JEE MainMathematics

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+4 marks−1 if incorrectNumericalPrevious-year question

Determining Composition and Inverse of Polynomial Functions

Q: How do we find f(g(x)) from its inverse function?

If h^(-1)(x) = y, then h(y) = x. Replacing y with x after solving gives the original function h(x) = (f ° g)(x).

Q: Can we directly compare coefficients of a x^b and 2 x^3?

Yes, because the equality a x^b + ac - 3 = 2x^3 + 7 must hold for all x ∈ ℝ and a, b, c are natural numbers.

For some a, b, c ∈ ℕ, let f(x) = ax - 3 and g(x) = x^b + c, x ∈ ℝ. If (f ° g)^(-1)(x) = ((x-7)/2)^(1/3), then (f ° g)(ac) + (g ° f)(b) is equal to

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Step-by-step solutionView

Correct answer

The value of $(f \circ g) (a c) + (g \circ f) (b)$ is 2039.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

$a, b, c \in N$ , $f (x) = a x - 3$ , $g (x) = x^{b} + c$ , and $(f \circ g)^{- 1} (x) = {(\frac{x - 7}{2})}^{1 / 3}$ .

02Goal

Find the value of $(f \circ g) (a c) + (g \circ f) (b)$ .

03Approach

Invert $(f \circ g)^{- 1} (x)$ to find the explicit formula for $(f \circ g) (x)$ . Then equate it to $f (g (x)) = a (x^{b} + c) - 3$ to determine the natural numbers $a, b, c$ . Finally, evaluate $(f \circ g) (a c)$ and $(g \circ f) (b)$ and compute their sum.

04Execute

Let $y = (f \circ g)^{- 1} (x) = {(\frac{x - 7}{2})}^{1 / 3}$ . Inverting gives $y^{3} = \frac{x - 7}{2} ⟹ x = 2 y^{3} + 7$ . Hence, $(f \circ g) (x) = 2 x^{3} + 7$ .

05Execute

Compute $f (g (x))$ from definitions: $f (g (x)) = a (x^{b} + c) - 3 = a x^{b} + a c - 3$ . Equating this to $2 x^{3} + 7$ for all $x$ : $a x^{b} + (a c - 3) = 2 x^{3} + 7$ . By comparing coefficients and exponents: $a = 2$ , $b = 3$ , and $a c - 3 = 7 ⟹ 2 c = 10 ⟹ c = 5$ . Since $a, b, c \in N$ , these values are valid. Then $a c = 2 \times 5 = 10$ .

06Execute

Calculate the two terms: First, $(f \circ g) (a c) = (f \circ g) (10) = 2 (10)^{3} + 7 = 2 (1000) + 7 = 2007$ . Second, evaluate $(g \circ f) (b) = g (f (3))$ . Since $f (3) = a (3) - 3 = 2 (3) - 3 = 3$ , we have $g (f (3)) = g (3) = 3^{b} + c = 3^{3} + 5 = 27 + 5 = 32$ .

07Execute

Sum the two values: $(f \circ g) (a c) + (g \circ f) (b) = 2007 + 32 = 2039$ .

✓Verify

Check: $f (x) = 2 x - 3$ , $g (x) = x^{3} + 5$ . $(f \circ g) (x) = 2 (x^{3} + 5) - 3 = 2 x^{3} + 7$ . Its inverse is $y = 2 x^{3} + 7 ⟹ x = {(\frac{y - 7}{2})}^{1 / 3}$ , which matches the stem exactly.

Hints that build this answer step by step

What is the explicit expression for the composition $(f \circ g) (x)$ in terms of $a, b,$ and $c$ ?
$(f \circ g) (x) = a (x^{b} + c) - 3 = a x^{b} + a c - 3$
Given $(f \circ g)^{- 1} (x) = {(\frac{x - 7}{2})}^{1 / 3}$ , what are the values of the natural numbers $a, b,$ and $c$ ?
$a = 2, b = 3, c = 5$
Using $a = 2, b = 3, c = 5$ , what is the numerical value of $(f \circ g) (a c) + (g \circ f) (b)$ ?
2039

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

Quick checks

Students also ask

How do we find

f (g (x))

from its inverse function?

If $h^{- 1} (x) = y$ , then $h (y) = x$ . Replacing $y$ with $x$ after solving gives the original function $h (x) = (f \circ g) (x)$ .

Can we directly compare coefficients of

a x^{b}

and

2 x^{3}

Yes, because the equality $a x^{b} + a c - 3 = 2 x^{3} + 7$ must hold for all $x \in R$ and $a, b, c$ are natural numbers.

Answer

The value of $(f \circ g) (a c) + (g \circ f) (b)$ is 2039.

Why each option works or fails

Step-by-step solution

Given: $a, b, c \in N$ , $f (x) = a x - 3$ , $g (x) = x^{b} + c$ , and $(f \circ g)^{- 1} (x) = {(\frac{x - 7}{2})}^{1 / 3}$ .
Goal: Find the value of $(f \circ g) (a c) + (g \circ f) (b)$ .
Approach: Invert $(f \circ g)^{- 1} (x)$ to find the explicit formula for $(f \circ g) (x)$ . Then equate it to $f (g (x)) = a (x^{b} + c) - 3$ to determine the natural numbers $a, b, c$ . Finally, evaluate $(f \circ g) (a c)$ and $(g \circ f) (b)$ and compute their sum.
Execute: Let $y = (f \circ g)^{- 1} (x) = {(\frac{x - 7}{2})}^{1 / 3}$ . Inverting gives $y^{3} = \frac{x - 7}{2} ⟹ x = 2 y^{3} + 7$ . Hence, $(f \circ g) (x) = 2 x^{3} + 7$ .
Execute: Compute $f (g (x))$ from definitions: $f (g (x)) = a (x^{b} + c) - 3 = a x^{b} + a c - 3$ . Equating this to $2 x^{3} + 7$ for all $x$ : $a x^{b} + (a c - 3) = 2 x^{3} + 7$ . By comparing coefficients and exponents: $a = 2$ , $b = 3$ , and $a c - 3 = 7 ⟹ 2 c = 10 ⟹ c = 5$ . Since $a, b, c \in N$ , these values are valid. Then $a c = 2 \times 5 = 10$ .
Execute: Calculate the two terms: First, $(f \circ g) (a c) = (f \circ g) (10) = 2 (10)^{3} + 7 = 2 (1000) + 7 = 2007$ . Second, evaluate $(g \circ f) (b) = g (f (3))$ . Since $f (g (x)) = a (x^{b} + c) - 3 = a x^{b} + a c - 3$ , we have $g (f (3)) = g (3) = 3^{b} + c = 3^{3} + 5 = 27 + 5 = 32$ .
Execute: Sum the two values: $(f \circ g) (a c) + (g \circ f) (b) = 2007 + 32 = 2039$ .
Verify: Check: $f (x) = 2 x - 3$ , $g (x) = x^{3} + 5$ . $(f \circ g) (x) = 2 (x^{3} + 5) - 3 = 2 x^{3} + 7$ . Its inverse is $y = 2 x^{3} + 7 ⟹ x = {(\frac{y - 7}{2})}^{1 / 3}$ , which matches the stem exactly.

Determining Composition and Inverse of Polynomial Functions

What feels right?

The value of $(f \circ g) (a c) + (g \circ f) (b)$ is 2039.

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

07Execute

✓Verify

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