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11 Nov 2021, 05:15
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D
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The operation is defined as all nonzero x and y by \(x\text{@}y = xy + \frac{x}{y}\). What is the value of \((((a\text{@}1)\text{@}1)\text{@}1)\text{@}1\) ?
(A) 2a
(B) 4a
(C) 8a
(D) 16a
(E) 32a
(A) 2a
(B) 4a
(C) 8a
(D) 16a
(E) 32a
11 Nov 2021, 06:24
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Bunuel
The operation is defined as all nonzero x and y by \(x\text{@}y = xy + \frac{x}{y}\). What is the value of \((((a\text{@}1)\text{@}1)\text{@}1)\text{@}1\) ?
(A) 2a
(B) 4a
(C) 8a
(D) 16a
(E) 32a
(A) 2a
(B) 4a
(C) 8a
(D) 16a
(E) 32a
a 1 means y=1
When y=1, \(x\text{@}1 = x*1 + \frac{x}{1}=x+x=2x\), that is the variable on left doubles in every operation.
So, \((((a\text{@}1)\text{@}1)\text{@}1)\text{@}1\)
\(((2a\text{@}1)\text{@}1)\text{@}1\)
\(((4a)\text{@}1)\text{@}1\)
\((8a)\text{@}1\)
\(16a\)
D
01 Jul 2022, 00:38
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Given that \(x\text{@}y = xy + \frac{x}{y}\) for all nonzero x and y and we need to find the value of \((((a\text{@}1)\text{@}1)\text{@}1)\text{@}1\)
Let's learn how to find the value of \(a\text{@}1\) first
To find \(a\text{@}1\) we need to compare what is before and after \(\text{@}\) in \(a\text{@}1\) and \(x\text{@}y\)
=> We need to substitute x with a and y with 1 in \(x\text{@}y = xy + \frac{x}{y}\)
=> \(a\text{@}1 = a*1 + \frac{a}{1}\) = a+a = 2a
Now, lets find the value of \((((a\text{@}1)\text{@}1)\text{@}1)\text{@}1\)
=> \((((a\text{@}1)\text{@}1)\text{@}1)\text{@}1\) = \(((2a\text{@}1)\text{@}1)\text{@}1\)
Now, \(2a\text{@}1 = 2a*1 + \frac{2a}{1}\) = 2a+2a = 4a
=> \((((a\text{@}1)\text{@}1)\text{@}1)\text{@}1\) = \((4a\text{@}1)\text{@}1\)
Now, \(4a\text{@}1 = 4a*1 + \frac{4a}{1}\) = 4a+4a = 8a
=> \((((a\text{@}1)\text{@}1)\text{@}1)\text{@}1\) = \(8a\text{@}1\) = 16a (following previous pattern)
So, Answer will be D
Hope it helps!
Watch the following video to learn the Basics of Functions and Custom Characters
Let's learn how to find the value of \(a\text{@}1\) first
To find \(a\text{@}1\) we need to compare what is before and after \(\text{@}\) in \(a\text{@}1\) and \(x\text{@}y\)
=> We need to substitute x with a and y with 1 in \(x\text{@}y = xy + \frac{x}{y}\)
=> \(a\text{@}1 = a*1 + \frac{a}{1}\) = a+a = 2a
Now, lets find the value of \((((a\text{@}1)\text{@}1)\text{@}1)\text{@}1\)
=> \((((a\text{@}1)\text{@}1)\text{@}1)\text{@}1\) = \(((2a\text{@}1)\text{@}1)\text{@}1\)
Now, \(2a\text{@}1 = 2a*1 + \frac{2a}{1}\) = 2a+2a = 4a
=> \((((a\text{@}1)\text{@}1)\text{@}1)\text{@}1\) = \((4a\text{@}1)\text{@}1\)
Now, \(4a\text{@}1 = 4a*1 + \frac{4a}{1}\) = 4a+4a = 8a
=> \((((a\text{@}1)\text{@}1)\text{@}1)\text{@}1\) = \(8a\text{@}1\) = 16a (following previous pattern)
So, Answer will be D
Hope it helps!
Watch the following video to learn the Basics of Functions and Custom Characters
