Question 6
The function f (x) = |x| + |x β 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 π(π₯)= |π₯|+|π₯β1|
Here, we have 2 critical points
i.e. π₯ = 0, and π = 1
So, our intervals will be
When πβ€π
When 0<π<π
When πβ₯π
When πβ€π
π(π₯)= |π₯|+ |π₯β1|.
Here, both will be negative
π(π₯)=(βπ₯)+(β(π₯β1))
π(π₯)=βπ₯β(π₯β1)
π(π₯)=βπ₯βπ₯+1
" " π(π)=βππ+π
When 0<π<π
π(π₯)= |π₯| +|π₯β1|.
Here, x will be positive, but (x - 1) will be negative
π(π₯)=(π₯)+(β(π₯β1))
π(π₯)=π₯β(π₯β1)
π(π₯)=π₯βπ₯+1
" " π(π)=π
When πβ₯π
π(π₯)= |π₯|+|π₯β1|.
Here, both will be positive
π(π₯)=π₯+(π₯β1)
π(π₯)=π₯+π₯β1
" " π(π)=ππβπ
Thus, our function becomes
π(π)={β(βππ+π" " ππ πβ€π@ π ππ π<π<π@ππβπ ππ πβ₯π)β€
Since, from options we need to check continuity of the function when x = 0 and x = 1
Letβs check continuity
Case 1 : When x = 0
f(x) is continuous at π₯ =0
if L.H.L = R.H.L = π(0)
if (π₯π’π¦)β¬(π±βπ^β ) π(π)=(π₯π’π¦)β¬(π±βπ^+ ) " " π(π)= π(π)
LHL at x β 0
(πππ)β¬(π±βπ^β ) f(x) = limβ¬(hβ0) f(0 β h)
= (πππ)β¬(π‘βπ) f (β h)
= limβ¬(hβ0) (β2(ββ))+1
= limβ¬(hβ0) (2β)+1
= 2(0) + 1
= 0 + 1
= 1
RHL at x β 0
(πππ)β¬(π±βγβπγ^+ ) f(x) = limβ¬(hβ0) f(0 + h)
= (πππ)β¬(π‘βπ) f(h)
= limβ¬(hβ0) 1
= 1
Case 2 : When x = 1
f(x) is continuous at π₯=1
if L.H.L = R.H.L = π(1)
if limβ¬(xβ1^β ) π(π₯)=limβ¬(xβ1^+ ) " " π(π₯)=π(1)
LHL at x β 1
limβ¬(xβ1^β ) f(x) = limβ¬(hβ0) f(1 β h)
= limβ¬(hβ0) 1
= 1
RHL at x β 0
limβ¬(xβ1^+ ) f(x) = limβ¬(hβ0) f(1 + h)
= limβ¬(hβ0) 2(1+β)β1
= limβ¬(hβ0) 2+2ββ1
= 2 + 2(0) β 1
=2 β 1
= 1
& π(1) = 2(1)β1
π(π) = 1
Hence, L.H.L = R.H.L = π(1)
β΄ 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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