TS ICET 23 May 2019 Shift 1 Paper

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Question : 74 of 200

Marks: +1, -0

If x ↓ y = (x + y) +

(x-y)^2

and x ↑ y =

(x+y)^3-(x-y)^3

then the value

{(0 ↓ 1) ↑ (0 ↓ 1)}/{(1 ↑ 2) ↓ (1 ↑ 2)}

=

$8/7$
${15}/7$
${14}/9$
${13}/8$

Solution:

(0 ↓ 1)=(0+1)+(0-1)^{2}=1+1=2

(1 ↑ 2)=(1+2)^{3}-(1-2)^{3}=27+1=28

\;{(0 ↓ 1)↑(0↓ 1)}/{(1 ↑ 2) ↓(1 ↑ 2)}=\;{(2 ↑ 2)}/{(28 ↓ 28)}

2 ↑ 2=(2+2)^{3}-(2-2)^{3}=64

28 ↓ 28=(28+28)+(28-28)^{2}=56

\;{(2 ↑ 2)}/{(28 ↓ 28)}=\;{64}/{56}=\;{8}/{7}

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