#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# F(x)=integrate [f(t)]dt from 0 to x where x>0 and [.] denotes greatest int. func. .Then,is f(x) continuous but not diff. for x=1,2,3,........ If yes,how?

Jitender Singh IIT Delhi
6 years ago
Ans:
f(t) should be given. f(t) = t
$F(x) = \int_{0}^{x}[t]dt$
Apply the newton leibniz’s theorm here, we have,
$F'(x) = [x].1 - [0].0 = [x]$
$F'(x) = f(x)$
$f(x) = [x]$
To check the continuity at integer ‘a’
LHL = RHL
$\lim_{x\rightarrow a^{-}}[x] = \lim_{x\rightarrow a^{+}}[x]$
$LHL = \lim_{x\rightarrow a^{-}}[x] = a-1$
$RHL = \lim_{x\rightarrow a^{+}}[x] = a$
So f(x) is not continuous at integer ‘a’. So it will definitely will not be differentiable at integer ‘a’.
Thanks & Regards
Jitender Singh
IIT Delhi