Algebra: Functions Test-5 – Wordpandit
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Algebra: Functions Test-5

Question 1
If x and y are real numbers, the functions are defined as f(x, y) = I x + Y I, F (x, y) = - f (x, y) and G (x, y) = - F (x, y). Now with the help of this information answer the following questions: If Y = which of the following will give x2 as the final value
A
f(x, y) G (x, y) 4
B
G (f (x, y)f(x, y)) F (x, y)/8
C
- F (x, y) G (x, y)/log216
D
– f(x, y) G (x, y) F (x, y)/F (3x, 3y)
Question 1 Explanation: 
F(x,y).G(x,y)=[|x+y|.|x+y|](2x.2x)log216=4x2[log216=log224=4]thereforeoptioncistherightanswer
Question 2
Using the relation answer the questions given below @ (A, B) = average of A and B /(A, B) = product of A and B, x (A, B) = the result when A is divided by B The sum of A and B is given by
A
\ (@ (A, B), 2)
B
@ ( \ (A, B), 2)
C
@ (X (A, B), 2)
D
none of these
Question 2 Explanation: 
Going through the various options
((A,B),2)=(A+B2,2)=A+B2×2=A+B
Question 3
Using the relation answer the questions given below @ (A, B) = average of A and B /(A, B) = product of A and B, x (A, B) = the result when A is divided by B The average of A, Band C is given by
A
@ (x (\ (@ (A, B), 2), C), 3)
B
\ (x (\ (@ (A, B)), C 2))
C
X (@ (\ (@ (A, B), 2), C,3))
D
X (\ (@ (\ (@ (A, B), 2), C), 2), 3)
Question 3 Explanation: 
(a) (x ( ( (A, B), 2), C), 3)=  (x ((A+B2,2),C), 3)= (x (A+B, C), 3)=(A+BC,3)(b)  (x ( ( (A, B)), C 2))=  (x((A+B2,C2)))=(x(A+B2×C2))(c)X((((A,B),2),C,3))=X(((A+B2,2),C,3))=X((A+B,C,3))=X(A+B+C2,3)=A+B+C6(d)X(((((A,B),2),C),2),3)=X(((((A,B),2),C),2),3)=X((((A+B2,2),C),2),3)=X(((A+B,C),2),3)=X((A+B+C2,2),3)=X((A+B+C,3))=A+B+C3
Question 4
x and yare non-zero real numbersf (x, y) = + (x +y)0.5,if (x +y)0.5is real otherwise = (x +y)2g (x, y) = (x +y)2 if (x + y)0.5is real, otherwise =- (x +y)For  which of the following is f (x, y) necessarily greater than g (x, y)option
A
x and y are positive
B
x and y are negative
C
x and y are greater than - 1
D
None of these
Question 4 Explanation: 
When both x and y are positive
f (x, y) = + (x +y)0.5g (x, y) = (x +y)2 so g(x,y)>f(x,y)
When both x and y are negative
f(x, y) = (x +y)2g (x, y) = -(x +y)
If   x+y>-1
g(x,y)>f(x,y)
When x and y are greater than -1. They can be both positive as well as in the range 0 to -1
so both a and b option cases arise.
Question 5
x and yare non-zero real numbers f (x, y) = + (x +y)0.5,if (x +y)0.5is real otherwise = (x +y)2 g (x, y) = (x +y)2 if (x + y)0.5is real, otherwise =- (x +y) Which of the following is necessarily false?
A
f(x,y) >g(x,y) for 0
B
f(x,y) > g (x,y) when x,y<-1
C
f (x, y) > g (x, y) for x, y > 1
D
None of these
Question 5 Explanation: 
In this case ,f (x, y) = + (x +y)0.5    g (x, y) = (x +y)2
This is true cause (x+y)<1 for which  (x +y)0.5>(x +y)2
In this case, f(x, y) = (x +y)2      g (x, y) = -(x +y)
This is also true cause for (x+y)<-2  (x +y)2>-(x+y)
In this case f (x, y) = + (x +y)0.5    g (x, y) = (x +y)2
This is false because for(x+y)>2 (x +y)2>(x +y)0.5
So the answer is c
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