JEE Main 2022 (27 July Shift 1)

LEVELJEE Advanced

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let a function $f : R \to R$ be defined as : $f (x) = {\int_{0}^{x} (5 - ∣ t - 3∣) d t, x^{2} + b x, x > 4 x \leq 4$ where $b \in R$ . If $f$ is continuous at $x = 4$ , then which of the following statements is NOT true ?

Q: Why do we split the integral at $t = 3$ instead of $t = 4$?

Great question! The absolute value function $|t - 3|$ is defined as $(t - 3)$ when $t \ge 3$ and $-(t - 3)$ when $t < 3$. Because the integrand changes its algebraic form at $t = 3$, we must split the integral there to correctly evaluate the area under the curve. If you integrate across $t = 3$ without splitting, you are essentially treating the absolute value as a single expression, which is mathematically incorrect.

Select Answer:

Progressive Hints

Hint 1

Think of continuity as a bridge. For the function to be continuous at

x = 4

, the 'left-side' path and the 'right-side' path must meet at the exact same elevation. What is the elevation of the left path at

x = 4

? Set that equal to the integral representing the right path.

Hint 2

The absolute value

∣ t - 3∣

is a mathematical chameleon. It changes its behavior at

t = 3

. To evaluate the integral

\int_{0}^{4} (5 - ∣ t - 3∣) d t

, you must split the domain into two distinct regions:

[0, 3]

and

[3, 4]

. If you don't, the integral will fail to capture the geometric reality of the function.

Hint 3

Start by setting the left-hand limit equal to the right-hand limit:

16 + 4 b = \int_{0}^{3} (5 - (3 - t)) d t + \int_{3}^{4} (5 - (t - 3)) d t

. Solve this for

b

first. Once you have

b

, you can find the derivative

f^{'} (x)

for both branches and test the monotonicity and differentiability.

Visualized Solution

Analyze Continuity at $x = 4$

Function $f (x)$ is continuous at $x = 4$ .
Value of function: $f (4) = 4^{2} + b (4) = 16 + 4 b$ .
Right Hand Limit: $lim_{x \to 4^{+}} f (x) = \int_{0}^{4} (5 - ∣ t - 3∣) d t$ .

Evaluate the Integral for RHL

Split the integral at $t = 3$ :
$\int_{0}^{4} (5 - ∣ t - 3∣) d t = \int_{0}^{3} (5 - (3 - t)) d t + \int_{3}^{4} (5 - (t - 3)) d t$
$= \int_{0}^{3} (2 + t) d t + \int_{3}^{4} (8 - t) d t$

Calculate the Integral Value

$= [2 t + \frac{t ^{2}}{2}]_{0}^{3} + [8 t - \frac{t ^{2}}{2}]_{3}^{4}$
$= (6 + 4.5) + (32 - 8 - (24 - 4.5))$
$= 10.5 + 4.5 = 15$ .

Solve for Constant $b$

Equating LHL and RHL for continuity:
$16 + 4 b = 15 ⟹ 4 b = - 1 ⟹ b = - \frac{1}{4}$ .
The function is: $f (x) = x^{2} - \frac{1}{4} x$ for $x \leq 4$ .

Find the Derivative $f^{'} (x)$ for $x \leq 4$

For $x < 4$ :
$f^{'} (x) = \frac{d}{d x} (x^{2} - \frac{1}{4} x) = 2 x - \frac{1}{4}$ .

Find the Derivative $f^{'} (x)$ for $x > 4$

For $x > 4$ :
$f^{'} (x) = \frac{d}{d x} \int_{0}^{x} (5 - ∣ t - 3∣) d t = 5 - ∣ x - 3∣$ .
Since $x > 4$ , $∣ x - 3∣ = x - 3$ , so $f^{'} (x) = 5 - (x - 3) = 8 - x$ .

Check Differentiability at $x = 4$

Left Hand Derivative: $f^{'} (4^{-}) = 2 (4) - \frac{1}{4} = \frac{31}{4}$ .
Right Hand Derivative: $f^{'} (4^{+}) = 8 - 4 = 4$ .
Since $f^{'} (4^{-}) \neq = f^{'} (4^{+})$ , $f$ is not differentiable at $x = 4$ . (Statement A is True)

Verify Option (B): $f^{'} (3) + f^{'} (5)$

Calculate $f^{'} (3)$ : $2 (3) - \frac{1}{4} = \frac{23}{4}$ .
Calculate $f^{'} (5)$ : $8 - 5 = 3$ .
Sum: $f^{'} (3) + f^{'} (5) = \frac{23}{4} + 3 = \frac{35}{4}$ . (Statement B is True)

Analyze Monotonicity for $x \leq 4$

For $x \leq 4$ , $f^{'} (x) = 2 x - \frac{1}{4}$ .
$f^{'} (x) = 0 ⟹ x = \frac{1}{8}$ .
$f^{'} (x) < 0$ for $x < \frac{1}{8}$ (Decreasing).
$f^{'} (x) > 0$ for $\frac{1}{8} < x \leq 4$ (Increasing).

Analyze Monotonicity for $x > 4$

For $x > 4$ , $f^{'} (x) = 8 - x$ .
$f^{'} (x) = 0 ⟹ x = 8$ .
$f^{'} (x) > 0$ for $4 < x < 8$ (Increasing).
$f^{'} (x) < 0$ for $x > 8$ (Decreasing).

Identify the False Statement

Combining results: $f$ is increasing in $(\frac{1}{8}, 8)$ .
$f$ is decreasing in $(- \infty, \frac{1}{8}) \cup (8, \infty)$ .
Statement (C) claims $f$ is increasing in $(- \infty, \frac{1}{8}) \cup (8, \infty)$ , which is NOT true.

Final Check: Local Minima at $x = 1/8$

At $x = \frac{1}{8}$ , $f^{'} (x)$ changes sign from negative to positive.
Therefore, $f$ has a local minima at $x = \frac{1}{8}$ . (Statement D is True)
The incorrect statement is (C).

00:00 / 00:00

The Sigma Insight: Relationship Between Continuity and Differentiability

Solution Diagram

Analyzing the Setup

We are examining a piecewise function

f (x)

defined as:

f (x) = {x^{2} + b x, \int_{0}^{x} (5 - ∣ t - 3∣) d t, x \leq 4 x > 4

To ensure the function is continuous at the junction

x = 4

, the Left-Hand Limit (LHL) must equal the Right-Hand Limit (RHL).

The Bridge of Continuity

First, we evaluate the left branch at

x = 4

f (4) = 4^{2} + b (4) = 16 + 4 b

Next, we calculate the RHL using the integral. Because of the absolute value

∣ t - 3∣

, we must split the integral at

t = 3

\int_{0}^{4} (5 - ∣ t - 3∣) d t = \int_{0}^{3} (5 - (3 - t)) d t + \int_{3}^{4} (5 - (t - 3)) d t

Simplifying the integrands, we obtain:

\int_{0}^{3} (2 + t) d t + \int_{3}^{4} (8 - t) d t

Evaluating these integrals:

[2 t + \frac{t ^{2}}{2}]_{0}^{3} + [8 t - \frac{t ^{2}}{2}]_{3}^{4}

The first part yields

6 + 4.5 = 10.5

, and the second part yields

(32 - 8) - (24 - 4.5) = 24 - 19.5 = 4.5

. The total sum is

10.5 + 4.5 = 15

Equating the LHL and RHL:

16 + 4 b = 15 \Rightarrow 4 b = - 1 \Rightarrow b = - \frac{1}{4}

The Derivative's Dance

To analyze the behavior of

f (x)

, we find the derivative

f^{'} (x)

for both intervals:

For

x < 4

f^{'} (x) = \frac{d}{d x} (x^{2} - \frac{1}{4} x) = 2 x - \frac{1}{4}

For

x > 4

, applying the Fundamental Theorem of Calculus:

f^{'} (x) = 5 - ∣ x - 3∣ = 5 - (x - 3) = 8 - x

Checking differentiability at

x = 4

f^{'} (4^{-}) = 2 (4) - 0.25 = 7.75

f^{'} (4^{+}) = 8 - 4 = 4

Since $7.75 eq 4$, the function is continuous but not differentiable at

x = 4

The Monotonicity Trap

We determine the intervals of increase and decrease by checking the sign of

f^{'} (x)

For

x < 4

f^{'} (x) = 2 x - \frac{1}{4}

. The function decreases for

x < \frac{1}{8}

and increases for

\frac{1}{8} < x < 4

For

x > 4

f^{'} (x) = 8 - x

. The function increases for

4 < x < 8

and decreases for

x > 8

Combining these results, the function

f (x)

is increasing on the interval

(\frac{1}{8}, 8)

and decreasing on

(- \infty, \frac{1}{8}) \cup (8, \infty)

Any statement claiming

f

is increasing on

(- \infty, \frac{1}{8}) \cup (8, \infty)

is false.

Common Traps & FAQs

Why do we split the integral at

t = 3

instead of

t = 4

Great question! The absolute value function

∣ t - 3∣

is defined as

(t - 3)

when

t \geq 3

and

- (t - 3)

when

t < 3

. Because the integrand changes its algebraic form at

t = 3

, we must split the integral there to correctly evaluate the area under the curve. If you integrate across

t = 3

without splitting, you are essentially treating the absolute value as a single expression, which is mathematically incorrect.

How do I know if the function is differentiable at

x = 4

Differentiability is a stricter condition than continuity. Even if a function is continuous (the graph is connected), it might have a 'sharp turn' or a 'kink' at the junction. To check this, you must calculate the Left-Hand Derivative (LHD) using the left branch and the Right-Hand Derivative (RHD) using the right branch. If

f^{'} (4^{-}) \neq = f^{'} (4^{+})

, the function is not differentiable at that point. In this problem, the slopes don't match, confirming the kink.

Why is the statement about monotonicity false?

Monotonicity is determined entirely by the sign of the derivative

f^{'} (x)

. We found that

f^{'} (x)

changes sign at

x = 1/8

and

x = 8

. By testing intervals, we discovered that the function is actually decreasing in the regions where the option claimed it was increasing. Always draw a sign chart for

f^{'} (x)

to visualize where the function is climbing (positive slope) and where it is falling (negative slope).

Select Answer:

Visualized Solution

Analyze Continuity at $x = 4$

Evaluate the Integral for RHL

Calculate the Integral Value

Solve for Constant $b$

Find the Derivative $f^{'} (x)$ for $x \leq 4$

Find the Derivative $f^{'} (x)$ for $x > 4$

Check Differentiability at $x = 4$

Verify Option (B): $f^{'} (3) + f^{'} (5)$

Analyze Monotonicity for $x \leq 4$

Analyze Monotonicity for $x > 4$

Identify the False Statement

Final Check: Local Minima at $x = 1/8$

The Sigma Insight: Relationship Between Continuity and Differentiability

Analyzing the Setup

The Bridge of Continuity

The Derivative's Dance

The Monotonicity Trap

Similar Questions

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

Let $f : [0, \infty) \to [0, 3]$ be a function defined by $f (x) = {max {sin t : 0 \leq t \leq x}, 2 + cos x, 0 \leq x \leq π x > π$ Then which of the following is true ?

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

The function $f (x) = {∣ x - 3∣, \frac{x ^{2}}{4} - \frac{3 x}{2} + \frac{13}{4}, x \geq 1 x < 1$ is

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

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

A function $f$ is defined on $[- 3, 3]$ as $f (x) = {min {∣ x ∣, 2 - x^{2}}, [∣ x ∣], - 2 \leq x \leq 2 2 < ∣ x ∣ \leq 3$ where $[x]$ denotes the greatest integer $\leq x$ . The number of points, where $f$ is not differentiable in $(- 3, 3)$ is

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

The function $f (x) = {\frac{π}{4} + tan^{- 1} x, \frac{1}{2} (∣ x ∣ - 1), ∣ x ∣ \leq 1 ∣ x ∣ > 1$ is:

Select Answer:

Visualized Solution

Analyze Continuity at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=4

Evaluate the Integral for RHL

Calculate the Integral Value

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

Find the Derivative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x) for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x≤4

Find the Derivative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x) for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x>4

Check Differentiability at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=4

Verify Option (B): .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(3)+f′(5)

Analyze Monotonicity for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x≤4

Analyze Monotonicity for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x>4

Identify the False Statement

Final Check: Local Minima at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=1/8

The Sigma Insight: Relationship Between Continuity and Differentiability

Analyzing the Setup

The Bridge of Continuity

The Derivative's Dance

The Monotonicity Trap

Similar Questions

.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:[0,∞)→[0,3] be a function defined by f(x)={max{sint:0≤t≤x},2+cosx,​0≤x≤πx>π​ Then which of the following is true ?

.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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The function f(x)={∣x−3∣,4x2​−23x​+413​,​x≥1x<1​ is

.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)=min{1,x2,x3}, then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The function f(x)={4π​+tan−1x,21​(∣x∣−1),​∣x∣≤1∣x∣>1​ is:

Analyze Continuity at $x = 4$

Solve for Constant $b$

Find the Derivative $f^{'} (x)$ for $x \leq 4$

Find the Derivative $f^{'} (x)$ for $x > 4$

Check Differentiability at $x = 4$

Verify Option (B): $f^{'} (3) + f^{'} (5)$

Analyze Monotonicity for $x \leq 4$

Analyze Monotonicity for $x > 4$

Final Check: Local Minima at $x = 1/8$

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

Let $f : [0, \infty) \to [0, 3]$ be a function defined by $f (x) = {max {sin t : 0 \leq t \leq x}, 2 + cos x, 0 \leq x \leq π x > π$ Then which of the following is true ?

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

The function $f (x) = {∣ x - 3∣, \frac{x ^{2}}{4} - \frac{3 x}{2} + \frac{13}{4}, x \geq 1 x < 1$ is

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

The function $f (x) = {\frac{π}{4} + tan^{- 1} x, \frac{1}{2} (∣ x ∣ - 1), ∣ x ∣ \leq 1 ∣ x ∣ > 1$ is: