If -x- stands for the greatest integer function then the value of -1-5-1-1000-1-5-1-2000 -1-5-999-1000- is-

Explanation for the correct option.[1/5+r/1000] will be equal to 1 for x=800.So, we can say that[1/5+r/1000]={[ 1 800≤ xNumber of terms in the interval 800≤ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

Mathematics

Fractional Part Function

If [x] stands...

Question

If $[x]$ stands for the greatest integer function then the value of $[\frac{1}{5} + \frac{1}{1000} + \frac{1}{5} + \frac{1}{2000} + . . . . . . + \frac{1}{5} + \frac{999}{1000}]$ is:

$199$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$201$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$202$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$200$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D
$200$

Explanation for the correct option.
$[\frac{1}{5} + \frac{r}{1000}]$ will be equal to $1$ for $x = 800$ .
So, we can say that
$[\frac{1}{5} + \frac{r}{1000}] = \{\begin{cases} 1 800 \leq x < 999 \\ 0 0 < x < 800 \end{cases}$
Number of terms in the interval $800 \leq x < 999$ is $200$ .
So,
$\begin{array}{rcl} [\frac{1}{5} + \frac{1}{1000} + \frac{1}{5} + \frac{1}{2000} + . . . . . . + \frac{1}{5} + \frac{999}{1000}] & = & 0 + (200 \times 1) \\ = & 200 \end{array}$
Hence, option D is correct.

Suggest Corrections

Similar questions

Q. Consider the sequence

a_{n}

given by

a_{1} = \frac{1}{2}, a_{n} + 1 = a_{n}^{2} + a_{n}

Let

S_{n} = \frac{1}{a_{1} + 1} + \frac{1}{a_{2} + 1} + . . . . . . . . . . . + \frac{1}{a_{n} + 1}

then find the value of

[S_{2012}]

, where [.] denotes greatest integer function.

$l o g_{e} (n + 1) - l o g_{e} (n - 1) = 4 a [(\frac{1}{n}) + (\frac{1}{3 n^{3}}) + (\frac{1}{5 n^{5}}) + . . . \infty]$ Find $8 a$ .

Q. Among

\frac{1}{12}

\frac{1}{15}

and

\frac{1}{17}

, is the greatest fraction

Q. If

α

β

γ

are the roots of

x^{3} + a x^{2} + b = 0

b \neq 0

then the determinant

Δ

, where

Δ = ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{1}{α} & \frac{1}{β} & \frac{1}{γ} \frac{1}{β} & \frac{1}{γ} & \frac{1}{α} \frac{1}{γ} & \frac{1}{α} & \frac{1}{β} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣

equals

Q. If

α

and

β

are the zeroes of the quadratic polynomial

f (x) = x^{2} - (\sqrt{5} - 1) x - (\sqrt{5} + 1)

, then the value of

\frac{1}{α^{2}} + \frac{1}{β^{2}}

is ____________.

Join BYJU'S Learning Program

If [x] stands for the greatest integer function then the value of 15+11000+15+12000+......+15+9991000 is:

The correct option is D 200Explanation for the correct option.15+r1000 will be equal to 1 for x=800.So, we can say that15+r1000=1800≤x<99900<x<800Number of terms in the interval 800≤x<999 is 200.So, 15+11000+15+12000+......+15+9991000=0+(200×1)=200Hence, option D is correct.

If $[x]$ stands for the greatest integer function then the value of $[\frac{1}{5} + \frac{1}{1000} + \frac{1}{5} + \frac{1}{2000} + . . . . . . + \frac{1}{5} + \frac{999}{1000}]$ is: