The sum of the maximum and minimum values of the function f(x)=|5 x-7|+[x²+2 x] in the interval [5/4, 2], where [t] is the greatest integer ≤ t, is ______________.

Solution

Maths sir · 4 steps

Both parts are increasing after x = 7/5, so look at the corner x = 7/5 and the end x = 2.

1. Range of the bracket. On [5/4, 2], x² + 2x increases from 65/16 = 4.0625 to 8. So [x²+2x] steps up through 4, 5, 6, 7 and becomes 8 only at x = 2.
2. Find the minimum. |5x-7| is smallest, equal to 0, at x = 7/5 = 1.4. There x² + 2x = 1.96 + 2.8 = 4.76, so [x²+2x] = 4, its smallest possible value. Both parts are at their lowest together, so the minimum is 0 + 4 = 4.
3. Find the maximum. For x > 1.4, both |5x-7| and [x²+2x] only increase, so the largest value comes at x = 2. The left end gives only f(5/4) = 0.75 + 4 = 4.75.

   f(2) = |10 - 7| + [4 + 4] = 3 + 8 = 11

4. So. The maximum is 11 and the minimum is 4, so their sum is 15.

The common slip: 14. Taking the maximum as 10, the value just before x = 2, misses the jump of [x²+2x] to 8 at x = 2 and gives 14.

The idea to keep. When a function is a sum of pieces, find where each piece is smallest or largest. Then check the corners of the modulus and the interval ends, where the greatest integer can jump.

Answer from the NTA final answer key. Solution by PrepWiser.