Ex 5.1 ,1 - Chapter 5 Class 12 Continuity and Differentiability
Last updated at Feb. 14, 2025 by Teachoo
Ex 5.1
Ex 5.1 ,1
You are here
You are here
Transcript
Ex 5.1, 1 Prove that the function π (π₯) = 5π₯ β 3 is continuous at π₯ = 0, at π₯ = β3 and at π₯ = 5 Given π(π₯)= 5π₯ β3 At π=π f(x) is continuous at x = 0 if (π₯π’π¦)β¬(π±βπ) π(π) = π(π) (π₯π’π¦)β¬(π±βπ) π(π) "= " limβ¬(xβ0) " "(5π₯β3) Putting x = 0 = 5(0) β 3 = β3 π(π) = 5(0) β 3 = 0 β 3 = β3 Since L.H.S = R.H.S Hence, f is continuous at π = π At x = β3 f(x) is continuous at x = β3 if ( lim)β¬(xββ3) π(π₯)= π(β3) Since, L.H.S = R.H.S Hence, f is continuous at π =β3 (π₯π’π¦)β¬(π±βπ) π(π) "= " limβ¬(xβ3) " "(5π₯β3) Putting x = β3 = 5(β3) β 3 = β18 π(βπ) = 5(β3) β 3 = β15 β 3 = β18 At π =π f(x) is continuous at x = 5 if ( lim)β¬(xβ5) π(π₯)= π(5) Since, L.H.S = R.H.S Hence, f is continuous at x = 5 Thus, the function is continuous at x = 0, x = β3, x = 5 (πππ)β¬(πβπ) π(π) "= " (πππ)β¬(π₯β5) " "(5π₯β3) Putting x = 5 = 5(5) β 3 = 22 π(π) = 5(5) β 3 = 25 β 3 = 22
