GATE CS 1993 Q01.6 — Maxima and Minima, Mean Value Theorem, Integration

We analyze the convergence of each improper integral:

• Option A: Near $x = 0$, we have $\sin x \approx x$ and $1 - \cos x \approx \frac{x^2}{2}$. Thus, $$\frac{\sin x}{1-\cos x} \approx \frac{x}{x^2/2} = \frac{2}{x}$$ Since $\int_{0}^{1} \frac{1}{x} dx$ diverges, this integral diverges.
• Option B: We apply Dirichlet's Test for improper integrals of the form $\int_{a}^{\infty} f(x)g(x) dx$:
  1. $g(x) = \frac{1}{1+x}$ is positive, monotonically decreasing, and $\lim_{x \to \infty} g(x) = 0$.
  2. The partial integral of $f(x) = \cos x$ is bounded: $\left| \int_{0}^{t} \cos x \, dx \right| = |\sin t| \le 1$ for all $t \ge 0$.
  Therefore, the integral converges.
• Option C: Evaluating the integral directly: $$\int_{0}^{\infty} \frac{x}{1+x^2} dx = \left[ \frac{1}{2} \ln(1+x^2) \right]_{0}^{\infty} = \infty$$ Thus, it diverges.
• Option D: Near $x = 0$, we have $1 - \cos x \approx \frac{x^2}{2}$. The integrand simplifies to: $$\frac{1-\cos x}{\frac{x^5}{2}} \approx \frac{x^2/2}{x^5/2} = \frac{1}{x^3}$$ Since $\int_{0}^{1} \frac{1}{x^3} dx$ diverges (by $p$-test with $p = 3 \ge 1$), this integral diverges.