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Question 32 : Let f(x) = 2x – 5 and g(x) = 7 – 2x. Then |f(x) + g(x)| = |f(x)| + |g(x)| if and only if

1. $\frac{5}{2}$ < x < $\frac{7}{2}$
2. x ≤ $\frac{5}{2}$ or x ≥ $\frac{7}{2}$
3. x < $\frac{5}{2}$ or x ≥ $\frac{7}{2}$
4. $\frac{5}{2}$ ≤ x ≤ $\frac{7}{2}$

Video Explanation

Explanatory Answer

Method of solving this CAT Question from Functions

Given that f(x) = 2x – 5 and g (x) = 7 – 2x.
Consider the LHS,
|f(x) + g(x)| = |2x - 5 + 7 – 2x|
|2| = 2
So, RHS should also be 2.

On considering the RHS, |f(x)| + |g(x)|, we get
$\frac{5}{2}$ and $\frac{7}{2}$ are the two pivotal points for which the values of f(x) and g(x) goes to zero.

We have three ranges,
1) x < $\frac{5}{2}$ or
2) $\frac{5}{2}$ < x < $\frac{7}{2}$ or
3) x > $\frac{7}{2}$.
When x lies between 0 and $\frac{5}{2}$,
|f(x)| + |g(x)| = 12 – 4x
When x lies between $\frac{5}{2}$ < x < $\frac{7}{2}$,
|f(x)| + |g(x)| = 2x – 5 + 7 – 2x = 2
When x is greater than $\frac{7}{2}$
|f(x)| + |g(x)| = 2x – 5 + 2x – 7 = 4x – 12
Hence, out of these 3 possibilities, only $\frac{5}{2}$ < x < $\frac{7}{2}$ is equal to 2. So, this has to be our answer. But, is that so?
Let us consider when x = $\frac{5}{2}$,
⟹ 12 – 4x = 12 – 4($\frac{5}{2}$) = 2
Similarly, when x = $\frac{7}{2}$,
⟹ 4x – 12 = 4($\frac{7}{2}$) - 12 = 2
The value of |f(x)| + |g(x)| is equal to 2 even at x = $\frac{5}{2}$ and x = $\frac{7}{2}$.
Hence, the range is $\frac{5}{2}$ ≤ x ≤ $\frac{7}{2}$

The question is "Let f(x) = 2x – 5 and g(x) = 7 – 2x. Then |f(x) + g(x)| = |f(x)| + |g(x)| if and only if"

Hence, the answer is $\frac{5}{2}$ ≤ x ≤ $\frac{7}{2}$

Choice D is the correct answer

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