Q.

Number of positive integers satisfying (x+2)x2−2x+1−4+3x−x2≥ 0 is

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a

0

b

1

c

2

d

3

answer is B.

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Detailed Solution

(x+2)x2−2x+1−x2+3x−4≥0⇒ (x+2)(x−1)2x2−3x+4≤0 but  x2−3x+4>0∀x∈R  since  ∆ =9-4·1·4=-7<0    ∴ (x+2)(x−1)2≤0⇒ x∈(−∞,−2]∪{1}  therefore only 1 is the positive integral solution
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Number of positive integers satisfying (x+2)x2−2x+1−4+3x−x2≥ 0 is