Let the sum of the maximum and the minimum values of the function f(x)=(2 x²-3 x+8)/(2 x²+3 x+8) be m/n, where gcd(m, n)=1. Then m+n is equal to :

Solution

Maths sir · 4 steps

Set y = f(x), rewrite it as a quadratic in x, and demand real roots. That pins down the range of y.

1. Denominator never vanishes. For 2x² + 3x + 8 the discriminant is 9 - 64 < 0, so the denominator is always positive. f is defined for every real x.
2. Quadratic in x. Put y = f(x) and cross-multiply: y(2x² + 3x + 8) = 2x² - 3x + 8. This gives 2(y-1)x² + 3(y+1)x + 8(y-1) = 0. A value y is attained only if this has a real root x. The case y = 1 is fine too, since f(0) = 1.

   9(y+1)² - 64(y-1)² ≥ 0 ⇒ (11 - 5y)(11y - 5) ≥ 0 ⇒ 5/11 ≤ y ≤ 11/5

3. Check the ends. f(2) = 10/22 = 5/11 and f(-2) = 22/10 = 11/5. Both ends are actually reached, so the minimum is 5/11 and the maximum is 11/5.
4. So. The sum is 5/11 + 11/5 = (25 + 121)/55 = 146/55. Since gcd(146, 55) = 1, we get m + n = 146 + 55 = 201, option (c).

The idea to keep. For a ratio of two quadratics, set y = f(x) and require the discriminant in x to be ≥ 0. The endpoints of that y-interval are the max and min, with no calculus needed.

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