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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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