Dear Student, x-1+x-2>4Case 1: x≤1⇒x-1≤0⇒x-1=-x-1Also, x≤1 4⇒-2x+3>4⇒3-4>2x⇒-1>2x⇒x 0 and x-2 4⇒1>4not possible Case 3: x≥2⇒x-2≥0⇒x-2=x-2Also x≥2>1⇒x-1>0⇒x-1=x-1Thus, x-1+x-2>4⇒2x-3>4⇒2x>4+3⇒2x>7⇒x>72⇒x∈72,∞Hence x∈-∞,-12∪72,∞ Hope this information will clear your doubts about topic Regards

sir

Dear Student,

|x - 1| + |x - 2| > 4 C a s e 1 : x \leq 1 \Rightarrow x - 1 \leq 0 \Rightarrow |x - 1| = - (x - 1) A l s o, x \leq 1 < 2 \Rightarrow x - 2 < 0 \Rightarrow |x - 2| = - (x - 2) T h u s, - (x - 1) - (x - 2) > 4 \Rightarrow - 2 x + 3 > 4 \Rightarrow 3 - 4 > 2 x \Rightarrow - 1 > 2 x \Rightarrow x < - \frac{1}{2} \Rightarrow x \in (- \infty, - \frac{1}{2}) C a s e 2 : 1 < x < 2 \Rightarrow x - 1 > 0 a n d x - 2 < 0 \Rightarrow |x - 1| = (x - 1) a n d |x - 2| = - (x - 2) T h u s, (x - 1) - (x - 2) > 4 \Rightarrow 1 > 4 n o t p o s s i b l e . C a s e 3 : x \geq 2 \Rightarrow x - 2 \geq 0 \Rightarrow |x - 2| = (x - 2) A l s o x \geq 2 > 1 \Rightarrow x - 1 > 0 \Rightarrow |x - 1| = (x - 1) T h u s, (x - 1) + (x - 2) > 4 \Rightarrow 2 x - 3 > 4 \Rightarrow 2 x > 4 + 3 \Rightarrow 2 x > 7 \Rightarrow x > \frac{7}{2} \Rightarrow x \in (\frac{7}{2}, \infty) H e n c e x \in (- \infty, - \frac{1}{2}) \cup (\frac{7}{2}, \infty) .

Hope this information will clear your doubts about topic.

Regards

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​sir

sir