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Let g(x) = f(f(x)) where f(x) = { 1 + x ...

Let `g(x) = f(f(x))` where `f(x) = { 1 + x ; 0 <=x<=2} and f(x) = {3 - x; 2 < x <= 3}` then the number of points of discontinuity of `g(x)` in [0,3] is :

A

0

B

1

C

2

D

3

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To solve the problem, we need to analyze the function \( g(x) = f(f(x)) \) where the function \( f(x) \) is defined piecewise: \[ f(x) = \begin{cases} 1 + x & \text{if } 0 \leq x \leq 2 \\ 3 - x & \text{if } 2 < x \leq 3 \end{cases} \] ### Step 1: Determine the values of \( f(x) \) 1. For \( 0 \leq x \leq 2 \): - \( f(x) = 1 + x \) - The range of \( f(x) \) when \( x \) varies from 0 to 2 is: - \( f(0) = 1 + 0 = 1 \) - \( f(2) = 1 + 2 = 3 \) - Therefore, \( f(x) \) maps \( [0, 2] \) to \( [1, 3] \). 2. For \( 2 < x \leq 3 \): - \( f(x) = 3 - x \) - The range of \( f(x) \) when \( x \) varies from 2 to 3 is: - \( f(2) = 3 - 2 = 1 \) - \( f(3) = 3 - 3 = 0 \) - Therefore, \( f(x) \) maps \( (2, 3] \) to \( (0, 1] \). ### Step 2: Determine the composition \( g(x) = f(f(x)) \) We need to evaluate \( g(x) \) for different intervals based on the piecewise definition of \( f(x) \). 1. **For \( 0 \leq x \leq 1 \)**: - \( f(x) = 1 + x \) which ranges from 1 to 2. - Therefore, \( g(x) = f(f(x)) = f(1 + x) \). - Since \( 1 + x \) is in the range \( [1, 2] \), we use the first case of \( f \): \[ g(x) = f(1 + x) = 1 + (1 + x) = 2 + x \] 2. **For \( 1 < x \leq 2 \)**: - \( f(x) = 1 + x \) which ranges from 2 to 3. - Therefore, \( g(x) = f(f(x)) = f(1 + x) \). - Since \( 1 + x \) is in the range \( [2, 3] \), we use the second case of \( f \): \[ g(x) = f(1 + x) = 3 - (1 + x) = 2 - x \] 3. **For \( 2 < x \leq 3 \)**: - \( f(x) = 3 - x \) which ranges from 1 to 0. - Therefore, \( g(x) = f(f(x)) = f(3 - x) \). - Since \( 3 - x \) is in the range \( [0, 1] \), we use the first case of \( f \): \[ g(x) = f(3 - x) = 1 + (3 - x) = 4 - x \] ### Step 3: Compile the piecewise definition of \( g(x) \) Thus, we have: \[ g(x) = \begin{cases} 2 + x & \text{if } 0 \leq x \leq 1 \\ 2 - x & \text{if } 1 < x \leq 2 \\ 4 - x & \text{if } 2 < x \leq 3 \end{cases} \] ### Step 4: Check for points of discontinuity 1. **At \( x = 1 \)**: - \( g(1) = 2 + 1 = 3 \) - Left-hand limit as \( x \to 1^- \): \( g(1^-) = 2 + 1 = 3 \) - Right-hand limit as \( x \to 1^+ \): \( g(1^+) = 2 - 1 = 1 \) - Since \( g(1^-) \neq g(1^+) \), \( g(x) \) is discontinuous at \( x = 1 \). 2. **At \( x = 2 \)**: - \( g(2) = 2 - 2 = 0 \) - Left-hand limit as \( x \to 2^- \): \( g(2^-) = 2 - 2 = 0 \) - Right-hand limit as \( x \to 2^+ \): \( g(2^+) = 4 - 2 = 2 \) - Since \( g(2^-) \neq g(2^+) \), \( g(x) \) is discontinuous at \( x = 2 \). ### Conclusion The points of discontinuity of \( g(x) \) in the interval \([0, 3]\) are at \( x = 1 \) and \( x = 2 \). Therefore, the number of points of discontinuity is **2**.
To solve the problem, we need to analyze the function \( g(x) = f(f(x)) \) where the function \( f(x) \) is defined piecewise: \[ f(x) = \begin{cases} 1 + x & \text{if } 0 \leq x \leq 2 \\ 3 - x & \text{if } 2 < x \leq 3 \end{cases} ...
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