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If `ax^2 + bx+ c = 0`, `a!=0`, `a, b, c in R` has distinct real roots in ( 1,2) then a and 5a + 2b + c have (A) same sign (B) opposite sign (C) not determined (D) none of theseA. of same typeB. of opposite typeC. undeterminedD. none of these |
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Answer» Correct Answer - A Let `alpha , beta` be the roots of `ax^(2)+bx+c = 0`. Then, `alpha + beta = -(b)/(a) and alpha beta = (c)/(a)` Now, `a(5a+2b+c) = a^(2)(5+2(b)/(a)+(c)/(a))` `rArr" "a(5a+2b+c)=a^(2){5-2(alpha+beta)+alpha beta}` `rArr" "a(5a+2b+c)=a^(2){alpha beta - 2(alpha + beta)+4 + 1}` `rArr" "a(5a+2b+c)=a^(2){(alpha-2)(beta-2)+1} gt 0` `rArr" "a(5a + 2b + c) gt 0" "[because alpha, beta lt 2 therefore (alpha-2)(beta-2) gt 0]` Signs of and 5a + 2b + c are same. |
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