Let \alpha, \beta be the distinct roots of the equation x^2-(t^2-5 t+6) x+1=0, t \in \mathbbR and a_n=\alpha^n+\beta^n . Then the minimum value of \fraca_2023+a_2025a_2024 is

Quadratic Equation and Inequalities PYQ — JEE Main 2024 · 6 Apr · Shift 1 · Q41 with Solution

View written solutionFree

Correct answer: B

Given quadratic and basic root relations

The equation is $x^2-(t^2-5t+6)x+1=0.$ Let its distinct roots be $\alpha,\beta$ . Then by Vieta,

\qquad \alpha\beta=1.$$ Also, $$a_n=\alpha^n+\beta^n.$$ --- 2. **Simplify the required expression** We need the minimum value of $$\frac{a_{2023}+a_{2025}}{a_{2024}}.

Now, $a_{2023}+a_{2025}=\alpha^{2023}+\beta^{2023}+\alpha^{2025}+\beta^{2025}.$ Group terms: $=\alpha^{2023}(1+\alpha^2)+\beta^{2023}(1+\beta^2).$

A better way is to use $\alpha\beta=1$ . Since $\beta=\frac1\alpha$ , $a_n=\alpha^n+\alpha^{-n}.$ Then

=(\alpha^{n-1}+\alpha^{-(n-1)})+(\alpha^{n+1}+\alpha^{-(n+1)}).$$ Factor: $$=\alpha^{n-1}(1+\alpha^2)+\alpha^{-(n+1)}(1+\alpha^2)$$ $$=(\alpha+\alpha^{-1})(\alpha^n+\alpha^{-n}).$$ Hence, $$a_{n-1}+a_{n+1}=(\alpha+\beta)a_n.$$ Taking $n=2024$, $$a_{2023}+a_{2025}=(\alpha+\beta)a_{2024}.$$ Therefore, $$\frac{a_{2023}+a_{2025}}{a_{2024}}=\alpha+\beta=t^2-5t+6.$$ So the problem reduces to finding the minimum of $$t^2-5t+6.$$ --- 3. **Minimize the quadratic in $t$** Complete the square: $$t^2-5t+6=\left(t-\frac52\right)^2+6-\frac{25}{4} =\left(t-\frac52\right)^2-\frac14.$$ Thus the minimum value is $$-\frac14,$$ achieved at $$t=\frac52.$$ --- 4. **Check distinct roots condition** The roots are distinct if discriminant $>0$: $$ (\alpha+\beta)^2-4\alpha\beta >0 \implies (t^2-5t+6)^2-4>0.$$ At the minimizing value $t=\frac52$, $$\alpha+\beta=-\frac14,$$ and discriminant becomes $$\left(-\frac14\right)^2-4<0,$$ so the quadratic would not have real roots. But the question only says $\alpha,\beta$ are distinct roots of the given equation, not necessarily real. Distinct complex roots are allowed. Distinctness for the quadratic in $x$ requires $$ (t^2-5t+6)^2-4\neq 0,$$ which is true at $t=\frac52$ because $$\frac{1}{16}-4\neq 0.$$ So $t=\frac52$ is valid. Also, $a_{2024}\neq 0$ at this value: if $\alpha=e^{i\theta},\beta=e^{-i\theta}$ with $$\alpha+\beta=2\cos\theta=-\frac14,$$ then $$a_{2024}=2\cos(2024\theta),$$ which is not forced to be zero by the given condition, and in fact the identity already shows the ratio equals $\alpha+\beta$ whenever defined. For the valid minimizing parameter, the expression evaluates to $-\frac14$. --- 5. **Option check** - A: $-\frac12$ ❌ - B: $-\frac14$ ✅ - C: $\frac14$ ❌ - D: $\frac12$ ❌ Therefore, the correct answer is $$\boxed{-\frac14}.$$

More from Quadratic Equation and Inequalities

Let $x_{1}, x_{2}, x_{3}, x_{4}$ be the solution of the equation $4 x^{4} + 8 x^{3} - 17 x^{2} - 12 x + 9 = 0$ and $(4 + x_{1}^{2}) (4 + x_{2}^{2}) (4 + x_{3}^{2}) (4 + x_{4}^{2}) = \frac{125}{16} m$ . Then the value of $m$ is …2024 · Numerical
Let $α, β$ be roots of $x^{2} + 2 x - 8 = 0$ . If $U_{n} = α^{n} + β^{n}$ , then $\frac{U _{10} + 2 U _{9}}{2 U _{8}}$ is equal to .2024 · Numerical
The sum of all the solutions of the equation $(8)^{2 x} - 16 \cdot (8)^{x} + 48 = 0$ is :2024 · MCQ
The number of distinct real roots of the equation $∣ x + 1∣∣ x + 3∣ - 4∣ x + 2∣ + 5 = 0$ , is 2024 · Numerical
Let $α, β$ be the roots of the equation $x^{2} + 22 x - 1 = 0$ . The quadratic equation, whose roots are $α^{4} + β^{4}$ and $\frac{1}{10} (α^{6} + β^{6})$ , is:2024 · MCQ
Let $α, β; α > β$ , be the roots of the equation $x^{2} - 2 x - 3 = 0$ . Let $P_{n} = α^{n} - β^{n}, n \in N$ . Then $(113 - 102) P_{10} + (112 + 10) P_{11} - 11 P_{12}$ …2024 · MCQ
If $α, β$ are the roots of the equation, $x^{2} - x - 1 = 0$ and $S_{n} = 2023 α^{n} + 2024 β^{n}$ , then :2024 · MCQ
Let the set $C = {(x, y) ∣ x^{2} - 2^{y} = 2023, x, y \in N}$ . Then $\sum_{(x, y) \in C} (x + y)$ is equal to .2024 · Numerical

View written solutionFree

Correct answer: B

Given quadratic and basic root relations

The equation is $x^2-(t^2-5t+6)x+1=0.$ Let its distinct roots be $\alpha,\beta$ . Then by Vieta,

\qquad \alpha\beta=1.$$ Also, $$a_n=\alpha^n+\beta^n.$$ --- 2. **Simplify the required expression** We need the minimum value of $$\frac{a_{2023}+a_{2025}}{a_{2024}}.

Now, $a_{2023}+a_{2025}=\alpha^{2023}+\beta^{2023}+\alpha^{2025}+\beta^{2025}.$ Group terms: $=\alpha^{2023}(1+\alpha^2)+\beta^{2023}(1+\beta^2).$

A better way is to use $\alpha\beta=1$ . Since $\beta=\frac1\alpha$ , $a_n=\alpha^n+\alpha^{-n}.$ Then

=(\alpha^{n-1}+\alpha^{-(n-1)})+(\alpha^{n+1}+\alpha^{-(n+1)}).$$ Factor: $$=\alpha^{n-1}(1+\alpha^2)+\alpha^{-(n+1)}(1+\alpha^2)$$ $$=(\alpha+\alpha^{-1})(\alpha^n+\alpha^{-n}).$$ Hence, $$a_{n-1}+a_{n+1}=(\alpha+\beta)a_n.$$ Taking $n=2024$, $$a_{2023}+a_{2025}=(\alpha+\beta)a_{2024}.$$ Therefore, $$\frac{a_{2023}+a_{2025}}{a_{2024}}=\alpha+\beta=t^2-5t+6.$$ So the problem reduces to finding the minimum of $$t^2-5t+6.$$ --- 3. **Minimize the quadratic in $t$** Complete the square: $$t^2-5t+6=\left(t-\frac52\right)^2+6-\frac{25}{4} =\left(t-\frac52\right)^2-\frac14.$$ Thus the minimum value is $$-\frac14,$$ achieved at $$t=\frac52.$$ --- 4. **Check distinct roots condition** The roots are distinct if discriminant $>0$: $$ (\alpha+\beta)^2-4\alpha\beta >0 \implies (t^2-5t+6)^2-4>0.$$ At the minimizing value $t=\frac52$, $$\alpha+\beta=-\frac14,$$ and discriminant becomes $$\left(-\frac14\right)^2-4<0,$$ so the quadratic would not have real roots. But the question only says $\alpha,\beta$ are distinct roots of the given equation, not necessarily real. Distinct complex roots are allowed. Distinctness for the quadratic in $x$ requires $$ (t^2-5t+6)^2-4\neq 0,$$ which is true at $t=\frac52$ because $$\frac{1}{16}-4\neq 0.$$ So $t=\frac52$ is valid. Also, $a_{2024}\neq 0$ at this value: if $\alpha=e^{i\theta},\beta=e^{-i\theta}$ with $$\alpha+\beta=2\cos\theta=-\frac14,$$ then $$a_{2024}=2\cos(2024\theta),$$ which is not forced to be zero by the given condition, and in fact the identity already shows the ratio equals $\alpha+\beta$ whenever defined. For the valid minimizing parameter, the expression evaluates to $-\frac14$. --- 5. **Option check** - A: $-\frac12$ ❌ - B: $-\frac14$ ✅ - C: $\frac14$ ❌ - D: $\frac12$ ❌ Therefore, the correct answer is $$\boxed{-\frac14}.$$

Quadratic Equation and Inequalities question

Quadratic Equation and Inequalities question

More from Quadratic Equation and Inequalities

Quadratic Equation and Inequalities question

Quadratic Equation and Inequalities question

More from Quadratic Equation and Inequalities