For example, consider the following counterexample: Let's define a continuous function f(x) = x on the interval [0, 1] and a discontinuous function g(x) = 1/x on the same interval. The function f(x) = x is continuous, and g(x) = 1/x is discontinuous at x = 0. Now, if we take the product of these functions, h(x) = f(x) * g(x) = x * (1/x) = 1, we obtain a constant function h(x) = 1. The function h(x) = 1 is continuous everywhere on the interval [0, 1]. Therefore, in this particular case, the product of a continuous function f(x) = x and a discontinuous function g(x) = 1/x results in a continuous function h(x) = 1. This counterexample demonstrates that the statement is not universally true. The continuity of the product function depends on the specific functions involved and their behavior.
Sir in Q(2) option B that you chose to be correct is clearly incorrect. Option C or D is correct. The product may be continuous or discontinuous. The "Not Sure" option could mean "I don't know if the product is continuous or discontinuous. Option D may be mean that "no proper option has been given". Sir kindly think it over and let me know what conclusion you have reached.
I have doubt in your q no 2 because we have x as continuous and 1/x discontinuous but the product is 1 which is continuous . So i think the right option is c
Yes bro. But what about option D? The product may be continuous or discontinuous. The "Not Sure" option could mean "I don't know if the product is continuous or discontinuous.
For example, consider the following counterexample:
Let's define a continuous function f(x) = x on the interval [0, 1] and a discontinuous function g(x) = 1/x on the same interval. The function f(x) = x is continuous, and g(x) = 1/x is discontinuous at x = 0.
Now, if we take the product of these functions, h(x) = f(x) * g(x) = x * (1/x) = 1, we obtain a constant function h(x) = 1. The function h(x) = 1 is continuous everywhere on the interval [0, 1].
Therefore, in this particular case, the product of a continuous function f(x) = x and a discontinuous function g(x) = 1/x results in a continuous function h(x) = 1.
This counterexample demonstrates that the statement is not universally true. The continuity of the product function depends on the specific functions involved and their behavior.
❤
great sharing sir
Helpful
Great
Sir padagogy ky liyee kon si books use kary plz
waw good nice video
Assalam.o.Alaikum Sir
Sir Ppsc and fpsc keliye jo interview questions hoty han un pay bhi koi video upload kar den......
Keep it up...
Nice
sir google py past pprs ni mill rhy
kia app provide kr skty hn
Full watch
💐💐
Good teaching
Thank you
Sir in Q(2) option B that you chose to be correct is clearly incorrect. Option C or D is correct. The product may be continuous or discontinuous. The "Not Sure" option could mean "I don't know if the product is continuous or discontinuous. Option D may be mean that "no proper option has been given". Sir kindly think it over and let me know what conclusion you have reached.
Product of continues and discontinues function may not be discontinues
f(x)=x and g(x)=1/x on [0,1]
f.g=x.1/x=1 what about constant function ??
@@arfananjum2705 Right. Constant functions are always continuous.
sir next part please
Website pr classical mechanics mention ni hai to kia usmy sy aey ga?
Yes, may come
I have doubt in your q no 2 because we have x as continuous and 1/x discontinuous but the product is 1 which is continuous .
So i think the right option is c
Yes bro. But what about option D? The product may be continuous or discontinuous. The "Not Sure" option could mean "I don't know if the product is continuous or discontinuous.
Product of function is defined as composition of function. This is not a product you used,
Sir Kia mujhy 2017 aur 2019 k past papers ml skty hain fpsc mathematics lecturer k?
April ma apply Kya tha uska test kb expected h?
Between, 15 August ~ 30 September
Sir kia saray past paper pdf me mil skty hn
Sir tyari krwa den fpsc ki plzzz
Englis wala part kaha se prepara kry?
2020 ka syallabus to different ha is bar jo fpsc ka Est k ley ha syallabus .
Sir please share this paper in pdf form.