Let R be the set of all real numbers. Let f: | vedclass

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(A) Given $f(x) = \begin{cases} 2x-5 & x < -3 \ x+2 & -3 \leq x < 5 \ 3x+1 & x \geq 5 \end{cases}$
$(A) f(-5)+f(0)+f(-1) = (2(-5)

Vedclass · Accepted Answer

$A-IV, B-V, C-III, D-I$

Vedclass · Answer

(A) Given $f(x) = \begin{cases} 2x-5 & x < -3 \ x+2 & -3 \leq x < 5 \ 3x+1 & x \geq 5 \end{cases}$
$(A) f(-5)+f(0)+f(-1) = (2(-5)-5) + (0+2) + (-1+2) = -15 + 2 + 1 = -12$. Thus $(A) ightarrow (IV)$.
$(B) f(f(5)+10f(-3)) = f((3(5)+1) + 10(-3+2)) = f(16 - 10) = f(6) = 3(6)+1 = 19$. Thus $(B) ightarrow (V)$.
$(C) f(f(-4)) = f(2(-4)-5) = f(-13) = 2(-13)-5 = -31$. Thus $(C) ightarrow (III)$.
$(D) f(f(f(1))) = f(f(1+2)) = f(f(3)) = f(3+2) = f(5) = 3(5)+1 = 16$. Thus $(D) ightarrow (I)$.
Therefore, the correct match is $A-IV, B-V, C-III, D-I$.

List-$I$	List-$II$
$(A) f(-5)+f(0)+f(-1)$	$(I) 16$
$(B) f(f(5)+10f(-3))$	$(II) 40$
$(C) f(f(-4))$	$(III) -31$
$(D) f(f(f(1)))$	$(IV) -12$
	$(V) 19$

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