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Bunuel
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Let f(x) = 2x – 5 and g(x) = 7 – 2x. Then |f(x) + g(x)| = |f(x)| + |g( 03 Apr 2020, 05:26
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D
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63% (02:25) correct 37% (02:26) wrong based on 501 sessions

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Let \(f(x) = 2x – 5\) and \(g(x) = 7 – 2x\). Then \(|f(x) + g(x)| = |f(x)| + |g(x)|\) if and only if


A. \(\frac{5}{2} < x < \frac{7}{2}\)

B. \(x ≤ \frac{5}{2}\) or \(x ≥ \frac{7}{2}\)

C. \(x < \frac{5}{2}\) or \(x ≥ 7/2\)

D. \(\frac{5}{2} ≤ x ≤ \frac{7}{2}\)

E. \(\frac{5}{2} ≤ x\)


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D
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Let f(x) = 2x – 5 and g(x) = 7 – 2x. Then |f(x) + g(x)| = |f(x)| + |g( 03 Apr 2020, 06:38
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Bunuel
Let \(f(x) = 2x – 5\) and \(g(x) = 7 – 2x\). Then \(|f(x) + g(x)| = |f(x)| + |g(x)|\) if and only if

A. \(\frac{5}{2} < x < \frac{7}{2}\)
B. \(x ≤ \frac{5}{2}\) or \(x ≥ \frac{7}{2}\)
C. \(x < \frac{5}{2}\) or \(x ≥ 7/2\)
D. \(\frac{5}{2} ≤ x ≤ \frac{7}{2}\)
E. \(\frac{5}{2} ≤ x\)
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\(|f(x) + g(x)| = |f(x)| + |g(x)|\)
--> \(|2x - 5 + 7 - 2x| = |2x - 5| + |7 - 2x|\)
--> \(|2| = |2x - 5| + |7 - 2x|\)
--> \(|2x - 5| + |7 - 2x| = 2\)

The only case possible is when \(|2x - 5| = +(2x - 5)\) and \(|7 - 2x| = +(7 - 2x)\)
--> \(|2x - 5| ≥ 0\) & \(|7 - 2x| ≥ 0\)
--> \(2x - 5 ≥ 0\) & \(7 - 2x ≥ 0\)
--> \(2x ≥ 5\) & \(2x ≤ 7\)
--> \(5 ≤ 2x ≤ 7\)
--> \(\frac{5}{2} ≤ x ≤ \frac{7}{2}\)

Option D
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Let f(x) = 2x – 5 and g(x) = 7 – 2x. Then |f(x) + g(x)| = |f(x)| + |g( 04 Apr 2020, 05:40
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Let's put x=2.5
LHS =RHS in the equation given both sides are giving value 2.
Option B,D and E can be the answers now,
Lets put x=4
LHS = 2
RHS= 4
Option B and E are out.
Hence option D is the answer

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Let f(x) = 2x – 5 and g(x) = 7 – 2x. Then |f(x) + g(x)| = |f(x)| + |g( 27 Aug 2022, 11:39
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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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Let f(x) = 2x – 5 and g(x) = 7 – 2x. Then |f(x) + g(x)| = |f(x)| + |g( 09 Dec 2024, 04:57
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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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Let f(x) = 2x – 5 and g(x) = 7 – 2x. Then |f(x) + g(x)| = |f(x)| + |g( 25 Jun 2026, 15:17
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