The number of negative values of a for which the inequation 4x2+2(2a+1)2x2+4a2−3>0 is satisfied for any x is
A
∞
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B
0
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C
1
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D
None of these
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Solution
The correct option is A∞ 4x2+2(2a+1)2x2+4a2−3>0 Put 2x2=t t2+2(2a+1)t+4a2−3>0 ⇒D<0 ⇒4(2a+1)2−4(4a2−3)<0 ⇒a+1<0 ⇒a<−1 ⇒a∈(−∞,−1) Hence, infinite values of a are possible.