Let a1,a2,a3,…,an be the integers which are not included in the range of f(x)=40(x−5)(x+1), where ai>ai+1 for each i=1,2,3,…,n. If n∑i=15−aiCi−1=xCy, then x+y can be
A
14
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B
21
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C
16
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D
15
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Solution
The correct options are A14 C16 y=40x2−4x−5 ⇒yx2−4yx−5y−40=0 Since x is real, ∴D≥0 and y≠0 16y2+4y(5y+40)≥0 ⇒9y2+40y≥0 ⇒y(9y+40)≥0 ∴y∈(−∞,−409]∪(0,∞) as y≠0 Hence, Rf=(−∞,−409]∪(0,∞) Integers which are not included in the range are −4,−3,−2,−1,0 ∴a1=0,a2=−1,a3=−2,a4=−3,a5=−4