IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

ishita247

In (0,1), f(x)= [3x²+1] , where [x] stands for the greatest integer not exceeding x, is

(a) continuous

(b) continuous except at one point

(c) continuous except at two points

(d) continuous except at three points

akhil_o

at 0, f(x)=1
at 1, f(x)=4

f(x)=[g(x)]
since g(x) is continuous throughout,,
it takes all values from 1 to 4
ponts of discontinuity=integral points
ie points when
f(x)=2,3

so ans is (c)

anchal_1991

i also think ans is c
u can make intervals like
0<=3x²+1<1
nd 1<=3x²+1<2
nd 2<=3x²+1<3
nd 3<=3x²+1<4
then u will get values of x
upto wich it gives 0,1,2,3

ishita247

well wt i thot of was

dt on putting x=1/2

we get the ans as 1

so 1 shld also b included in d range of d function

thus der should b 3 points of discontinuity

i.e 1.2.3

hw du verify dt???

akhil_o

see...
the function is increasing...and at 0 it is just 1...
and keeps increasing till f(x)=2
so f(x) is continuous at 1 from 0 to 1/root(3), where it just becomes 2 so it is discontinuous
so the 1 is not considered here
it is only 2,3

ishita247

all ryt thnx a ton

u rock!!!

spideyunlimited

In (0,1), f(x)= [3x^2+1]

Now here f(x) will be 1 as long as x^2 < 1/3 because when x^2 becomes 1/3, f(x) becomes [ 1 + 1] = 2.
so x = (1 / root 3) is one point at which it is discontinuous as next integer is obtained.

similary at x^2 = 2/3, f(x) = [2 + 1] = 3
x = root2 / root 3 is another point of discontinuity.

and at x = 1, f(x) = [3 + 1] = 4
so x= 1 is another point of discontinuity. but 1 is not within the given interval

THUS POINTS OF DISCONTINUITY ARE x = 1/root3, root2/root3 (not 1 but)

Thus 2 points of discontinuity.!

anchal_1991

it is open interval

ishita247

but spideyy

1 is not included in d domain v mentioned

see its (0,1)

open interval

so 2 points of discontinuity

spideyunlimited

arre i corrected it as soon as i posted it lol

neeraj_agarwal_1990

making its graph would be the best option to go for...

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