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Bunuel
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If the operation ∆ is defined by a ∆ b = (b - a)^2/a^2 for all numbers 03 Jan 2016, 12:55
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E
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15% (low)

Question Stats:

76% (01:12) correct 24% (01:35) wrong based on 128 sessions

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If the operation ∆ is defined by a ∆ b = (b - a)^2/a^2 for all numbers a and b, and a ≠ 0, then − 1 ∆ (1 ∆ − 1) =

A. −1
B. 0
C. 1
D. 9
E. 25
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E
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If the operation ∆ is defined by a ∆ b = (b - a)^2/a^2 for all numbers 04 Jan 2016, 07:23
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Bunuel
If the operation ∆ is defined by a ∆ b = (b - a^2)/a^2 for all numbers a and b, and a ≠ 0, then − 1 ∆ (1 ∆ − 1) =

A. −1
B. 0
C. 1
D. 9
E. 25

Hi Bunuel ,
i don't find the answer as a choice..
a ∆ b = (b - a^2)/a^2..
− 1 ∆ (1 ∆ − 1) ...
first we solve what is there in bracket.. (1 ∆ − 1) .
here b=-1 and a=1..
so (b - a^2)/a^2 = (-1 - 1^2)/1^2 =-2/1=-2..
next is -1 ∆ -2..
now b=-2 and a=-1..
so (b - a^2)/a^2 = (-2 - (-1)^2)/(-1)^2 =-3/1=-3....
-3 is not there in choices but???

either, there is a problem of bracket or operation..
pl recheck..
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If the operation ∆ is defined by a ∆ b = (b - a)^2/a^2 for all numbers 04 Jan 2016, 08:29
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Bunuel
If the operation ∆ is defined by a ∆ b = (b - a^2)/a^2 for all numbers a and b, and a ≠ 0, then − 1 ∆ (1 ∆ − 1) =

A. −1
B. 0
C. 1
D. 9
E. 25

Hi Bunuel ,
i don't find the answer as a choice..
a ∆ b = (b - a^2)/a^2..
− 1 ∆ (1 ∆ − 1) ...
first we solve what is there in bracket.. (1 ∆ − 1) .
here b=-1 and a=1..
so (b - a^2)/a^2 = (-1 - 1^2)/1^2 =-2/1=-2..
next is -1 ∆ -2..
now b=-2 and a=-1..
so (b - a^2)/a^2 = (-2 - (-1)^2)/(-1)^2 =-3/1=-3....
-3 is not there in choices but???

either, there is a problem of bracket or operation..
pl recheck..

Edited. Thank you very much!
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If the operation ∆ is defined by a ∆ b = (b - a)^2/a^2 for all numbers 04 Jan 2016, 10:36
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Answer is E,25

On solving the inner bracket we get 4.
then we get equation as -1delta4 where b=4 and a=-1

On solving this we get 25 as answer
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If the operation ∆ is defined by a ∆ b = (b - a)^2/a^2 for all numbers 08 Aug 2022, 09:29
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Given that a ∆ b = \(\frac{(b - a)^2}{a^2}\) and we need to find the value of − 1 ∆ (1 ∆ − 1)

Lets start by finding the value of 1 ∆ − 1

To find 1 ∆ − 1 we need to compare what is before and after ∆ in 1 ∆ − 1 and a ∆ b

=> We need to substitute a with 1 and b with -1 in a ∆ b = \(\frac{(b - a)^2}{a^2}\) to get the value of 1 ∆ − 1

=> 1 ∆ -1 = \(\frac{(-1 - 1)^2}{1^2}\) = \(\frac{(-2)^2}{1}\) = 4

=> − 1 ∆ (1 ∆ − 1) = − 1 ∆ 4

Similarly, -1 ∆ 4 = \(\frac{(4 - (-1))^2}{(-1)^2}\) = \(\frac{(5)^2}{1}\) = 25

So, Answer will be E
Hope it helps!

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