JEE Main 2025 April

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Animated Solution for Mathematics - Differentiation: Let $f : R \to R$ be a function defined by $f (x) = ∣∣ x + 2∣ - 2∣ x ∣∣$ . If $m$ is the number of points of local minima and $n$ is the number of points of local maxima of $f$ , then $m + n$ is

Q: Why do we split the intervals at $x = -2$ and $x = 0$?

The absolute value function $|u|$ changes its behavior exactly when the expression $u$ changes sign. For $f(x) = ||x + 2| - 2|x||$, the inner expressions are $x + 2$ and $x$. These change sign at $x = -2$ and $x = 0$, respectively. By splitting the real line at these points, we ensure that in each interval, the expressions inside the absolute values have a constant sign, allowing us to remove the modulus signs algebraically.

Q: Do I need to use derivatives to find the local minima and maxima?

Not necessarily! While derivatives are a powerful tool, they are often overkill for piecewise linear functions. Since the function is composed of straight-line segments, the local extrema will always occur at the 'corners' (the points where the slope changes). You can identify these points by sketching the graph or by observing where the function changes from decreasing to increasing (minima) or increasing to decreasing (maxima).

Q: Why is $x = -2$ not a local minimum or maximum?

Great observation! If you look at the graph, at $x = -2$, the function is continuous, but the slope changes. However, for a point to be a local minimum, the function must be smaller than its neighbors, and for a local maximum, it must be larger. At $x = -2$, the function value is $f(-2) = ||-2 + 2| - 2|-2|| = |0 - 4| = 4$. If you check the values slightly to the left and right, you will see that the function is decreasing on both sides of $x = -2$. Since it doesn't change direction (from decreasing to increasing or vice versa), it is not an extremum.

Select Answer:

Progressive Hints

Hint 1

The nested absolute value function

f (x) = ∣∣ x + 2∣ - 2∣ x ∣∣

might look complex, but it is simply a piecewise linear function. The key is to identify the critical points where the expressions inside the absolute values change sign. These points are

x = - 2

and

x = 0

Hint 2

To solve this, you must analyze the function in three distinct intervals:

x < - 2

- 2 \leq x < 0

, and

x \geq 0

. In each interval, remove the absolute value signs by considering the sign of the expressions inside. Remember that

∣ a ∣ = a

a \geq 0

and

∣ a ∣ = - a

a < 0

Hint 3

Once you have the piecewise definition, sketch the graph. The local minima and maxima are the 'corners' or vertices of this graph. For example, in the interval

- 2 \leq x < 0

, the function simplifies to

∣3 x + 2∣

, which has a corner at

x = - \frac{2}{3}

. Find all such corners to count the local extrema.

Visualized Solution

Analyzing the Function $f (x)$

Function: $f (x) = ∣∣ x + 2∣ - 2∣ x ∣∣$
Identify critical points for the inner absolute values:
$x + 2 = 0 ⟹ x = - 2$
$x = 0$
The real line is divided into: $(- \infty, - 2)$ , $[- 2, 0)$ , and $[0, \infty)$ .

Case 1: $x < - 2$

For $x < - 2$ :
$∣ x + 2∣ = - (x + 2)$ and $∣ x ∣ = - x$
$f (x) = ∣ - (x + 2) - 2 (- x) ∣$
$f (x) = ∣ - x - 2 + 2 x ∣ = ∣ x - 2∣$
Since $x < - 2$ , $x - 2 < 0$ , so $f (x) = - (x - 2) = 2 - x$

Case 2: $- 2 \leq x < 0$

For $- 2 \leq x < 0$ :
$∣ x + 2∣ = x + 2$ and $∣ x ∣ = - x$
$f (x) = ∣ (x + 2) - 2 (- x) ∣ = ∣3 x + 2∣$
The expression $3 x + 2$ changes sign at $x = - \frac{2}{3}$ .

Subcase 2a: $- 2 \leq x < - \frac{2}{3}$

For $- 2 \leq x < - \frac{2}{3}$ :
$3 x + 2 < 0 ⟹ f (x) = - (3 x + 2) = - 3 x - 2$

Subcase 2b: -\frac{2}{3} \le x < 0

For $- \frac{2}{3} \leq x < 0$ :
$3 x + 2 \geq 0 ⟹ f (x) = 3 x + 2$

Case 3: $x \geq 0$

For $x \geq 0$ :
$∣ x + 2∣ = x + 2$ and $∣ x ∣ = x$
$f (x) = ∣ (x + 2) - 2 x ∣ = ∣2 - x ∣$
The expression $2 - x$ changes sign at $x = 2$ .

Subcase 3a: $0 \leq x < 2$

For $0 \leq x < 2$ :
$2 - x > 0 ⟹ f (x) = 2 - x$

Subcase 3b: $x \geq 2$

For $x \geq 2$ :
$2 - x \leq 0 ⟹ f (x) = - (2 - x) = x - 2$

Identifying Local Minima ( $m$ )

Local minima occur at the sharp "V" shaped dips in the graph.
From the graph, the dips are at $x = - \frac{2}{3}$ and $x = 2$ .
Number of local minima, $m = 2$ .

Identifying Local Maxima ( $n$ )

Local maxima occur at the sharp peaks in the graph.
From the graph, there is one peak at $x = 0$ .
Number of local maxima, $n = 1$ .

Final Calculation: $m + n$

We found $m = 2$ and $n = 1$ .
Calculate the sum: $m + n = 2 + 1 = 3$ .
The final answer is $3$ .

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The Sigma Insight: Maxima and Minima

Solution Diagram

Analyzing the Setup

The function is defined as

f (x) = ∣∣ x + 2∣ - 2∣ x ∣∣

. To simplify this expression, we must identify the critical points where the expressions inside the absolute value bars change sign.

Setting the inner expressions to zero, we find:

x + 2 = 0 \Rightarrow x = - 2

x = 0 \Rightarrow x = 0

These critical points divide the real number line into three distinct intervals:

x < - 2

- 2 \leq x < 0

, and

x \geq 0

The Three Realms

Realm 1: $x < - 2$ In this region,

x + 2 < 0

and

x < 0

. The function becomes:

f (x) = ∣ - (x + 2) - 2 (- x) ∣ = ∣ - x - 2 + 2 x ∣ = ∣ x - 2∣

Since

x < - 2

, the term

x - 2

is negative. Thus,

f (x) = - (x - 2) = 2 - x

Realm 2: $- 2 \leq x < 0$ Here,

x + 2 \geq 0

and

x < 0

. The function becomes:

f (x) = ∣ (x + 2) - 2 (- x) ∣ = ∣3 x + 2∣

This expression changes sign at

x = - \frac{2}{3}

. We split this into two sub-realms: 1. For

- 2 \leq x < - \frac{2}{3}

f (x) = - (3 x + 2) = - 3 x - 2

. 2. For

- \frac{2}{3} \leq x < 0

f (x) = 3 x + 2

Realm 3: $x \geq 0$ In this region,

x + 2 > 0

and

x \geq 0

. The function becomes:

f (x) = ∣ (x + 2) - 2 x ∣ = ∣2 - x ∣

This expression changes sign at

x = 2

, leading to two further sub-realms: 1. For

0 \leq x < 2

f (x) = 2 - x

. 2. For

x \geq 2

f (x) = x - 2

The Visual Stage

By analyzing the piecewise segments, we can map the behavior of the function. The graph exhibits sharp corners (cusps) at the points where the function transitions between these linear segments.

The local minima (valleys) occur at: 1.

x = - \frac{2}{3}

x = 2

Thus, the number of local minima is

m = 2

The local maximum (peak) occurs at: 1.

x = 0

Thus, the number of local maxima is

n = 1

Final Calculation

To find the final result, we sum the number of local minima and local maxima:

m + n = 2 + 1 = 3

The final answer is 3.

Common Traps & FAQs

Why do we split the intervals at

x = - 2

and

x = 0

The absolute value function

∣ u ∣

changes its behavior exactly when the expression

u

changes sign. For

f (x) = ∣∣ x + 2∣ - 2∣ x ∣∣

, the inner expressions are

x + 2

and

x

. These change sign at

x = - 2

and

x = 0

, respectively. By splitting the real line at these points, we ensure that in each interval, the expressions inside the absolute values have a constant sign, allowing us to remove the modulus signs algebraically.

Do I need to use derivatives to find the local minima and maxima?

Not necessarily! While derivatives are a powerful tool, they are often overkill for piecewise linear functions. Since the function is composed of straight-line segments, the local extrema will always occur at the 'corners' (the points where the slope changes). You can identify these points by sketching the graph or by observing where the function changes from decreasing to increasing (minima) or increasing to decreasing (maxima).

Why is

x = - 2

not a local minimum or maximum?

Great observation! If you look at the graph, at

x = - 2

, the function is continuous, but the slope changes. However, for a point to be a local minimum, the function must be smaller than its neighbors, and for a local maximum, it must be larger. At

x = - 2

, the function value is

f (- 2) = ∣∣ - 2 + 2∣ - 2∣ - 2∣∣ = ∣0 - 4∣ = 4

. If you check the values slightly to the left and right, you will see that the function is decreasing on both sides of

x = - 2

. Since it doesn't change direction (from decreasing to increasing or vice versa), it is not an extremum.

Animated Solution for Mathematics - Differentiation: Let $f : R \to R$ be a function defined by $f (x) = ∣∣ x + 2∣ - 2∣ x ∣∣$ . If $m$ is the number of points of local minima and $n$ is the number of points of local maxima of $f$ , then $m + n$ is

Select Answer:

Visualized Solution

Analyzing the Function $f (x)$

Case 1: $x < - 2$

Case 2: $- 2 \leq x < 0$

Subcase 2a: $- 2 \leq x < - \frac{2}{3}$

Subcase 2b: -\frac{2}{3} \le x < 0

Case 3: $x \geq 0$

Subcase 3a: $0 \leq x < 2$

Subcase 3b: $x \geq 2$

Identifying Local Minima ( $m$ )

Identifying Local Maxima ( $n$ )

Final Calculation: $m + n$

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Three Realms

The Visual Stage

Final Calculation

Similar Questions

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Select Answer:

Visualized Solution

Analyzing the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x<−2

Case 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } −2≤x<0

Subcase 2a: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } −2≤x<−32​

Subcase 2b: -\frac{2}{3} \le x < 0

Case 3: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x≥0

Subcase 3a: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 0≤x<2

Subcase 3b: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x≥2

Identifying Local Minima ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m)

Identifying Local Maxima ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n)

Final Calculation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m+n

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Three Realms

The Visual Stage

Final Calculation

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Analyzing the Function $f (x)$

Case 1: $x < - 2$

Case 2: $- 2 \leq x < 0$

Subcase 2a: $- 2 \leq x < - \frac{2}{3}$

Case 3: $x \geq 0$

Subcase 3a: $0 \leq x < 2$

Subcase 3b: $x \geq 2$

Identifying Local Minima ( $m$ )

Identifying Local Maxima ( $n$ )

Final Calculation: $m + n$

Let $f : R \to R$ be defined as $f (x) = ∣ x ∣ + ∣ x^{2} - 1∣$ . The total number of points at which $f$ attains either a local maximum or a local minimum is

If m and n respectively are the number of local maximum and local minimum points of the function $f (x) = \int_{0}^{x^{2}} \frac{t ^{2} - 5 t + 4}{2 + e ^{t}} d t$ , then the ordered pair (m, n) is equal to

The total number of local maxima and local minima of the function $f (x) = {(2 + x)^{3}, x^{2/3}, - 3 < x \leq - 1 - 1 < x < 2$ is

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The function $f (x) = 2 x + 3 (x)^{\frac{2}{3}}, x \in R$ , has

Let $f (x) = {∣ x ∣, 1, for 0 < ∣ x ∣ \leq 2 for x = 0$ then at $x = 0, f$ has

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