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Question

# The value of a for which the equation (1âˆ’a2)x2+2axâˆ’1=0 has roots belonging to (0,1) is

A
a>1+52
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B
a>2
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C
1+52<a<2
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D
a>2
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Solution

## The correct option is B a>2Let f(x)=(1−a2)x2+2ax−1. then f(x)=0 has roots between 0 and 1 if(1) D≥0(2) (1−a2)f(0)>0 and (1−a2)f(1)>0Now, D≥0⇒4a2+4(1−a2)>0 which is always true.(1−a2)f(0)>0⇒−(1−a2)>0⇒a2−1>0⇒a<−1 and a>1 ...(i)and (1−a2)f(1)>0⇒(1−a2)(2a−a2)>0⇒a(a−1)(a+1)(a−2)>0⇒a<−1 or a>2 or 0<a<1 ...(ii)From (i) and (ii) we geta<−1 and a>2

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