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Grade 12Differential Calculus

see attachment ...and please tell the method which is used in solution please reply anyone ….i heard lot about this website

Question image for see attachment ...and please tell the method which
Profile image of milind
11 Years agoGrade 12
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1 Answer

Profile image of Jitender Singh
11 Years ago
Ans:

Hello student,
Please find the answer to your question below

?L = [\frac{3}{4}]+[\frac{3}{4}+\frac{1}{100}]+[\frac{3}{4}+\frac{2}{100}]+............+[\frac{3}{4}+\frac{99}{100}]L = [\frac{3}{4}]+[\frac{3}{4}+\frac{1}{100}]+[\frac{3}{4}+\frac{2}{100}]+..........+[\frac{3}{4}+\frac{99}{100}]
[t] = 0, 0 <=t < 1
[t] = 1, 1 <= t < 2
Now looking at the problem in hand
[\frac{3}{4}+\frac{x}{100}] = 0, 0 <= x < 25
[\frac{3}{4}+\frac{x}{100}] = 0, 25 <= x < 99
There are total 100 term in L out of which 25 terms are zero. Rest all the terms are equal to 1. So the sum is 75.

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