If f x -lim n - r -1 n rx -1-3 -5 -2 r -1- then -03- f x - d x - x - is equal to where - denotes greatest integer functionA- 0B- 1C- 2D- 3

The correct option is B 1 T_r = 1/2 [x/1.3.5...(2r 1) x/1.3.5....(2r 1)(2r+1)] ∑^n_r=1 T_r = 1/2 [(x/1 x/1.3) + (x/1.3 x/1.35)+......] = 1/2 [x x/1.3. ...

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Standard XII

Mathematics

Monotonically Increasing Functions

If f x =lim n...

Question

If $f (x) = l i m n \to \infty \sum_{r - = 1}^{n} \frac{r x}{1.3.5 \dots (2 r + 1)}$ , then $\int_{0}^{3} [f (x)] d (x - [x])$ is equal to (where [.] denotes greatest integer function)

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Solution

The correct option is B
1

$T_{r} = \frac{1}{2} [\frac{x}{1.3.5... (2 r - 1)} - \frac{x}{1.3.5.... (2 r - 1) (2 r + 1)}] \sum_{r = 1}^{n} T_{r} = \frac{1}{2} [(\frac{x}{1} - \frac{x}{1.3}) + (\frac{x}{1.3} - \frac{x}{} 1.35) + . . . . . .] = \frac{1}{2} [x - \frac{x}{1.3.5.... (2 n + 1)}] L t n \to \infty \sum_{r = 1}^{n} T_{r} = \frac{x}{2}$

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If f(x)=limn→∞∑nr−=1rx1.3.5…(2r+1), then ∫30[f(x)]d(x−[x]) is equal to (where [.] denotes greatest integer function)

The correct option is B 1 Tr=12[x1.3.5...(2r−1)−x1.3.5....(2r−1)(2r+1)]∑nr=1Tr=12[(x1−x1.3)+(x1.3−x1.35)+......]=12[x−x1.3.5....(2n+1)]Ltn→∞∑nr=1Tr=x2

If $f (x) = l i m n \to \infty \sum_{r - = 1}^{n} \frac{r x}{1.3.5 \dots (2 r + 1)}$ , then $\int_{0}^{3} [f (x)] d (x - [x])$ is equal to (where [.] denotes greatest integer function)