Q.2. Consider the quadratic equation . Let be the minimum value of the product of its roots and be

Q: What is the correct answer for JEE Main Mathematics Q2?

The correct answer is C. Sol. Let $P = n^2 - 2n + 2 = (n-1)^2 + 1$. The minimum value of $P$ is $1$ (when $n=1$). The quadratic equation is $Px^2 - 3x + P^2 = 0$. Product of roots $P_r = \frac{P^2}{P} = P$. So, $\alpha = \min(P_r) = \min(P) = 1$. Sum of roots $S_r = \frac{3}{P}$. To maximize $S_r$, $P$ must be minimum. So, $\beta = \max(S_r) = \frac{3}{\min(P)} = \frac{3}{1} = 3$. The G.P. has first term $a = \alpha = 1$

Q: Q.2. Consider the quadratic equation . Let be the minimum value of the product of its roots and be the maximum value of the sum of its roots. Then the sum of the first six terms of the G.P., whose…

The correct answer is C. Sol. Let $P = n^2 - 2n + 2 = (n-1)^2 + 1$. The minimum value of $P$ is $1$ (when $n=1$). The quadratic equation is $Px^2 - 3x + P^2 = 0$. Product of roots $P_r = \frac{P^2}{P} = P$. So, $\alpha = \min(P_r) = \min(P) = 1$. Sum of roots $S_r = \frac{3}{P}$. To maximize $S_r$, $P$ must be minimum. So, $\beta = \max(S_r) = \frac{3}{\min(P)} = \frac{3}{1} = 3$. The G.P. has first term $a = \alpha = 1$

Q: What are the options for this JEE Main Mathematics question?

A. $\frac{61}{37}$ · B. $\frac{121}{81}$ · C. $\frac{364}{243}$ · D. $\frac{1093}{729}$

Question

Q.2. Consider the quadratic equation . Let  be the minimum value of the product of its roots and  be the maximum value of the sum of its roots. Then the sum of the first six terms of the G.P., whose first term is  and the common ratio is  is:

FirstInTest · Accepted Answer

Sol. Let $P = n^2 - 2n + 2 = (n-1)^2 + 1$. The minimum value of $P$ is $1$ (when $n=1$). The quadratic equation is $Px^2 - 3x + P^2 = 0$. Product of roots $P_r = \frac{P^2}{P} = P$. So, $\alpha = \min(P_r) = \min(P) = 1$. Sum of roots $S_r = \frac{3}{P}$. To maximize $S_r$, $P$ must be minimum. So, $\beta = \max(S_r) = \frac{3}{\min(P)} = \frac{3}{1} = 3$. The G.P. has first term $a = \alpha = 1$ and common ratio $r = \frac{\alpha}{\beta} = \frac{1}{3}$. The sum of the first six terms is $S_6 = a\frac{1-r^6}{1-r} = 1 \cdot \frac{1 - (\frac{1}{3})^6}{1 - \frac{1}{3}} = \frac{1 - \frac{1}{729}}{\frac{2}{3}} = \frac{\frac{728}{729}}{\frac{2}{3}} = \frac{728}{729} \cdot \frac{3}{2} = \frac{364}{243}$.

Answer

$\frac{61}{37}$

Answer

$\frac{121}{81}$

Answer

$\frac{364}{243}$

Answer

$\frac{1093}{729}$

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