Ex 5.1
Ex 5.1 ,2
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
Last updated at April 16, 2024 by Teachoo
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