Let f : R → R be a function defined by : f(x) = { max_{t ≤ x} {t³ - 3t} ; x ≤ 2; x² + 2x - 6 ; 2

Q: Wait, why does $f(x) = 2$ for $x \in (-1, 2]$? I thought it should follow $t^3 - 3t$.

That is the beauty of the 'running maximum' definition! For $x > -1$, the function $g(t) = t^3 - 3t$ starts decreasing from its local maximum of $2$ at $t = -1$. Since the running maximum 'remembers' the highest value encountered so far, it stays locked at $2$ until the curve $g(t)$ rises back up to $2$ at $x = 2$. It is like a hiker who marks their highest altitude reached; they don't lower their record just because they walk downhill.

Q: Do I really need to check continuity at $x=4$ and $x=5$? Can't I just check the derivatives?

Never skip the continuity check! If a function is discontinuous at a point, it is impossible for it to be differentiable there. At $x=4$ and $x=5$, the function jumps, creating a gap. A sharp corner (where derivatives don't match) is one thing, but a jump is a complete break. Always check continuity first—it saves you from wasting time on derivatives where they don't exist.

Q: Why did we split the integral at $x=-1$?

We split the integral because the algebraic definition of $f(x)$ changes at that point. For $x \le -1$, $f(x)$ is defined by the cubic $x^3 - 3x$. For $x > -1$, the running maximum 'locks' the value at $2$. To calculate the area under the curve, we must integrate the specific function active in each region. Splitting the integral allows us to handle these two distinct behaviors separately.

Question

Let $f : \mathbb{R} 	o \mathbb{R}$ be a function defined by : $f(x) = \begin{cases} \max_{t \le x} \{t^3 - 3t\} & ; x \le 2 \ x^2 + 2x - 6 & ; 2 < x < 3 \ [x - 3] + 9 & ; 3 \le x \le 5 \ 2x + 1 & ; x > 5 \end{cases}$ Where $[t]$ is the greatest integer less than or equal to $t$. Let $m$ be the number of points where $f$ is not differentiable and $I = \int_{-2}^2 f(x) dx$. Then the ordered pair $(m, I)$ is equal to :

Options:

(A) $(3, \frac{27}{4})$

(B) $(3, \frac{23}{4})$

(C) $(4, \frac{27}{4})$

(D) $(4, \frac{23}{4})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are going to dissect a problem that might look like a chaotic mess of piecewise definitions, but is actually a beautifully structured puzzle. We are dealing with a function $f(x)$ defined by a 'running maximum.'

This is a classic JEE concept that tests your ability to visualize how a function evolves over time. Let's peel back the layers.

### Decoding the Cubic

First, let's look at the region $x \le 2$. The definition $f(x) = \max_{t \le x} \{t^3 - 3t\}$ is not just a formula; it is a memory. Imagine you are tracking the altitude of a climber.

The climber starts at $t = -2$ and moves forward. The 'running maximum' is the highest altitude they have reached *so far*. Let $g(t) = t^3 - 3t$.

To understand its behavior, we find its critical points:
$$g'(t) = 3t^2 - 3 = 3(t-1)(t+1)$$

We have a local maximum at $t = -1$ with $g(-1) = 2$ and a local minimum at $t = 1$ with $g(1) = -2$. For $x \le -1$, the function is climbing, so the running maximum is just $g(x)$.

But once we pass $x = -1$, the curve dips. The running maximum, however, refuses to drop. It stays at the peak value of $2$ until the curve $g(x)$ climbs back up to $2$ at $x = 2$.

Thus, for $x \in (-1, 2]$, our function is a flat, constant line: $f(x) = 2$.

### The Differentiability Gauntlet

Now, let's hunt for the points of non-differentiability, $m$. We check the boundaries: $x = 2, 3, 4,$ and $5$.

At $x = 2$, the left limit is $2$ and the right limit (from the parabola $x^2 + 2x - 6$) is $2^2 + 2(2) - 6 = 2$. It is continuous!

However, the left derivative is $0$ (derivative of a constant), and the right derivative is $2x + 2$, which is $6$ at $x = 2$. Since $0 
eq 6$, we have a sharp corner. That is our first point.

At $x = 3$, the left limit is $3^2 + 2(3) - 6 = 9$, and the right limit (from the greatest integer function $[x-3] + 9$) is $[3-3] + 9 = 9$. Continuous again!

The left derivative is $2(3) + 2 = 8$, and the right derivative is $0$. Another sharp corner! At $x = 4$ and $x = 5$, the function jumps. A jump means discontinuity, and discontinuity implies non-differentiability.

Counting them all up, we have **$m = 4$** points of non-differentiability.

### The Integral

Finally, we calculate $I = \int_{-2}^2 f(x) dx$. Because the definition of $f(x)$ changes at $x = -1$, we split the integral:
$$I = \int_{-2}^{-1} (x^3 - 3x) dx + \int_{-1}^2 2 dx$$

The first part is a standard integral:
$$\left[ \frac{x^4}{4} - \frac{3x^2}{2} ight]_{-2}^{-1}$$

Plugging in the limits, we get:
$$\left( \frac{1}{4} - \frac{3}{2} ight) - \left( \frac{16}{4} - \frac{12}{2} ight) = -\frac{5}{4} - (-2) = \frac{3}{4}$$

The second part is just the area of a rectangle with width $3$ and height $2$, which is $6$. Adding them together:
$$I = \frac{3}{4} + 6 = \frac{27}{4}$$

We have found our pair $(m, I) = (4, \frac{27}{4})$. You have just conquered a complex piecewise function by breaking it down into its fundamental geometric and algebraic parts. Keep this analytical mindset, and no JEE problem will ever be too daunting!

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Visualized Solution

$U n d er s t an d in g t h e P i ece w i se F u n c t i o n f (x)$

$A na l y z in g g (t) = t^{3} - 3 t$

$S im pl i f y in g f (x) f or x \leq 2$

$D e f inin g f (x) f or x > 2$

$C h ec k in g D i f f er e n t iabi l i t y a t x = 2$

$C h ec k in g D i f f er e n t iabi l i t y a t x = 3$

$C h ec k in g x = 4 an d x = 5$

$S e tt in g u pt h e I n t e g r a l I$

$C a l c u l a t in g t h e F i r s tP a r t o f t h e I n t e g r a l$

$C a l c u l a t in g t h e S eco n d P a r t an d F ina l S u m$

$C o n c l u s i o nan d F ina l A n s w er$

The Sigma Insight: Relationship Between Continuity and Differentiability

Analyzing the Setup

Decoding the Cubic

The Differentiability Gauntlet

The Integral

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Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzingg(t)=t3−3t

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifyingf(x)forx≤2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Definingf(x)forx>2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CheckingDifferentiabilityatx=2

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Checkingx=4andx=5

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

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

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

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