JEE Main 2023 (25 January Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Functions: For some $a, b, c \in N$ , let $f (x) = a x - 3$ and $g (x) = x^{b} + c, x \in R$ . If $(f o g)^{- 1} (x) = (\frac{x - 7}{2})^{1/3}$ , then $(f o g) (a c) + (g o f) (b)$ is equal to

Q: Why does the order of composition matter so much in the final step? Why is $(f \circ g)(ac)$ different from $(g \circ f)(b)$?

Composition is not commutative! $f(g(x))$ is fundamentally different from $g(f(x))$. In the first case, $g$ acts on $x$ first, then $f$ acts on the result. In the second, $f$ acts first. Always visualize the 'flow' of the variable $x$ through the functions. Never assume $f(g(x)) = g(f(x))$ unless the functions are inverses of each other!

Enter Numerical Value:

Progressive Hints

Hint 1

The Spark: Don't get lost in the algebra yet. Remember the fundamental definition of an inverse function: if

y = (f \circ g) (x)

, then

x = (f \circ g)^{- 1} (y)

. By setting the given inverse expression equal to

y

and solving for

x

, you can 'undo' the inverse and reveal the original composite function

f (g (x))

in its raw, polynomial form.

Hint 2

The Arsenal: Once you have the polynomial form of

(f \circ g) (x)

, you must compare it to the algebraic expansion of

f (g (x))

. Remember the Principle of Identity for polynomials: if two polynomials are equal for all

x

, their coefficients for each power of

x

must be identical. This is your key to unlocking

a

b

, and

c

Hint 3

The Execution: Start by setting

(\frac{x - 7}{2})^{1/3} = y

. Cube both sides to get

\frac{x - 7}{2} = y^{3}

, then isolate

x

to find

(f \circ g) (y) = 2 y^{3} + 7

. Now, equate this to

f (g (x)) = a (x^{b} + c) - 3

and match the coefficients.

Visualized Solution

Understanding the Given Functions

Given: $f (x) = a x - 3$ and $g (x) = x^{b} + c$
Constants $a, b, c \in N$

The Composite Function and its Inverse

The composite function is $(f \circ g) (x)$
We are given its inverse: $(f \circ g)^{- 1} (x) = (\frac{x - 7}{2})^{\frac{1}{3}}$

Finding $(f \circ g) (x)$ from its Inverse

Let $y = (f \circ g)^{- 1} (x) = (\frac{x - 7}{2})^{\frac{1}{3}}$
To find the original function, we need to express $x$ in terms of $y$ .

Inverting the Inverse

Cubing both sides: $y^{3} = \frac{x - 7}{2}$
Multiply by 2: $2 y^{3} = x - 7$
Rearrange for $x$ : $x = 2 y^{3} + 7$

Expressing $(f \circ g) (x)$ Algebraically

Now, let's find $(f \circ g) (x)$ using the given definitions.
$(f \circ g) (x) = f (g (x))$
Substitute $g (x)$ : $f (x^{b} + c)$

Expanding the Composition

$f (x) = a x - 3$
$f (x^{b} + c) = a (x^{b} + c) - 3$
Expand: $(f \circ g) (x) = a x^{b} + a c - 3$

Comparing the Two Expressions

We have two forms for $(f \circ g) (x)$ :
Form 1: $2 x^{3} + 7$
Form 2: $a x^{b} + a c - 3$
Equating them: $a x^{b} + a c - 3 = 2 x^{3} + 7$

Extracting Constants $a$ and $b$

Compare coefficients of the variable term:
$a x^{b} = 2 x^{3}$
Therefore, $a = 2$ and $b = 3$

Solving for the Constant $c$

Compare the constant terms:
$a c - 3 = 7$
Substitute $a = 2$ : $2 c - 3 = 7$
$2 c = 10 ⟹ c = 5$

Verifying Natural Number Constraint

Values found: $a = 2, b = 3, c = 5$
Check constraint: $a, b, c \in N$
$2, 3, 5$ are all natural numbers. Condition satisfied!

Calculating $(f \circ g) (a c)$

We need to find $(f \circ g) (a c) + (g \circ f) (b)$
First part: $(f \circ g) (a c)$
Calculate $a c = 2 \times 5 = 10$
We know $(f \circ g) (x) = 2 x^{3} + 7$

Evaluating the First Term

Substitute $x = 10$ into $(f \circ g) (x)$ :
$(f \circ g) (10) = 2 (10)^{3} + 7$
$= 2 (1000) + 7 = 2007$

Finding $(g \circ f) (x)$

Second part: $(g \circ f) (b)$
$(g \circ f) (x) = g (f (x)) = (f (x))^{b} + c$
Substitute $f (x) = 2 x - 3$ , $b = 3$ , $c = 5$ :
$(g \circ f) (x) = (2 x - 3)^{3} + 5$

Evaluating the Second Term

We need $(g \circ f) (b)$ where $b = 3$
Substitute $x = 3$ :
$(g \circ f) (3) = (2 (3) - 3)^{3} + 5$
$= (6 - 3)^{3} + 5 = 3^{3} + 5$
$= 27 + 5 = 32$

Final Summation

Final expression: $(f \circ g) (a c) + (g \circ f) (b)$
Substitute the calculated values:
$2007 + 32 = 2039$
Final Answer: 2039

00:00 / 00:00

The Sigma Insight: Inverse of a Function

Solution Diagram

Analyzing the Setup

We are given two functions,

f (x) = a x - 3

and

g (x) = x^{b} + c

, where

a, b, c \in N

. We are also provided with the inverse of their composition:

(f \circ g)^{- 1} (x) = (\frac{x - 7}{2})^{1/3}

Our goal is to determine the value of

(f \circ g) (a c) + (g \circ f) (b)

The Inverse Mystery

To find the original composite function

(f \circ g) (x)

, we must reverse the inverse operation. Let

y = (f \circ g)^{- 1} (x)

, which implies

x = (f \circ g) (y)

Setting

y = (\frac{x - 7}{2})^{1/3}

, we cube both sides to obtain:

y^{3} = \frac{x - 7}{2}

Solving for

x

, we multiply by

2

and add

7

x = 2 y^{3} + 7

Replacing

y

with

x

to return to standard notation, we find the composite function:

(f \circ g) (x) = 2 x^{3} + 7

The Algebraic Mirror

We know that

(f \circ g) (x) = f (g (x))

. Substituting the definitions of

f

and

g

(f \circ g) (x) = f (x^{b} + c) = a (x^{b} + c) - 3 = a x^{b} + a c - 3

We now equate this to our derived expression

2 x^{3} + 7

. By comparing the coefficients and powers, we observe:

a x^{b} = 2 x^{3} \Rightarrow a = 2, b = 3

Equating the constant terms:

a c - 3 = 7

Substituting

a = 2

into the equation:

2 c - 3 = 7 \Rightarrow 2 c = 10 \Rightarrow c = 5

Thus, the constants are $a = 2$ , $b = 3$ , and $c = 5$ .

Final Calculation

First, we calculate

(f \circ g) (a c)

. Since

a c = 2 \times 5 = 10

(f \circ g) (10) = 2 (10)^{3} + 7 = 2 (1000) + 7 = 2007

Next, we calculate

(g \circ f) (b) = (g \circ f) (3)

. Given

(g \circ f) (x) = g (f (x)) = (f (x))^{b} + c

(g \circ f) (x) = (2 x - 3)^{3} + 5

Substituting

x = 3

(g \circ f) (3) = (2 (3) - 3)^{3} + 5 = (6 - 3)^{3} + 5 = 3^{3} + 5 = 27 + 5 = 32

Summing the two results:

2007 + 32 = 2039

The final value is $2039$ .

Common Traps & FAQs

Wait, why can't we just find

f^{- 1} (x)

and

g^{- 1} (x)

separately and compose them? Is

(f \circ g)^{- 1}

the same as

f^{- 1} \circ g^{- 1}

That is a classic trap! Remember the 'socks and shoes' rule: if you put on socks then shoes, to undo it, you must take off shoes then socks. So,

(f \circ g)^{- 1} = g^{- 1} \circ f^{- 1}

. Trying to compose them in the wrong order will lead you down a path of unnecessary complexity. Stick to the definition of the inverse as a whole function—it's much safer and more direct.

How can we be sure that

a = 2

and

b = 3

are the only possible values? Could there be other solutions?

Excellent question! Because we are dealing with polynomials, the coefficients are unique. If

a x^{b} + (a c - 3) = 2 x^{3} + 7

, the only way this holds for all

x

is if the power

b

matches the power

3

, and the coefficient

a

matches

2

. Since

a, b, c

are natural numbers, we don't have to worry about complex or non-integer solutions. The structure of the polynomial forces these values to be unique.

Why does the order of composition matter so much in the final step? Why is

(f \circ g) (a c)

different from

(g \circ f) (b)

Composition is not commutative!

f (g (x))

is fundamentally different from

g (f (x))

. In the first case,

g

acts on

x

first, then

f

acts on the result. In the second,

f

acts first. Always visualize the 'flow' of the variable

x

through the functions. Never assume

f (g (x)) = g (f (x))

unless the functions are inverses of each other!

Animated Solution for Mathematics - Functions: For some $a, b, c \in N$ , let $f (x) = a x - 3$ and $g (x) = x^{b} + c, x \in R$ . If $(f o g)^{- 1} (x) = (\frac{x - 7}{2})^{1/3}$ , then $(f o g) (a c) + (g o f) (b)$ is equal to

Enter Numerical Value:

Visualized Solution

Understanding the Given Functions

The Composite Function and its Inverse

Finding $(f \circ g) (x)$ from its Inverse

Inverting the Inverse

Expressing $(f \circ g) (x)$ Algebraically

Expanding the Composition

Comparing the Two Expressions

Extracting Constants $a$ and $b$

Solving for the Constant $c$

Verifying Natural Number Constraint

Calculating $(f \circ g) (a c)$

Evaluating the First Term

Finding $(g \circ f) (x)$

Evaluating the Second Term

Final Summation

The Sigma Insight: Inverse of a Function

Analyzing the Setup

The Inverse Mystery

The Algebraic Mirror

Final Calculation

Similar Questions

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Animated Solution for Mathematics - Functions: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For some a,b,c∈N, let f(x)=ax−3 and g(x)=xb+c,x∈R. If (fog)−1(x)=(2x−7​)1/3, then (fog)(ac)+(gof)(b) is equal to

Enter Numerical Value:

Visualized Solution

Understanding the Given Functions

The Composite Function and its Inverse

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (f∘g)(x) from its Inverse

Inverting the Inverse

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

Expanding the Composition

Comparing the Two Expressions

Extracting Constants .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

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

Verifying Natural Number Constraint

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

Evaluating the First Term

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

Evaluating the Second Term

Final Summation

The Sigma Insight: Inverse of a Function

Analyzing the Setup

The Inverse Mystery

The Algebraic Mirror

Final Calculation

Similar Questions

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Animated Solution for Mathematics - Functions: For some $a, b, c \in N$ , let $f (x) = a x - 3$ and $g (x) = x^{b} + c, x \in R$ . If $(f o g)^{- 1} (x) = (\frac{x - 7}{2})^{1/3}$ , then $(f o g) (a c) + (g o f) (b)$ is equal to

Finding $(f \circ g) (x)$ from its Inverse

Expressing $(f \circ g) (x)$ Algebraically

Extracting Constants $a$ and $b$

Solving for the Constant $c$

Calculating $(f \circ g) (a c)$

Finding $(g \circ f) (x)$

Let $f : N \to Y$ be a function defined as $f (x) = 4 x + 3$ where $Y = {y \in N : y = 4 x + 3 for some x \in N}$ . Show that $f$ is invertible and its inverse is

If the function $f : [1, \infty) \to [1, \infty)$ is defined by $f (x) = 2^{x (x - 1)}$ , then $f^{- 1} (x)$ is

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If $f : [1, \infty) \to [2, \infty)$ is given by $f (x) = x + 1/ x$ , then $f^{- 1} (x)$ equals

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Let $f (x) = (x + 1)^{2} - 1, x \geq - 1$ . Then the set ${x : f (x) = f^{- 1} (x)}$ is

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