University:
University of Cambridge
Course:
0580 IGCSE | Mathematics
Academic year:

2023
Views:

290

Pages:

5
Author:

Shahbaz A.

Q5 Calculus and analytic geometry shahbaz ahmed August 2024 Introduction to the relevant concepts and skills The Left Hand Limit 1 limx→c− f (x) = L For values of x < c ,as x is getting close to c, the corresponding values of f(x) are getting close to L. The Right Hand Limit limx→c+ f (x) = M For values of x > c ,as x is getting close to c, the corresponding values of f(x) are getting close to M. Criterion for existence of Limit of a Function limx→c f (x) = L ⇔ limx→c− f (x) = limx→c+ f (x) 2 Criterion for Continuity of a Function at a Number A function is said to be continuous at a number ”c” ⇔ (i) f(c) is defined (ii) limx→c f (x) exists, (iii)) limx→c f (x)=f(c) Q5 (i) Find the values m and n,so that given function is continuous at x=3.     mx i f x < 3     f (x) = n i f x = 3       −2x + 9 i f x > 3 Solution limx−→3− f (x) = limx−→3− [mx] limx−→3− f (x) = limx−→3− [3m] Similarly limx−→3+ f (x) = limx−→3+ [−2x + 9] limx−→3− f (x) = limx−→3− [−2(3) + 9] = 3 limx−→3 f (x) exist ⇔ 3 limx−→3− f (x) = limx−→3+ f (x) 3m = 3 m=1 =⇒ limx−→3 f (x) = 3 For continuity limx−→3 f (x) = f (3) 3=n So required values are m=1,n=3 (ii) f (x) =    mx i f x < 3   x2 i f x ≥ 3 Solution limx−→3− f (x) = limx−→3− [mx] limx−→3− f (x) = limx−→3− [3m] Similarly limx−→3+ f (x) = limx−→3+ [x2 ] limx−→3− f (x) = limx−→3− [9] = 9 4 limx−→3 f (x) exist ⇔ limx−→3− f (x) = limx−→3+ f (x) 3m=9 m=3 =⇒ limx−→3 f (x) = 9 For continuity limx−→3 f (x) = f (3) 9=9 Hence condition for continuity is satisfied. m=3 5