The values of a for which the quadratic equation 3x2+2(a2+1)x+(a2−3a+2)=0 has roots of opposite sign, is
A
a∈(1,2)
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B
a∈(2,∞)
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C
a∈(1,∞)
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D
a∈(2,3)
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Solution
The correct option is Aa∈(1,2) Given equation is 3x2+2(a2+1)x+(a2−3a+2)=0 It is a upward opening parabola As the roots are of opposite sign, so f(0)<0⇒a2−3a+2<0⇒(a−1)(a−2)<0∴a∈(1,2)