If x is real, then the value of {x^2} - 6x + | vedclass

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

Question

(a) Let $y = {x^2} - 6x + 13 \Rightarrow {x^2} - 6x + 13 - y = 0$

Its discriminant $D \ge 0 \Rightarrow 36 - 4(13 - y) \ge 0$

==>$36 - 52 + 4

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(a) Let $y = {x^2} - 6x + 13 \Rightarrow {x^2} - 6x + 13 - y = 0$

Its discriminant $D \ge 0 \Rightarrow 36 - 4(13 - y) \ge 0$

==>$36 - 52 + 4y \ge 0 \Rightarrow 4y \ge 16 \Rightarrow y \ge 4$

Hence $y$ is not less than $4.$

Aliter : ${x^2} - 6x + 13 = {(x - 3)^2} + 4$

Obviously the minimum value is $4.$

If $x$ is real, then the value of ${x^2} - 6x + 13$ will not be less than

Similar Questions

Let $f: R -\{0\} \rightarrow(-\infty, 1)$ be a polynomial of degree $2$ , satisfying $f( x ) f\left(\frac{1}{ x }\right)=f( x )+f\left(\frac{1}{ x }\right)$. If $f(K)=-2 K$, then the sum of squares of all possible values of $K$ is :

The number of solutions of the equation $\left(\frac{9}{x}-\frac{9}{\sqrt{x}}+2\right)\left(\frac{2}{x}-\frac{7}{\sqrt{x}}+3\right)=0$ is :

Suppose that $x$ and $y$ are positive number with $xy = \frac{1}{9};\,x\left( {y + 1} \right) = \frac{7}{9};\,y\left( {x + 1} \right) = \frac{5}{{18}}$ . The value of $(x + 1) (y + 1)$ equals

The number of integers $a$ in the interval $[1,2014]$ for which the system of equations $x+y=a$, $\frac{x^2}{x-1}+\frac{y^2}{y-1}=4$ has finitely many solutions is

If the set of all $a \in R$, for which the equation $2 x^2+$ $(a-5) x+15=3 a$ has no real root, is the interval $(\alpha, \beta)$, and $X=\{x \in Z: \alpha < x < \beta\}$, then $\sum_{x \in X} x^2$ is equal to