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Question

The complete set of values of x for which the inequality logx(4x+565x)<1 holds good, is

A
(1,65)
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B
(0,1)
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C
(12,1)
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D
(0,1)(1,65)
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Solution

The correct option is C (12,1)
For logx(4x+565x)<1 to be defined,
4x+565x>0, x>0, x1x(54,65), x>0, x1x(0,65){1}

Case 1: When x(0,1)
logx4x+565x<14x+565x>x14x+565x1x>04x2+10x6x(65x)>02(2x1)(x+3)x(5x6)<0x(3,0)(12,65)
x(12,1) (1) as x(0,1)

Case 2: When x(1,65)
logx4x+565x<1
4x+565x<x1
4x+565x1x<0
4x2+10x6x(65x)<0
2(2x1)(x+3)x(5x6)>0
x(,3)(0,12)(65,)
xϕ (2) as x(1,65)

From (1) and (2),
x(12,1)

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