If α > β > 0 are the roots of the equation a x²+b x+1=0, and lim x → 1/α((1-cos (x²+b x+a))/2(1-α x)²)^(1/2)=1/k(1/β-1/α), then k is equal to :

Solution

Maths sir · 4 steps

x²+bx+a has roots 1/α, 1/β, so factorise it and cancel 1-α x.

1. Reciprocal roots. Put x → 1/x in ax²+bx+1=0 and multiply by x² to get x²+bx+a=0. Its roots are 1/α and 1/β, so x²+bx+a = (x-1/α)(x-1/β).
2. Use half angle. Call t = x²+bx+a, so t → 0 as x → 1/α. Since 1-cos t = 2 sin² t/2, the expression becomes (|sin t/2|)/(|1-α x|). Near 0, sin t/2 ≈ t/2, so it behaves like |t|/(2|1-α x|).
3. Cancel the factor. Write 1-α x = -α(x-1/α). The factor (x-1/α) cancels, leaving 1/2α|x-1/β|. Since α > β > 0, 1/β > 1/α, so the modulus opens as 1/β-1/α.

   lim(x → 1/α) 1/2α|x-1/β| = 1/2α(1/β-1/α)

4. So. Comparing with 1/k(1/β-1/α) gives k = 2α, option (d).

The idea to keep. Swapping the leading and constant coefficients of a quadratic gives reciprocal roots. With 1-cos t over a square, write it as 2 sin² t/2 and cancel the common factor.

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