Ex 5.1, 33 - Examine that sin |x| is continuous - Class 12 CBSE

Ex 5.1, 33 - Chapter 5 Class 12 Continuity and Differentiability - Part 2

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Ex 5.1, 33 Examine that sin⁑| π‘₯ | is a continuous function.𝑓(π‘₯) = γ€–sin 〗⁑|π‘₯| Let π’ˆ(𝒙) = sin⁑π‘₯ & 𝒉(𝒙) = |π‘₯| Now, π’ˆπ’π’‰ (𝒙) = g(β„Ž(π‘₯)) = 𝑔(|π‘₯|) = γ€–sin 〗⁑|π‘₯| = 𝒇(𝒙) Hence, 𝑓(π‘₯) = π‘”π‘œβ„Ž (π‘₯) We know that π’ˆ(𝒙) = sin⁑π‘₯ is continuous as sin x is always continuous & 𝒉(𝒙) = |π‘₯| is continuous as modulus function is continuous Hence, 𝑔(π‘₯) & β„Ž(π‘₯) are both continuous . We know that If two function of 𝑔(π‘₯) & β„Ž(π‘₯) both continuous, then their composition π’ˆπ’π’‰(𝒙) is also continuous Hence, 𝒇(𝒙) is continuous . Given 𝑓(π‘₯)= |π‘₯| – |π‘₯+1|. Here, we have 2 critical points x = 0 and x + 1 = 0 i.e. x = 0, and x = βˆ’1 So, our intervals will be When π’™β‰€βˆ’πŸ When βˆ’πŸ<𝒙<𝟎 When 𝒙β‰₯𝟎 When π’™β‰€βˆ’πŸ 𝑓(π‘₯)= |π‘₯| – |π‘₯+1|. Here, both will be negative 𝑓(π‘₯)=(βˆ’π‘₯) –(βˆ’(π‘₯+1)) 𝑓(π‘₯)=βˆ’π‘₯+(π‘₯+1) " " 𝒇(𝒙)=𝟏 When βˆ’πŸ<π’™β‰€πŸŽ 𝑓(π‘₯)= |π‘₯| – |π‘₯+1|. Here, x will be negative, but (x + 1) will be positive 𝑓(π‘₯)=(βˆ’π‘₯) –(π‘₯+1) 𝑓(π‘₯)=βˆ’π‘₯βˆ’π‘₯βˆ’1 " " 𝒇(𝒙)=βˆ’πŸπ’™βˆ’πŸ |π‘₯| = {β–ˆ(π‘₯, π‘₯ β‰₯0@βˆ’π‘₯, π‘₯<0)─ |π‘₯+1| = {β–ˆ((π‘₯+1) , π‘₯+1β‰₯0@βˆ’(π‘₯+1) π‘₯+1<0)─ = {β–ˆ((π‘₯+1) , π‘₯β‰₯βˆ’1@βˆ’(π‘₯+1) π‘₯<1)─ When 𝒙β‰₯𝟎 𝑓(π‘₯)= |π‘₯| – |π‘₯+1|. Here, both will be positive 𝑓(π‘₯)=π‘₯ –(π‘₯+1) 𝑓(π‘₯)=π‘₯βˆ’π‘₯βˆ’1 " " 𝒇(𝒙)=βˆ’πŸ Thus, our function becomes 𝒇(𝒙)={β–ˆ(𝟏 π’Šπ’‡ π’™β‰€βˆ’πŸ@βˆ’πŸπ’™βˆ’πŸ π’Šπ’‡ βˆ’πŸ<𝒙<𝟎@βˆ’πŸ π’Šπ’‡ 𝒙β‰₯𝟎)─ Since we need to find continuity at of the function We check continuity for different values of x When x < βˆ’1 When x = βˆ’1 When βˆ’1 < x < 0 When x = 0 When x > 0 Checking continuity Case 1 : When x < βˆ’1 For x < βˆ’1, f(x) = 1 Since this constant It is continuous ∴ f(x) is continuous for x < βˆ’1 Case 2 : When x = βˆ’1 f(x) is continuous at π‘₯ =βˆ’1 if L.H.L = R.H.L = 𝑓(βˆ’1) if lim┬(xβ†’γ€–βˆ’1γ€—^βˆ’ ) 𝑓(π‘₯)=lim┬(xβ†’γ€–βˆ’1γ€—^+ ) " " 𝑓(π‘₯)= 𝑓(βˆ’1) Since there are two different functions on the left & right of βˆ’1, we take LHL & RHL . LHL at x β†’ βˆ’1 lim┬(xβ†’γ€–βˆ’1γ€—^βˆ’ ) f(x) = lim┬(hβ†’0) f(βˆ’1 βˆ’ h) = lim┬(hβ†’0) 1 = 1 RHL at x β†’ 0 lim┬(xβ†’γ€–βˆ’1γ€—^+ ) f(x) = lim┬(hβ†’0) f(βˆ’1 + h) = lim┬(hβ†’0) (βˆ’2(βˆ’1+β„Ž))βˆ’1 = lim┬(hβ†’0) (2βˆ’2β„Ž)βˆ’1 = (2 βˆ’ 2(0)) βˆ’ 1 = 2 βˆ’ 0 βˆ’ 1 = 1 & 𝑓(0) = βˆ’1 Hence, L.H.L = R.H.L = 𝑓(0) ∴ f is continuous at x = 0 Case 5: When x > 0 For x > 0, f(x) = βˆ’1 Since this constant It is continuous ∴ f(x) is continuous for x > 0 Since there is no point of discontinuity Therefore, f is continuous for all x ∈ R

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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.