Question

If x is real, then the value of ${x^2} - 6x + 13$ will not be less than
A. 4
B. 6
C. 7
D. 8

Answer 1

Hint: Equate the above expression to y. Use the fact that the discriminant must be greater than or equal to 0 if the roots of the equation are real. Find the range of y and the minimum value of the expression can be found using this range.

Formula used: Discriminant of the standard quadratic equation $a{x^2} + bx + c = 0$ is ${b^2} - 4ac$

Complete step by step solution:
Let $y = {x^2} - 6x + 13$
${x^2} - 6x + 13 - y = 0$
It is given to us that x is real. Therefore, the discriminant must be greater than or equal to 0.
(Discriminant of the standard quadratic equation $a{x^2} + bx + c = 0$ is ${b^2} - 4ac$).
Therefore, ${\left( { - 6} \right)^2} - 4\left( 1 \right)\left( {13 - y} \right) \geqslant 0$
$36 - 52 + 4y \geqslant 0$
$4y \geqslant 16$
Dividing both sides of the inequality by 4,
$y \geqslant 4$
$y \in [4,\infty )$
The minimum value of y is 4.
Therefore, the correct answer is option A. 4

Note: This question can also be solved using applications of derivatives.
Let $f(x) = {x^2} - 6x + 13$.
Now find the first derivative of $f(x)$ and equate it to 0 to get the critical point.
$f'(x) = 2x - 6 = 0$
$x = 3$
Find the second derivative at the critical point to check if it is the point of minima or point of maxima.
$f''(x) = 2 > 0$. Since $f''(3) > 0$, we will get the minimum value of $f(x)$ at $x = 3$.
$f(3) = {3^2} - 6(3) + 13 = 9 - 18 + 13 = 4$. Since 4 is the minimum value of $f(x)$, ${x^2} - 6x + 13$ cannot be less than 4.