JEE Main 2025 (January)

LEVELJEE Main

Animated Solution for Mathematics - Differentiation: Let $f : [0, 3] \to A$ be defined by $f (x) = 2 x^{3} - 15 x^{2} + 36 x + 7$ and $g : [0, \infty) \to B$ be defined by $g (x) = \frac{x ^{2025}}{x ^{2025} + 1} .$ If both the functions are onto and $S = {x \in Z : x \in A \cup B}$ , then $n (S)$ is equal to:

Q: Why is the range of $g(x)$ not $[0, 1]$?

Great question! Look at the function $g(x) = \frac{x^{2025}}{x^{2025}+1}$. At $x=0$, $g(0)=0$. As $x$ approaches infinity, the fraction $\frac{1}{x^{2025}+1}$ approaches $0$, so $g(x)$ approaches $1$. However, $g(x)$ never actually reaches $1$ for any finite $x$. Thus, the range is $[0, 1)$, not $[0, 1]$.

Q: How do I count the number of integers in $[7, 35]$?

A common mistake is to just do $35-7$, which is $28$. But you must include both endpoints! The correct formula for the number of integers in $[a, b]$ is $b-a+1$. So, $35-7+1 = 29$.

Select Answer:

Progressive Hints

Hint 1

The Spark: The term 'onto' is your best friend here. It tells you that the codomain is exactly the range. Your mission is to find the range of two distinct functions,

f (x)

and

g (x)

, and then find the number of integers in their union. Don't let the complexity of the expressions intimidate you; focus on the boundaries.

Hint 2

The Arsenal: For the cubic function

f (x)

, you need to find the absolute extrema on the closed interval

[0, 3]

. This requires finding the derivative

f^{'} (x)

, identifying critical points, and evaluating the function at those points and the endpoints. For the rational function

g (x)

, analyze its behavior as

x

ranges from

0

\infty

. Rewrite it to see the horizontal asymptote.

Hint 3

The Execution: Start by calculating

f (0)

f (2)

, and

f (3)

to determine the range

A

. Then, rewrite

g (x)

1 - \frac{1}{x ^{2025} + 1}

to determine the range

B

. Finally, identify the integers in the union

A \cup B

and count them carefully.

Visualized Solution

Objective: Find Range of $f (x)$ and $g (x)$

Given $f : [0, 3] \to A$ is onto, so $A = Range of f$ .
Given $g : [0, \infty) \to B$ is onto, so $B = Range of g$ .
We need to find $n (S)$ where $S = {x \in Z : x \in A \cup B}$ .

Differentiating $f (x)$

$f (x) = 2 x^{3} - 15 x^{2} + 36 x + 7$
Differentiating with respect to $x$ :
$f^{'} (x) = 6 x^{2} - 30 x + 36$

Finding Critical Points

$f^{'} (x) = 6 (x^{2} - 5 x + 6)$
$f^{'} (x) = 6 (x - 2) (x - 3)$
For critical points, $f^{'} (x) = 0 \Rightarrow x = 2, 3$ .

Evaluating $f (x)$ at $x = 0$

At $x = 0$ :
$f (0) = 2 (0)^{3} - 15 (0)^{2} + 36 (0) + 7 = 7$

Evaluating $f (x)$ at $x = 2$

At $x = 2$ :
$f (2) = 2 (8) - 15 (4) + 36 (2) + 7$
$f (2) = 16 - 60 + 72 + 7 = 35$

Evaluating $f (x)$ at $x = 3$

At $x = 3$ :
$f (3) = 2 (27) - 15 (9) + 36 (3) + 7$
$f (3) = 54 - 135 + 108 + 7 = 34$

Determining Range $A$

Comparing values: $f (0) = 7, f (2) = 35, f (3) = 34$ .
Minimum value $= 7$ , Maximum value $= 35$ .
Range $A = [7, 35]$ .

Analyzing Function $g (x)$

$g (x) = \frac{x ^{2025}}{x ^{2025} + 1}$ for $x \in [0, \infty)$
Rewriting: $g (x) = \frac{x ^{2025} + 1 - 1}{x ^{2025} + 1} = 1 - \frac{1}{x ^{2025} + 1}$

Determining Range $B$

At $x = 0$ , $g (0) = 1 - \frac{1}{0 + 1} = 0$ .
As $x \to \infty$ , $g (x) \to 1 - 0 = 1$ .
Since $g (x)$ is strictly increasing for $x \geq 0$ ,
Range $B = [0, 1)$ .

Finding the Union $A \cup B$

$A \cup B = [0, 1) \cup [7, 35]$
We need to find integers $x$ such that $x \in [0, 1)$ or $x \in [7, 35]$ .

Identifying Integers in $S$

Integers in $[0, 1)$ : ${0}$ (Total 1 integer)
Integers in $[7, 35]$ : ${7, 8, 9, \dots, 35}$

Final Count $n (S)$

Number of integers in $[7, 35] = 35 - 7 + 1 = 29$
Total number of integers $n (S) = 1 + 29 = 30$
Final Answer: 30

00:00 / 00:00

The Sigma Insight: Maxima and Minima

Solution Diagram

Analyzing the Cubic Function $f (x)$

We are given the function

f (x) = 2 x^{3} - 15 x^{2} + 36 x + 7

defined on the interval

[0, 3]

. Since

f (x)

is a continuous polynomial, we determine its range by evaluating the function at its critical points and the boundaries of the interval.

First, we compute the derivative to locate the critical points:

f^{'} (x) = 6 x^{2} - 30 x + 36

Setting the derivative to zero, we solve for

x

6 (x^{2} - 5 x + 6) = 0

6 (x - 2) (x - 3) = 0

The critical points are

x = 2

and

x = 3

. We now evaluate

f (x)

at the endpoints

x = 0, 3

and the critical point

x = 2

f (0) = 2 (0)^{3} - 15 (0)^{2} + 36 (0) + 7 = 7

f (2) = 2 (8) - 15 (4) + 36 (2) + 7 = 16 - 60 + 72 + 7 = 35

f (3) = 2 (27) - 15 (9) + 36 (3) + 7 = 54 - 135 + 108 + 7 = 34

Comparing these values, the minimum value is

7

and the maximum value is

35

. Thus, the range

A

[7, 35]

Decoding the Rational Function $g (x)$

Next, we examine

g (x) = \frac{x ^{2025}}{x ^{2025} + 1}

for

x \in [0, \infty)

. We simplify the expression to understand its behavior:

g (x) = \frac{x ^{2025} + 1 - 1}{x ^{2025} + 1} = 1 - \frac{1}{x ^{2025} + 1}

At the lower bound

x = 0

, we find:

g (0) = 1 - \frac{1}{0 + 1} = 0

x \to \infty

, the term

\frac{1}{x ^{2025} + 1}

approaches

0

, meaning

g (x)

approaches

1

. Since the function is strictly increasing on the given domain, the range

B

is the interval

[0, 1)

The Final Count

Uniting the Sets

We have identified the two sets as

A = [7, 35]

and

B = [0, 1)

. We seek the number of integers in their union

S = A \cup B

The integers contained in the interval

B = [0, 1)

consist solely of the integer

{0}

. The integers contained in the interval

A = [7, 35]

are

{7, 8, 9, \dots, 35}

To count the number of integers in

A

, we use the formula

n = b - a + 1

35 - 7 + 1 = 29

Including the integer

0

from set

B

, the total number of integers in the union is:

29 + 1 = 30

The total number of integers in the union is 30.

Common Traps & FAQs

Why is the range of

g (x)

not

[0, 1]

Great question! Look at the function

g (x) = \frac{x ^{2025}}{x ^{2025} + 1}

. At

x = 0

g (0) = 0

. As

x

approaches infinity, the fraction

\frac{1}{x ^{2025} + 1}

approaches

0

, so

g (x)

approaches

1

. However,

g (x)

never actually reaches

1

for any finite

x

. Thus, the range is

[0, 1)

, not

[0, 1]

Do I really need to check the endpoints for

f (x)

Absolutely! When finding the range of a continuous function on a closed interval

[a, b]

, the absolute maximum or minimum can occur at the critical points *or* at the endpoints. Skipping the endpoints is a classic trap that will lead you to the wrong range.

How do I count the number of integers in

[7, 35]

A common mistake is to just do

35 - 7

, which is

28

. But you must include both endpoints! The correct formula for the number of integers in

[a, b]

b - a + 1

. So,

35 - 7 + 1 = 29

Select Answer:

Visualized Solution

Objective: Find Range of $f (x)$ and $g (x)$

Differentiating $f (x)$

Finding Critical Points

Evaluating $f (x)$ at $x = 0$

Evaluating $f (x)$ at $x = 2$

Evaluating $f (x)$ at $x = 3$

Determining Range $A$

Analyzing Function $g (x)$

Determining Range $B$

Finding the Union $A \cup B$

Identifying Integers in $S$

Final Count $n (S)$

The Sigma Insight: Maxima and Minima

Analyzing the Cubic Function $f (x)$

Decoding the Rational Function $g (x)$

The Final Count

Similar Questions

Let $f (x) = x^{2025} - x^{2000}, x \in [0, 1]$ and the minimum value of the function $f (x)$ in the interval $[0, 1]$ be $(80)^{80} (n)^{- 81}$ . Then $n$ is equal to

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Let $f (x) = ∣ (x - 1) (x^{2} - 2 x - 3) ∣ + x - 3, x \in R$ . If m and M are respectively the number of points of local minimum and local maximum of f in the interval (0, 4), then m + M is equal to ______

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Let $f : R \to R$ be a function defined by $f (x) = (x - 3)^{n_{1}} (x - 5)^{n_{2}}, n_{1}, n_{2} \in N$ . Then, which of the following is NOT true?

Select Answer:

Visualized Solution

Objective: Find Range of .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)

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

Finding Critical Points

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Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x) at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=3

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

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

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

Finding the Union .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A∪B

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

Final Count .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n(S)

The Sigma Insight: Maxima and Minima

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

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

The Final Count

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=x2025−x2000,x∈[0,1] and the minimum value of the function f(x) in the interval [0,1] be (80)80(n)−81. Then n is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=∫0x​t(t2−9t+20)dt, 1≤x≤5 If the range of f is [α,β], then 4(α+β) equals

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be a function defined by f(x)=(x−3)n1​(x−5)n2​,n1​,n2​∈N. Then, which of the following is NOT true?

Objective: Find Range of $f (x)$ and $g (x)$

Differentiating $f (x)$

Evaluating $f (x)$ at $x = 0$

Evaluating $f (x)$ at $x = 2$

Evaluating $f (x)$ at $x = 3$

Determining Range $A$

Analyzing Function $g (x)$

Determining Range $B$

Finding the Union $A \cup B$

Identifying Integers in $S$

Final Count $n (S)$

Analyzing the Cubic Function $f (x)$

Decoding the Rational Function $g (x)$

Let $f (x) = x^{2025} - x^{2000}, x \in [0, 1]$ and the minimum value of the function $f (x)$ in the interval $[0, 1]$ be $(80)^{80} (n)^{- 81}$ . Then $n$ is equal to

Let $f (x) = \int_{0}^{x} t (t^{2} - 9 t + 20) d t$ , $1 \leq x \leq 5$ If the range of f is $[α, β]$ , then $4 (α + β)$ equals

The maximum value of the function $f (x) = 2 x^{3} - 15 x^{2} + 36 x - 48$ on the set $A = {x ∣ x^{2} + 20 \leq 9 x}$ is

The total number of local maxima and local minima of the function $f (x) = {(2 + x)^{3}, x^{2/3}, - 3 < x \leq - 1 - 1 < x < 2$ is

Let $f : R \to R$ be a function defined by $f (x) = (x - 3)^{n_{1}} (x - 5)^{n_{2}}, n_{1}, n_{2} \in N$ . Then, which of the following is NOT true?