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Question

Equation 2x2−2(2a+1)x+a(a+1)=0 has one root less than a and other root greater than a, then which of the following is true

A
0<a<1
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B
1<a<0
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C
a>0
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D
a<1
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Solution

The correct options are
A 0<a<1
B a>0
C a<1
Quadratic equation : 2x22(2a+1)x+a(a+1)=0
Both roots are real, discriminant =D>0
Comparing with the standard quadratic equation Ax2+Bx+C=0, we get A=2>0.
Since, x=a lies between the two roots and A is positive, we have f(a)<0
2(a)22(2a+1)(a)+a(a+1)<0
a2a<0
a2+a>0
a(,1)(0,)
Hence, options (A), (C) and (D) are correct.

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