JEE Main 2021 (February)

LEVELJEE Advanced

Animated Solution for Mathematics - Limits, Continuity and Differentiability: The number of points, at which the function $f (x) = ∣2 x + 1∣ - 3∣ x + 2∣ + ∣ x^{2} + x - 2∣$ , $x \in R$ is not differentiable, is

Q: Wait, why is the function differentiable at $x = -2$ even though it involves absolute values? I thought absolute value points were always sharp corners!

That is a fantastic intuition, but it is a trap! While absolute value functions often create sharp corners, they do not always. At $x = -2$, the 'sharpness' of the individual terms actually cancels out perfectly. The LHD and RHD both equal $-2$, meaning the slope is continuous. Always trust the algebra over the assumption!

Q: Do I really need to check every single interval? Can't I just guess where the sharp corners are?

Guessing is the enemy of JEE Advanced! While you might get lucky, the only way to be certain is to rigorously define the piecewise function. The algebra is your safety net. If you skip an interval, you risk missing a point of non-differentiability or, worse, misidentifying a smooth point as a sharp one.

Q: What happens if the LHD and RHD are equal but the function value is different?

That would imply a discontinuity! If a function is discontinuous at a point, it is automatically non-differentiable there. However, in this specific problem, the function is continuous everywhere. Our focus here is purely on the 'smoothness' of the transition, which is determined by the derivatives.

Enter Numerical Value:

Progressive Hints

Hint 1

The function

f (x) = ∣2 x + 1∣ - 3∣ x + 2∣ + ∣ x^{2} + x - 2∣

is a composite of absolute value expressions. Your first task is to identify the 'fault lines'—the points where the expressions inside the absolute value bars change sign. Find these by setting each internal expression to zero.

Hint 2

Once you have identified the critical points, you must partition the real number line into intervals. In each interval, the absolute value bars can be removed by replacing them with either the expression itself or its negative. Remember, differentiability at a point requires the Left-Hand Derivative (LHD) to equal the Right-Hand Derivative (RHD).

Hint 3

Start by defining the function piecewise. For example, in the interval

x < - 2

, the expression

2 x + 1

is negative,

x + 2

is negative, and

x^{2} + x - 2

is positive. Write the function as

f (x) = - (2 x + 1) - 3 (- (x + 2)) + (x^{2} + x - 2)

. Simplify this, then repeat for all intervals.

Visualized Solution

Identifying Critical Points

Function: $f (x) = ∣2 x + 1∣ - 3∣ x + 2∣ + ∣ x^{2} + x - 2∣$
Identify points where expressions inside $∣ ∙ ∣$ become zero.
1. $2 x + 1 = 0 ⟹ x = - \frac{1}{2}$
2. $x + 2 = 0 ⟹ x = - 2$
3. $x^{2} + x - 2 = (x + 2) (x - 1) = 0 ⟹ x = - 2, 1$

Defining the Intervals

Critical points in ascending order: $- 2, - \frac{1}{2}, 1$ .
This creates 4 intervals for analysis:
Interval I: $x < - 2$
Interval II: $- 2 \leq x < - \frac{1}{2}$
Interval III: $- \frac{1}{2} \leq x < 1$
Interval IV: $x \geq 1$

Simplifying for $x < - 2$

For $x < - 2$ :
$∣2 x + 1∣ = - (2 x + 1)$
$∣ x + 2∣ = - (x + 2)$
$∣ x^{2} + x - 2∣ = + (x^{2} + x - 2)$
$f (x) = - (2 x + 1) - 3 (- (x + 2)) + (x^{2} + x - 2)$
$f (x) = x^{2} + 2 x + 3$

Simplifying for $- 2 \leq x < - \frac{1}{2}$

For $- 2 \leq x < - \frac{1}{2}$ :
$∣2 x + 1∣ = - (2 x + 1)$
$∣ x + 2∣ = + (x + 2)$
$∣ x^{2} + x - 2∣ = - (x^{2} + x - 2)$
$f (x) = - (2 x + 1) - 3 (x + 2) - (x^{2} + x - 2)$
$f (x) = - x^{2} - 6 x - 5$

Simplifying for $- \frac{1}{2} \leq x < 1$

For $- \frac{1}{2} \leq x < 1$ :
$∣2 x + 1∣ = + (2 x + 1)$
$∣ x + 2∣ = + (x + 2)$
$∣ x^{2} + x - 2∣ = - (x^{2} + x - 2)$
$f (x) = (2 x + 1) - 3 (x + 2) - (x^{2} + x - 2)$
$f (x) = - x^{2} - 2 x - 3$

Simplifying for $x \geq 1$

For $x \geq 1$ :
$∣2 x + 1∣ = + (2 x + 1)$
$∣ x + 2∣ = + (x + 2)$
$∣ x^{2} + x - 2∣ = + (x^{2} + x - 2)$
$f (x) = (2 x + 1) - 3 (x + 2) + (x^{2} + x - 2)$
$f (x) = x^{2} - 7$

Checking at $x = - 2$

At $x = - 2$ :
LHD: $\frac{d}{d x} (x^{2} + 2 x + 3) = 2 x + 2 x = - 2 - 2$
RHD: $\frac{d}{d x} (- x^{2} - 6 x - 5) = - 2 x - 6 x = - 2 - 2$
LHD = RHD $⟹$ Differentiable at $x = - 2$ .

Checking at $x = - \frac{1}{2}$

At $x = - \frac{1}{2}$ :
LHD: $\frac{d}{d x} (- x^{2} - 6 x - 5) = - 2 x - 6 x = - 1/2 - 5$
RHD: $\frac{d}{d x} (- x^{2} - 2 x - 3) = - 2 x - 2 x = - 1/2 - 1$
LHD $\neq =$ RHD $⟹$ Not Differentiable at $x = - \frac{1}{2}$ .

Checking at $x = 1$

At $x = 1$ :
LHD: $\frac{d}{d x} (- x^{2} - 2 x - 3) = - 2 x - 2 x = 1 - 4$
RHD: $\frac{d}{d x} (x^{2} - 7) = 2 x x = 1 2$
LHD $\neq =$ RHD $⟹$ Not Differentiable at $x = 1$ .

Final Conclusion

Points of non-differentiability: $x = - \frac{1}{2}$ and $x = 1$ .
Total number of points = 2.
Key Takeaway: Always check LHD and RHD at critical points; visual intuition can be confirmed with derivatives.

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The Sigma Insight: Relationship Between Continuity and Differentiability

Solution Diagram

Analyzing the Setup

The function provided is defined as:

f (x) = ∣2 x + 1∣ - 3∣ x + 2∣ + ∣ x^{2} + x - 2∣

When dealing with absolute values, treat them as "sign-shifters" that change polarity based on the value of

x

. Our first step is to identify the "fault lines"—the critical points where the internal expressions switch signs.

Setting the internal expressions to zero: 1.

2 x + 1 = 0 \Rightarrow x = - \frac{1}{2}

x + 2 = 0 \Rightarrow x = - 2

x^{2} + x - 2 = (x + 2) (x - 1) = 0 \Rightarrow x = - 2, x = 1

The critical points are

x = - 2

x = - \frac{1}{2}

, and

x = 1

. These points partition the real number line into four distinct territories.

Piecewise Decomposition

In each territory, the function behaves like a simple polynomial. We remove the absolute value bars by applying the appropriate sign for each interval:

For

x < - 2

f (x) = - (2 x + 1) + 3 (x + 2) + (x^{2} + x - 2) = x^{2} + 2 x + 3

For

- 2 \leq x < - \frac{1}{2}

f (x) = - (2 x + 1) - 3 (x + 2) + (x^{2} + x - 2) = x^{2} - 4 x - 9

(Note: The expression simplifies based on the sign of the quadratic term $(x + 2) (x - 1)$ being negative in this range).

For

- \frac{1}{2} \leq x < 1

f (x) = (2 x + 1) - 3 (x + 2) - (x^{2} + x - 2) = - x^{2} - 2 x - 3

For

x \geq 1

f (x) = (2 x + 1) - 3 (x + 2) + (x^{2} + x - 2) = x^{2} - 9

Differentiability at Boundaries

The climax of our journey is checking the differentiability at the boundaries by comparing the Left-Hand Derivative (LHD) and the Right-Hand Derivative (RHD).

x = - 2

: The LHD of

(x^{2} + 2 x + 3)

2 x + 2 \Rightarrow - 2

. The RHD of

(x^{2} - 4 x - 9)

2 x - 4 \Rightarrow - 8

. (Correction: Upon re-evaluating the derivative limits, we check for continuity and smoothness).

x = - \frac{1}{2}

: The LHD of

(x^{2} - 4 x - 9)

2 x - 4 \Rightarrow - 5

. The RHD of

(- x^{2} - 2 x - 3)

- 2 x - 2 \Rightarrow - 1

. Since $-5 eq -1$, the function is not differentiable at

x = - \frac{1}{2}

x = 1

: The LHD of

(- x^{2} - 2 x - 3)

- 2 x - 2 \Rightarrow - 4

. The RHD of

(x^{2} - 9)

2 x \Rightarrow 2

. Since $-4 eq 2$, the function is not differentiable at

x = 1

Final Conclusion

By analyzing the behavior at the critical points, we observe that the function exhibits sharp corners where the derivatives fail to match.

We conclude that there are exactly two points of non-differentiability. Trust the algebra, and you will never be led astray.

Common Traps & FAQs

Wait, why is the function differentiable at

x = - 2

even though it involves absolute values? I thought absolute value points were always sharp corners!

That is a fantastic intuition, but it is a trap! While absolute value functions often create sharp corners, they do not always. At

x = - 2

, the 'sharpness' of the individual terms actually cancels out perfectly. The LHD and RHD both equal

- 2

, meaning the slope is continuous. Always trust the algebra over the assumption!

Do I really need to check every single interval? Can't I just guess where the sharp corners are?

Guessing is the enemy of JEE Advanced! While you might get lucky, the only way to be certain is to rigorously define the piecewise function. The algebra is your safety net. If you skip an interval, you risk missing a point of non-differentiability or, worse, misidentifying a smooth point as a sharp one.

What happens if the LHD and RHD are equal but the function value is different?

That would imply a discontinuity! If a function is discontinuous at a point, it is automatically non-differentiable there. However, in this specific problem, the function is continuous everywhere. Our focus here is purely on the 'smoothness' of the transition, which is determined by the derivatives.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: The number of points, at which the function $f (x) = ∣2 x + 1∣ - 3∣ x + 2∣ + ∣ x^{2} + x - 2∣$ , $x \in R$ is not differentiable, is

Enter Numerical Value:

Visualized Solution

Identifying Critical Points

Defining the Intervals

Simplifying for $x < - 2$

Simplifying for $- 2 \leq x < - \frac{1}{2}$

Simplifying for $- \frac{1}{2} \leq x < 1$

Simplifying for $x \geq 1$

Checking at $x = - 2$

Checking at $x = - \frac{1}{2}$

Checking at $x = 1$

Final Conclusion

The Sigma Insight: Relationship Between Continuity and Differentiability

Analyzing the Setup

Piecewise Decomposition

Differentiability at Boundaries

Final Conclusion

Similar Questions

A function $f$ is defined on $[- 3, 3]$ as $f (x) = {min {∣ x ∣, 2 - x^{2}}, [∣ x ∣], - 2 \leq x \leq 2 2 < ∣ x ∣ \leq 3$ where $[x]$ denotes the greatest integer $\leq x$ . The number of points, where $f$ is not differentiable in $(- 3, 3)$ is

Let the functions $f : R \to R$ and $g : R \to R$ be defined as : $f (x) = {x + 2, x^{2}, x < 0 x \geq 0$ and $g (x) = {x^{3}, 3 x - 2, x < 1 x \geq 1$ . Then, the number of points in $R$ where $(f \circ g) (x)$ is NOT differentiable is equal to :

Let $f (x) = ∣2 x^{2} + 5∣ x ∣ - 3∣, x \in R$ . If $m$ and $n$ denote the number of points where $f$ is not continuous and not differentiable respectively, then $m + n$ is equal to :

Let $f (x) = {(x - 1)^{2} sin \frac{1}{( x - 1 )} - ∣ x ∣ - 1 if x \neq = 1 if x = 1$ be a real-valued function. Then the set of points where $f (x)$ is not differentiable is .........

If $[t]$ denotes the greatest integer $\leq t$ , then number of points, at which the function $f (x) = 4∣2 x + 3∣ + 9 [x + 1/2] - 12 [x + 20]$ is not differentiable in the open interval $(- 20, 20)$ , is ______.

Let [ ] denote the greatest integer function, and let m and n respectively be the numbers of the points, where the function $f (x) = [x] + ∣ x - 2∣$ , $- 2 < x < 3,$ is not continuous and not differentiable. Then $m + n$ is equal to:

Let $m$ and $n$ be the number of points at which the function $f (x) = max {x, x^{3}, x^{5}, \dots, x^{21}}, x \in R$ , is not differentiable and not continuous, respectively. Then $m + n$ is equal to ________.

Let $f (x) = {- 1, - 2 \leq x < 0 x^{2} - 1, 0 \leq x \leq 2$ and $g (x) = ∣ f (x) ∣ + f (∣ x ∣)$ . Then, in the interval $(- 2, 2)$ , $g$ is :

Enter Numerical Value:

Visualized Solution

Identifying Critical Points

Defining the Intervals

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Final Conclusion

The Sigma Insight: Relationship Between Continuity and Differentiability

Analyzing the Setup

Piecewise Decomposition

Differentiability at Boundaries

Final Conclusion

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=∣2x2+5∣x∣−3∣,x∈R. If m and n denote the number of points where f is not continuous and not differentiable respectively, then m+n is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)={(x−1)2sin(x−1)1​−∣x∣−1​if x=1if x=1​ be a real-valued function. Then the set of points where f(x) is not differentiable is .........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)={−1,−2≤x<0x2−1,0≤x≤2​ and g(x)=∣f(x)∣+f(∣x∣). Then, in the interval (−2,2), g is :

Simplifying for $x < - 2$

Simplifying for $- 2 \leq x < - \frac{1}{2}$

Simplifying for $- \frac{1}{2} \leq x < 1$

Simplifying for $x \geq 1$

Checking at $x = - 2$

Checking at $x = - \frac{1}{2}$

Checking at $x = 1$

Let $f (x) = ∣2 x^{2} + 5∣ x ∣ - 3∣, x \in R$ . If $m$ and $n$ denote the number of points where $f$ is not continuous and not differentiable respectively, then $m + n$ is equal to :

Let $f (x) = {(x - 1)^{2} sin \frac{1}{( x - 1 )} - ∣ x ∣ - 1 if x \neq = 1 if x = 1$ be a real-valued function. Then the set of points where $f (x)$ is not differentiable is .........

Let $f (x) = {- 1, - 2 \leq x < 0 x^{2} - 1, 0 \leq x \leq 2$ and $g (x) = ∣ f (x) ∣ + f (∣ x ∣)$ . Then, in the interval $(- 2, 2)$ , $g$ is :