1.

The equation `|x+1||x-1|=a^(2) - 2a - 3` can have real solutions for x, if a belongs toA. `(-oo, -1]uu[3, oo)`B. `[1 - sqrt(5), 1 + sqrt(5)]`C. `[1-sqrt(5), 1] uu [3, 1 + sqrt(5)]`D. none of these

Answer» Correct Answer - C
We have,
`|x+1||x-1|=a^(2)-2a - 3" "...(i)`
`rArr" "|x^(2)-1|=a^(2) - 2a -3`
`rArr" "x^(2) -1 = +-(a^(2) -2a - 3)`
`rArr" "x^(2) = a^(2) - 2a - 2, -a^(2) + 2a + 4`
Thus, for equation (i) to have real solutiions, we must have `(a^(2)-2a - 3 ge 0 and a^(2) - 2a - 2 ge 0)`
`or," "(a^(2)-3a-3 ge 0 and -a^(2) + 2a + 4 ge 0)`
`or," "((a-3)(a+1) ge 0 and (a-1-sqrt(3))(a-1+sqrt(3)) ge 0)`
`or," "((a-1+sqrt(5))(a-1-sqrt(5)) le 0 and (a-3)(a+1) ge 0)`
`rArr" "a in (-oo, -1] uu [3, oo) or, a in [1-sqrt(5,)1 ] uu [3, 1 + sqrt(5)]`


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