Let x_1, x_2, x_3 in mathbb{R} setminus {0},x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} = \frac{1}{x_1 + x_2 + x_3}$.This simplifies to $\frac{x_1 x_2 + x_2 x_3 + x_3 x_1}{x_1 x_2 x

Vedclass · Accepted Answer

all odd integers $n$

Vedclass · Answer

(B) Given $\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} = \frac{1}{x_1 + x_2 + x_3}$.This simplifies to $\frac{x_1 x_2 + x_2 x_3 + x_3 x_1}{x_1 x_2 x_3} = \frac{1}{x_1 + x_2 + x_3}$.$(x_1 + x_2 + x_3)(x_1 x_2 + x_2 x_3 + x_3 x_1) = x_1 x_2 x_3$.Let $x_1, x_2, x_3$ be roots of the cubic equation $t^3 + \alpha t^2 + \beta t + \gamma = 0$,where $\alpha = -(x_1 + x_2 + x_3)$,$\beta = x_1 x_2 + x_2 x_3 + x_3 x_1$,and $\gamma = -x_1 x_2 x_3$.The equation becomes $(-\alpha)(\beta) = -(-\gamma) \implies \gamma = \alpha \beta$.Substituting this into the cubic equation: $t^3 + \alpha t^2 + \beta t + \alpha \beta = 0$.$t^2(t + \alpha) + \beta(t + \alpha) = 0 \implies (t + \alpha)(t^2 + \beta) = 0$.Thus,the roots are $t = -\alpha, \sqrt{-\beta}, -\sqrt{-\beta}$.Let $x_1 = -\alpha$,$x_2 = k$,and $x_3 = -k$ where $k = \sqrt{-\beta}$.Then $x_1^n + x_2^n + x_3^n = x_1^n + k^n + (-k)^n$.If $n$ is an odd integer,$k^n + (-k)^n = 0$,so $x_1^n + x_2^n + x_3^n = x_1^n$.Also,$\frac{1}{x_1^n} + \frac{1}{x_2^n} + \frac{1}{x_3^n} = \frac{1}{x_1^n} + \frac{1}{k^n} + \frac{1}{(-k)^n} = \frac{1}{x_1^n} + 0 = \frac{1}{x_1^n}$.Thus,the equality holds for all odd integers $n$.

Let $x_1, x_2, x_3 \in \mathbb{R} \setminus \{0\}$,$x_1 + x_2 + x_3 \neq 0$ and $\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} = \frac{1}{x_1 + x_2 + x_3}$. Then $\frac{1}{x_1^n + x_2^n + x_3^n} = \frac{1}{x_1^n} + \frac{1}{x_2^n} + \frac{1}{x_3^n}$ holds good for:

Similar Questions

The solution of the equation $\sqrt{x + 10} + \sqrt{x - 2} = 6$ is

What is the number of real solutions for the equation $x^2 - 3|x| + 2 = 0$?

If one root of the equation $x^3-6x^2+3x+10=0$ is the average of the other two,then the sum of the fourth powers of the roots of the equation is

If $(3+2 \sqrt{2})^{x^2-4}+(3-2 \sqrt{2})^{x^2-4}=6$,then $x^4+x^2+5=$

One root of the given equation $2x^5 - 14x^4 + 31x^3 - 64x^2 + 19x + 130 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module