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GATE 2026 CE (CE2) – Question 36

Engineering Mathematics · Calculus: Limits, continuity, mean value theorems, maxima and minima, Taylor series · 2 marks · Multiple choice

Let

$$f(x) = \begin{vmatrix} x^3 & \sin x & \cos x \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{vmatrix}$$

where $p$ is a constant.

The value of $\frac{d^3}{dx^3} f(x)$ at $x = 0$ is

  1. $6p^3$
  2. $p + p^2$
  3. $p + p^3$
  4. independent of $p$

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Correct answer: (D) independent of $p$

Explanation

Expanding along the first row gives $f(x) = -p^3x^3 - 6p^3\sin x + (6p^2 + p)\cos x$. The third derivative is $f'''(x) = -6p^3 + 6p^3\cos x + (6p^2 + p)\sin x$. At $x = 0$ this is $-6p^3 + 6p^3 + 0 = 0$, so the value does not depend on $p$.