calculus expected qestions for jee

f(x)=[x]+[x+1/3]+[x+2/3],where [] is G.I.F and x belongs to R.find the number of points of discontinuty of f(x) in[-1,1]? and mentipon the points?draw the graph.??
plz anwer with full explantion.

number of points carry 1 mark
graph caary 2 marks
points cary 1 marks
making totla of 4 marks
plz anwer as soon as poosible

8 Answers

341
Hari Shankar ·

See Hermite Identity

1
ujjwalkalra kalra ·

whts tht???is it in sllybus of jee

62
Lokesh Verma ·

The name of the Identity may not be in syllabus but what prophet sir was trying to do was to give you a hint so that you could look into hermite identity and see its application.... [1] which is a more general result of what you have written...

As a starting point.. look at 3x

now [3x] will either be of form 3N or 3N+1 or 3N+2... now try the remaining...

11
sagnik sarkar ·

No of pts of discontinuity=6.
pts are:{-2/3,-1/3,0,1/3,2/3,1}

23
qwerty ·

f(x+1/3) =f(x) + 1

this can help in drawing the graph

1
jangra28192manoj jangra ·

answer is 7
pts are:{-2/3,-1/3,0,1/3,2/3,1}
hey sagnik correct ur solution

1
ujjwalkalra kalra ·

but u hv mentioned only 6 points in set..whts the 7th one??

1
jangra28192manoj jangra ·

-1 is the 7th point

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