If $I=\sum_{k=1}^{98} \int_{k}^{k+1} \frac{k+1}{x(x+1)} dx$, then

JEE Advanced · 2017 · 21 May 2017 · Paper 2 · Definite Integrals

I = \sum_{k = 1}^{98} \int_{k}^{k + 1} \frac{k + 1}{x ( x + 1 )} d x

, then

Answer: (B), (D)

For

x \in [k, k + 1]

, we have:

\frac{1}{x + 1} < \frac{k + 1}{x ( x + 1 )} < \frac{1}{x}

Integrating from

k

k + 1

\int_{k}^{k + 1} \frac{d x}{x + 1} < \int_{k}^{k + 1} \frac{k + 1}{x ( x + 1 )} d x < \int_{k}^{k + 1} \frac{d x}{x}

ln (k + 2) - ln (k + 1) < \int_{k}^{k + 1} \frac{k + 1}{x ( x + 1 )} d x < ln (k + 1) - ln k

Summing from

k = 1

98

k = 1 \sum 98 [ln (k + 2) - ln (k + 1)] < I < k = 1 \sum 98 [ln (k + 1) - ln k]

ln 100 - ln 2 < I < ln 99 - ln 1

ln 50 < I < ln 99

Since

ln 50 = ln \frac{100}{2} > ln \frac{49}{50} \cdot 50 = ln 49 > \frac{49}{50}

(for large arguments,

ln x > \frac{x - 1}{x}

), we have

I > \frac{49}{50}

and

I < ln 99

JEE Advanced · 2017 · 21 May 2017 · Paper 2 · One or more correct options