If lim x → 0 (eᵃˣ-cos (b x)-(cx e⁻ᶜˣ)/2)/(1-cos (2 x))=17, then 5 a²+b² is equal to

Solution

Maths sir · 3 steps

Expand everything up to x²: the x term must vanish, and the x² term gives you 17.

1. Expand the numerator. Use eᵃˣ = 1 + ax + a²x²/2, cos bx = 1 - b²x²/2 and (cx e⁻ᶜˣ)/2 = c/2(x - cx²) = cx/2 - c²x²/2, all up to x². The constants cancel, leaving (a - c/2)x + (a² + b² + c²)/2x².
2. Kill the x term. The denominator is 1 - cos 2x ≈ 2x². For a finite limit, the numerator cannot have an x term, so a = c/2, that is, c = 2a.

   lim(x → 0) ((a² + b² + 4a²)/2 x²)/2x² = (5a² + b²)/4 = 17

3. So. 5a² + b² = 4 × 17 = 68, which is option (b).

The idea to keep. For a limit of type 0/0, expand the top and bottom to the same power of x. Lower powers in the numerator must vanish, and the matching coefficients give the limit.

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