Animated Solution for Mathematics - Limits, Continuity and Differentiability: In the following [x] denotes the greatest integer less than or equal to x. Match the functions in Column I with the properties in Column II and indicate your answer by darkening the appropriate bubbles in the 4×4 matrix given in the ORS.
List-I
(P)
x∣x∣
(Q)
∣x∣
(R)
x+[x]
(S)
∣x−1∣+∣x+1∣
List-II
(1)
continuous in (−1,1)
(2)
differentiable in (−1,1)
(3)
strictly increasing in (−1,1)
(4)
not differentiable at least at one point in (−1,1)
Select Matching Pairs:
* Multiple Allowed
PMatches
QMatches
RMatches
SMatches
Visualized Solution
The Matrix Match Challenge
Objective: Match functions in Column I with properties in Column II for x∈(−1,1).
Properties to check:
(p) Continuous in (−1,1)
(q) Differentiable in (−1,1)
(r) Strictly increasing in (−1,1)
(s) Not differentiable at least at one point in (−1,1)
Analyzing f(x)=x∣x∣
f(x)=x∣x∣
For x≥0, ∣x∣=x⟹f(x)=x2
For x<0, ∣x∣=−x⟹f(x)=−x2
f(x)={−x2x2x<0x≥0
Continuity & Differentiability of x∣x∣
Continuity:limx→0−(−x2)=0 and limx→0+(x2)=0. Continuous everywhere in (−1,1).
Differentiability:f′(x)={−2x2xx<0x≥0
At x=0, LHD=0 and RHD=0. Differentiable everywhere in (−1,1).
Monotonicity of x∣x∣
f′(x)=2∣x∣
For all x∈(−1,1), f′(x)≥0.
f′(x)=0 only at discrete point x=0.
Therefore, f(x) is strictly increasing.
Match: (A) → (p, q, r)
Analyzing f(x)=∣x∣
f(x)=∣x∣
f(x)={−xxx<0x≥0
Continuity:limx→0∣x∣=0=f(0). Continuous in (−1,1).
Differentiability of ∣x∣
f′(x)={2−x−12x1x<0x>0
As x→0+, f′(x)→∞. As x→0−, f′(x)→−∞.
Vertical tangent at x=0⟹ Not differentiable at x=0.
Match: (B) → (p, s)
Analyzing f(x)=x+[x]
f(x)=x+[x] where [x] is the greatest integer function.
For x∈(−1,0), [x]=−1⟹f(x)=x−1
For x∈[0,1), [x]=0⟹f(x)=x
Discontinuity of x+[x]
At x=0:
limx→0−f(x)=0−1=−1
limx→0+f(x)=0
f(0)=0
LHL = RHL. Discontinuous at x=0.
Discontinuous ⟹ Not differentiable at x=0.
Monotonicity of x+[x]
In (−1,0), f(x)=x−1 (slope = 1, increasing)
In [0,1), f(x)=x (slope = 1, increasing)
At x=0, f(0)=0>limx→0−f(x)=−1.
The function strictly increases across the jump.
Match: (C) → (r, s)
Analyzing f(x)=∣x−1∣+∣x+1∣
f(x)=∣x−1∣+∣x+1∣
We are analyzing in the interval x∈(−1,1).
For any x in (−1,1):
x<1⟹x−1<0⟹∣x−1∣=−(x−1)=1−x
x>−1⟹x+1>0⟹∣x+1∣=x+1
Properties of f(x)=∣x−1∣+∣x+1∣
Substitute back: f(x)=(1−x)+(x+1)
f(x)=2 for all x∈(−1,1).
A constant function is continuous and differentiable everywhere.
It is not strictly increasing (slope is 0).
Match: (D) → (p, q)
Final Matching Matrix
(A) x∣x∣→ (p, q, r)
(B) ∣x∣→ (p, s)
(C) x+[x]→ (r, s)
(D) ∣x−1∣+∣x+1∣→ (p, q)
Key Takeaway: Always analyze piecewise functions at their critical points, especially when absolute values or step functions are involved.
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The Sigma Insight: Relationship Between Continuity and Differentiability
Solution Diagram
The Matrix Match Challenge
A Journey Through Functions
Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are dissecting the very anatomy of functions.
We have four distinct mathematical entities, and our goal is to map them to their fundamental behaviors—continuity, differentiability, and monotonicity—within the open interval (−1,1). This is a classic JEE Advanced challenge, and it requires us to look past the symbols and visualize the behavior of these curves.
Phase 1
The Smooth Operator (x∣x∣)
Let us begin with f(x)=x∣x∣. The absolute value function is the great divider; it changes its nature at x=0.
For x≥0, ∣x∣=x, so our function becomes f(x)=x2. For x<0, ∣x∣=−x, turning our function into f(x)=−x2.
When we look at the origin, the left-hand limit limx→0−(−x2)=0 and the right-hand limit limx→0+(x2)=0. They meet perfectly at f(0)=0, so the function is continuous.
The derivative f′(x) is −2x for x<0 and 2x for x≥0. At x=0, both sides yield 0, meaning the curve is smooth.
Since f′(x)=2∣x∣≥0 and only hits zero at a single point, the function is strictly increasing. Thus, f(x)=x∣x∣ is continuous, differentiable, and strictly increasing. A perfect match for (p, q, r).
Phase 2
The Cusp of Reality (∣x∣)
Next, we encounter f(x)=∣x∣. Again, we split the domain at x=0.
For x<0, we have −x, and for x≥0, we have x. As x approaches 0, the function value approaches 0, so it is continuous.
However, watch what happens when we differentiate. For x>0:
f′(x)=2x1
As x→0+, this derivative explodes to +∞. For x<0, f′(x)=−2−x1, which shoots to −∞.
This infinite slope at the origin is a 'vertical tangent' or a sharp cusp. The function is not differentiable at x=0. Match: (p, s).
Phase 3
The Staircase Jump (x+[x])
Now, we face the greatest integer function, [x]. In the interval (−1,0), [x]=−1, so f(x)=x−1.
In the interval [0,1), [x]=0, so f(x)=x. At x=0, the left-hand limit is −1, but the right-hand limit is 0.
There is a jump! A jump discontinuity means the function is not continuous, and by the fundamental laws of calculus, it cannot be differentiable.
However, look at the values: the function is always increasing in both segments, and the jump itself is an upward step. It is strictly increasing, despite the break. Match: (r, s).
Phase 4
The Hidden Constant (∣x−1∣+∣x+1∣)
Finally, we arrive at f(x)=∣x−1∣+∣x+1∣. In the domain x∈(−1,1), x is always less than 1, so ∣x−1∣=1−x.
Simultaneously, x is always greater than −1, so ∣x+1∣=x+1. When we add them:
f(x)=(1−x)+(x+1)=2
The variables x and −x cancel out, leaving us with a constant function f(x)=2.
A horizontal line is the epitome of smoothness—it is continuous and differentiable everywhere. It is not strictly increasing because the slope is zero. Match: (p, q).
The Takeaway
We have navigated the terrain. We saw how x∣x∣ smooths out, how ∣x∣ creates a cusp, how x+[x] jumps, and how ∣x−1∣+∣x+1∣ hides a constant value.
The secret to these problems is never to fear the piecewise definition. Embrace the critical points, analyze the limits, and the properties will reveal themselves.
