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Solution of x−2x+5>2\dfrac{x - 2}{x + 5} > 2x+5x−2​>2​ is:
Question

Solution of x−2x+5>2\dfrac{x - 2}{x + 5} > 2​ is:

A.

​x∈(5,12)x \in (5, 12)​​

B.

​x∈(−12,−7)x \in (-12, -7)​​

C.

​x∈(−12,−5)x \in (-12, -5)​​

D.

​x∈(−12,5)x \in (-12, 5)​​

Correct option is C

Given inequality:
​x−2x+5>2\dfrac{x - 2}{x + 5} > 2​​
Solution:
​x−2x+5−2>0 =>x−2−2(x+5)x+5>0 =>x−2−2x−10x+5>0 =>−x−12x+5>0\dfrac{x - 2}{x + 5} - 2 > 0 \\ \ \\\Rightarrow \dfrac{x - 2 - 2(x + 5)}{x + 5} > 0 \\ \ \\\Rightarrow \dfrac{x - 2 - 2x - 10}{x + 5} > 0 \\ \ \\\Rightarrow \dfrac{-x - 12}{x + 5} > 0​​
Multiply numerator and denominator by -1 to simplify:

​=>x+12x+5<0\Rightarrow \dfrac{x + 12}{x + 5} < 0​​
Find critical points
Set numerator and denominator equal to 0.
x + 12 = 0 =>x=−12 \Rightarrow x = -12​​
x + 5 = 0 =>x=−5 \Rightarrow x = -5​​
These divide the number line into intervals:
​(−∞,−12),(−12,−5),(−5,∞)(-\infty, -12), (-12, -5), (-5, \infty)​​
Sign analysis of f(x)=x+12x+5: f(x) = \dfrac{x + 12}{x + 5}:​​
For x > -5:
Choose  x=0=>f(0)=125>0x = 0 \Rightarrow f(0) = \dfrac{12}{5} > 0​ (Positive)
For −12<x<−5:-12 < x < -5:​​
Choose x=−10=>f(−10)=−10+12−10+5=2−5<0 x = -10 \Rightarrow f(-10) = \dfrac{-10 + 12}{-10 + 5} = \dfrac{2}{-5} < 0 ​ (Negative)
For x < -12:
Choose 
x=−13=>f(−13)=−13+12−13+5=−1−8>0x = -13 \Rightarrow f(-13) = \dfrac{-13 + 12}{-13 + 5} = \dfrac{-1}{-8} > 0​ (Positive)

We need x+12x+5<0\dfrac{x + 12}{x + 5} < 0​, which occurs for
​−12<x<−5-12 < x < -5​​
Therefore, the solution is:
​x∈(−12,−5)x \in (-12, -5)​​
\textbf{Correct answer: (c) x∈(−12,−5) x \in (-12, -5)​​

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