JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

For some $a, b, c \in \mathbb{N}$ , let $f(x) = ax - 3$ and $g(x) = x^b + c, x \in \mathbb{R}$ . If $(f \circ g)^{-1}(x) = \left( \frac{x-7}{2} \right)^{\frac{1}{3}}$ then $(f \circ g)(ac) + (g \circ f)b$ is equal to

Choose the correct answer:

Explanation

Solution:

1. Find $(f \circ g)(x)$ :

(f \circ g)(x) = f(g(x)) = a(x^b + c) - 3 = ax^b + ac - 3

2. Find the inverse $(f \circ g)^{-1}(x)$ :

Let $y = ax^b + ac - 3$ . Solve for $x$ :

y + 3 - ac = ax^b

x^b = \frac{y + 3 - ac}{a}

x = \left( \frac{y + 3 - ac}{a} \right)^{\frac{1}{b}}

So, $(f \circ g)^{-1}(x) = \left( \frac{x + 3 - ac}{a} \right)^{\frac{1}{b}}$

3. Compare with the given inverse:

Given: $(f \circ g)^{-1}(x) = \left( \frac{x - 7}{2} \right)^{\frac{1}{3}}$

By comparison:

$b = 3$
$a = 2$
$3 - ac = -7 \implies ac = 10$

Since $a = 2$ , then $2c = 10 \implies c = 5$ .

So, $a=2, b=3, c=5$ .

4. Calculate the required value:

$ac = 2 \times 5 = 10$
$(f \circ g)(10) = 2(10^3) + 10 - 3 = 2000 + 7 = 2007$
$(g \circ f)(b) = (g \circ f)(3) = g(f(3)) = g(2(3) - 3) = g(3) = 3^3 + 5 = 27 + 5 = 32$

Final Result:

(f \circ g)(ac) + (g \circ f)b = 2007 + 32 = 2039

Choose the correct answer:

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