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A. function f: [1, 6) → R is defined as follows  \( f(x) = \begin{cases} 1 + x, & \quad \text{} 1≤x<2 \text{ }\\ 2x-1, & \quad \text{ } 2≤x<4 \text{ }\\3x^2-10,& \quad \text{ } 4≤x<6 \end{cases}\)

(Here, [1, 6) = {x ∈ R : 1 < x < 6})

Find the value of

(i) f(5)

(ii) f(3)

(iii) f(1)

(iv) f(2) – f(4)

(v) 2f(5) – 3f(1).

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(i) Let us find f(5). Since 5 lies between 4 and 6, we have to use f(x) = 3x2 – 10.

Thus, f(5) = 3(52) – 10 = 65.

(ii) To find f(3), note that 3 lies between 2 and 4.

So, we use f(x) = 2x – 1 to calculate f(3).

Thus, f(3) = 2(3) – 1 = 5.

(iii) Let us find f(1).

Now, 1 is in the interval 1 < x < 2

Thus, we have to use f(x) = 1 + x to obtain f(1) = 1 + 1 = 2.

(iv) f(2) – f(4)

Now, 2 is in the interval 2 < x < 4 and so, we use f(x) = 2x – 1.

Thus, f(2) = 2(2) -1 = 3.

Also, 4 is in the interval 4 < x < 6. Thus, we use f(x) = 3x2 – 10

Therefore, f(4) = 3(4 ) – 10 = 3(16) – 10 = 48 – 10 = 38.

Hence, f(2) – f(4) = 3 – 38 = -35.

(v) To calculate 2 f(5) – 3f(1), we shall make use of the values that we have already calculated in (i) and (iii).

Thus, 2f(5) – 3f(1) = 2(65) – 3(2) = 130 – 6 – 124.

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