Q.3. Let and be two distinct roots of the equation . Let the sets {a: and are the eccentricities

Q: What is the correct answer for JEE Main Mathematics Q3?

The correct answer is C. Sol. For the quadratic equation $x^2 - ax + 2 = 0$, the roots are $e_1$ and $e_2$. By Vieta's formulas: $e_1 + e_2 = a$ and $e_1 e_2 = 2$. Case 1: Both $e_1$ and $e_2$ are eccentricities of hyperbolas. This means $e_1 > 1$ and $e_2 > 1$. Since $e_1 e_2 = 2$, we can write $e_2 = \frac{2}{e_1}$. So, $e_1 > 1$ and $\frac{2}{e_1} > 1 \Rightarrow e_1 < 2$. Thus, $e_1 \in (1,2)$. The sum $a = e_1 + e_2

Q: Q.3. Let and be two distinct roots of the equation . Let the sets {a: and are the eccentricities of hyperbolas} = (, ), and {a: and are the eccentricities of an ellipse and a hyperbola,…

The correct answer is C. Sol. For the quadratic equation $x^2 - ax + 2 = 0$, the roots are $e_1$ and $e_2$. By Vieta's formulas: $e_1 + e_2 = a$ and $e_1 e_2 = 2$. Case 1: Both $e_1$ and $e_2$ are eccentricities of hyperbolas. This means $e_1 > 1$ and $e_2 > 1$. Since $e_1 e_2 = 2$, we can write $e_2 = \frac{2}{e_1}$. So, $e_1 > 1$ and $\frac{2}{e_1} > 1 \Rightarrow e_1 < 2$. Thus, $e_1 \in (1,2)$. The sum $a = e_1 + e_2

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

A. 18 · B. 22 · C. 26 · D. 34

Question

Q.3. Let  and  be two distinct roots of the equation . Let the sets {a:  and  are the eccentricities of hyperbolas} = (, ), and {a:  and  are the eccentricities of an ellipse and a hyperbola, respectively} = (, ). The  is equal to:

FirstInTest · Accepted Answer

Sol. For the quadratic equation $x^2 - ax + 2 = 0$, the roots are $e_1$ and $e_2$. By Vieta's formulas: $e_1 + e_2 = a$ and $e_1 e_2 = 2$. Case 1: Both $e_1$ and $e_2$ are eccentricities of hyperbolas. This means $e_1 > 1$ and $e_2 > 1$. Since $e_1 e_2 = 2$, we can write $e_2 = \frac{2}{e_1}$. So, $e_1 > 1$ and $\frac{2}{e_1} > 1 \Rightarrow e_1 < 2$. Thus, $e_1 \in (1,2)$. The sum $a = e_1 + e_2 = e_1 + \frac{2}{e_1}$. To find the range of $a$, we can analyze the function $f(x) = x + \frac{2}{x}$ for $x \in (1,2)$. $f'(x) = 1 - \frac{2}{x^2}$. $f'(x) = 0 \Rightarrow x^2 = 2 \Rightarrow x = \sqrt{2}$. The minimum value occurs at $x = \sqrt{2}$, so $a_{min} = \sqrt{2} + \frac{2}{\sqrt{2}} = 2\sqrt{2}$. As $e_1 o 1^+$, $a o 1+2 = 3$. As $e_1 o 2^-$, $a o 2+1 = 3$. So, $a \in [2\sqrt{2}, 3)$. Therefore, $(\alpha, \beta) = (2\sqrt{2}, 3)$, which means $\alpha = 2\sqrt{2}$ and $\beta = 3$. Case 2: One eccentricity is for an ellipse and the other for a hyperbola. Let $e_1$ be for an ellipse and $e_2$ for a hyperbola. So, $0 < e_1 < 1$ and $e_2 > 1$. Again, $e_1 e_2 = 2 \Rightarrow e_2 = \frac{2}{e_1}$. Since $0 < e_1 < 1$, then $e_2 = \frac{2}{e_1}$ will be greater than 2 ($e_2 > 2$). The sum $a = e_1 + e_2 = e_1 + \frac{2}{e_1}$. As $e_1 o 0^+$, $a o \infty$. As $e_1 o 1^-$, $a o 1+2 = 3$. So, $a \in (3, \infty)$. Therefore, $(\gamma, \infty) = (3, \infty)$, which means $\gamma = 3$. Finally, $\alpha^2 + \beta^2 + \gamma^2 = (2\sqrt{2})^2 + 3^2 + 3^2 = 8 + 9 + 9 = 26$.

Answer

18

Answer

22

Answer

26

Answer

34

Related Questionsin Mathematics

More from JEE Main

Browse Engineering Questions by Subject

Practice More JEE Main Questions