The modulus equation ||x−1|+a|=4 (a∈R) can have a real solution for x . If 'a' lies in the interval

| | x − 1 | + a | = 4 | x − 1 | + a = ± 4 | x − 1 | = ± 4 − a x − 1 > 0 ∴ ± 4 − a ≥ 0 4 − a > 0 and − ( 4 + a ) > 0 a < 4 and a < − 4 ( − ∞ , 4 ] and ( − ∞ , − 4 ]

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The modulus equation $| | x - 1 | + a | = 4 (a \in R)$ can have a real solution for $x$ . If $' a'$ lies in the interval

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(4, \infty)

[- 4, 4]

(- \infty, - 4]

(- \infty, 4]

answer is A, C.

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Detailed Solution

| | x - 1 | + a | = 4

$| x - 1 | + a = \pm 4$

$| x - 1 | = \pm 4 - a$

$x - 1 > 0$

$∴ \pm 4 - a \geq 0$

$4 - a > 0 and - (4 + a) > 0$

$a < 4 and a < - 4$

$(- \infty, 4] and (- \infty, - 4]$

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