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Given: Let \(f(x) = 2x – 5\) and \(g(x) = 7 – 2x\).

Asked: Then \(|f(x) + g(x)| = |f(x)| + |g(x)|\) if and only if

|f(x) + g(x)| = |2x-5 + 7-2x| = 2
|f(x)| + |g(x)| = |2x-5| + |7-2x|

Case 1: x>=3.5
|f(x)| + |g(x)| = |2x-5| + |7-2x| = 2x- 5 + 2x - 7 = 4x - 12 = 2
x = 3.5

Case 2: 2.5<x<3.5
|f(x)| + |g(x)| = |2x-5| + |7-2x| = 2x- 5 + 7-2x = 2

Case 3: x<=2.5
|f(x)| + |g(x)| = |2x-5| + |7-2x| = 5-2x + 7-2x = 12 - 4x = 2
x = 2.5


IMO D
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↧↧↧ Detailed Video Solution to the Problem ↧↧↧




\(f(x) = 2x – 5\) and \(g(x) = 7 – 2x\)

\(|f(x) + g(x)| = |f(x)| + |g(x)|\)

=> | 2x - 5 + 7 - 2x | = | 2x - 5 | + | 7 - 2x |
=> | 2 | = 2 = | 2x - 5 | + | 7 - 2x |

Now, this will be true only when both the absolute values open as it is to give us
| 2x - 5 | + | 7 - 2x | = 2x - 5 + 7 - 2x + 2

This will happen only when the values inside both the absolute value is non-negative
[ Watch this video to MASTER Absolute Value ]

=> 2x - 5 ≥ 0 and 7 - 2x ≥ 0
=> x ≥ \(\frac{5}{2}\) and x ≤ \(\frac{7}{2}\)
=> \(\frac{5}{2}\) ≤ x ≤ \(\frac{7}{2}\)

So, Answer will be D
Hope it helps!

Watch the following video to MASTER Functions and Custom Characters

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