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Question

# The value of k for which the equation 3x2+2x(k2+1)+k2−3k+2=0 has roots of opposite signs, lies in the interval

A
(1,2)
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B
(32,2)
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C
(,0)
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D
(,1)
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Solution

## The correct option is A (1,2)Given quadratic equation 3x2+2x(k2+1)+k2−3k+2=0 Now, for the equation ax2+bx+c=0 to have roots of opposite signs a⋅c<0 Comparing the equations, we get: a=3, c=k2−3k+2 ∴ for 3x2+2x(k2+1)+k2−3k+2=0 to have roots of opposite signs, 3(k2−3k+2)<0⇒k2−3k+2<0⇒(k−1)(k−2)<0 Using wavy curve method we find the required range of k ∴(k−1)(k−2)<0⇒k∈(1,2)

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