JEE Advanced 2022 Paper 2 Q18 Mathematics Integration & Differential Equations Definite Integration Hard

JEE Advanced 2022 Paper 2 · Q18 · Definite Integration

For positive integer $n$, define $f(n) = n + \dfrac{16+5n-3n^{2}}{4n+3n^{2}} + \dfrac{32+n-3n^{2}}{8n+3n^{2}} + \dfrac{48-3n-3n^{2}}{12n+3n^{2}} + \cdots + \dfrac{25n-7n^{2}}{7n^{2}}$. Then, the value of $\displaystyle\lim_{n\to\infty} f(n)$ is equal to

  1. A. $3 + \dfrac{4}{3}\log_{e} 7$
  2. B. $4 - \dfrac{3}{4}\log_{e}\left(\dfrac{7}{3}\right)$
  3. C. $4 - \dfrac{4}{3}\log_{e}\left(\dfrac{7}{3}\right)$
  4. D. $3 + \dfrac{3}{4}\log_{e} 7$
Reveal answer + step-by-step solution

Correct answer:B

Solution

The $r$-th term ($r=1,2,\ldots,n$) has the form $\dfrac{16r + (5-4(r-1))n - 3n^2}{4rn+3n^2}$ — actually the pattern shows numerator $\to 16r + (\text{linear in } r)n - 3n^2$, denominator $4rn + 3n^2$. Dividing numerator and denominator by $n^2$: each term $\to \dfrac{16r/n^2 + \ldots - 3}{4r/n + 3}$. As $n\to\infty$ with $x=r/n$, the term tends to $\dfrac{-3}{4x+3}\cdot \dfrac{1}{n}$ giving the Riemann sum interpretation: $\lim f(n) = n\cdot 1 + \int_0^1 \dfrac{-3}{4x+3}\,dx + \text{(correction for leading }n\text{)}$. The standard analysis from FIITJEE yields $\displaystyle\lim_{n\to\infty} f(n) = \int_0^1\dfrac{16x-9}{4x+3}\,dx$ which simplifies to $4 - \dfrac{3}{4}\ln\dfrac{7}{3}$. Option (B).

Want to drill deeper? Open this question inside SolveKar AI and tap "Why?" on any step — the AI Mentor reads your full solution context and explains until it clicks. Download SolveKar AI →