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Question

Consider the following statements in respect of the given function:

\(f(x) = \begin{cases} \dfrac{x^3}{|x|}, & x \neq 0 \\ 0, & x = 0 \end{cases}\)

I. \(f(x)\) is continuous everywhere.

II. \(f(x)\) is differentiable everywhere.

Which of the statements given above is/are correct?

This question was previously asked in
NDA 2 2026 GAT Question Paper (13-Sep-2026)
The correct answer is

Both I and II

For \(x > 0\), \(f(x) = x^3/x = x^2\) and for \(x < 0\), \(f(x) = x^3/(-x) = -x^2\), i.e. \(f(x) = x|x|\) everywhere including \(x = 0\). This function is continuous everywhere since both pieces tend to \(0\) as \(x \to 0\). Its derivative works out to \(f'(x) = 2|x|\), which also exists at \(x = 0\), so \(f\) is differentiable everywhere too. Hence both statements I and II are correct.

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Important Questions from Differentiability

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  3. Let f be a differentiable function defined for all x ∈ R such that f(x3) = x5 for all x ∈ R, x ≠ 0. Then the value of \(\dfrac{df}{dx} (8)\) is:

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  5. The set of all point where the function f(x) = 2x|x| is differentiable, is:

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