JEE Advanced 2017 · Paper 2 · Mathematics Q14

MathematicsMultiple correct+4 / -2Question 50 of 54

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

I > \log_e 99

I < \log_e 99

I < \frac{49}{50}

I > \frac{49}{50}

Answer and worked solution

Official answer: B, D

Bound the integrand. For $x\in(k,k+1)$ ,
$\frac1{x+1}<\frac{k+1}{x(x+1)}<\frac1x.$

Sum over k.
$\int_1^{99}\frac{dx}{x+1}<I<\int_1^{99}\frac{dx}x\implies\ln50<I<\ln99.$

Conclusion. $I<\ln 99$ , so (B) is true. Also $I>\ln50\approx3.9>\frac{49}{50}$ , so (D) is true.

(B), (D)

Topics:Properties of definite integrals

JEE Advanced 2017 · Paper 2, Mathematics question 14