Let a be the sum of all coefficients in the expansion of (1-2 x+2 x²)²⁰²³(3-4 x²+2 x³)²⁰²⁴ and b=lim x → 0((∫₀ˣ (log (1+t))/(t²⁰²⁴+1) d t)/x²). If the equation c x²+d x+e=0 and 2 b x²+a x+4=0 have a common root, where c, d, e ∈ ℝ, then d: c: e equals

Solution

Maths sir · 5 steps

Find a and b first. The quadratic has no real roots, so sharing one root means sharing both.

1. Find a. The sum of all coefficients is the value at x = 1. That gives a = (1 - 2 + 2)²⁰²³(3 - 4 + 2)²⁰²⁴ = 1²⁰²³ × 1²⁰²⁴ = 1.
2. Find b. The limit has the form 0/0, so use L'Hôpital's rule. Differentiate the integral with the fundamental theorem of calculus and differentiate x² to get 2x.

   b = lim(x → 0) log(1+x)/((x²⁰²⁴+1) · 2x) = 1/1 × 1/2 = 1/2

3. The given quadratic. With a = 1 and b = 1/2, the equation 2bx² + ax + 4 = 0 becomes x² + x + 4 = 0. Its discriminant is 1 - 16 = -15 < 0, so both roots are non-real.
4. Both roots shared. Non-real roots of a quadratic with real coefficients come in conjugate pairs. If cx² + dx + e = 0 (real c, d, e) shares one root, it must also share the conjugate, so its coefficients are proportional: c/1 = d/1 = e/4.
5. So. c : d : e = 1 : 1 : 4. Written in the asked order, this is d : c : e = 1 : 1 : 4, which is option (b).

The idea to keep. When a real quadratic has a negative discriminant, a common root with another real quadratic forces a common pair of roots. The coefficients are then proportional.

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