PYQs / GATE EE / 2018 / Set 1 / Q21 GATE EE 2018 Set 1 — Question 21 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. MCQ +1 / -0.33 Easy Calculus Engineering Mathematics
Engineering Mathematics → Calculus
Last updated 5 September 2026
Question Let
f f f be a real-valued function of a real variable defined as
f ( x ) = x 2 f(x) = x^2 f ( x ) = x 2 for
x ≥ 0 x \geq 0 x ≥ 0 , and
f ( x ) = − x 2 f(x) = -x^2 f ( x ) = − x 2 for
x < 0 x < 0 x < 0 . Which one of the following statements is true?
Correct answer (D) f(x) is differentiable but its first derivative is not differentiable at x = 0.
Solution Given the function:
f ( x ) = { x 2 , x ≥ 0 − x 2 , x < 0 f(x) = \begin{cases} x^2, & x \geq 0 \\ -x^2, & x < 0 \end{cases} f ( x ) = { x 2 , − x 2 , x ≥ 0 x < 0 1. Continuity at x = 0 x = 0 x = 0 lim x → 0 + f ( x ) = lim x → 0 + x 2 = 0 \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x^2 = 0 lim x → 0 + f ( x ) = lim x → 0 + x 2 = 0 lim x → 0 − f ( x ) = lim x → 0 − ( − x 2 ) = 0 \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-x^2) = 0 lim x → 0 − f ( x ) = lim x → 0 − ( − x 2 ) = 0 f ( 0 ) = 0 2 = 0 f(0) = 0^2 = 0 f ( 0 ) = 0 2 = 0 Since
lim x → 0 + f ( x ) = lim x → 0 − f ( x ) = f ( 0 ) \lim_{x \to 0^+} f(x) = \lim_{x \to 0^-} f(x) = f(0) lim x → 0 + f ( x ) = lim x → 0 − f ( x ) = f ( 0 ) ,
f ( x ) f(x) f ( x ) is continuous at
x = 0 x = 0 x = 0 .
2. Differentiability at x = 0 x = 0 x = 0 f ′ ( 0 + ) = lim h → 0 + f ( h ) − f ( 0 ) h = lim h → 0 + h 2 − 0 h = lim h → 0 + h = 0 f'(0^+) = \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{h^2 - 0}{h} = \lim_{h \to 0^+} h = 0 f ′ ( 0 + ) = lim h → 0 + h f ( h ) − f ( 0 ) = lim h → 0 + h h 2 − 0 = lim h → 0 + h = 0 f ′ ( 0 − ) = lim h → 0 − f ( h ) − f ( 0 ) h = lim h → 0 − − h 2 − 0 h = lim h → 0 − ( − h ) = 0 f'(0^-) = \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{-h^2 - 0}{h} = \lim_{h \to 0^-} (-h) = 0 f ′ ( 0 − ) = lim h → 0 − h f ( h ) − f ( 0 ) = lim h → 0 − h − h 2 − 0 = lim h → 0 − ( − h ) = 0 Since
f ′ ( 0 + ) = f ′ ( 0 − ) = 0 f'(0^+) = f'(0^-) = 0 f ′ ( 0 + ) = f ′ ( 0 − ) = 0 ,
f ( x ) f(x) f ( x ) is differentiable at
x = 0 x = 0 x = 0 and
f ′ ( 0 ) = 0 f'(0) = 0 f ′ ( 0 ) = 0 .
3. Continuity of f ′ ( x ) f'(x) f ′ ( x ) x = 0 x = 0 x = 0 For
x > 0 x > 0 x > 0 ,
f ′ ( x ) = 2 x f'(x) = 2x f ′ ( x ) = 2 x For
x < 0 x < 0 x < 0 ,
f ′ ( x ) = − 2 x f'(x) = -2x f ′ ( x ) = − 2 x So,
f ′ ( x ) = 2 ∣ x ∣ f'(x) = 2|x| f ′ ( x ) = 2∣ x ∣ for all
x x x .
lim x → 0 f ′ ( x ) = lim x → 0 2 ∣ x ∣ = 0 = f ′ ( 0 ) \lim_{x \to 0} f'(x) = \lim_{x \to 0} 2|x| = 0 = f'(0) lim x → 0 f ′ ( x ) = lim x → 0 2∣ x ∣ = 0 = f ′ ( 0 ) Thus,
f ′ ( x ) f'(x) f ′ ( x ) is continuous at
x = 0 x = 0 x = 0 .
4. Differentiability of f ′ ( x ) f'(x) f ′ ( x ) x = 0 x = 0 x = 0 The derivative of
f ′ ( x ) = 2 ∣ x ∣ f'(x) = 2|x| f ′ ( x ) = 2∣ x ∣ at
x = 0 x = 0 x = 0 is:
lim h → 0 + f ′ ( h ) − f ′ ( 0 ) h = lim h → 0 + 2 h − 0 h = 2 \lim_{h \to 0^+} \frac{f'(h) - f'(0)}{h} = \lim_{h \to 0^+} \frac{2h - 0}{h} = 2 lim h → 0 + h f ′ ( h ) − f ′ ( 0 ) = lim h → 0 + h 2 h − 0 = 2 lim h → 0 − f ′ ( h ) − f ′ ( 0 ) h = lim h → 0 − − 2 h − 0 h = − 2 \lim_{h \to 0^-} \frac{f'(h) - f'(0)}{h} = \lim_{h \to 0^-} \frac{-2h - 0}{h} = -2 lim h → 0 − h f ′ ( h ) − f ′ ( 0 ) = lim h → 0 − h − 2 h − 0 = − 2 Since the left-hand and right-hand derivatives of
f ′ ( x ) f'(x) f ′ ( x ) at
x = 0 x = 0 x = 0 are not equal (
2 ≠ − 2 2 \neq -2 2 = − 2 ),
f ′ ( x ) f'(x) f ′ ( x ) is not differentiable at
x = 0 x = 0 x = 0 .
Conclusion:
f ( x ) f(x) f ( x ) is differentiable at
x = 0 x = 0 x = 0 , but its first derivative
f ′ ( x ) f'(x) f ′ ( x ) is not differentiable at
x = 0 x = 0 x = 0 . This matches option (D).
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Correct answer (D) f(x) is differentiable but its first derivative is not differentiable at x = 0.
Solution Given the function:
f ( x ) = { x 2 , x ≥ 0 − x 2 , x < 0 f(x) = \begin{cases} x^2, & x \geq 0 \\ -x^2, & x < 0 \end{cases} f ( x ) = { x 2 , − x 2 , x ≥ 0 x < 0 1. Continuity at x = 0 x = 0 x = 0 lim x → 0 + f ( x ) = lim x → 0 + x 2 = 0 \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x^2 = 0 lim x → 0 + f ( x ) = lim x → 0 + x 2 = 0 lim x → 0 − f ( x ) = lim x → 0 − ( − x 2 ) = 0 \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-x^2) = 0 lim x → 0 − f ( x ) = lim x → 0 − ( − x 2 ) = 0 f ( 0 ) = 0 2 = 0 f(0) = 0^2 = 0 f ( 0 ) = 0 2 = 0 Since
lim x → 0 + f ( x ) = lim x → 0 − f ( x ) = f ( 0 ) \lim_{x \to 0^+} f(x) = \lim_{x \to 0^-} f(x) = f(0) lim x → 0 + f ( x ) = lim x → 0 − f ( x ) = f ( 0 ) ,
f ( x ) f(x) f ( x ) is continuous at
x = 0 x = 0 x = 0 .
2. Differentiability at x = 0 x = 0 x = 0 f ′ ( 0 + ) = lim h → 0 + f ( h ) − f ( 0 ) h = lim h → 0 + h 2 − 0 h = lim h → 0 + h = 0 f'(0^+) = \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{h^2 - 0}{h} = \lim_{h \to 0^+} h = 0 f ′ ( 0 + ) = lim h → 0 + h f ( h ) − f ( 0 ) = lim h → 0 + h h 2 − 0 = lim h → 0 + h = 0 f ′ ( 0 − ) = lim h → 0 − f ( h ) − f ( 0 ) h = lim h → 0 − − h 2 − 0 h = lim h → 0 − ( − h ) = 0 f'(0^-) = \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{-h^2 - 0}{h} = \lim_{h \to 0^-} (-h) = 0 f ′ ( 0 − ) = lim h → 0 − h f ( h ) − f ( 0 ) = lim h → 0 − h − h 2 − 0 = lim h → 0 − ( − h ) = 0 Since
f ′ ( 0 + ) = f ′ ( 0 − ) = 0 f'(0^+) = f'(0^-) = 0 f ′ ( 0 + ) = f ′ ( 0 − ) = 0 ,
f ( x ) f(x) f ( x ) is differentiable at
x = 0 x = 0 x = 0 and
f ′ ( 0 ) = 0 f'(0) = 0 f ′ ( 0 ) = 0 .
3. Continuity of f ′ ( x ) f'(x) f ′ ( x ) x = 0 x = 0 x = 0 For
x > 0 x > 0 x > 0 ,
f ′ ( x ) = 2 x f'(x) = 2x f ′ ( x ) = 2 x For
x < 0 x < 0 x < 0 ,
f ′ ( x ) = − 2 x f'(x) = -2x f ′ ( x ) = − 2 x So,
f ′ ( x ) = 2 ∣ x ∣ f'(x) = 2|x| f ′ ( x ) = 2∣ x ∣ for all
x x x .
lim x → 0 f ′ ( x ) = lim x → 0 2 ∣ x ∣ = 0 = f ′ ( 0 ) \lim_{x \to 0} f'(x) = \lim_{x \to 0} 2|x| = 0 = f'(0) lim x → 0 f ′ ( x ) = lim x → 0 2∣ x ∣ = 0 = f ′ ( 0 ) Thus,
f ′ ( x ) f'(x) f ′ ( x ) is continuous at
x = 0 x = 0 x = 0 .
4. Differentiability of f ′ ( x ) f'(x) f ′ ( x ) x = 0 x = 0 x = 0 The derivative of
f ′ ( x ) = 2 ∣ x ∣ f'(x) = 2|x| f ′ ( x ) = 2∣ x ∣ at
x = 0 x = 0 x = 0 is:
lim h → 0 + f ′ ( h ) − f ′ ( 0 ) h = lim h → 0 + 2 h − 0 h = 2 \lim_{h \to 0^+} \frac{f'(h) - f'(0)}{h} = \lim_{h \to 0^+} \frac{2h - 0}{h} = 2 lim h → 0 + h f ′ ( h ) − f ′ ( 0 ) = lim h → 0 + h 2 h − 0 = 2 lim h → 0 − f ′ ( h ) − f ′ ( 0 ) h = lim h → 0 − − 2 h − 0 h = − 2 \lim_{h \to 0^-} \frac{f'(h) - f'(0)}{h} = \lim_{h \to 0^-} \frac{-2h - 0}{h} = -2 lim h → 0 − h f ′ ( h ) − f ′ ( 0 ) = lim h → 0 − h − 2 h − 0 = − 2 Since the left-hand and right-hand derivatives of
f ′ ( x ) f'(x) f ′ ( x ) at
x = 0 x = 0 x = 0 are not equal (
2 ≠ − 2 2 \neq -2 2 = − 2 ),
f ′ ( x ) f'(x) f ′ ( x ) is not differentiable at
x = 0 x = 0 x = 0 .
Conclusion:
f ( x ) f(x) f ( x ) is differentiable at
x = 0 x = 0 x = 0 , but its first derivative
f ′ ( x ) f'(x) f ′ ( x ) is not differentiable at
x = 0 x = 0 x = 0 . This matches option (D).
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