If x is real, then the maximum and minimum va | vedclass

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

Question

(c) Let $y = \frac{{{x^2} - 3x + 4}}{{{x^2} + 3x + 4}}$

==> $(y - 1){x^2} + 3(y + 1)x + 4(y - 1) = 0$

For $x$ is real $D \ge 0$

==> $9{

Vedclass · Accepted Answer

$7,\frac{1}{7}$

Vedclass · Answer

(c) Let $y = \frac{{{x^2} - 3x + 4}}{{{x^2} + 3x + 4}}$

==> $(y - 1){x^2} + 3(y + 1)x + 4(y - 1) = 0$

For $x$ is real $D \ge 0$

==> $9{(y + 1)^2} - 16{(y - 1)^2} \ge 0$

==> $ - 7{y^2} + 50y - 7 \ge 0$==>$7{y^2} - 50y + 7 \le 0$

==> $(y - 7)(7y - 1) \le 0$

Now, the product of two factors is negative if one in $ - ve$and one in $ + ve$.

Case I : $(y - 7) \ge 0$ and $(7y - 1) \le 0$

==> $y \ge 7$and $y \le \frac{1}{7}$. But it is impossible

Case II : $(y - 7) \le 0$and $(7y - 1) \ge 0$

==> $y \le 7$and $y \ge \frac{1}{7} \Rightarrow \frac{1}{7} \le y \le 7$

Hence maximum value is $7$ and minimum value is $\frac{1}{7}$

If $x$ is real, then the maximum and minimum values of the expression $\frac{{{x^2} - 3x + 4}}{{{x^2} + 3x + 4}}$ will be

Similar Questions

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

If $\log _{(3 x-1)}(x-2)=\log _{\left(9 x^2-6 x+1\right)}\left(2 x^2-10 x-2\right)$, then $x$ equals

Let $\alpha, \beta, \gamma$ be the three roots of the equation $x ^3+ bx + c =0$. If $\beta \gamma=1=-\alpha$, then $b^3+2 c^3-3 \alpha^3-6 \beta^3-8 \gamma^3$ is equal to $......$.

The set of all $a \in R$ for which the equation $x | x -1|+| x +2|+a=0$ has exactly one real root is:

Let $a, b, c$ be non-zero real numbers such that $a+b+c=0$, let $q=a^2+b^2+c^2$ and $r=a^4+b^4+c^4$. Then,