# CAT 2017 Question Paper | Quants Slot 2

###### CAT Previous Year Paper | CAT Functions Questions | Question 32

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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 ≤ $\frac{7}{2}$ ## Best CAT Online Coaching Try upto 40 hours for free Learn from the best! #### 2IIM : Best Online CAT Coaching. ### Video Explanation ## Best CAT Coaching in Chennai #### CAT Coaching in Chennai - CAT 2022Limited Seats Available - Register Now! ### 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"

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