Ex 5.1

Chapter 5 Class 12 Continuity and Differentiability
Serial order wise

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Ex 5.1, 24 Determine if f defined by π(π₯)={β( π₯2 sinβ‘γ1/π₯γ, ππ π₯β [email protected]&0, ππ π₯=0)β€ is a continuous function? Since we need to find continuity at of the function We check continuity for different values of x When x β  0 When x = 0 Case 1 : When x β  0 For x β  0, f(x) = π₯2 sinβ‘γ1/π₯γ Since x2 is continuous and sinβ‘γ1/π₯γ is continuous So, π₯2 sinβ‘γ1/π₯γ is continuous β΄ f(x) is continuous for x β  0 Case 2 : When x = 0 f(x) is continuous at π₯ =0 if L.H.L = R.H.L = π(0) if limβ¬(xβ0^β ) π(π₯)=limβ¬(xβ0^+ ) " " π(π₯)= π(0) Since there are two different functions on the left & right of 0, we take LHL & RHL . LHL at x β 0 limβ¬(xβ0^β ) f(x) = limβ¬(hβ0) f(0 β h) = limβ¬(hβ0) f(βh) = limβ¬(hβ0) (ββ)^2 sinβ‘γ1/((ββ))γ = limβ¬(hβ0) β^2 π = 02 .π = 0 We know that β 1 β€ sin ΞΈ β€ 1 β β 1β€γsin γβ‘γ1/((ββ))γβ€ 1 β΄ γsin γβ‘γ1/((ββ))γ is a finite value Let γsin γβ‘γ1/((ββ))γ = k RHL at x β 0 limβ¬(xβ0^+ ) f(x) = limβ¬(hβ0) f(0 + h) = limβ¬(hβ0) f(h) = limβ¬(hβ0) β^2 sinβ‘γ1/βγ = limβ¬(hβ0) β^2 π = 02 .π = 0 We know that β 1 β€ sin ΞΈ β€ 1 β β 1β€γsin γβ‘γ1/βγβ€ 1 β΄ γsin γβ‘γ1/βγ is a finite value Let γsin γβ‘γ1/βγ = k And, f(0) = 0 Hence, L.H.L = R.H.L = π(0) β΄ f is continuous at x=0 Hence, π(π) is continuous for all real number