The set of real values of x satisfying log _{ | vedclass

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

Question

(B) Given the inequality: $\log _{1/2}(x^2 - 6x + 12) \ge -2$ $(i)$For the logarithm to be defined,the argument must be positive: $x^2 - 6x + 12 > 0$.

Vedclass · Accepted Answer

$[ 2 , 4 ]$

Vedclass · Answer

(B) Given the inequality: $\log _{1/2}(x^2 - 6x + 12) \ge -2$ $(i)$For the logarithm to be defined,the argument must be positive: $x^2 - 6x + 12 > 0$.Since the discriminant $D = (-6)^2 - 4(1)(12) = 36 - 48 = -12  0$ for all $x \in \mathbb{R}$.Since the base of the logarithm is $1/2$ (which is between $0$ and $1$),the inequality sign reverses when we remove the logarithm:$x^2 - 6x + 12 \le (1/2)^{-2}$$x^2 - 6x + 12 \le 4$$x^2 - 6x + 8 \le 0$$(x - 2)(x - 4) \le 0$Solving the quadratic inequality,we get $2 \le x \le 4$.Thus,the solution set is $x \in [2, 4]$.

The set of real values of $x$ satisfying $\log _{1/2}(x^2 - 6x + 12) \ge -2$ is

Similar Questions

If $\log_2 x + \log_x 2 = \frac{10}{3} = \log_2 y + \log_y 2$ and $x \neq y$,then $x + y = $

If $A = \log _2 \log _2 \log _4 256 + 2 \log _{\sqrt{2}} 2$,then $A$ is equal to

The number of solutions of the equation $\log_{\sqrt{3}}(x^{3} - 1) = \log_{\sqrt{3}}(x - 1) + 2$ is:

The solution to the equation ${\log _7}{\log _5}(\sqrt {{x^2} + 5 + x} ) = 0$ is:

The solution to the equation ${9^x} - {2^{x + 1/2}} = {2^{x + 3/2}} - {3^{2x - 1}}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module