If x is real,then the minimum value of y = fr | vedclass

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

Question

(B) Let $y = \frac{x^2-x+1}{x^2+x+1}$.Multiplying both sides by $(x^2+x+1)$,we get $y(x^2+x+1) = x^2-x+1$.Rearranging the terms as a quadratic in $x$:

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(B) Let $y = \frac{x^2-x+1}{x^2+x+1}$.Multiplying both sides by $(x^2+x+1)$,we get $y(x^2+x+1) = x^2-x+1$.Rearranging the terms as a quadratic in $x$:$(y-1)x^2 + (y+1)x + (y-1) = 0$.Since $x$ is real,the discriminant $D \geq 0$.$D = (y+1)^2 - 4(y-1)^2 \geq 0$.$(y+1)^2 - [2(y-1)]^2 \geq 0$.Using $a^2-b^2 = (a-b)(a+b)$:$[(y+1) - 2(y-1)][(y+1) + 2(y-1)] \geq 0$.$(-y+3)(3y-1) \geq 0$.$(y-3)(3y-1) \leq 0$.This inequality holds for $\frac{1}{3} \leq y \leq 3$.Thus,the minimum value of $y$ is $\frac{1}{3}$.

If $x$ is real,then the minimum value of $y = \frac{x^2-x+1}{x^2+x+1}$ is

Similar Questions

Let $f(x) = x^2 + ax + b$,where $a, b \in R$. If $f(x) = 0$ has all its roots imaginary,then the roots of $f(x) + f'(x) + f''(x) = 0$ are

Suppose $m, n$ are positive integers such that $6^m + 2^{m+n} \cdot 3^w + 2^n = 332$. The value of the expression $m^2 + mn + n^2$ is

If ${b_1}{b_2} = 2({c_1} + {c_2})$,then at least one of the equations ${x^2} + {b_1}x + {c_1} = 0$ and ${x^2} + {b_2}x + {c_2} = 0$ has

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

The roots of the equation $(p - q)x^2 + (q - r)x + (r - p) = 0$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module