arthham

JEE Advanced 2021 · Continuity and Differentiability

Let S=(0,1)∪(1,2)∪(3,4)S = (0, 1) \cup (1, 2) \cup (3, 4) and T={0,1,2,3}T = \{0, 1, 2, 3\}. Then which of the following statements is(are) true?

Answers: A. There are infinitely many functions from SS to TT · C. The number of continuous functions from SS to TT is at most 120120 · D. Every continuous function from SS to TT is differentiable

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S=(0,1)∪(1,2)∪(3,4)S=(0,1)\cup(1,2)\cup(3,4) T={0,1,2,3}T=\{0,1,2,3\} (1) infinitely many functions S→T(1)\ \text{infinitely many functions } S\to T (2) infinitely many strictly increasing S→T(2)\ \text{infinitely many strictly increasing } S\to T (3) #{continuous}≤120;(4) every continuous is differentiable(3)\ \#\{\text{continuous}\}\le 120;\quad (4)\ \text{every continuous is differentiable} S has 3 components: I1=(0,1), I2=(1,2), I3=(3,4)S \text{ has } 3 \text{ components: } I_1=(0,1),\ I_2=(1,2),\ I_3=(3,4) ∣T∣=4,T finite (discrete)|T|=4,\quad T \text{ finite (discrete)} (1) #{f:S→T}=4∣S∣,∣S∣=∞(1)\ \#\{f:S\to T\}=4^{|S|},\quad |S|=\infty ⇒(1) TRUE\Rightarrow (1)\ \textbf{TRUE} (2) x<y⇒f(x)<f(y),f(x)∈T(2)\ x<y \Rightarrow f(x)<f(y),\quad f(x)\in T an increasing chain needs distinct values in T\text{an increasing chain needs distinct values in } T x1<⋯<x5⇒f(x1)<⋯<f(x5)x_1<\dots<x_5 \Rightarrow f(x_1)<\dots<f(x_5) needs 5 distinct values, but ∣T∣=4\text{needs } 5 \text{ distinct values, but } |T|=4 #{increasing}=0 ⇒(2) FALSE\#\{\text{increasing}\}=0\ \Rightarrow (2)\ \textbf{FALSE} continuous⇒f constant on each connected component\text{continuous} \Rightarrow f \text{ constant on each connected component} connected image of connected set is connected\text{connected image of connected set is connected} I1,I2,I3→4 choices each independentlyI_1,I_2,I_3 \to 4 \text{ choices each independently} #{continuous}=43=64\#\{\text{continuous}\}=4^{3}=64 64≤120 ⇒(3) TRUE64\le 120\ \Rightarrow (3)\ \textbf{TRUE} (4) f constant on each component⇒f′=0(4)\ f \text{ constant on each component} \Rightarrow f'=0 S open⇒every point interior⇒(4) TRUES \text{ open} \Rightarrow \text{every point interior} \Rightarrow (4)\ \textbf{TRUE} TRUE: (1), (3), (4)\text{TRUE: }(1),\ (3),\ (4) (1), (3), (4)\boxed{(1),\ (3),\ (4)}

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