Check Full Chapter Explained - Continuity and Differentiability - Continuity and Differentiability Class 12




Last updated at Jan. 2, 2020 by Teachoo
Check Full Chapter Explained - Continuity and Differentiability - Continuity and Differentiability Class 12
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) limβ¬(xβ0) π(π₯) "= " limβ¬(xβ0) " "(5π₯β3) Putting x = 0 = 5(0) β 3 = β3 π(0) = 5(0) β 3 = 0 β 3 = β3 Since L.H.S = R.H.S β΄ limβ¬(xβ0) π(π₯) = π(0) 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 β΄ limβ¬(xββ3) π(π₯) = π(β3) Hence, f is continuous at π =β3 limβ¬(xβ3) π(π₯) "= " limβ¬(xβ3) " "(5π₯β3) Putting x = β3 = 5(β3) β 3 = β18 π(β3) = 5(β3) β 3 = β15 β 3 = β18 Since, L.H.S = R.H.S β΄ limβ¬(xββ3) π(π₯) = π(β3) Hence, f is continuous at π =β3 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 β΄ ( lim)β¬(xβ5) π(π₯)= π(5) (πππ)β¬(π₯β5) π(π₯) "= " (πππ)β¬(π₯β5) " "(5π₯β3) Putting x = 5 = 5(5) β 3 = 22 π(5) = 5(5) β 3 = 25 β 3 = 22 Hence, f is continuous at π =π Thus the function is continuous at π =π, at π =βπ & at π =π
Ex 5.1
Ex 5.1 ,2
Ex 5.1 ,3 Important
Ex 5.1 ,4
Ex 5.1 ,5 Important
Ex 5.1 ,6
Ex 5.1 ,7 Important
Ex 5.1 ,8
Ex 5.1, 9 Important
Ex 5.1, 10
Ex 5.1, 11
Ex 5.1, 12
Ex 5.1, 13
Ex 5.1, 14
Ex 5.1, 15 Important
Ex 5.1, 16
Ex 5.1, 17 Important
Ex 5.1, 18 Important
Ex 5.1, 19 Important
Ex 5.1, 20
Ex 5.1, 21
Ex 5.1, 22
Ex 5.1, 23
Ex 5.1, 24 Important
Ex 5.1, 25
Ex 5.1, 26 Important
Ex 5.1, 27
Ex 5.1, 28 Important
Ex 5.1, 29
Ex 5.1, 30 Important
Ex 5.1, 31
Ex 5.1, 32
Ex 5.1, 33
Ex 5.1, 34 Important
About the Author