Let a>0 be a root of the equation 2 x²+x-2=0. If lim x → 1/a 16(1-cos (2+x-2 x²))/(1-a x)²=α+β √17, where α, β ∈ Z, then α+β is equal to _________.

Solution

Maths sir · 4 steps

At x = 1/a the cosine's argument becomes zero, so use 1-cos θ ≈ θ²/2 and factor.

1. Spot the zero. Put x = 1/a: 2 + 1/a - 2/a² = (2a² + a - 2)/a² = 0. So 1/a is a root of 2 + x - 2x² = 0, and the limit has the form 0/0.
2. Use the standard limit. Since (1-cos θ)/θ² → 1/2 as θ → 0, the limit equals 8 · (2+x-2x²)²/(1-ax)². The roots of 2+x-2x²=0 are the reciprocals 1/a and 1/b of the roots of 2x²+x-2=0. So 2+x-2x² = -2(x-1/a)(x-1/b) and 1-ax = -a(x-1/a).

   L = 8 · 4(x-1/a)²(x-1/b)²/a²(x-1/a)² → 32/a²(1/a-1/b)²

3. Put in values. 1/a, 1/b are the roots of 2x² - x - 2 = 0, so 1/a = (1+√17)/4 (positive, because a>0) and 1/b = (1-√17)/4. Then (1/a-1/b)² = 17/4 and 1/a² = (18 + 2√17)/16 = (9+√17)/8.
4. So. L = 32 × 17/4 × (9+√17)/8 = 153 + 17√17. So α = 153, β = 17, and α + β = 170.

The common slip: 340. Treating 1-cos θ as θ² and dropping the 1/2 doubles everything to 306 + 34√17, giving 340.

The idea to keep. When a 1-cos(something) limit is 0/0, replace it by 1/2(something)², then factor so the vanishing factor cancels with the denominator.

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