Ex 5.1, 3 (c) - Chapter 5 Class 12 Continuity and Differentiability
Last updated at Dec. 16, 2024 by Teachoo
Ex 5.1
Ex 5.1 ,1
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Transcript
Ex 5.1, 3 Examine the following functions for continuity. (c) f (x) = (π₯^(2 )β 25 )/(π₯ + 5), x β β5 f (x) = (π₯^(2 )β 25 )/(π₯ + 5) Putting x = β5 f (β5) = (γ(β5)γ^(2 )β 25 )/(β5 + 5) = (25β 25 )/(β5 + 5) = 0/0 = Undefined Hence, f(x) is not defined at x = β5 So, we check for continuity at all points except β5 Let c be any real number except β5. f is continuous at π₯ = π if (π₯π’π¦)β¬(π±βπ) π(π) = π(π) LHS (π₯π’π¦)β¬(π±βπ) π(π) = limβ¬(xβπ) (π₯^2β 25)/(π₯ + 5) = limβ¬(xβπ) ((π₯ β 5) (π₯ + 5))/(π₯ + 5) = limβ¬(xβπ) π₯β5 Putting x = c = c β 5 RHS f (c) = (π^(2 )β 25 )/(π + 5) = ((π β 5)(π + 5))/((π + 5)) = c β 5 Let c be any real number except β5. f is continuous at π₯ = π if (π₯π’π¦)β¬(π±βπ) π(π) = π(π) Since, L.H.S = R.H.S β΄ Function is continuous at x = c (except β5) Thus, we can write that f is continuous for all real numbers except β5 β΄ f is continuous at each π± β R β {βπ}
