The set of real values of `x` for which `log_(2x + 3) x^(2)ltlog_(2x+3)(2x + 3)` is `(a,b) uu (b,c) uu (c,d)` then
A
`2a + b = c + d + 1`
B
`2a = 3b`
C
`c + d = 3`
D
`b + d = 2`
Text Solution
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The correct Answer is:
B, C, D
Case - `1` If `2x + 3 gt 1 rArr x gt -1 …..(1)` then `x^(2) lt 2x + 3` `x^(2) - 2x - 3 lt 0` `x in (-1, 3) …(2)` `rArr (1) cap (2) = (-1, 3)` Case `-2` `0 lt 2x + 3 lt 1 rArr (-3)/(2) lt x lt 1 …..(3)` then `x^(2) gt 2x + 3` `rArr x lt -1 cup x gt 3 ......(4)` `rArr (3) cap (4) (-3)/(2) lt x lt -1` but by definition of log `x^(2) gt 0` `rArr x lt 0 cup gt 0, x ne 0` so final `((-3)/(2), -1) cup (-1, 0) cup(0, 3)` `rArr a = (-3)/(2), b = -1, c = 0, d = 3`
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