NCERT Exemplar - MCQs

Chapter 5 Class 12 Continuity and Differentiability
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## (D) continuous at x = 0 but not at x = 1.

This question is similar to Ex 5.1, 34 - Chapter 5 Class 12 - Continuity and Differentiability

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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₯β1| Here, we have 2 critical points i.e. π₯ = 0, and π = 1 So, our intervals will be When πβ€π When 0<π<π When πβ₯π When πβ€π π(π₯)= |π₯|+ |π₯β1|. Here, both will be negative π(π₯)=(βπ₯)+(β(π₯β1)) π(π₯)=βπ₯β(π₯β1) π(π₯)=βπ₯βπ₯+1 " " π(π)=βππ+π When 0<π<π π(π₯)= |π₯| +|π₯β1|. Here, x will be positive, but (x - 1) will be negative π(π₯)=(π₯)+(β(π₯β1)) π(π₯)=π₯β(π₯β1) π(π₯)=π₯βπ₯+1 " " π(π)=π When πβ₯π π(π₯)= |π₯|+|π₯β1|. Here, both will be positive π(π₯)=π₯+(π₯β1) π(π₯)=π₯+π₯β1 " " π(π)=ππβπ Thus, our function becomes π(π)={β(βππ+π" " ππ πβ€π@ π ππ π<π<π@ππβπ ππ πβ₯π)β€ Since, from options we need to check continuity of the function when x = 0 and x = 1 Letβs 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±βπ^β ) π(π)=(π₯π’π¦)β¬(π±βπ^+ ) " " π(π)= π(π) LHL at x β 0 (πππ)β¬(π±βπ^β ) f(x) = limβ¬(hβ0) f(0 β h) = (πππ)β¬(π‘βπ) f (β h) = limβ¬(hβ0) (β2(ββ))+1 = limβ¬(hβ0) (2β)+1 = 2(0) + 1 = 0 + 1 = 1 RHL at x β 0 (πππ)β¬(π±βγβπγ^+ ) f(x) = limβ¬(hβ0) f(0 + h) = (πππ)β¬(π‘βπ) f(h) = limβ¬(hβ0) 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¬(xβ1^β ) π(π₯)=limβ¬(xβ1^+ ) " " π(π₯)=π(1) LHL at x β 1 limβ¬(xβ1^β ) f(x) = limβ¬(hβ0) f(1 β h) = limβ¬(hβ0) 1 = 1 RHL at x β 0 limβ¬(xβ1^+ ) f(x) = limβ¬(hβ0) f(1 + h) = limβ¬(hβ0) 2(1+β)β1 = limβ¬(hβ0) 2+2ββ1 = 2 + 2(0) β 1 =2 β 1 = 1 & π(1) = 2(1)β1 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)