Does the condition $f(t) \ge 1$ actually matter?

It matters immensely! If $f(a)$ were 0, the slopes might match, and we could have differentiability. But because $f(t) \ge 1$, the slope of the integral curve is at least 1. It forces a mismatch with the slope of 0 on the outside, guaranteeing a sharp corner at both boundaries.

JEE Advanced 2014

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: 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

Q: Why can't we just say it's differentiable because $f(t)$ is continuous?

Ah, the classic trap! $f(t)$ being continuous makes the *integral* continuous, but differentiability is about the *slope*. Even if the road is continuous, if you turn the steering wheel instantly from straight to a curve, you create a sharp corner. That is exactly what happens here; the slope changes abruptly.

Select Answer:

* Multiple Correct

Progressive Hints

Hint 1

Visualize the function as a traveler: first resting on the x-axis, then climbing a hill defined by the integral, and finally resting on a plateau. The question of differentiability is simply asking: does the traveler change speed smoothly, or is there a sudden jerk?

Hint 2

Recall that for a function to be differentiable at a point, the slope from the left must equal the slope from the right. Use the Fundamental Theorem of Calculus to find the derivative of the integral term:

\frac{d}{d x} \int_{a}^{x} f (t) d t = f (x)

Hint 3

Calculate the derivatives at the boundaries. At

x = a

, compare the slope of the flat line (0) with the slope of the integral curve (

f (a)

). At

x = b

, compare the slope of the integral curve (

f (b)

) with the slope of the flat plateau (0). Since

f (t) \geq 1

, what does this tell you about the equality of these slopes?

Visualized Solution

Visualizing the Function $g (x)$

We are given a continuous function $f : [a, b] \to [1, \infty)$ .
The function $g (x)$ is defined piecewise across three distinct intervals.
Let's set up our coordinate system with boundary points $a$ and $b$ on the x-axis.

Region 1: $x < a$

For the interval $x < a$ , the function is defined as $g (x) = 0$ .
This represents a flat horizontal line lying exactly on the x-axis.

Region 2: $a \leq x \leq b$

For $a \leq x \leq b$ , the function is defined as $g (x) = \int_{a}^{x} f (t) d t$ .
Since $f (t) \geq 1$ , the integrand is strictly positive, meaning $g (x)$ is strictly increasing.

Region 3: $x > b$

For $x > b$ , the function is defined as $g (x) = \int_{a}^{b} f (t) d t$ .
This is a constant value, representing a flat horizontal line starting from $x = b$ .

Continuity at $x = a$

Let's check the continuity at $x = a$ :
Left-Hand Limit (LHL): $lim_{x \to a^{-}} g (x) = 0$ .
Right-Hand Limit (RHL): $lim_{x \to a^{+}} g (x) = \int_{a}^{a} f (t) d t = 0$ .
Since $LHL = RHL = g (a) = 0$ , $g (x)$ is continuous at $x = a$ .

Continuity at $x = b$

Let's check the continuity at $x = b$ :
Left-Hand Limit (LHL): $lim_{x \to b^{-}} g (x) = \int_{a}^{b} f (t) d t$ .
Right-Hand Limit (RHL): $lim_{x \to b^{+}} g (x) = \int_{a}^{b} f (t) d t$ .
Since $LHL = RHL = g (b)$ , $g (x)$ is continuous at $x = b$ .

Differentiability at $x = a$ : Left-Hand Derivative

To check differentiability at $x = a$ , we find the Left-Hand Derivative (LHD):
$LHD = \frac{d}{d x} (0) = 0$ .

Differentiability at $x = a$ : Right-Hand Derivative

Now we find the Right-Hand Derivative (RHD) at $x = a$ :
Using the Leibniz Rule: $\frac{d}{d x} \int_{a}^{x} f (t) d t = f (x)$ .
At $x = a$ , $RHD = f (a)$ .

Non-Differentiability at $x = a$

Since $f (t) \geq 1$ , we have $f (a) \geq 1$ .
Therefore, $LHD = 0 \neq = f (a) = RHD$ .
This mismatch in slopes creates a sharp corner at $x = a$ , so $g (x)$ is not differentiable at $x = a$ .

Differentiability at $x = b$ : Left-Hand Derivative

Now let's check differentiability at $x = b$ :
$LHD = \frac{d}{d x} \int_{a}^{x} f (t) d t = f (x)$ .
At $x = b$ , $LHD = f (b) \geq 1$ .

Differentiability at $x = b$ : Right-Hand Derivative

The Right-Hand Derivative (RHD) at $x = b$ is:
$RHD = \frac{d}{d x} (constant) = 0$ .

Non-Differentiability at $x = b$

Since $f (b) \geq 1$ , we have $LHD = f (b) \neq = 0 = RHD$ .
This mismatch in slopes creates another sharp corner at $x = b$ .
Thus, $g (x)$ is not differentiable at $x = b$ .

Final Conclusion

The function $g (x)$ is continuous everywhere on $R$ .
However, $g (x)$ is not differentiable at $x = a$ and $x = b$ .
Correct Options: (a) and (c).

00:00 / 00:00

The Sigma Insight: Relationship Between Continuity and Differentiability

Solution Diagram

Analyzing the Setup

Imagine you are standing on a vast, flat plain. You are looking at a function

g (x)

that behaves like a traveler on a journey. This function is a story told in three distinct acts.

In the first act, for

x < a

, our traveler is resting on the x-axis. The function is defined as

g (x) = 0

. It is a perfectly flat, horizontal line with no movement and no slope.

The Climb

At the boundary

x = a

, the traveler begins to climb. For the interval

a \leq x \leq b

, the function is defined as:

g (x) = \int_{a}^{x} f (t) d t

This is the heart of the problem, representing the accumulation of area under the curve

f (t)

. Because we are given that

f (t) \geq 1

, this is a steady, upward climb.

The slope of this path is determined by the Fundamental Theorem of Calculus:

\frac{d}{d x} \int_{a}^{x} f (t) d t = f (x)

Since

f (t) \geq 1

, our traveler is moving upward with a slope of at least

1

The Plateau

Finally, we reach

x = b

. The climbing stops, and the function becomes:

g (x) = \int_{a}^{b} f (t) d t

This is a constant value—a plateau. The traveler stops climbing and walks along a flat, horizontal line once more, where the slope is

0

Continuity and Differentiability

Now, we must ask the critical questions: Is this journey smooth?

Continuity is straightforward to verify. At

x = a

, the left-hand limit is

0

, and the right-hand limit is

\int_{a}^{a} f (t) d t = 0

. They match.

x = b

, the left-hand limit is

\int_{a}^{b} f (t) d t

, and the right-hand limit is the constant

\int_{a}^{b} f (t) d t

. These also match, confirming the function is continuous everywhere.

However, differentiability reveals where the sharp corners hide. At

x = a

, the slope from the left is

0

, while the slope from the right is

f (a)

. Since

f (a) \geq 1

, the slopes are not equal, creating a sharp corner.

x = b

, the slope from the left is

f (b)

, and the slope from the right is

0

. Again, because

f (b) \geq 1

, the slopes are not equal, resulting in another sharp corner.

We have discovered that while our function is continuous, it is not differentiable at the boundaries. It is a beautiful example of how a function can be perfectly connected but fundamentally "broken" in its smoothness.

Common Traps & FAQs

Wait, why is the function continuous at

a

and

b

? I thought piecewise functions usually have jumps.

That is a brilliant observation, but don't let the 'piecewise' label intimidate you! Continuity simply means the graph doesn't break. At

x = a

, the left side is 0, and the right side is

\int_{a}^{a} f (t) d t

, which is also 0. They meet perfectly! It is like two roads joining seamlessly without a gap.

Why can't we just say it's differentiable because

f (t)

is continuous?

Ah, the classic trap!

f (t)

being continuous makes the *integral* continuous, but differentiability is about the *slope*. Even if the road is continuous, if you turn the steering wheel instantly from straight to a curve, you create a sharp corner. That is exactly what happens here; the slope changes abruptly.

Does the condition

f (t) \geq 1

actually matter?

It matters immensely! If

f (a)

were 0, the slopes might match, and we could have differentiability. But because

f (t) \geq 1

, the slope of the integral curve is at least 1. It forces a mismatch with the slope of 0 on the outside, guaranteeing a sharp corner at both boundaries.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: 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

Select Answer:

Visualized Solution

Visualizing the Function $g (x)$

Region 1: $x < a$

Region 2: $a \leq x \leq b$

Region 3: $x > b$

Continuity at $x = a$

Continuity at $x = b$

Differentiability at $x = a$ : Left-Hand Derivative

Differentiability at $x = a$ : Right-Hand Derivative

Non-Differentiability at $x = a$

Differentiability at $x = b$ : Left-Hand Derivative

Differentiability at $x = b$ : Right-Hand Derivative

Non-Differentiability at $x = b$

Final Conclusion

The Sigma Insight: Relationship Between Continuity and Differentiability

Analyzing the Setup

The Climb

The Plateau

Continuity and Differentiability

Similar Questions

Consider the function $f : (0, 2) \to R$ defined by $f (x) = \frac{x}{2} + \frac{2}{x}$ and the function $g (x)$ defined by $g (x) = {min {f (t)}, \frac{3}{2} + x, 0 < t \leq x and 0 < x \leq 1 1 < x < 2$ . Then

If $f (x) = {\int_{0}^{x} (5 + ∣1 - t ∣) d t, 5 x + 1, x > 2 x \leq 2$ , then

Let $f (x) = {0, x^{2}, x < 0 x \geq 0$ then for all $x$

Let $f (x) = {x^{2} sin (\frac{1}{x}) 0, x \neq = 0, x = 0$ ; Then at $x = 0$

Let $g (x) = x f (x)$ , where $f (x) = {x sin \frac{1}{x}, 0, x \neq = 0 x = 0$ . At $x = 0$

If $f (x) = ⎩ ⎨ ⎧ - x - \frac{π}{2}, - cos x, x - 1, ln x, x \leq - \frac{π}{2} - \frac{π}{2} < x \leq 0 0 < x \leq 1 x > 1$

If $f (x) = min {1, x^{2}, x^{3}}$ , then

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 :

If $f (x) = {x e^{- (\frac{1}{∣ x ∣} + \frac{1}{x})}, 0, x \neq = 0 x = 0$ then $f (x)$ is

Let $f : [- 1/2, 2] \to R$ and $g : [- 1/2, 2] \to R$ be functions defined by $f (x) = [x^{2} - 3]$ and $g (x) = ∣ x ∣ f (x) + ∣4 x - 7∣ f (x)$ , where $[y]$ denotes the greatest integer less than or equal to $y$ . Then

Select Answer:

Visualized Solution

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

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

Region 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a≤x≤b

Region 3: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x>b

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

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

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

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

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

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

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

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

Final Conclusion

The Sigma Insight: Relationship Between Continuity and Differentiability

Analyzing the Setup

The Climb

The Plateau

Continuity and Differentiability

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Consider the function f:(0,2)→R defined by f(x)=2x​+x2​ and the function g(x) defined by g(x)={min{f(t)},23​+x,​0<t≤x and 0<x≤11<x<2​. Then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)={∫0x​(5+∣1−t∣)dt,5x+1,​x>2x≤2​, then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)={0,x2,​x<0x≥0​ then for all x

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)={x2sin(x1​)0​,x=0,x=0​; Then at x=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let g(x)=xf(x), where f(x)={xsinx1​,0,​x=0x=0​. At x=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=⎩⎨⎧​−x−2π​,−cosx,x−1,lnx,​x≤−2π​−2π​<x≤00<x≤1x>1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=min{1,x2,x3}, then

.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 :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)={xe−(∣x∣1​+x1​),0,​x=0x=0​ then f(x) is

Visualizing the Function $g (x)$

Region 1: $x < a$

Region 2: $a \leq x \leq b$

Region 3: $x > b$

Continuity at $x = a$

Continuity at $x = b$

Differentiability at $x = a$ : Left-Hand Derivative

Differentiability at $x = a$ : Right-Hand Derivative

Non-Differentiability at $x = a$

Differentiability at $x = b$ : Left-Hand Derivative

Differentiability at $x = b$ : Right-Hand Derivative

Non-Differentiability at $x = b$

Consider the function $f : (0, 2) \to R$ defined by $f (x) = \frac{x}{2} + \frac{2}{x}$ and the function $g (x)$ defined by $g (x) = {min {f (t)}, \frac{3}{2} + x, 0 < t \leq x and 0 < x \leq 1 1 < x < 2$ . Then

If $f (x) = {\int_{0}^{x} (5 + ∣1 - t ∣) d t, 5 x + 1, x > 2 x \leq 2$ , then

Let $f (x) = {0, x^{2}, x < 0 x \geq 0$ then for all $x$

Let $f (x) = {x^{2} sin (\frac{1}{x}) 0, x \neq = 0, x = 0$ ; Then at $x = 0$

Let $g (x) = x f (x)$ , where $f (x) = {x sin \frac{1}{x}, 0, x \neq = 0 x = 0$ . At $x = 0$

If $f (x) = ⎩ ⎨ ⎧ - x - \frac{π}{2}, - cos x, x - 1, ln x, x \leq - \frac{π}{2} - \frac{π}{2} < x \leq 0 0 < x \leq 1 x > 1$

If $f (x) = min {1, x^{2}, x^{3}}$ , then

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 :

If $f (x) = {x e^{- (\frac{1}{∣ x ∣} + \frac{1}{x})}, 0, x \neq = 0 x = 0$ then $f (x)$ is