GATE DA 2024 | Question 37

MCQ · 2 marks · −0.66

Let \(f: \mathbb{R} \to \mathbb{R}\) be a function. Note: \(\mathbb{R}\) denotes the set of real numbers.

\[f(x) = \begin{cases} -x, & \text{if } x \lt -2 \\ ax^{2} + bx + c, & \text{if } x \in [-2,2] \\ x, & \text{if } x \gt 2 \end{cases}\]

Which ONE of the following choices gives the values of \(a, b, c\) that make the function \(f\) continuous and differentiable?

  1. Option A: \(a = \dfrac{1}{4}, b = 0, c = 1\) (correct)
  2. Option B: \(a = \dfrac{1}{2}, b = 0, c = 0\)
  3. Option C: \(a = 0, b = 0, c = 0\)
  4. Option D: \(a = 1, b = 1, c = -4\)

Answer

(A)

Explanation

Let \(q(x) = ax^{2} + bx + c\). Both outer pieces equal 2 at the joins, with slope \(-1\) on the left and \(+1\) on the right.

  • Continuity: \(q(-2) = 4a - 2b + c = 2\) and \(q(2) = 4a + 2b + c = 2\).
  • Differentiability: \(q'(-2) = -4a + b = -1\) and \(q'(2) = 4a + b = 1\).

Adding the slope equations gives \(b = 0\), so \(a = \frac{1}{4}\), and then \(1 + c = 2\) gives \(c = 1\). The middle piece \(\frac{x^{2}}{4} + 1\) meets both lines smoothly.

Option B, \(\frac{x^{2}}{2}\), reaches the value 2 at both joins but with slopes \(-2\) and \(2\), so it is continuous but not differentiable. Options C and D are not even continuous at \(x = -2\).

Answer: A