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 a real number,the discriminant $D$ must be greater than or equal to $0$.$D = (k + 1)^2 - 4(k - 1)(k - 1) \ge 0$.$(k + 1)^2 - 4(k - 1)^2 \ge 0$.Using the identity $a^2 - b^2 = (a - b)(a + b)$:$((k + 1) - 2(k - 1))((k + 1) + 2(k - 1)) \ge 0$.$(k + 1 - 2k + 2)(k + 1 + 2k - 2) \ge 0$.$(-k + 3)(3k - 1) \ge 0$.Multiplying by $-1$ reverses the inequality sign:$(k - 3)(3k - 1) \le 0$.The roots are $k = 3$ and $k = \frac{1}{3}$.Thus,the inequality holds for $\frac{1}{3} \le k \le 3$.

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

Similar Questions

If the roots of $x^2 - 7x + 6 = 0$ are $\alpha$ and $\beta$,then $\frac{1}{\alpha} + \frac{1}{\beta} = $

Construct a quadratic equation whose roots have the sum $= 6$ and product $= -16$.

${x \in R : |x - 2| = x^2} = $

If the equation $\frac{1}{x} + \frac{1}{x - 1} + \frac{1}{x - 2} = 3x^3$ has $k$ real roots,then $k$ is equal to -

If $x+y+z=0,$ then find the value of $\frac{(x+y)(y+z)(z+x)}{x y z}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module