The set of real values of x that satisfy the | vedclass

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

Question

(B) Given the inequality: ${\log _{1/2}}({x^2} - 6x + 12) \ge - 2$. Since the base of the logarithm is $1/2$ (which is between $0$ and $1$),the inequa

Vedclass · Accepted Answer

$[2, 4]$

Vedclass · Answer

(B) Given the inequality: ${\log _{1/2}}({x^2} - 6x + 12) \ge - 2$. Since the base of the logarithm is $1/2$ (which is between $0$ and $1$),the inequality sign reverses when we remove the logarithm: ${x^2} - 6x + 12 \le {(1/2)^{-2}}$. ${x^2} - 6x + 12 \le 4$. ${x^2} - 6x + 8 \le 0$. Factoring the quadratic: $(x - 2)(x - 4) \le 0$. The solution to this inequality is $x \in [2, 4]$. We must also ensure the argument of the logarithm is positive: ${x^2} - 6x + 12 > 0$. The discriminant $D = {(-6)^2} - 4(1)(12) = 36 - 48 = -12 Since the leading coefficient is positive and $D Thus,the solution set is $[2, 4]$.

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

Similar Questions

If $\log 2=a, \log 3=b, \log 7=c$ and $6^x=7^{x+4}$, then $x$ is equal to:

Let $x$ and $y$ be real numbers such that $x > 2y > 0$ and $2 \log (x - 2y) = \log x + \log y$. Then,the possible value$(s)$ of $\frac{x}{y}$ is/are:

The value of $\log_{20} 3$ lies in which interval?

The sum of the roots of the equation $x+1-2 \log _{2}\left(3+2^{x}\right)+2 \log _{4}\left(10-2^{-x}\right)=0$ is:

Number of solution$(s)$ of the equation $\ln(1 + \sin^2 x) = 1 - \ln(5 + x^2)$ is -

Vedclass Test Series

Exam Paper Generator

Online Exam Module