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DUE: 18 JUNE 2026 -+
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Question 1
o Answer.~ False
0 Exp1anation: A simp e counterexample dlsproves th is Let a. = 9 a • d b. = 11 1 •
a LHS = D (!J, + 16) = ,J25 = .5
1
o R.HS = D 9) + D (16) = Jg yl6 = 3 + 4 = 7
a S f'lce 5 -:j: 7, he stateme "l't does not hold iurnversa □ y.
(1. 2) If s -/:- t, where s E and f' is a fjunctfon then it fol~lows that J(s ) # f (t) ,.
o .Answer~ False
o Exp1a1nation: This • escribes the co:ndition for a1on.e--w-one (injeetive) • uoctio11, but it is no·tt t ue
for all functions. For ai many-to-one functron like f(x) = · 2. dis.t inct :inputs can yield the same
output. For instance, i = - 2 and t ·= 2, then s =/- .t , but J ( - . ) = J 2') = 4.
(.1.3) If J and g are fiuncUons tJh .n .f (ag( .~ == 1) g( / (ax · •~
o .A.11sweli.! False
o Ex:p.ilaination: Function composition LS,g;enera11y non-commutartive. and constants cannot be
a1rbitrarfly moved between the inner and outer a erations. Fo •exampte. le·t / { 2:) = :E 2',
g x) = and a=
a LH = , ( g( ,) ) = , (2:EJ = (2 r~ = ; .2
o RHS = g(/(2. )) = g((2 ~) = g( x~) = :l
o Howeve. ~f we change the unctrons to f( ·) = , • 1 and 9( ·), = ~:
a LH = / (2:r) = 2x + 1
a H..H = g ( J(2 -) = g{ 2 · 1) = 2x 1 (This s, . if.ic setup. matches. but if we
chedkg(/{ax)} pmperly.: R,H' = g(2(:r + 1)) = 2x + .Since2:r + 1 =t- 2x -r 2,
it is fake in general).