Question:

If f:RRf: \mathbb{R} \rightarrow \mathbb{R} be defined by f(x)={ 2xif x>3x2if 1<x33xif x1f(x) = \begin{cases}  2x & \text{if } x > 3 \\ x^2 & \text{if } 1 < x \leq 3 \\ 3x & \text{if } x \leq 1 \end{cases}. Then f(1)+f(2)+f(4)f(-1) + f (2) + f(4) is

Updated On: Apr 20, 2024
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The Correct Option is B

Solution and Explanation

First, let's calculate f(1):f(-1):
Since x1x ≤ 1, we use the third part of the function definition:
f(1)=3(1)=3f(-1) = 3(-1) = -3

Next, let's calculate f(2):f(2):
Since 1<x31 < x ≤ 3, we use the second part of the function definition:
f(2)=(2)2=4f(2) = (2)^2 = 4

Finally, let's calculate f(4):f(4):
Since x>3x > 3, we use the first part of the function definition:
f(4)=2(4)=8f(4) = 2(4) = 8

Now, we can compute f(1)+f(2)+f(4):f(-1) + f(2) + f(4):
3+4+8=9-3 + 4 + 8 = 9

Therefore, f(1)+f(2)+f(4)f(-1) + f(2) + f(4) is equal to 9 (option B).

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