GATE 2024 DA Q.37: solution and answer

GATE 2024 · DAQ.37 · 2 marks · MCQ

Let $f : R \to R$ be a function. Note: $R$ denotes the set of real numbers.

f (x) = ⎩ ⎨ ⎧ - x, a x^{2} + b x + c, x, if x < - 2 if x \in [- 2, 2] if x > 2

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

(A)
$a = \frac{1}{4}, b = 0, c = 1$
(B)
$a = \frac{1}{2}, b = 0, c = 0$
(C)
$a = 0, b = 0, c = 0$
(D)
$a = 1, b = 1, c = - 4$

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Answer and solution

ANSWER

Option (A): $a = \frac{1}{4}, b = 0, c = 1$ .

Official answer: ASource: Official answer key, GATE 2024 DA (IISc Bengaluru)

The outer pieces are $- x$ for $x < - 2$ and $x$ for $x > 2$ . Both are differentiable, so only the joins at $x = \pm 2$ need checking. The middle piece is $q (x) = a x^{2} + b x + c$ , with $q^{'} (x) = 2 a x + b$ .
Continuity at $x = - 2$ : the left limit is $- (- 2) = 2$ , so $q (- 2) = 4 a - 2 b + c = 2$ .
Continuity at $x = 2$ : the right limit is $2$ , so $q (2) = 4 a + 2 b + c = 2$ .
Subtracting the first equation from the second gives $4 b = 0$ , so $b = 0$ . Then $4 a + c = 2$ .
Differentiability at $x = - 2$ : the left derivative is $\frac{d}{d x} (- x) = - 1$ and the right derivative is $q^{'} (- 2) = - 4 a + b$ . Setting them equal gives $- 4 a + b = - 1$ . With $b = 0$ this gives $a = \frac{1}{4}$ .
Differentiability at $x = 2$ : the right derivative is $\frac{d}{d x} (x) = 1$ and the left derivative is $q^{'} (2) = 4 a + b$ . Setting them equal gives $4 a + b = 1$ . With $b = 0$ this again gives $a = \frac{1}{4}$ , so the two conditions agree.
From $4 a + c = 2$ with $a = \frac{1}{4}$ : $c = 2 - 1 = 1$ .
Hence $a = \frac{1}{4}, b = 0, c = 1$ , which is option (A). Check: $q (x) = \frac{1}{4} x^{2} + 1$ gives $q (\pm 2) = 2$ , matching the outer pieces. Its slope $q^{'} (x) = \frac{x}{2}$ is $- 1$ at $x = - 2$ and $+ 1$ at $x = 2$ , matching the slopes of $- x$ and $x$ .
Options (B), (C) and (D) fail. (B) has $a = \frac{1}{2}$ , which violates $a = \frac{1}{4}$ . (C) gives $q \equiv 0$ , so $q (2) = 0 \neq = 2$ . (D) gives $q (- 2) = 4 - 2 - 4 = - 2 \neq = 2$ .

KEY IDEA

Differentiable implies continuous. At each junction, match both the function value and the derivative of the middle quadratic with those of the outer line. This gives four linear equations in $a, b, c$ (two of them consistent).

WATCH OUT FOR

Imposing only continuity. Continuity alone gives $b = 0$ and $4 a + c = 2$ , which leaves a whole family of solutions, and options (B) and (C) are tempting for that reason. The derivative conditions fix $a = \frac{1}{4}$ .

From the GATE 2024 DA paper. Every question of this paper · Download the full paper · All 6 Limits & continuity questions, with notes

GATE 2024 DA Q.37: solution and answer

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