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



Last updated at March 10, 2021 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) (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
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