Question 6
The function f (x) = |x| + |x – 1| is
(A) continuous at x = 0 as well as at x = 1.
(B) continuous at x = 1 but not at x = 0.
(C) discontinuous at x = 0 as well as at x = 1.
(D) continuous at x = 0 but not at x = 1.
Given 𝑓(𝑥)= |𝑥|+|𝑥−1|
Here, we have 2 critical points
i.e. 𝑥 = 0, and 𝒙 = 1
So, our intervals will be
When 𝒙≤𝟎
When 0<𝒙<𝟏
When 𝒙≥𝟏
When 𝒙≤𝟎
𝑓(𝑥)= |𝑥|+ |𝑥−1|.
Here, both will be negative
𝑓(𝑥)=(−𝑥)+(−(𝑥−1))
𝑓(𝑥)=−𝑥−(𝑥−1)
𝑓(𝑥)=−𝑥−𝑥+1
" " 𝒇(𝒙)=−𝟐𝒙+𝟏
When 0<𝒙<𝟏
𝑓(𝑥)= |𝑥| +|𝑥−1|.
Here, x will be positive, but (x - 1) will be negative
𝑓(𝑥)=(𝑥)+(−(𝑥−1))
𝑓(𝑥)=𝑥−(𝑥−1)
𝑓(𝑥)=𝑥−𝑥+1
" " 𝒇(𝒙)=𝟏
When 𝒙≥𝟏
𝑓(𝑥)= |𝑥|+|𝑥−1|.
Here, both will be positive
𝑓(𝑥)=𝑥+(𝑥−1)
𝑓(𝑥)=𝑥+𝑥−1
" " 𝒇(𝒙)=𝟐𝒙−𝟏
Thus, our function becomes
𝒇(𝒙)={█(−𝟐𝒙+𝟏" " 𝒊𝒇 𝒙≤𝟎@ 𝟏 𝒊𝒇 𝟎<𝒙<𝟏@𝟐𝒙−𝟏 𝒊𝒇 𝒙≥𝟏)┤
Since, from options we need to check continuity of the function when x = 0 and x = 1
Let’s check continuity
Case 1 : When x = 0
f(x) is continuous at 𝑥 =0
if L.H.L = R.H.L = 𝑓(0)
if (𝐥𝐢𝐦)┬(𝐱→𝟎^− ) 𝒇(𝒙)=(𝐥𝐢𝐦)┬(𝐱→𝟎^+ ) " " 𝒇(𝒙)= 𝒇(𝟎)
LHL at x → 0
(𝒍𝒊𝒎)┬(𝐱→𝟎^− ) f(x) = lim┬(h→0) f(0 − h)
= (𝒍𝒊𝒎)┬(𝐡→𝟎) f (− h)
= lim┬(h→0) (−2(−ℎ))+1
= lim┬(h→0) (2ℎ)+1
= 2(0) + 1
= 0 + 1
= 1
RHL at x → 0
(𝒍𝒊𝒎)┬(𝐱→〖−𝟏〗^+ ) f(x) = lim┬(h→0) f(0 + h)
= (𝒍𝒊𝒎)┬(𝐡→𝟎) f(h)
= lim┬(h→0) 1
= 1
Case 2 : When x = 1
f(x) is continuous at 𝑥=1
if L.H.L = R.H.L = 𝑓(1)
if lim┬(x→1^− ) 𝑓(𝑥)=lim┬(x→1^+ ) " " 𝑓(𝑥)=𝑓(1)
LHL at x → 1
lim┬(x→1^− ) f(x) = lim┬(h→0) f(1 − h)
= lim┬(h→0) 1
= 1
RHL at x → 0
lim┬(x→1^+ ) f(x) = lim┬(h→0) f(1 + h)
= lim┬(h→0) 2(1+ℎ)−1
= lim┬(h→0) 2+2ℎ−1
= 2 + 2(0) – 1
=2 − 1
= 1
& 𝑓(1) = 2(1)−1
𝒇(𝟏) = 1
Hence, L.H.L = R.H.L = 𝑓(1)
∴ f is continuous at x = 1
Therefore, f is continuous at x = 0 as well as at x = 1.
So, the correct answer is (A)

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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