Fun with inequalities-3

Algebra Level 2

log ⁡ 3 ∣ x 2 − 4 x ∣ + 3 x 2 + ∣ x − 5 ∣ ≥ 0 \log _{ 3 }{ \frac { \left| { x }^{ 2 }-4x \right| +3 }{ { x }^{ 2 }+\left| x-5 \right| } } \ge 0

Solve the following for x x .

Try Fun with inequalities-1 and Fun with inequalities-2
x ∈ ( − ∞ , − 2 3 ] ∪ [ 1 2 , 2 ] x\in \left (-\infty ,-\frac { 2 }{ 3 } \right ]\cup \left [\frac { 1 }{ 2 } ,2 \right ] x ∈ ( − ∞ , 2 3 ] ∪ [ − 1 2 , 2 ] x\in \left (-\infty ,\frac { 2 }{ 3 } \right ]\cup \left [\frac { -1 }{ 2 } ,2 \right ] None of the above x ∈ [ − 2 3 , 1 2 ] ∪ [ 2 , ∞ ) x\in \left [\frac { -2 }{ 3 } ,\frac { 1 }{ 2 } \right ]\cup \left[2,\infty \right )

View wiki
Yash Choudhary by Yash Choudhary , 22, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

For log ⁡ 3 ∣ x 2 − 4 x ∣ + 3 x 2 + ∣ x − 5 ∣ ≥ 0 ⇒ ∣ x 2 − 4 x ∣ + 3 x 2 + ∣ x − 5 ∣ ≥ 1 \log_3 {\frac{|x^2-4x|+3}{x^2+|x-5|}} \ge 0\quad \Rightarrow \dfrac{|x^2-4x|+3}{x^2+|x-5|} \ge 1

⇒ x 2 + ∣ x − 5 ∣ − ∣ x 2 − 4 x ∣ − 3 ≤ 0 \Rightarrow x^2+|x-5|-|x^2-4x|-3 \le 0

For x < 0 x<0 :

⇒ x 2 − x + 5 − x 2 + 4 x − 3 = 3 x + 2 ≤ 0 ⇒ x ∈ ( − ∞ , − 2 3 ] \Rightarrow x^2-x+5-x^2+4x-3 = 3x+2 \le 0\quad \Rightarrow x \in (-\infty, -\frac{2}{3}]

For 0 ≤ x < 4 0\le x < 4 :

⇒ x 2 − x + 5 + x 2 − 4 x − 3 = 2 x 2 − 5 x + 2 ≤ 0 \Rightarrow x^2-x+5+x^2-4x-3 = 2x^2-5x+2 \le 0 ⇒ ( 2 x − 1 ) ( x − 2 ) ≤ 0 ⇒ x ∈ [ 1 2 , 2 ] \Rightarrow (2x-1)(x-2) \le 0\quad \Rightarrow x \in [\frac{1}{2},2]

For 4 ≤ x < 5 4\le x < 5 : ⇒ x 2 − x + 5 − x 2 + 4 x − 3 = 3 x + 2 > 0 \quad \Rightarrow x^2-x+5-x^2+4x-3 = 3x+2 > 0

⇒ ∣ x 2 − 4 x ∣ + 3 x 2 + ∣ x − 5 ∣ < 1 \Rightarrow \dfrac{|x^2-4x|+3}{x^2+|x-5|} < 1

For x ≥ 5 x \ge 5 : ⇒ x 2 + x − 5 − x 2 + 4 x − 3 = 5 x − 8 > 0 \quad \Rightarrow x^2+x-5-x^2+4x-3 = 5x-8 > 0

⇒ ∣ x 2 − 4 x ∣ + 3 x 2 + ∣ x − 5 ∣ < 1 \Rightarrow \dfrac{|x^2-4x|+3}{x^2+|x-5|} < 1

Therefore, log ⁡ 3 ∣ x 2 − 4 x ∣ + 3 x 2 + ∣ x − 5 ∣ ≥ 0 \log_3 {\frac{|x^2-4x|+3}{x^2+|x-5|}} \ge 0 when x ∈ ( − ∞ , − 2 3 ] ∪ [ 1 2 , 2 ] \boxed{x \in (-\infty, -\frac{2}{3}] \cup [\frac{1}{2}, 2]}

A plot of log ⁡ 3 ∣ x 2 − 4 x ∣ + 3 x 2 + ∣ x − 5 ∣ \log_3 {\frac{|x^2-4x|+3}{x^2+|x-5|}} clearly show the answer.

17 Helpful 0 Interesting 2 Brilliant 0 Confused

That's a very nice solution sir. Exactly how I did it.

Yash Choudhary - 6 years, 2 months ago

I like your solution, but could you elaborate more on how you knew to evaluate for x< 0 , 0<x<4, and 4<x<5 and x>5 ?

That was the information I was missing to solve the problem. Good explanation :)

Oli Hohman - 6 years, 1 month ago

Log in to reply

These are the values of x x where discontinuity starts due to the absolute value function ∣ ∣ | \quad | . You can see that in the graph. The values come from ∣ x 2 − 4 x ∣ |x^2-4x| for x < 0 , 0 ≤ x < 4 x<0, 0 \le x < 4 and ∣ x − 5 ∣ |x - 5| for 4 ≤ x < 5 , x ≥ 5 4 \le x < 5, x \ge 5 . For ease of calculation, we need to replace the original equation with ∣ ∣ |\quad | to one without at these values of x x .

For example,

∣ x − 5 ∣ = { x − 5 for x ≥ 5 e.g. : ∣ 5.5 − 5 ∣ = 5.5 − 5 = 0.5 5 − x for x < 5 e.g. : ∣ 5 − 4.5 ∣ = 5 − 4.5 = 0.5 |x -5| = \begin{cases} x - 5 \text{ for } x \ge 5 \text{ e.g.}: |5.5-5| = 5.5 - 5 = 0.5 \\ 5 - x \text{ for } x < 5 \text{ e.g.}: |5-4.5| = 5 - 4.5 = 0.5 \end{cases}

Chew-Seong Cheong - 6 years, 1 month ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...