Get live Maths 1-on-1 Classs - Class 6 to 12
Ex 5.1 ,2 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)
Last updated at March 22, 2023 by Teachoo
Ex 5.1
Ex 5.1 ,1
Ex 5.1 ,2 You are here
Ex 5.1, 3 (a)
Ex 5.1, 3 (b)
Ex 5.1, 3 (c) Important
Ex 5.1, 3 (d) 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 Important
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 (i) Important
Ex 5.1, 22 (ii)
Ex 5.1, 22 (iii)
Ex 5.1, 22 (iv) Important
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
Transcript
Ex 5.1, 2 Examine the continuity of the function f (x) = 2x2 β 1 at x = 3. π(π₯) is continuous at x = 3 if limβ¬(xβ3) π(π₯) = π(3) Since, L.H.S = R.H.S Hence, f is continuous at π =3 (π₯π’π¦)β¬(π±βπ) π(π) "= " limβ¬(xβ3) " "(2π₯2β1) Putting π₯ = 3 = 2(3)2 β 1 = 2 Γ 9 β 1 = 17 π(π) = 2(3)2 β 1 = 2 Γ 9 β 1 = 18β1 = 17
