First let us find the range where Min (-(x – 1)^{2} + 2, (x – 2)^{2} + 1) is – (x – 1)^{2} + 2.
In other words, in which range is – (x – 1)^{2} + 2 < (x – 2)^{2} + 1.
–(x^{2} – 2x +1) + 2 < x^{2} – 4x + 4 + 1
0 < 2x^{2} – 6x + 4
x^{2} – 3x + 2 > 0
(x – 1) (x – 2) > 0
=> x > 2 or x < 1
So, for x ∈ (1, 2) , f(x) = (x – 2)^{2} + 1
And f(x) = –(x – 1)^{2} + 2 elsewhere.
Let us also compute f(1) and f(2)
f(1) = 2, f(2) = 1
For x ∈ (-∞, 1), f(x) = –(x – 1)^{2} + 2
f(1) = 2
For x ∈ (1, 2), f(x) = (x – 2)^{2} + 1
f(2) = 1
For x ∈ (2, ∞), f(x) = –(x – 1)^{2} + 2
For x < 1 and x > 2, f(x) is -(square) + 2 and so is less than 2.
When x lies between 1 and 2, the maximum value it can take is 2. f(1) = 2 is the highest value f(x) can take.
As a simple rule of thumb, the best way to approach this question is to solve the two expressions. This gives us the meeting points of the two curves. One of the two meeting points should be the maximum value.
Answer choice (b)
Correct Answer: 2