If -x- denotes the greatest integers -x- then find-23-23-199-23-299-23-9899-

[23]=0 [23+3399]=[23+13]=[33]=1 ∴[23+3499]=1 [23+3599]=1 · · · [23+9899]=1 ∴ Adding them, we get 66. ...

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Standard XII

Mathematics

Set Builder Form

If [x] denote...

Question

If $[x]$ denotes the greatest integers $\leq x,$ then find

$[\frac{2}{3}] + [(\frac{2}{3}) + (\frac{1}{99})] + [(\frac{2}{3}) + (\frac{2}{99})] + \dots \dots + [(\frac{2}{3}) + (\frac{98}{99})]$

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Solution

The correct option is A
66

$[\frac{2}{3}] = 0$
$[\frac{2}{3} + \frac{33}{99}] = [\frac{2}{3} + \frac{1}{3}] = [\frac{3}{3}] = 1$
$∴ [\frac{2}{3} + \frac{34}{99}] = 1$
$[\frac{2}{3} + \frac{35}{99}] = 1$
$\cdot$
$\cdot$
$\cdot$
$[\frac{2}{3} + \frac{98}{99}] = 1$
$∴$ Adding them, we get $66.$

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If [x] denotes the greatest integers ≤x, then find [23]+[(23)+(199)]+[(23)+(299)]+……+[(23)+(9899)]

The correct option is A 66 [23]=0 [23+3399]=[23+13]=[33]=1 ∴[23+3499]=1 [23+3599]=1 ⋅ ⋅ ⋅ [23+9899]=1 ∴ Adding them, we get 66.

If $[x]$ denotes the greatest integers $\leq x,$ then find

$[\frac{2}{3}] + [(\frac{2}{3}) + (\frac{1}{99})] + [(\frac{2}{3}) + (\frac{2}{99})] + \dots \dots + [(\frac{2}{3}) + (\frac{98}{99})]$

The correct option is A
66

$[\frac{2}{3}] = 0$
$[\frac{2}{3} + \frac{33}{99}] = [\frac{2}{3} + \frac{1}{3}] = [\frac{3}{3}] = 1$
$∴ [\frac{2}{3} + \frac{34}{99}] = 1$
$[\frac{2}{3} + \frac{35}{99}] = 1$
$\cdot$
$\cdot$
$\cdot$
$[\frac{2}{3} + \frac{98}{99}] = 1$
$∴$ Adding them, we get $66.$