-1)(b+1) (b+2)≥0 (b-1) (b+1) (b+2)≥0 b≠-1, -2 b∈(-2, -1)∪[1, ∞)

Let ⨍ (x)=-x3+b3-b2+b-1b2+3b+2,0≤x<1 2x-31≤x≤3If ⨍ (x) attains minimum at x=1, then b∈

LHL ≥ ⨍ (1) -1+b3-b2+b-1b2+3b+2≥-1 b2(b-1)+(b-1)(b+1) (b+2)≥0 (b-1) (b+1) (b+2)≥0 b≠-1, -2 b∈(-2, -1)∪[1, ∞)

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Let $⨍ (x)= \{\begin{cases} - x^{3} + \frac{b^{3} - b^{2} + b - 1}{b^{2} + 3 b + 2}, & 0 \leq x < 1 \\ 2 x - 3 & 1 \leq x \leq 3 \end{cases}$
If $⨍ (x)$ attains minimum at $x = 1$ , then $b \in$

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$(0, \infty)$

$(- 2, - 1) \cup [1, \infty)$

$(- \infty, - 2) \cup [- 1, 1]$

answer is D.

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Detailed Solution

$LHL \geq ⨍ (1) - 1 + \frac{b^{3} - b^{2} + b - 1}{b^{2} + 3 b + 2} \geq - 1 \frac{b^{2} (b - 1) + (b - 1)}{(b + 1) (b + 2)} \geq 0 (b - 1) (b + 1) (b + 2) \geq 0 b \neq - 1, - 2$

$b \in (- 2, - 1) \cup [1, \infty)$

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Let ⨍ (x)=-x3+b3-b2+b-1b2+3b+2,0≤x<1 2x-31≤x≤3If ⨍ (x) attains minimum at x=1, then b∈

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Let $⨍ (x)= \{\begin{cases} - x^{3} + \frac{b^{3} - b^{2} + b - 1}{b^{2} + 3 b + 2}, & 0 \leq x < 1 \\ 2 x - 3 & 1 \leq x \leq 3 \end{cases}$
If $⨍ (x)$ attains minimum at $x = 1$ , then $b \in$