JEE Main 2024 (31 Jan Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Differentiation: If the function $f : (- \infty, - 1] \to (a, b]$ defined by $f (x) = e^{x^{3} - 3 x + 1}$ is one-one and onto, then the distance of the point $P (2 b + 4, a + 2)$ from the line $x + e^{- 3} y = 4$ is :

Q: I got $a=0$ and $b=e^3$. Is the range $(0, e^3]$ or $[0, e^3)$?

Look closely at the domain $(-\infty, -1]$. As $x \to -\infty$, the function approaches $0$ but never touches it (hence the open parenthesis). At $x = -1$, the function is defined, so we include $e^3$ (hence the closed bracket). Always respect the boundaries of your domain!

Select Answer:

Progressive Hints

Hint 1

To unlock the secrets of

a

and

b

, you must first realize that for a function to be 'onto', its range must equal its codomain. Since the domain is

(- \infty, - 1]

, your first mission is to determine the behavior of

f (x)

on this specific interval. Is it climbing or falling?

Hint 2

Don't guess! Use the derivative

f^{'} (x) = 3 e^{x^{3} - 3 x + 1} (x^{2} - 1)

. Analyze the sign of

(x^{2} - 1)

within the domain

(- \infty, - 1]

. If the derivative is always positive, the function is strictly increasing, which makes finding the range a simple matter of evaluating the endpoints.

Hint 3

Once you have

a

and

b

, you are dealing with a standard coordinate geometry problem. For the line

x + e^{- 3} y = 4

, multiply the entire equation by

e^{3}

to clear the negative exponent. Now, apply the perpendicular distance formula:

d = \frac{∣ A x _{1} + B y _{1} + C ∣}{A ^{2} + B ^{2}}

Visualized Solution

Introduction to Function $f (x)$

Function: $f (x) = e^{x^{3} - 3 x + 1}$
Domain: $D_{f} = (- \infty, - 1]$
Codomain: $C_{f} = (a, b]$
Since $f$ is one-one and onto, Range of $f = (a, b]$

Differentiating the Function

To find the range, we check the monotonicity using $f^{'} (x)$
Apply Chain Rule: $f^{'} (x) = e^{x^{3} - 3 x + 1} \cdot \frac{d}{d x} (x^{3} - 3 x + 1)$
$f^{'} (x) = e^{x^{3} - 3 x + 1} \cdot (3 x^{2} - 3)$
$f^{'} (x) = 3 e^{x^{3} - 3 x + 1} (x^{2} - 1)$

Analyzing Monotonicity

$f^{'} (x) = 3 e^{x^{3} - 3 x + 1} (x - 1) (x + 1)$
For $x \leq - 1$ , $(x - 1) < 0$ and $(x + 1) \leq 0$
Thus, $(x - 1) (x + 1) \geq 0$
Since $e^{g (x)} > 0$ , we have $f^{'} (x) \geq 0$ for all $x \in (- \infty, - 1]$

Finding the Lower Bound $a$

As $x \to - \infty$ , the exponent $(x^{3} - 3 x + 1) \to - \infty$
Lower bound $a = lim_{x \to - \infty} e^{x^{3} - 3 x + 1} = 0$
So, $a = 0$

Finding the Upper Bound $b$

Upper bound occurs at $x = - 1$
$b = f (- 1) = e^{(- 1)^{3} - 3 (- 1) + 1}$
$b = e^{- 1 + 3 + 1} = e^{3}$
Range of $f = (0, e^{3}] = (a, b]$

Coordinates of Point $P$

Given point $P (2 b + 4, a + 2)$
Substitute $a = 0$ and $b = e^{3}$ :
$P (2 e^{3} + 4, 0 + 2) = P (2 e^{3} + 4, 2)$

Simplifying the Line Equation

Line equation: $x + e^{- 3} y = 4$
Multiply by $e^{3}$ to simplify:
$e^{3} x + y = 4 e^{3}$
Standard form: $e^{3} x + y - 4 e^{3} = 0$

Applying the Distance Formula

Distance $d = \frac{∣ A x _{1} + B y _{1} + C ∣}{A ^{2} + B ^{2}}$
Substitute $A = e^{3}, B = 1, C = - 4 e^{3}$ and $P (2 e^{3} + 4, 2)$ :
$d = \frac{∣ e ^{3} ( 2 e ^{3} + 4 ) + 1 ( 2 ) - 4 e ^{3} ∣}{( e ^{3} ) ^{2} + 1 ^{2}}$

Simplifying the Numerator

Numerator: $∣2 e^{6} + 4 e^{3} + 2 - 4 e^{3} ∣$
Cancel $4 e^{3}$ and $- 4 e^{3}$ :
Numerator: $∣2 e^{6} + 2∣ = 2 (e^{6} + 1)$

Final Calculation

Distance $d = \frac{2 ( e ^{6} + 1 )}{e ^{6} + 1}$
Using $\frac{x}{x} = x$ :
$d = 2 e^{6} + 1$

00:00 / 00:00

The Sigma Insight: Monotonicity

Solution Diagram

Analyzing the Function Behavior

We are given the function

f (x) = e^{x^{3} - 3 x + 1}

defined on the domain

(- \infty, - 1]

. To determine the range

(a, b]

, we first analyze the monotonicity of the function using its derivative.

Applying the chain rule, we find:

f^{'} (x) = e^{x^{3} - 3 x + 1} \cdot \frac{d}{d x} (x^{3} - 3 x + 1) = 3 e^{x^{3} - 3 x + 1} (x^{2} - 1)

For any

x \in (- \infty, - 1]

, the term

(x^{2} - 1)

is non-negative. Since the exponential term

e^{x^{3} - 3 x + 1}

is always positive, the derivative

f^{'} (x) \geq 0

throughout the domain. This confirms that the function is strictly increasing.

Determining the Range

Because the function is strictly increasing, the minimum value occurs as

x \to - \infty

and the maximum value occurs at the right boundary

x = - 1

x \to - \infty

, the exponent

(x^{3} - 3 x + 1) \to - \infty

, which implies:

a = x \to - \infty lim e^{x^{3} - 3 x + 1} = 0

At the upper boundary

x = - 1

b = f (- 1) = e^{(- 1)^{3} - 3 (- 1) + 1} = e^{- 1 + 3 + 1} = e^{3}

Thus, the range of the function is

(0, e^{3}]

, identifying our constants as

a = 0

and

b = e^{3}

Geometry and Distance Calculation

We are given a point

P (2 b + 4, a + 2)

. Substituting our values for

a

and

b

, the coordinates of

P

are:

P (2 e^{3} + 4, 2)

We seek the perpendicular distance from

P

to the line

x + e^{- 3} y = 4

. Multiplying by

e^{3}

to clear the fraction, the equation of the line becomes:

e^{3} x + y - 4 e^{3} = 0

Using the perpendicular distance formula

d = \frac{∣ A x _{1} + B y _{1} + C ∣}{A ^{2} + B ^{2}}

, we substitute

A = e^{3}

B = 1

C = - 4 e^{3}

x_{1} = 2 e^{3} + 4

, and

y_{1} = 2

d = \frac{∣ e ^{3} ( 2 e ^{3} + 4 ) + 1 ( 2 ) - 4 e ^{3} ∣}{( e ^{3} ) ^{2} + 1 ^{2}}

Final Calculation

Expanding the numerator, we observe the cancellation of terms:

∣2 e^{6} + 4 e^{3} + 2 - 4 e^{3} ∣ = ∣2 e^{6} + 2∣ = 2 (e^{6} + 1)

The denominator is

e^{6} + 1

. Therefore, the distance is:

d = \frac{2 ( e ^{6} + 1 )}{e ^{6} + 1}

Simplifying the expression, we arrive at the final result: $d = 2 e^{6} + 1$

Common Traps & FAQs

Wait, why do we assume the function is increasing? What if it has a local maximum inside the interval?

That is a brilliant question! In JEE Advanced, never assume. We checked the derivative

f^{'} (x) = 3 e^{x^{3} - 3 x + 1} (x^{2} - 1)

. For

x \leq - 1

, both

(x - 1)

and

(x + 1)

are negative (or zero at

- 1

), so their product is positive. Because the derivative is non-negative, the function is strictly increasing. No local maxima exist in this interval!

Why did we multiply the line equation by

e^{3}

It's not strictly necessary, but it is a mark of a seasoned problem solver. Working with

e^{- 3}

in the distance formula makes the arithmetic messy and prone to errors. Multiplying by

e^{3}

clears the fraction and gives us integer-like coefficients, making the final cancellation of terms much more obvious.

I got

a = 0

and

b = e^{3}

. Is the range

(0, e^{3}]

[0, e^{3})

Look closely at the domain

(- \infty, - 1]

. As

x \to - \infty

, the function approaches

0

but never touches it (hence the open parenthesis). At

x = - 1

, the function is defined, so we include

e^{3}

(hence the closed bracket). Always respect the boundaries of your domain!

Animated Solution for Mathematics - Differentiation: If the function $f : (- \infty, - 1] \to (a, b]$ defined by $f (x) = e^{x^{3} - 3 x + 1}$ is one-one and onto, then the distance of the point $P (2 b + 4, a + 2)$ from the line $x + e^{- 3} y = 4$ is :

Select Answer:

Visualized Solution

Introduction to Function $f (x)$

Differentiating the Function

Analyzing Monotonicity

Finding the Lower Bound $a$

Finding the Upper Bound $b$

Coordinates of Point $P$

Simplifying the Line Equation

Applying the Distance Formula

Simplifying the Numerator

Final Calculation

The Sigma Insight: Monotonicity

Analyzing the Function Behavior

Determining the Range

Geometry and Distance Calculation

Final Calculation

Similar Questions

Let $f : R \to R$ be defined by $f (x) = \frac{x ^{2} - 3 x - 6}{x ^{2} + 2 x + 4}$ . Then which of the following statements is (are) TRUE ?

The function defined by $f (x) = (x + 2) e^{- x}$ is

Let $f (x) = 2 cos^{- 1} x + 4 cot^{- 1} x - 3 x^{2} - 2 x + 10, x \in [- 1, 1]$ . If [a, b] is the range of the function then 4a - b is equal to:

Let $f : R \to R$ be defined as $f (x) = {- \frac{4}{3} x^{3} + 2 x^{2} + 3 x 3 x e^{x}, x > 0, x \leq 0$ . Then $f$ is increasing function in the interval

Let $f : (0, 1) \to R$ be a function defined by $f (x) = \frac{1}{1 - e ^{- x}}$ , and $g (x) = (f (- x) - f (x))$ . Consider two statements (I) $g$ is an increasing function in $(0, 1)$ (II) $g$ is one-one in $(0, 1)$ Then,

Let $f : R \to R$ and $g : R \to R$ be two functions defined by $f (x) = lo g_{e} (x^{2} + 1) - e^{- x} + 1$ and $g (x) = \frac{1 - 2 e ^{2 x}}{e ^{x}}$ . Then, for which of the following range of $α$ , the inequality $f (g (\frac{( α - 1 ) ^{2}}{3})) > f (g (α - \frac{5}{3}))$ holds?

If the function $f$ given by $f (x) = x^{3} - 3 (a - 2) x^{2} + 3 a x + 7$ , for some $a \in R$ is increasing in $(0, 1]$ and decreasing in $[1, 5)$ , then a root of the equation $\frac{f ( x ) - 14}{( x - 1 ) ^{2}} = 0 (x \neq = 1)$ is :

If $f (x) = \int_{x^{2}}^{x^{2} + 1} e^{- t^{2}} d t$ , then $f (x)$ increases in

Select Answer:

Visualized Solution

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

Differentiating the Function

Analyzing Monotonicity

Finding the Lower Bound .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Finding the Upper Bound .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Coordinates of Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Simplifying the Line Equation

Applying the Distance Formula

Simplifying the Numerator

Final Calculation

The Sigma Insight: Monotonicity

Analyzing the Function Behavior

Determining the Range

Geometry and Distance Calculation

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be defined by f(x)=x2+2x+4x2−3x−6​. Then which of the following statements is (are) TRUE ?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The function defined by f(x)=(x+2)e−x is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=2cos−1x+4cot−1x−3x2−2x+10,x∈[−1,1]. If [a, b] is the range of the function then 4a - b is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be defined as f(x)={−34​x3+2x2+3x3xex​,x>0,x≤0​. Then f is increasing function in the interval

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=∫x2x2+1​e−t2dt, then f(x) increases in

Introduction to Function $f (x)$

Finding the Lower Bound $a$

Finding the Upper Bound $b$

Coordinates of Point $P$

Let $f : R \to R$ be defined by $f (x) = \frac{x ^{2} - 3 x - 6}{x ^{2} + 2 x + 4}$ . Then which of the following statements is (are) TRUE ?

The function defined by $f (x) = (x + 2) e^{- x}$ is

Let $f (x) = 2 cos^{- 1} x + 4 cot^{- 1} x - 3 x^{2} - 2 x + 10, x \in [- 1, 1]$ . If [a, b] is the range of the function then 4a - b is equal to:

Let $f : R \to R$ be defined as $f (x) = {- \frac{4}{3} x^{3} + 2 x^{2} + 3 x 3 x e^{x}, x > 0, x \leq 0$ . Then $f$ is increasing function in the interval

If $f (x) = \int_{x^{2}}^{x^{2} + 1} e^{- t^{2}} d t$ , then $f (x)$ increases in