GATE DA Calculus and Optimization Previous Year Questions

GATE DA 2024 | Question 50

MSQ · 2 marks

Consider the function \(f: \mathbb{R} \to \mathbb{R}\) where \(\mathbb{R}\) is the set of all real numbers.

\[ f(x) = \frac{x^{4}}{4} - \frac{2x^{3}}{3} - \frac{3x^{2}}{2} + 1 \]

Which of the following statements is/are TRUE?

\(x = 0\) is a local maximum of \(f\) \(x = 3\) is a local minimum of \(f\) \(x = -1\) is a local maximum of \(f\) \(x = 0\) is a local minimum of \(f\)

Explanation

Answer: (A), (B)

Differentiate and factor:

\[ f'(x) = x^{3} - 2x^{2} - 3x = x(x - 3)(x + 1) \]

so the critical points are \(x = -1, 0, 3\). Test each with the second derivative \(f''(x) = 3x^{2} - 4x - 3\):

\(f''(0) = -3 < 0\): a local maximum at \(x = 0\), so A holds and D fails.
\(f''(3) = 27 - 12 - 3 = 12 > 0\): a local minimum at \(x = 3\), so B holds.
\(f''(-1) = 3 + 4 - 3 = 4 > 0\): \(x = -1\) is a local minimum, not a maximum, so C fails.

Answer: A, B

GATE DA 2024 | Question 60

NAT · 2 marks

Evaluate the following limit:

\[ \lim_{x \to 0} \frac{\ln((x^{2}+1)\cos x)}{x^{2}} = \]

______.

Your answer

Explanation

Answer: 0.5

The logarithm of a product splits into a sum:

\[ \frac{\ln((x^{2}+1)\cos x)}{x^{2}} = \frac{\ln(1 + x^{2})}{x^{2}} + \frac{\ln(\cos x)}{x^{2}} \]

Since \(\ln(1 + t) \approx t\) for small \(t\), the first term tends to 1.
Since \(\cos x \approx 1 - \frac{x^{2}}{2}\), we get \(\ln(\cos x) \approx -\frac{x^{2}}{2}\), so the second term tends to \(-\frac{1}{2}\).

The limit is \(1 - \frac{1}{2} = 0.5\). L'Hôpital's rule agrees: differentiating the top and the bottom gives \(\frac{2x/(1 + x^{2}) - \tan x}{2x}\), which also tends to \(1 - \frac{1}{2}\).

Answer: 0.5