Let (2,3) be the largest open interval in which the function f(x)=2 log ₑ(x-2)-x²+a x+1 is strictly increasing and (b, c) be the largest open interval, in which the function g(x)=(x-1)³(x+2-a)² is strictly decreasing. Then 100(a+b-c) is equal to :

Solution

Maths sir · 5 steps

The end x = 3 of the increasing interval must be a root of f'(x) = 0, and that root fixes a.

1. Derivative of f. The domain is x > 2. Here f'(x) = 2/(x-2) - 2x + a. Multiply f'(x) > 0 by (x-2) > 0 and rearrange. The condition becomes 2x² - (4+a)x + 2a - 2 < 0.
2. Find a. The interval ends at x = 3, so the quadratic must be zero there: 18 - 3(4+a) + 2a - 2 = 4 - a = 0, so a = 4. Check: 2x² - 8x + 6 = 2(x-1)(x-3) < 0 on (1, 3), and with x > 2 this gives (2, 3).
3. Derivative of g. With a = 4, g(x) = (x-1)³(x-2)². Use the product rule and take out the common factor (x-1)²(x-2).

   g'(x) = (x-1)²(x-2)[3(x-2) + 2(x-1)] = (x-1)²(x-2)(5x-8)

4. Sign of g'. (x-1)² ≥ 0, so g' < 0 exactly when (x-2)(5x-8) < 0, that is on (8/5, 2). So b = 8/5 and c = 2.
5. So. a + b - c = 4 + 1.6 - 2 = 3.6, and 100(a + b - c) = 360, option (a).

The idea to keep. When a question gives you the exact interval of monotonicity, its endpoints are where f' changes sign. Put an endpoint into f' = 0 to find the unknown constant, then check the whole interval.

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