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Functions — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question

Let

f: \mathbb{R} \to \mathbb{R}

be defined as

f(x) = x^3 + x + 1

1 \leq x \leq 2

. The graph of

y = g(x)

is the reflection of the graph of

y = f(x)

through the line

y = x

. If the domain of

g(x)

[a, b]

, then

|a-b|

is.

Solution

Answer: 8

Step 1: Find the range of

f

f'(x) = 3x^2+1 > 0

, so

f

is strictly increasing on

[1,2]

f(1) = 1+1+1 = 3, \quad f(2) = 8+2+1 = 11

Step 2: Relate domain of

g

to range of

f

Since

g

is the reflection of

f

through

y = x

g = f^{-1}

. The domain of

g

equals the range of

f

, which is

[3, 11]

a = 3, \quad b = 11 \implies |a-b| = 8

Answer: 8

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