If a>0 and the equation ax2+bx+c=0 has two real roots α and β such that |α|≤1,|β|≤1, then
A
a+b+c≥0
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B
a−b+c≥0
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C
a+|b|+c≥0
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D
a−c≥0
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Solution
The correct options are Aa+b+c≥0 Ba−b+c≥0 Ca+|b|+c≥0 Da−c≥0 As both the roots lie in the interval [-1,1], we have y(−1)≥0,y(1)≥0⇒a−b+c≥0,a+b+c≥0⇒a+|b|+c≥0Also,αβ≤|αβ|=|α||β|≤1⇒ca≤1orc≤aora−c≥0