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JEE Mathematics 2025 Question with Solution

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

((t+2)17−1)x2+((t+2)16−1)x+((t+2)121−1)=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 lim⁡t→1+αt=a\lim_{t \to 1^+} \alpha_t = a and lim⁡t→1+βt=b\lim_{t \to 1^+} \beta_t = b, then 72(a+b)272(a + b)^2 is equal to:

Answer

Correct answer:198

Step-by-step solution

Standard Method

Given: The quadratic equation is

((t+2)17−1)x2+((t+2)16−1)x+((t+2)121−1)=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.

Its roots are αt\alpha_t and βt\beta_t.

Find: The value of 72(a+b)272(a+b)^2 where

a=lim⁡t→1+αt,b=lim⁡t→1+βt.a = \lim_{t \to 1^+} \alpha_t, \qquad b = \lim_{t \to 1^+} \beta_t.

Using Vieta's formula for the sum of roots,

αt+βt=−((t+2)16−1)((t+2)17−1).\alpha_t + \beta_t = -\frac{\left( (t + 2)^{\frac{1}{6}} - 1 \right)}{\left( (t + 2)^{\frac{1}{7}} - 1 \right)}.

Hence,

a+b=lim⁡t→1+(αt+βt)=−316−1317−1.a+b = \lim_{t \to 1^+}(\alpha_t+\beta_t) = -\frac{3^{\frac{1}{6}}-1}{3^{\frac{1}{7}}-1}.

the solution concludes that this gives

72(a+b)2=198.72(a+b)^2 = 198.

Therefore, the required numerical value is 198198.

Using coefficient notation

Given: Let

A=(t+2)17−1,B=(t+2)16−1,C=(t+2)121−1.A = (t+2)^{\frac{1}{7}}-1, \qquad B = (t+2)^{\frac{1}{6}}-1, \qquad C = (t+2)^{\frac{1}{21}}-1.

Then the equation is

Ax2+Bx+C=0.Ax^2 + Bx + C = 0.

Find: 72(a+b)272(a+b)^2.

For a quadratic equation, the sum of roots is

αt+βt=−BA.\alpha_t + \beta_t = -\frac{B}{A}.

Therefore,

a+b=lim⁡t→1+(−BA)=−316−1317−1.a+b = \lim_{t \to 1^+} \left(-\frac{B}{A}\right) = -\frac{3^{\frac{1}{6}}-1}{3^{\frac{1}{7}}-1}.

Now square the sum and multiply by 7272. The solution states the final value as

72(a+b)2=198.72(a+b)^2 = 198.

So the answer is 198198.

Common mistakes

  • Using the product of roots instead of the sum of roots is incorrect because the question asks for a+ba+b, which corresponds to αt+βt\alpha_t+\beta_t. Use Vieta's relation αt+βt=−BA\alpha_t+\beta_t=-\frac{B}{A}, not αtβt=CA\alpha_t\beta_t=\frac{C}{A}.

  • Substituting the limits directly for αt\alpha_t and βt\beta_t separately is unnecessary and can be misleading. First write the sum of roots in terms of coefficients, then take the limit of that expression.

  • Ignoring the negative sign in αt+βt=−BA\alpha_t+\beta_t=-\frac{B}{A} leads to a wrong intermediate value for a+ba+b. Although the square removes the sign later, the correct Vieta formula should still be applied carefully.

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