Let a function $g : [0, 4] \to \mathbb{R}$ be defined as $g(x) = \begin{cases} \max_{0 \le t \le x} \{t^3 - 6t^2 + 9t - 3\}, & 0 \le x \le 3 \\ 4 - x, & 3 < x \le 4 \end{cases}$ then the number of points in the interval $(0, 4)$ where $g(x)$ is NOT differentiable, is

### Analyzing the Cubic Landscape To understand the behavior of the function $g(x) = \max_{0 \le t \le x} \{f(t)\}$, we first analyze the inner function $f(t) = t^3 - 6t^2 + 9t - 3$. We determine the critical points by calculating the derivative: $$f'(t) = 3t^2 - 12t + 9$$ Setting the derivative to zero, we obtain: $$3(t^2 - 4t + 3) = 0 \Rightarrow 3(t-1)(t-3) = 0$$ The critical points are located at $t=1$ and $t=3$. Evaluating the function at these points and the boundary $t=0$, we find: $f(0) = -3$, $f(1) = 1$, and $f(3) = -3$. The function climbs from $-3$ to $1$ on the interval $[0, 1]$ and descends from $1$ to $-3$ on the interval $[1, 3]$. ### Constructing the High-Water Mark We now define $g(x)$ based on the running maximum of $f(t)$. For $x \in [0, 1]$, the function $f(t)$ is strictly increasing. Therefore, the maximum value is the current value: $$g(x) = x^3 - 6x^2 + 9x - 3$$ For $x \in (1, 3]$, the function $f(t)$ decreases from $1$ to $-3$. Since $g(x)$ tracks the maximum value encountered, it remains constant at the peak value: $$g(x) = 1$$ For $x \in (3, 4]$, the function is defined by the linear descent: $$g(x) = 4 - x$$ ### The Test of Smoothness We examine the differentiability of the piecewise function $g(x)$ at the transition points $x=1$ and $x=3$. At $x=1$: The left-hand derivative is $f'(1) = 3(1)^2 - 12(1) + 9 = 0$. The right-hand derivative is the derivative of the constant $1$, which is $0$. Since the derivatives match, the function is **differentiable at $x=1$**. At $x=3$: The left-hand derivative is $0$ (the derivative of the constant $1$). The right-hand derivative is the derivative of $4-x$, which is $-1$. Since $0 \neq -1$, the function is **not differentiable at $x=3$**. Consequently, there is exactly **$1$ point of non-differentiability** in the interval $(0, 4)$.

JEE Main 2021 (20 July Shift 2)

LEVELJEE Advanced

Animated Solution for Mathematics - Differentiation: Let a function $g : [0, 4] \to R$ be defined as $g (x) = {max_{0 \leq t \leq x} {t^{3} - 6 t^{2} + 9 t - 3}, 4 - x, 0 \leq x \leq 3 3 < x \leq 4$ then the number of points in the interval $(0, 4)$ where $g (x)$ is NOT differentiable, is

Q: Wait, why is $g(x) = 1$ for $x \in [1, 3]$? Shouldn't it just be $f(x)$?

That is a fantastic question! Remember, $g(x)$ is defined as the *maximum* value of $f(t)$ on the interval $[0, x]$. Once $x$ passes $1$, the function $f(t)$ starts to decrease. However, $g(x)$ is a 'running maximum'—it remembers the highest value it has seen so far. Since $f(1) = 1$ is the highest point reached in the interval $[0, 3]$, $g(x)$ stays at $1$ because no value of $f(t)$ for $t > 1$ can exceed that peak.

Q: Why is it differentiable at $x=1$ if it's a transition point?

Differentiability is about smoothness. At $x=1$, the left-hand side is the cubic function $f(x)$, and the right-hand side is the constant function $1$. If you calculate the derivative of the cubic at $x=1$, you get $0$. The derivative of the constant $1$ is also $0$. Since the slopes match (both are $0$), the transition is perfectly smooth, making it differentiable!

Q: Can we just look at the graph to find non-differentiable points?

Visual intuition is a great starting point, but in JEE Advanced, you must verify it analytically. A graph might look smooth but have a subtle cusp. Always calculate the LHD and RHD at the transition points. At $x=3$, the LHD is $0$ (from the constant $1$) and the RHD is $-1$ (from the line $4-x$). Since $0 \neq -1$, the sharp corner is mathematically confirmed.

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1: Visualize

g (x)

as a 'high-water mark' tracker. For any

x

g (x)

is the maximum value the cubic function

f (t) = t^{3} - 6 t^{2} + 9 t - 3

has attained at any point

t

in the interval

[0, x]

. Think of it as a record-keeping function that only updates when a new peak is reached.

Hint 2

Hint 2: To determine differentiability, you must first find the explicit piecewise definition of

g (x)

. Remember that differentiability at a point requires two things: continuity at that point and the equality of the left-hand derivative (LHD) and the right-hand derivative (RHD).

Hint 3

Hint 3: Start by finding the critical points of

f (t)

to see where it rises and falls. For

x \in [0, 1]

g (x) = f (x)

. For

x \in [1, 3]

, since

f (t)

is decreasing, the maximum value is 'locked' at

f (1) = 1

. Use this to write the full piecewise function and then test the transition points

x = 1

and

x = 3

Visualized Solution

Defining the inner function $f (t)$

Let $f (t) = t^{3} - 6 t^{2} + 9 t - 3$ for $t \in [0, 3]$ .
The function $g (x)$ is defined as the maximum value of $f (t)$ in the interval $[0, x]$ .

Finding critical points of $f (t)$

To find where $f (t)$ increases or decreases, we find its derivative.
$f^{'} (t) = \frac{d}{d t} (t^{3} - 6 t^{2} + 9 t - 3)$
$f^{'} (t) = 3 t^{2} - 12 t + 9$

Solving for critical points

Set $f^{'} (t) = 0$ to find critical points.
$3 (t^{2} - 4 t + 3) = 0$
$3 (t - 1) (t - 3) = 0$
Critical points are at $t = 1$ and $t = 3$ .

Evaluating $f (t)$ at key points

At $t = 0$ : $f (0) = - 3$
At $t = 1$ : $f (1) = 1^{3} - 6 (1)^{2} + 9 (1) - 3 = 1$
At $t = 3$ : $f (3) = 3^{3} - 6 (3)^{2} + 9 (3) - 3 = - 3$

Analyzing the shape of $f (t)$

For $t \in [0, 1]$ , $f^{'} (t) > 0$ , so $f (t)$ is strictly increasing.
For $t \in [1, 3]$ , $f^{'} (t) < 0$ , so $f (t)$ is strictly decreasing.

Constructing $g (x)$ for $x \in [0, 1]$

$g (x) = max_{0 \leq t \leq x} f (t)$
Since $f (t)$ is increasing on $[0, 1]$ , its maximum on $[0, x]$ occurs at the right endpoint $t = x$ .
Therefore, $g (x) = f (x) = x^{3} - 6 x^{2} + 9 x - 3$ for $0 \leq x \leq 1$ .

Constructing $g (x)$ for $x \in [1, 3]$

For $x \in [1, 3]$ , $f (t)$ starts decreasing after $t = 1$ .
The maximum value achieved so far remains at $t = 1$ , which is $f (1) = 1$ .
Therefore, $g (x) = 1$ for $1 < x \leq 3$ .

Defining $g (x)$ for $x \in (3, 4]$

The problem explicitly defines $g (x) = 4 - x$ for $3 < x \leq 4$ .
This is a straight line with a slope of $- 1$ .

The complete piecewise definition of $g (x)$

$g (x) = ⎩ ⎨ ⎧ x^{3} - 6 x^{2} + 9 x - 3, 1, 4 - x, 0 \leq x \leq 1 1 < x \leq 3 3 < x \leq 4$

Checking differentiability at $x = 1$

Left Hand Derivative (LHD) at $x = 1$ : $\frac{d}{d x} (x^{3} - 6 x^{2} + 9 x - 3) = 3 x^{2} - 12 x + 9$ . At $x = 1$ , LHD $= 0$ .
Right Hand Derivative (RHD) at $x = 1$ : $\frac{d}{d x} (1) = 0$ .
Since LHD $=$ RHD $= 0$ , $g (x)$ is differentiable at $x = 1$ .

Checking differentiability at $x = 3$

Left Hand Derivative (LHD) at $x = 3$ : $\frac{d}{d x} (1) = 0$ .
Right Hand Derivative (RHD) at $x = 3$ : $\frac{d}{d x} (4 - x) = - 1$ .
Since LHD $\neq =$ RHD, $g (x)$ is NOT differentiable at $x = 3$ .

Final Conclusion

The function $g (x)$ is continuous everywhere in $(0, 4)$ .
It has a sharp corner only at $x = 3$ .
Therefore, there is exactly 1 point of non-differentiability.

00:00 / 00:00

The Sigma Insight: Differentiability of a Function

Solution Diagram

Analyzing the Cubic Landscape

To understand the behavior of the function

g (x) = max_{0 \leq t \leq x} {f (t)}

, we first analyze the inner function

f (t) = t^{3} - 6 t^{2} + 9 t - 3

We determine the critical points by calculating the derivative:

f^{'} (t) = 3 t^{2} - 12 t + 9

Setting the derivative to zero, we obtain:

3 (t^{2} - 4 t + 3) = 0 \Rightarrow 3 (t - 1) (t - 3) = 0

The critical points are located at

t = 1

and

t = 3

. Evaluating the function at these points and the boundary

t = 0

, we find:

f (0) = - 3

f (1) = 1

, and

f (3) = - 3

The function climbs from

- 3

1

on the interval

[0, 1]

and descends from

1

- 3

on the interval

[1, 3]

Constructing the High-Water Mark

We now define

g (x)

based on the running maximum of

f (t)

For

x \in [0, 1]

, the function

f (t)

is strictly increasing. Therefore, the maximum value is the current value:

g (x) = x^{3} - 6 x^{2} + 9 x - 3

For

x \in (1, 3]

, the function

f (t)

decreases from

1

- 3

. Since

g (x)

tracks the maximum value encountered, it remains constant at the peak value:

g (x) = 1

For

x \in (3, 4]

, the function is defined by the linear descent:

g (x) = 4 - x

The Test of Smoothness

We examine the differentiability of the piecewise function

g (x)

at the transition points

x = 1

and

x = 3

x = 1

: The left-hand derivative is

f^{'} (1) = 3 (1)^{2} - 12 (1) + 9 = 0

. The right-hand derivative is the derivative of the constant

1

, which is

0

. Since the derivatives match, the function is differentiable at $x = 1$ .

x = 3

: The left-hand derivative is

0

(the derivative of the constant

1

). The right-hand derivative is the derivative of

4 - x

, which is

- 1

. Since $0 eq -1$, the function is not differentiable at $x = 3$ .

Consequently, there is exactly $1$ point of non-differentiability in the interval

(0, 4)

Common Traps & FAQs

Wait, why is

g (x) = 1

for

x \in [1, 3]

? Shouldn't it just be

f (x)

That is a fantastic question! Remember,

g (x)

is defined as the *maximum* value of

f (t)

on the interval

[0, x]

. Once

x

passes

1

, the function

f (t)

starts to decrease. However,

g (x)

is a 'running maximum'—it remembers the highest value it has seen so far. Since

f (1) = 1

is the highest point reached in the interval

[0, 3]

g (x)

stays at

1

because no value of

f (t)

for

t > 1

can exceed that peak.

Why is it differentiable at

x = 1

if it's a transition point?

Differentiability is about smoothness. At

x = 1

, the left-hand side is the cubic function

f (x)

, and the right-hand side is the constant function

1

. If you calculate the derivative of the cubic at

x = 1

, you get

0

. The derivative of the constant

1

is also

0

. Since the slopes match (both are

0

), the transition is perfectly smooth, making it differentiable!

Can we just look at the graph to find non-differentiable points?

Visual intuition is a great starting point, but in JEE Advanced, you must verify it analytically. A graph might look smooth but have a subtle cusp. Always calculate the LHD and RHD at the transition points. At

x = 3

, the LHD is

0

(from the constant

1

) and the RHD is

- 1

(from the line

4 - x

). Since

0 \neq = - 1

, the sharp corner is mathematically confirmed.

Animated Solution for Mathematics - Differentiation: Let a function $g : [0, 4] \to R$ be defined as $g (x) = {max_{0 \leq t \leq x} {t^{3} - 6 t^{2} + 9 t - 3}, 4 - x, 0 \leq x \leq 3 3 < x \leq 4$ then the number of points in the interval $(0, 4)$ where $g (x)$ is NOT differentiable, is

Enter Numerical Value:

Visualized Solution

Defining the inner function $f (t)$

Finding critical points of $f (t)$

Solving for critical points

Evaluating $f (t)$ at key points

Analyzing the shape of $f (t)$

Constructing $g (x)$ for $x \in [0, 1]$

Constructing $g (x)$ for $x \in [1, 3]$

Defining $g (x)$ for $x \in (3, 4]$

The complete piecewise definition of $g (x)$

Checking differentiability at $x = 1$

Checking differentiability at $x = 3$

Final Conclusion

The Sigma Insight: Differentiability of a Function

Analyzing the Cubic Landscape

Constructing the High-Water Mark

The Test of Smoothness

Similar Questions

Let $f : R \to R$ be a function defined by $f (x) = max {x, x^{3}}$ . The set of all points where $f (x)$ is NOT differentiable is

The number of points, where the function $f : R \to R$ , $f (x) = ∣ x - 1∣ cos ∣ x - 2∣ sin ∣ x - 1∣ + (x - 3) ∣ x^{2} - 5 x + 4∣$ , is NOT differentiable, is :

The function $f (x) = ∣ x^{2} - 2 x - 3∣ \cdot e^{∣9 x^{2} - 12 x + 4∣}$ is not differentiable at exactly:

Let $f (x) = {∣4 x^{2} - 8 x + 5∣, [4 x^{2} - 8 x + 5], if 8 x^{2} - 6 x + 1 \geq 0 if 8 x^{2} - 6 x + 1 < 0$ where $[α]$ denotes the greatest integer less than or equal to $α$ . Then the number of points in $R$ where $f$ is not differentiable is ______.

Let $f (x) = {max {∣ x ∣, x^{2}}, 8 - 2∣ x ∣, ∣ x ∣ \leq 2 2 < ∣ x ∣ \leq 4$ . Let S be the set of points in the interval (-4, 4) at which f is not differentiable. Then S:

Let $f : R \to R$ and $g : R \to R$ be respectively given by $f (x) = ∣ x ∣ + 1$ and $g (x) = x^{2} + 1$ . Define $h : R \to R$ by $h (x) = {max {f (x), g (x)} min {f (x), g (x)} if x \leq 0, if x > 0.$ The number of points at which $h (x)$ is not differentiable is .........

Let $a \in Z$ and $[t]$ be the greatest integer $\leq t$ , then the number of points, where the function $f (x) = [a + 13 sin x]$ , $x \in (0, π)$ is not differentiable, is ________.

Let $f : R \to R$ be a function defined by $f (x) = max {x, x^{2}}$ . Let $S$ denote the set of all points in $R$ , where $f$ is not differentiable. Then :

Let $f : (- 1, 1) \to R$ be a function defined by $f (x) = max {- ∣ x ∣, - 1 - x^{2}}$ . If K be the set of all points at which f is not differentiable, then K has exactly :

Let $f : R \to R$ be differentiable at $c \in R$ and $f (c) = 0$ . If $g (x) = ∣ f (x) ∣$ , then at $x = c$ , $g$ is :

Enter Numerical Value:

Visualized Solution

Defining the inner function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(t)

Finding critical points of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(t)

Solving for critical points

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(t) at key points

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

Constructing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x) for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x∈[0,1]

Constructing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x) for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x∈[1,3]

Defining .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x) for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x∈(3,4]

The complete piecewise definition of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

Checking differentiability at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=1

Checking differentiability at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=3

Final Conclusion

The Sigma Insight: Differentiability of a Function

Analyzing the Cubic Landscape

Constructing the High-Water Mark

The Test of Smoothness

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be a function defined by f(x)=max{x,x3}. The set of all 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; } The number of points, where the function f:R→R, f(x)=∣x−1∣cos∣x−2∣sin∣x−1∣+(x−3)∣x2−5x+4∣, is NOT differentiable, is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The function f(x)=∣x2−2x−3∣⋅e∣9x2−12x+4∣ is not differentiable at exactly:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)={max{∣x∣,x2},8−2∣x∣,​∣x∣≤22<∣x∣≤4​. Let S be the set of points in the interval (-4, 4) at which f is not differentiable. Then S:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a∈Z and [t] be the greatest integer ≤t, then the number of points, where the function f(x)=[a+13sinx], x∈(0,π) is not differentiable, is ________.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be a function defined by f(x)=max{x,x2}. Let S denote the set of all points in R, where f is not differentiable. Then :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:(−1,1)→R be a function defined by f(x)=max{−∣x∣,−1−x2​}. If K be the set of all points at which f is not differentiable, then K has exactly :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be differentiable at c∈R and f(c)=0. If g(x)=∣f(x)∣, then at x=c, g is :

Defining the inner function $f (t)$

Finding critical points of $f (t)$

Evaluating $f (t)$ at key points

Analyzing the shape of $f (t)$

Constructing $g (x)$ for $x \in [0, 1]$

Constructing $g (x)$ for $x \in [1, 3]$

Defining $g (x)$ for $x \in (3, 4]$

The complete piecewise definition of $g (x)$

Checking differentiability at $x = 1$

Checking differentiability at $x = 3$

Let $f : R \to R$ be a function defined by $f (x) = max {x, x^{3}}$ . The set of all points where $f (x)$ is NOT differentiable is

The number of points, where the function $f : R \to R$ , $f (x) = ∣ x - 1∣ cos ∣ x - 2∣ sin ∣ x - 1∣ + (x - 3) ∣ x^{2} - 5 x + 4∣$ , is NOT differentiable, is :

The function $f (x) = ∣ x^{2} - 2 x - 3∣ \cdot e^{∣9 x^{2} - 12 x + 4∣}$ is not differentiable at exactly:

Let $f (x) = {max {∣ x ∣, x^{2}}, 8 - 2∣ x ∣, ∣ x ∣ \leq 2 2 < ∣ x ∣ \leq 4$ . Let S be the set of points in the interval (-4, 4) at which f is not differentiable. Then S:

Let $a \in Z$ and $[t]$ be the greatest integer $\leq t$ , then the number of points, where the function $f (x) = [a + 13 sin x]$ , $x \in (0, π)$ is not differentiable, is ________.

Let $f : R \to R$ be a function defined by $f (x) = max {x, x^{2}}$ . Let $S$ denote the set of all points in $R$ , where $f$ is not differentiable. Then :

Let $f : (- 1, 1) \to R$ be a function defined by $f (x) = max {- ∣ x ∣, - 1 - x^{2}}$ . If K be the set of all points at which f is not differentiable, then K has exactly :

Let $f : R \to R$ be differentiable at $c \in R$ and $f (c) = 0$ . If $g (x) = ∣ f (x) ∣$ , then at $x = c$ , $g$ is :