NVAMediumJEE Main 2025 · 7 April, Shift 2Limits

Mathematics Question from JEE Main 2025 · 7 April, Shift 2

For t>1t > -1, let αt\alpha_t and βt\beta_t be the roots of the equation

((t+2)171)x2+((t+2)161)x+((t+2)1211)=0.\left( (t + 2)^{\frac{1}{7}} - 1 \right)x^2 + \left( (t + 2)^{\frac{1}{6}} - 1 \right)x + \left( (t + 2)^{\frac{1}{21}} - 1 \right) = 0.

If limt1+αt=a\lim_{t \to -1^+} \alpha_t = a and limt1+βt=b\lim_{t \to -1^+} \beta_t = b, then 72(a+b)272(a + b)^2 is equal to:

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