If x is real and k = frac{x^2 - x + 1}{x^2 + | vedclass

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

Question

(A) Given $k = \frac{x^2 - x + 1}{x^2 + x + 1}$.Rearranging the terms,we get $k(x^2 + x + 1) = x^2 - x + 1$.$kx^2 + kx + k = x^2 - x + 1$.$(k - 1)x^2

Vedclass · Accepted Answer

$\frac{1}{3} \le k \le 3$

Vedclass · Answer

(A) Given $k = \frac{x^2 - x + 1}{x^2 + x + 1}$.Rearranging the terms,we get $k(x^2 + x + 1) = x^2 - x + 1$.$kx^2 + kx + k = x^2 - x + 1$.$(k - 1)x^2 + (k + 1)x + (k - 1) = 0$.Since $x$ is real,the discriminant $D \ge 0$.$D = (k + 1)^2 - 4(k - 1)(k - 1) \ge 0$.$(k + 1)^2 - 4(k - 1)^2 \ge 0$.Using $a^2 - b^2 = (a - b)(a + b)$,we get $((k + 1) - 2(k - 1))((k + 1) + 2(k - 1)) \ge 0$.$(-k + 3)(3k - 1) \ge 0$.Multiplying by $-1$,we get $(k - 3)(3k - 1) \le 0$.The roots are $k = \frac{1}{3}$ and $k = 3$.Thus,$\frac{1}{3} \le k \le 3$.

If $x$ is real and $k = \frac{x^2 - x + 1}{x^2 + x + 1}$,then

Similar Questions

If $x+\frac{1}{x}=a$ and $x^2+\frac{1}{x^3}=b$,then $x^3+\frac{1}{x^2}$ is

For $x \in R$,the number of real roots of the equation $3x^2 - 4|x^2 - 1| + x - 1 = 0$ is:

Let $\tan A$ and $\tan B$, where $A, B \in (-\frac{\pi}{2}, \frac{\pi}{2})$, be the roots of the quadratic equation $x^2 - 2x - 5 = 0$. Then $20 \sin^2(\frac{A+B}{2})$ is equal to:

If $x$ is real,then the value of $\frac{x^2 + 34x - 71}{x^2 + 2x - 7}$ does not lie between

Let $[r]$ denote the greatest integer not exceeding $r$. The roots of the equation $3 x^2 + 6 x + 5 + \alpha (x^2 + 2 x + 2) = 0$ are complex numbers whenever $\alpha > L$ or $\alpha < M$. If $(L - M)$ is minimum,then the greatest value of $[r]$ such that $L y^2 + M y + r < 0$ for all $y \in R$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module