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Question

If 1+log5(x2+1)log5(ax2+4x+a) for all xR, then a can be equal to


  1. 3
  2. 4
  3. 2
  4. 32


Solution

The correct option is A 3

We have, log55(x2+1)log5(ax2+4x+a)
5(x2+1)ax2+4x+a,  xR
x2(5a)4x+(5a)0,  xR

5a>0
a<5     (1)
and D0
164(5a)20
(2(5a))(2+5a)0
(a3)(a7)0
a(,3][7,)     (2)

(1)(2), we get
a(,3]     (3)

But for the domain of log5(ax2+4x+a),
we must have ax2+4x+a>0,  xR
a>0 and 164a2<0
a>0 and a(,2)(2,)
a(2,)     (4)

(3)(4), we get
a(2,3]

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