GATE 2018 EE Q.11: solution and answer

GATE 2018 · EEQ.11 · 1 mark · MCQ

Let $f$ be a real-valued function of a real variable defined as $f (x) = x^{2}$ for $x \geq 0$ , and $f (x) = - x^{2}$ for $x < 0$ . Which one of the following statements is true?

(A)
$f (x)$ is discontinuous at $x = 0$ .
(B)
$f (x)$ is continuous but not differentiable at $x = 0$ .
(C)
$f (x)$ is differentiable but its first derivative is not continuous at $x = 0$ .
(D)
$f (x)$ is differentiable but its first derivative is not differentiable at $x = 0$ .

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Answer and solution

ANSWER

(D) $f (x)$ is differentiable but its first derivative is not differentiable at $x = 0$ .

Official answer: DSource: Official answer key, GATE 2018 EE (IIT Guwahati)

The two branches combine into one formula, $f (x) = x ∣ x ∣$ . For $x \geq 0$ , $x \cdot x = x^{2}$ , and for $x < 0$ , $x \cdot (- x) = - x^{2}$ .
Continuity at $0$ : $lim_{x \to 0^{-}} (- x^{2}) = 0 = lim_{x \to 0^{+}} x^{2} = f (0)$ , so $f$ is continuous at $0$ . This rules out (A).
Differentiability at $0$ , from the definition of the derivative: $f^{'} (0) = h \to 0 lim \frac{f ( h ) - f ( 0 )}{h} = h \to 0 lim \frac{h ∣ h ∣}{h} = h \to 0 lim ∣ h ∣ = 0.$ The limit exists, so $f$ is differentiable at $0$ . This rules out (B).
The first derivative everywhere: $f^{'} (x) = 2 x$ for $x > 0$ , $f^{'} (x) = - 2 x$ for $x < 0$ , and $f^{'} (0) = 0$ . Taken together, $f^{'} (x) = 2∣ x ∣$ .
Continuity of $f^{'}$ at $0$ : $lim_{x \to 0} 2∣ x ∣ = 0 = f^{'} (0)$ , so $f^{'}$ is continuous at $0$ . This rules out (C).
Differentiability of $f^{'}$ at $0$ , from the one-sided derivatives: $h \to 0^{+} lim \frac{2∣ h ∣ - 0}{h} = h \to 0^{+} lim \frac{2 h}{h} = 2, h \to 0^{-} lim \frac{2∣ h ∣ - 0}{h} = h \to 0^{-} lim \frac{- 2 h}{h} = - 2.$ The two sides differ, so $f^{'}$ is not differentiable at $0$ . The graph of $f^{'} (x) = 2∣ x ∣$ has a corner there.
So $f$ is differentiable and $f^{'}$ is continuous at $x = 0$ , but $f^{'}$ is not differentiable there. The answer is (D).

KEY IDEA

$f (x) = x ∣ x ∣$ . Its derivative is $f^{'} (x) = 2∣ x ∣$ , which is continuous at $0$ but has a corner there. So $f$ has a continuous first derivative, but no second derivative exists at $0$ : the one-sided second derivatives are $+ 2$ and $- 2$ . When a function is defined in pieces, test each level of smoothness at the joint separately with one-sided limits.

WATCH OUT FOR

Mixing up $x ∣ x ∣$ with $∣ x ∣$ and choosing (B). Another mistake is to see that the branch derivatives $2 x$ and $- 2 x$ come from different formulas and conclude that $f^{'}$ jumps, which gives (C). In fact both branches give $0$ at $x = 0$ , so $f^{'}$ is continuous there. What fails is the next level: the slopes of $f^{'}$ are $+ 2$ and $- 2$ .

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GATE 2018 EE Q.11: solution and answer

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