Differentiability and Mean Value Theorems Questions and Answers

Q: Let \(g: \mathbb{R} \to \mathbb{R}\) be a function defined by \(g(x) = \int_{0}^{x^{3}} (t^{2} - 9t + 8)\,dt\). Which one of the following statements is NOT true?

The statement that is NOT true is \(g\) has a local minimum at \(x = 1\); \(g\) has a local maximum there.By the fundamental theorem of calculus and the chain rule:\[ \begin{aligned} g'(x) &= (x^{6} - 9x^{3} + 8) \cdot 3x^{2} \\ &= 3x^{2}(x^{3} - 1)(x^{3} - 8) \end{aligned} \]So the critical points are \(x = 0, 1, 2\), which makes the third statement true. Near \(x = 1\) the factor \(x^{3} - 8\) is negative while \(x^{3} - 1\) changes from negative to positive, so \(g'\) changes from positive to negative: a local maximum, not a minimum.Near \(x = 0\), \(g'(x) \approx 24x^{2} \geq 0\), so \(g'\) has a local minimum of 0 there and \(g''\) changes sign: \(x = 0\) is a point of inflection. Since \(g'(0) = g'(1) = g'(2) = 0\), Rolle's theorem puts a zero of \(g''\) in each of \((0,1)\) and \((1,2)\), so \(g'' = 0\) has at least two solutions.

Q: Let \(f: \mathbb{R} \setminus \{0\} \to \mathbb{R}\) and \(g: \mathbb{R} \setminus \{0\} \to \mathbb{R}\) be two bounded and differentiable functions. Define \(F: \mathbb{R}^{2} \to \mathbb{R}\) by\[ F(x, y) = \begin{cases} x f(y) + y^{2} g(x), & xy \neq 0, \\ 0, & xy = 0. \end{cases} \]Consider the following statements.P: \(F\) is continuous at \((0, 0)\).Q: Both the partial derivatives \(\frac{\partial F}{\partial x}\) and \(\frac{\partial F}{\partial y}\) exist at \((1, 0)\).Then

P is correct and Q is NOT correct.P: let \(|f| \leq M\) and \(|g| \leq M\). Then \(|F(x, y)| \leq M|x| + My^{2} \to 0\) as \((x, y) \to (0, 0)\), and \(F(0,0) = 0\), so \(F\) is continuous at the origin.Q: on the line \(y = 0\), \(F(x, 0) = 0\) for all \(x\), so \(\frac{\partial F}{\partial x}(1, 0) = 0\) exists. But\[ \begin{aligned} \frac{\partial F}{\partial y}(1, 0) &= \lim_{y \to 0} \frac{F(1, y) - 0}{y} \\ &= \lim_{y \to 0} \frac{f(y) + y^{2}g(1)}{y} \end{aligned} \]which exists only if \(\frac{f(y)}{y}\) has a limit. A bounded differentiable \(f\) on \(\mathbb{R} \setminus \{0\}\) need not tend to 0 at 0; with \(f(y) = 1\) the quotient \(\frac{1}{y}\) blows up. So both partial derivatives need not exist, and Q fails.

Q: Let :\[ \begin{aligned} f(x) &= e^{-1/x} \quad \text{if } x \gt 0 \\ &= 0 \quad \text{if } x \le 0, \end{aligned} \]

f(x) is \(C^{\infty}\) but not analytic.For \(x \gt 0\) each derivative has the form \(f^{(k)}(x) = p_{k}(1/x)\,e^{-1/x}\) for a polynomial \(p_{k}\). As \(x \to 0^{+}\), \(e^{-1/x}\) goes to 0 faster than any power of \(1/x\) grows, so every derivative tends to 0. On the left all derivatives are 0. Hence \(f^{(k)}(0) = 0\) for every \(k\) and f is infinitely differentiable.But then the Taylor series at 0 is identically zero, while \(f(x) \gt 0\) for every \(x \gt 0\). The series does not represent f on any neighbourhood of 0, so f is not analytic there.This is the classical example separating smooth from analytic. It rules out the three options that stop at continuity or a finite number of derivatives.

Q: The function \(f(x) = |x|^{3}, x \in \mathbf{R}\) is :

\(f(x) = |x|^{3}\) is of class \(c^{2}\) but not \(c^{3}\).Write \(f(x) = x^{3}\) for \(x \ge 0\) and \(-x^{3}\) for \(x \lt 0\). Then\[ f'(x) = 3x|x|, \qquad f''(x) = 6|x| \]Both are continuous everywhere, including at 0 (where each equals 0), so f is twice continuously differentiable. But \(f''(x) = 6|x|\) has a corner at 0: its one-sided derivatives are \(+6\) and \(-6\), so \(f'''(0)\) does not exist and f is not of class \(c^{3}\).'Continuous but not differentiable' and 'differentiable but not \(c^{1}\)' stop too early: the cube flattens the corner of \(|x|\) twice over.

Q: The maximum value of \(\dfrac{\log x}{x}\), \(x \gt 0\) is :

The maximum value is \(\dfrac{1}{e}\).Let \(g(x) = \dfrac{\log x}{x}\). Then\[ g'(x) = \frac{1 - \log x}{x^{2}} \]which is positive for \(x \lt e\) and negative for \(x \gt e\). So g rises to a maximum at \(x = e\) and falls after it, and the maximum value is \(g(e) = \dfrac{\log e}{e} = \dfrac{1}{e}\).\(e\) is where the maximum occurs, not its value. \(e + 1/e\) and \(e - 1/e\) are not values of g at all: since \(\log x \lt x\), \(g(x) \lt 1\) everywhere, while both those numbers exceed 2.

Q: Consider the function :\[ f(x) = \begin{cases} x^{2}\sin(1/x), & \text{if } 0 \lt x \le 1 \\ 0, & \text{if } x = 0 \end{cases} \]Then :

\(f\) is of bounded variation.At 0, \(|f(x)/x| = |x \sin(1/x)| \le |x| \to 0\), so \(f'(0) = 0\). For \(0 \lt x \le 1\),\[ f'(x) = 2x\sin(1/x) - \cos(1/x) \]so \(|f'(x)| \le 2 + 1 = 3\) on [0, 1]. A function differentiable on a closed interval with a bounded derivative is Lipschitz (mean value theorem), and every Lipschitz function on [a, b] is of bounded variation.So \(f\) is continuous and differentiable at 0, ruling out (A) and (B), and (C) is false because \(f\) is of bounded variation. (With \(x\sin(1/x)\) instead, the function would be continuous but not of bounded variation; the square makes the difference.)

Let \(f: \mathbb{R} \to \mathbb{R}\) be a twice differentiable function such that \(f(n) = 1\) for all \(n \in \mathbb{Z}\).

Which ONE of the following statements is necessarily TRUE?

A.\(f'(n) = 0\) for all \(n \in \mathbb{Z}\).
B.\(f'(x) = 0\) for infinitely many \(x \in \mathbb{R} \setminus \mathbb{Z}\).
C.\(f''(n) = 0\) for all \(n \in \mathbb{Z}\).
D.\(f''(x) = 0\) for infinitely many \(x \in \mathbb{R} \setminus \mathbb{Z}\).

Exam Year(s)

IIT JAM - 2026,

Answer: B

The necessarily true statement is \(f'(x) = 0\) for infinitely many \(x \in \mathbb{R} \setminus \mathbb{Z}\).

For each integer \(n\), \(f(n) = f(n+1) = 1\), so Rolle's theorem gives a point \(c_n\) strictly between \(n\) and \(n+1\) with \(f'(c_n) = 0\). Each \(c_n\) is a non-integer, and they lie in different intervals, so there are infinitely many.

The other three can fail. With \(f(x) = 1 + \sin(\pi x)\), \(f(n) = 1\) but \(f'(n) = \pi\cos(n\pi) = \pm\pi \neq 0\), so the first option fails; and \(f''(x) = -\pi^{2}\sin(\pi x)\) vanishes only at integers, so the last option fails too. With \(f(x) = 1 + \sin^{2}(\pi x)\), \(f''(x) = 2\pi^{2}\cos(2\pi x)\), which equals \(2\pi^{2} \neq 0\) at every integer, so \(f''(n) = 0\) is not forced.

Discuss in Forum

Let \(f: \mathbb{R} \to \mathbb{R}\) be a function that has derivatives of all orders.

Which ONE of the following statements is FALSE?

A.If \(f\) is a polynomial in \(x\) of degree at most 5, then \(f^{(6)}(x) = 0\) for all \(x \in \mathbb{R}\).
B.If \(f^{(6)}(x) = 0\) for all \(x \in \mathbb{R}\), then \(f\) is a polynomial in \(x\) of degree at most 5.
C.If \(f^{(k)}(a) = 0\) for all \(1 \leq k \leq 5\) and \(f^{(6)}(a) \gt 0\) for some \(a \in \mathbb{R}\), then \(f(x)\) has a local minimum at \(a\).
D.If \(f^{(k)}(a) = 0\) for all \(1 \leq k \leq 6\) and \(f^{(7)}(a) \lt 0\) for some \(a \in \mathbb{R}\), then \(f(x)\) has a local maximum at \(a\).

Exam Year(s)

IIT JAM - 2026,

Answer: D

The FALSE statement is the last one: \(f^{(k)}(a) = 0\) for \(1 \leq k \leq 6\) and \(f^{(7)}(a) \lt 0\) does not give a local maximum.

The higher derivative test looks at the first derivative that does not vanish at \(a\). If its order is even, \(f\) has an extremum at \(a\): a minimum when that derivative is positive, a maximum when negative. If its order is odd, \(f\) has no extremum at \(a\). Here the first non-zero derivative is of order 7, which is odd. The model case \(f(x) = -(x-a)^{7}\) is strictly decreasing through \(a\), so it has neither a maximum nor a minimum there.

The other statements are true. Differentiating a polynomial of degree at most 5 six times gives 0. Conversely, integrating \(f^{(6)} = 0\) six times gives a polynomial of degree at most 5. And a first non-zero derivative of even order 6 that is positive gives a local minimum, as for \((x-a)^{6}\).

Discuss in Forum

Let \(g: \mathbb{R} \to \mathbb{R}\) be a function defined by \(g(x) = \int_{0}^{x^{3}} (t^{2} - 9t + 8)\,dt\). Which one of the following statements is NOT true?

A.\(g\) has a local minimum at \(x = 1\)
B.\(g\) has a point of inflection at \(x = 0\)
C.\(x = 0, x = 1, x = 2\) are critical points of \(g\)
D.\(g''(x) = 0\) has at least two distinct solutions, where \(g''\) denotes the second order derivative of \(g\)

Exam Year(s)

IIT JAM - 2026,

Answer: A

The statement that is NOT true is \(g\) has a local minimum at \(x = 1\); \(g\) has a local maximum there.

By the fundamental theorem of calculus and the chain rule:

\[ \begin{aligned} g'(x) &= (x^{6} - 9x^{3} + 8) \cdot 3x^{2} \\ &= 3x^{2}(x^{3} - 1)(x^{3} - 8) \end{aligned} \]

So the critical points are \(x = 0, 1, 2\), which makes the third statement true. Near \(x = 1\) the factor \(x^{3} - 8\) is negative while \(x^{3} - 1\) changes from negative to positive, so \(g'\) changes from positive to negative: a local maximum, not a minimum.

Near \(x = 0\), \(g'(x) \approx 24x^{2} \geq 0\), so \(g'\) has a local minimum of 0 there and \(g''\) changes sign: \(x = 0\) is a point of inflection. Since \(g'(0) = g'(1) = g'(2) = 0\), Rolle's theorem puts a zero of \(g''\) in each of \((0,1)\) and \((1,2)\), so \(g'' = 0\) has at least two solutions.

Discuss in Forum

Let \(f: \mathbb{R} \setminus \{0\} \to \mathbb{R}\) and \(g: \mathbb{R} \setminus \{0\} \to \mathbb{R}\) be two bounded and differentiable functions. Define \(F: \mathbb{R}^{2} \to \mathbb{R}\) by

\[ F(x, y) = \begin{cases} x f(y) + y^{2} g(x), & xy \neq 0, \\ 0, & xy = 0. \end{cases} \]

Consider the following statements.

P: \(F\) is continuous at \((0, 0)\).

Q: Both the partial derivatives \(\frac{\partial F}{\partial x}\) and \(\frac{\partial F}{\partial y}\) exist at \((1, 0)\).

Then

A.P is correct and Q is NOT correct
B.P is NOT correct and Q is correct
C.Both P and Q are correct
D.Neither P nor Q is correct

Exam Year(s)

IIT JAM - 2026,

Answer: A

P is correct and Q is NOT correct.

P: let \(|f| \leq M\) and \(|g| \leq M\). Then \(|F(x, y)| \leq M|x| + My^{2} \to 0\) as \((x, y) \to (0, 0)\), and \(F(0,0) = 0\), so \(F\) is continuous at the origin.

Q: on the line \(y = 0\), \(F(x, 0) = 0\) for all \(x\), so \(\frac{\partial F}{\partial x}(1, 0) = 0\) exists. But

\[ \begin{aligned} \frac{\partial F}{\partial y}(1, 0) &= \lim_{y \to 0} \frac{F(1, y) - 0}{y} \\ &= \lim_{y \to 0} \frac{f(y) + y^{2}g(1)}{y} \end{aligned} \]

which exists only if \(\frac{f(y)}{y}\) has a limit. A bounded differentiable \(f\) on \(\mathbb{R} \setminus \{0\}\) need not tend to 0 at 0; with \(f(y) = 1\) the quotient \(\frac{1}{y}\) blows up. So both partial derivatives need not exist, and Q fails.

Discuss in Forum

Let :

\[ \begin{aligned} f(x) &= e^{-1/x} \quad \text{if } x \gt 0 \\ &= 0 \quad \text{if } x \le 0, \end{aligned} \]

A.\(f(x)\) is not continuous
B.\(f(x)\) is continuous but not \(C^{1}\)
C.\(f(x)\) is \(C^{1}\) but not \(C^{\infty}\)
D.\(f(x)\) is \(C^{\infty}\) but not analytic

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

f(x) is \(C^{\infty}\) but not analytic.

For \(x \gt 0\) each derivative has the form \(f^{(k)}(x) = p_{k}(1/x)\,e^{-1/x}\) for a polynomial \(p_{k}\). As \(x \to 0^{+}\), \(e^{-1/x}\) goes to 0 faster than any power of \(1/x\) grows, so every derivative tends to 0. On the left all derivatives are 0. Hence \(f^{(k)}(0) = 0\) for every \(k\) and f is infinitely differentiable.

But then the Taylor series at 0 is identically zero, while \(f(x) \gt 0\) for every \(x \gt 0\). The series does not represent f on any neighbourhood of 0, so f is not analytic there.

This is the classical example separating smooth from analytic. It rules out the three options that stop at continuity or a finite number of derivatives.

Discuss in Forum

The function \(f(x) = |x|^{3}, x \in \mathbf{R}\) is :

A.continuous but not differentiable
B.differentiable but not \(c^{1}\)
C.of class \(c^{2}\)
D.of class \(c^{3}\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

\(f(x) = |x|^{3}\) is of class \(c^{2}\) but not \(c^{3}\).

Write \(f(x) = x^{3}\) for \(x \ge 0\) and \(-x^{3}\) for \(x \lt 0\). Then

\[ f'(x) = 3x|x|, \qquad f''(x) = 6|x| \]

Both are continuous everywhere, including at 0 (where each equals 0), so f is twice continuously differentiable. But \(f''(x) = 6|x|\) has a corner at 0: its one-sided derivatives are \(+6\) and \(-6\), so \(f'''(0)\) does not exist and f is not of class \(c^{3}\).

'Continuous but not differentiable' and 'differentiable but not \(c^{1}\)' stop too early: the cube flattens the corner of \(|x|\) twice over.

Discuss in Forum

The maximum value of \(\dfrac{\log x}{x}\), \(x \gt 0\) is :

A.\(e\)
B.\(\dfrac{1}{e}\)
C.\(e + \dfrac{1}{e}\)
D.\(\dfrac{-1}{e} + e\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The maximum value is \(\dfrac{1}{e}\).

Let \(g(x) = \dfrac{\log x}{x}\). Then

\[ g'(x) = \frac{1 - \log x}{x^{2}} \]

which is positive for \(x \lt e\) and negative for \(x \gt e\). So g rises to a maximum at \(x = e\) and falls after it, and the maximum value is \(g(e) = \dfrac{\log e}{e} = \dfrac{1}{e}\).

\(e\) is where the maximum occurs, not its value. \(e + 1/e\) and \(e - 1/e\) are not values of g at all: since \(\log x \lt x\), \(g(x) \lt 1\) everywhere, while both those numbers exceed 2.

Discuss in Forum

Consider the function :

\[ f(x) = \begin{cases} x^{2}\sin(1/x), & \text{if } 0 \lt x \le 1 \\ 0, & \text{if } x = 0 \end{cases} \]

Then :

A.\(f\) is not continuous at \(x = 0\)
B.\(f\) is continuous but not differentiable at \(x = 0\)
C.\(f\) is differentiable but not of bounded variation
D.\(f\) is of bounded variation

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

\(f\) is of bounded variation.

At 0, \(|f(x)/x| = |x \sin(1/x)| \le |x| \to 0\), so \(f'(0) = 0\). For \(0 \lt x \le 1\),

\[ f'(x) = 2x\sin(1/x) - \cos(1/x) \]

so \(|f'(x)| \le 2 + 1 = 3\) on [0, 1]. A function differentiable on a closed interval with a bounded derivative is Lipschitz (mean value theorem), and every Lipschitz function on [a, b] is of bounded variation.

So \(f\) is continuous and differentiable at 0, ruling out (A) and (B), and (C) is false because \(f\) is of bounded variation. (With \(x\sin(1/x)\) instead, the function would be continuous but not of bounded variation; the square makes the difference.)

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