JEE Main 2021 (16 March Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: 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 :

Select Answer:

Visualized Solution

Given Functions $f (x)$ and $g (x)$

$f (x) = {x + 2, x^{2}, x < 0 x \geq 0$
$g (x) = {x^{3}, 3 x - 2, x < 1 x \geq 1$

Defining the Composite Function $f (g (x))$

By definition of $f (x)$ :
$f (g (x)) = {g (x) + 2, (g (x))^{2}, g (x) < 0 g (x) \geq 0$

Analyzing the Condition $g (x) < 0$

Case 1: $g (x) < 0$
For $x < 1$ , $g (x) = x^{3} ⟹ x^{3} < 0 ⟹ x < 0$
For $x \geq 1$ , $g (x) = 3 x - 2 ⟹ 3 x - 2 < 0 ⟹ x < \frac{2}{3}$ (Rejected as $x \geq 1$ )
Conclusion: $g (x) < 0$ only for $x < 0$

Analyzing the Condition $g (x) \geq 0$

Case 2: $g (x) \geq 0$
This happens when $x \geq 0$ .
Sub-case (a): $0 \leq x < 1 ⟹ g (x) = x^{3}$
Sub-case (b): $x \geq 1 ⟹ g (x) = 3 x - 2$

Constructing the Piecewise $f (g (x))$

Combining all cases:
$(f \circ g) (x) = ⎩ ⎨ ⎧ x^{3} + 2, (x^{3})^{2} = x^{6}, (3 x - 2)^{2}, x < 0 0 \leq x < 1 x \geq 1$

Identifying Critical Points

The function $(f \circ g) (x)$ changes definition at $x = 0$ and $x = 1$ .
We must check continuity and differentiability at these critical points.

Continuity Check at $x = 0$

At $x = 0$ :
L.H.L. $= lim_{x \to 0^{-}} (x^{3} + 2) = 2$
R.H.L. $= lim_{x \to 0^{+}} (x^{6}) = 0$
Since L.H.L. $\neq =$ R.H.L., $(f \circ g) (x)$ is discontinuous at $x = 0$ .

Differentiability at $x = 0$

A function cannot be differentiable at a point where it is discontinuous.
Therefore, $(f \circ g) (x)$ is not differentiable at $x = 0$ .

Continuity Check at $x = 1$

At $x = 1$ :
L.H.L. $= lim_{x \to 1^{-}} (x^{6}) = 1$
R.H.L. $= lim_{x \to 1^{+}} (3 x - 2)^{2} = (3 (1) - 2)^{2} = 1$
$f (g (1)) = 1$
Since L.H.L. $=$ R.H.L. $= f (g (1))$ , it is continuous at $x = 1$ .

Differentiability Check at $x = 1$

Differentiating for $x > 0$ :
$(f \circ g)^{'} (x) = {6 x^{5}, 2 (3 x - 2) \cdot 3 = 6 (3 x - 2), 0 < x < 1 x > 1$
L.H.D. at $x = 1$ : $6 (1)^{5} = 6$
R.H.D. at $x = 1$ : $6 (3 (1) - 2) = 6$

Final Conclusion

Since L.H.D. $=$ R.H.D. $= 6$ , the function is differentiable at $x = 1$ .
Summary:
At $x = 0$ : Not differentiable.
At $x = 1$ : Differentiable.
Total number of points of non-differentiability = 1.

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The Sigma Insight: Relationship Between Continuity and Differentiability

Solution Diagram

Analyzing the Setup

We are tasked with finding the points of non-differentiability for the composite function

(f \circ g) (x)

. The given functions are defined as:

f (x) = {x + 2, x^{2}, x < 0 x \geq 0

g (x) = {x^{3}, 3 x - 2, x < 1 x \geq 1

Constructing the Composite Function

To build

(f \circ g) (x)

, we must analyze the output of

g (x)

relative to the domain constraints of

f (x)

For

x < 0

g (x) = x^{3} < 0

. Thus,

(f \circ g) (x) = f (g (x)) = g (x) + 2 = x^{3} + 2

For

0 \leq x < 1

g (x) = x^{3}

. Since

0 \leq x^{3} < 1

, the output is non-negative, so

(f \circ g) (x) = f (g (x)) = (g (x))^{2} = (x^{3})^{2} = x^{6}

For

x \geq 1

g (x) = 3 x - 2

. Since

3 x - 2 \geq 1

for

x \geq 1

, the output is positive, so

(f \circ g) (x) = f (g (x)) = (g (x))^{2} = (3 x - 2)^{2}

The composite function is:

(f \circ g) (x) = ⎩ ⎨ ⎧ x^{3} + 2, x^{6}, (3 x - 2)^{2}, x < 0 0 \leq x < 1 x \geq 1

Investigating Continuity at the Seams

Differentiability requires continuity. We check the "seams" at

x = 0

and

x = 1

x = 0

: The Left-Hand Limit (LHL) is

lim_{x \to 0^{-}} (x^{3} + 2) = 2

. The Right-Hand Limit (RHL) is

lim_{x \to 0^{+}} (x^{6}) = 0

Since $2 eq 0$, the function is discontinuous at $x = 0$ . Therefore, the function is not differentiable at

x = 0

Checking Differentiability at $x = 1$

x = 1

: The LHL is

lim_{x \to 1^{-}} (x^{6}) = 1

. The RHL is

lim_{x \to 1^{+}} (3 x - 2)^{2} = (3 (1) - 2)^{2} = 1

. The function is continuous at

x = 1

Now, we check the derivatives: The derivative of

x^{6}

6 x^{5}

. At

x = 1

, the slope is

6 (1)^{5} = 6

. The derivative of

(3 x - 2)^{2}

2 (3 x - 2) \cdot 3 = 6 (3 x - 2)

. At

x = 1

, the slope is

6 (3 (1) - 2) = 6

Since the left-hand derivative equals the right-hand derivative, the function is differentiable at $x = 1$ .

Final Conclusion

The only point where the function fails to be differentiable is $x = 0$ .

Common Traps & FAQs

Wait, why do we only check

x = 0

and

x = 1

? Shouldn't we check every single point?

That is a fantastic question! In a piecewise function, the 'danger zones' are only at the boundaries where the definition changes. Everywhere else, the function is just a simple polynomial (like

x^{3} + 2

x^{6}

), which is inherently smooth and differentiable. We only need to investigate the 'seams' where the pieces are stitched together.

If the function is discontinuous at

x = 0

, why do we even bother checking the derivative?

You are absolutely right to question that! It is a fundamental theorem: if a function is discontinuous at a point, it is automatically non-differentiable there. We check continuity first because it is a 'shortcut'—if it fails, we save ourselves the time of calculating derivatives. It is about working smart, not just hard.

Why did the derivative at

x = 1

turn out to be differentiable? I thought piecewise functions are usually non-differentiable at their boundaries.

That is a common trap! A piecewise function is only non-differentiable at a boundary if the 'slopes' (derivatives) from the left and right don't match. Sometimes, the two pieces meet so perfectly that the transition is smooth. In this case, both sides had a slope of

6

x = 1

, so the 'kink' in the graph vanished!

Select Answer:

Visualized Solution

Given Functions $f (x)$ and $g (x)$

Defining the Composite Function $f (g (x))$

Analyzing the Condition $g (x) < 0$

Analyzing the Condition $g (x) \geq 0$

Constructing the Piecewise $f (g (x))$

Identifying Critical Points

Continuity Check at $x = 0$

Differentiability at $x = 0$

Continuity Check at $x = 1$

Differentiability Check at $x = 1$

Final Conclusion

The Sigma Insight: Relationship Between Continuity and Differentiability

Analyzing the Setup

Constructing the Composite Function

Investigating Continuity at the Seams

Checking Differentiability at $x = 1$

Final Conclusion

Similar Questions

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Select Answer:

Visualized Solution

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

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

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

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

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

Identifying Critical Points

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

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

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

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

Final Conclusion

The Sigma Insight: Relationship Between Continuity and Differentiability

Analyzing the Setup

Constructing the Composite Function

Investigating Continuity at the Seams

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

Final Conclusion

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:[a,b]→[1,∞) be a continuous function and let g:R→R be defined as g(x)=⎩⎨⎧​0,∫ax​f(t)dt,∫ab​f(t)dt,​if x<a,if a≤x≤b;if x>b.​ then

Given Functions $f (x)$ and $g (x)$

Defining the Composite Function $f (g (x))$

Analyzing the Condition $g (x) < 0$

Analyzing the Condition $g (x) \geq 0$

Constructing the Piecewise $f (g (x))$

Continuity Check at $x = 0$

Differentiability at $x = 0$

Continuity Check at $x = 1$

Differentiability Check at $x = 1$

Checking Differentiability at $x = 1$

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

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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 $f : [a, b] \to [1, \infty)$ be a continuous function and let $g : R \to R$ be defined as $g (x) = ⎩ ⎨ ⎧ 0, \int_{a}^{x} f (t) d t, \int_{a}^{b} f (t) d t, if x < a, if a \leq x \leq b; if x > b .$ then