फलन `f(x) = |x - 1| + |x - 2|` के सांतत्य की विवेचना x = 1 और x = 2 पर कीजिए ।

Explanation

दिया गया है -
`f(x) = |x - 1| + |x - 2|`
`rArr" "f(x)={{:(-(x-1)-(x-2)",","यदि",x lt 1),((x-1)-(x-2)",","यदि",1 le x lt 2),((x-1)+(x-2)",","यदि",x ge 2):}`
`rArr" "f(x)={{:(-2x + 3",","यदि",x lt 1),(1",","यदि",1 le x lt 2),(2x-3",","यदि",x ge 2):}`
x = 1 पर सांतत्यता -
(i) `f(1)=1`
(ii) R.H.L. `=underset(x rarr 1^(+))(lim)f(x)=underset(h rarr 0)(lim)f(1+h)`
`=underset(h rarr 0)(lim)1," "[because x = 1 + h gt 1]`
= 1
(iii) L.H.L. `=underset(x rarr 1^(-))(lim)f(x)=underset(h rarr 0)(lim)f(1-h)`
`=underset(h rarr 0)(lim)-2(1-h)+3," "[because x = 1 - h lt 1]`
`=underset(h rarr 0)(lim)1 + 2h`
= 1
`therefore" ""R.H.L.=L.H.L."=f(1)`
अत: f(x) बिन्दु x = 1 पर संतत है ।
x = 2 पर सांतत्यता -
(i) `f(2)=2 xx 2 - 3 = 1`
(ii) R.H.L. `=underset(x rarr 2^(+))(lim)f(x)=underset(h rarr 0)(lim)f(2+h)`
`=underset(h rarr 0)(lim)2(2+h)-3," "[because x = 2 + h gt 0]`
`=underset(h rarr 0)(lim)1+2h`
`=1 + 0 = 1`
(iii) L.H.L. `= underset(x rarr 2^(-))(lim) f(x) = underset(h rarr 0)(lim)f(2-h)`
`=underset(h rarr 0)(lim)1," "[because x = 2 - h lt 2]`
= 1
`therefore" ""R.H.L.=L.H.L."=f(2)`
अत: f(x) बिन्दु x = 2 पर संतत है ।