The set of real values of x satisfying l o g 1 2 ( x 2 − 6 x + 12 ) ≥ − 2 is Options: A . &nbsp ( − ∞ , 2 ] B . &nbsp [ 2 , 4 ] C . &nbsp [ 4 , + ∞ ] D . &nbsp N o n e o f t h e s e

Answer: Option B : B l o g 1 2 ( x 2 − 6 x + 12 ) ≥ − 2 ...(i) For log to be defined, x 2 − 6 x + 12 > 0 ⇒ ( x − 3 ) 2 + 3 > 0 , which is true ∀ x ϵ R . From (i), x 2 − 6 x + 12 ≤ ( 1 2 ) − 2 ⇒ x 2 − 6 x + 12 ≤ 4 ⇒ x 2 − 6 x + 8 ≤ 0 ⇒ ( x − 2 ) ( x − 4 ) ≤ 0 ⇒ 2 ≤ x ≤ 4 ; ∴ x ϵ [ 2 , 4 ] .

The set of real values of x satisfying log12(x2−6x+12)≥−2

Discussion Forum : Inequalities Modulus And Logarithms

Question -

The set of real values of x satisfying $l o g_{\frac{1}{2}} (x^{2} - 6 x + 12) \geq - 2$ is

Options:

A .

(- \infty, 2]

B .

[2, 4]

C .

[4, + \infty]

D .

N o n e o f t h e s e

Answer: Option B
:
B

l o g_{\frac{1}{2}} (x^{2} - 6 x + 12) \geq - 2

...(i)
For log to be defined,

x^{2} - 6 x + 12 > 0

\Rightarrow (x - 3)^{2} + 3 > 0

, which is true

\forall x ϵ R .

From (i),

x^{2} - 6 x + 12 \leq (\frac{1}{2})^{- 2}

\Rightarrow x^{2} - 6 x + 12 \leq 4 \Rightarrow x^{2} - 6 x + 8 \leq 0 \Rightarrow (x - 2) (x - 4) \leq 0 \Rightarrow 2 \leq x \leq 4; ∴ x ϵ [2, 4] .

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Discussion Forum : Inequalities Modulus And Logarithms

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