For x ∈ R, Let [ x ] denotes the greatest integer ≤ x, then the sum of the series [- (1)/(3)] + [- (1)/(3) - (1)/(100)] + [- (1)/(3) - (2)/(100)] + . . . . . + [- (1)/(3) - (99)/(100)] is

Correct answer: D. - 133 Greatest integer function f(x) = [x] = \ array ll 0; 0 ≤ x Given series S = [- (1)/(3)] + [- (1)/(3) - (1)/(100)] + [- (1)/(3) - (2)/(100)] + [- (1)/(3) - (3)/(100)] + + [- (1)/(3) - (99)/(100)] General term T _ r = [- (1)/(3) - (r)/(100)] = \ array ll - 10 ≤ r ≤ 66 - 2r > 66 array ) ⇒ S = ∑ _ r = 0 ^ 66 (- 1) + ∑ _ r = 67 ^ 99 (- 2) = (- 67) + (- 2) × 33 = - 133

JEE Main 201912 AprilMorning Shift

Exam: JEE Main
Year: 2019
Date: 12 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Functions

JEE Main 2019 · 12 April · Morning Shift · Mathematics · Functions

For x ∈ R, Let [ x ] denotes the greatest integer ≤ x, then the sum...

For

x \in R,

Let

[ x ]

denotes the greatest integer

\le x,

then the sum of the series \left[- \dfrac{1}{3}\right] +

\left[- \dfrac{1}{3} - \dfrac{1}{100}\right] + \left[- \dfrac{1}{3} - \dfrac{2}{100}\right] + . . . . . + \left[- \dfrac{1}{3} - \dfrac{99}{100}\right]

Options

A $- 131$
B $- 153$
C $- 135$
D $- 133$

Correct answer

D. $- 133$

Step-by-step solution

Greatest integer function

f\left(x\right) = \left[x\right] = \left\{

\begin{array}{ll}0; 0 \le x < 1 \\ - 1; - 1 \le x < 0 \\ - 2; - 2 \le x < - 1\end{array}

\right)<br /><br />Given series<br />

S = \left[- \dfrac{1}{3}\right] + \left[- \dfrac{1}{3} - \dfrac{1}{100}\right] + \left[- \dfrac{1}{3} - \dfrac{2}{100}\right] + \left[- \dfrac{1}{3} - \dfrac{3}{100}\right] + \ldots + \left[- \dfrac{1}{3} - \dfrac{99}{100}\right]

<br />General term

{T}_{r} = \left[- \dfrac{1}{3} - \dfrac{r}{100}\right] = \left\{

\begin{array}{ll}- 10 \le r \le 66 \\ - 2r > 66\end{array}

\right)

⇒ S = \mathop{∑}_{r = 0}^{66}\left(- 1\right) +\mathop{∑}_{r = 67}^{99}\left(- 2\right) = \left(- 67\right) + \left(- 2\right) \times 33

= - 133

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