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## Algebra: Functions Test-5

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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 x

^{2}as the final valuef(x, y) G (x, y) 4 | |

G (f (x, y)f(x, y)) F (x, y)/8 | |

- F (x, y) G (x, y)/log _{2}16 | |

â€“ f(x, y) G (x, y) F (x, y)/F (3x, 3y) |

Question 1 Explanation:

$ \displaystyle \begin{array}{l}-F\,\left( x,\,\,y \right)\,.\,G\,\left( x,y \right)=-\left[ -\left| \,\,\,\,x+y\,\,\, \right|\,\,\,\,.\,\,\,\,\left| \,\,\,x+y\,\,\, \right|\,\, \right]\\\frac{-\left( -2x.2x \right)}{{{\log }_{2}}16}=4{{x}^{2}}\,\,\,\,\,\,\because \,[{{\log }_{2}}16={{\log }_{2}}{{2}^{4}}=4]\\therefore\,\,option\,\,c\,\,is\,\,the\,\,right\,\,answer\end{array}$

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, B), 2) | |

@ ( \ (A, B), 2) | |

@ (X (A, B), 2) | |

none of these |

Question 2 Explanation:

Going through the various options

$ \backslash \left( \left( A,B \right),2 \right)=\backslash \left( \frac{A+B}{2},2 \right)=\frac{A+B}{2}\times 2=A+B$

$ \backslash \left( \left( A,B \right),2 \right)=\backslash \left( \frac{A+B}{2},2 \right)=\frac{A+B}{2}\times 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

@ (x (\ (@ (A, B), 2), C), 3) | |

\ (x (\ (@ (A, B)), C 2)) | |

X (@ (\ (@ (A, B), 2), C,3)) | |

X (\ (@ (\ (@ (A, B), 2), C), 2), 3) |

Question 3 Explanation:

$ \displaystyle \begin{array}{l}(a)\,\,\text{ }\left( x\text{ }\left( \backslash \text{ }\left( \text{ }\left( A,\text{ }B \right),\text{ }2 \right),\text{ }C \right),\text{ }3 \right)=\text{ }\text{ }(x\text{ }(\backslash \left( \frac{A+B}{2},2 \right),C),\text{ }3)\\=\text{ }\left( x\text{ }\left( A+B,\text{ }C \right),\text{ }3 \right)\\=\left( \frac{A+B}{C},3 \right)\\\left( b \right)\text{ }\backslash \text{ }\left( x\text{ }\left( \backslash \text{ }\left( \text{ }\left( A,\text{ }B \right) \right),\text{ }C\text{ }2 \right) \right)\\=\text{ }\!\!\backslash\!\!\text{ }\left( x\left( \backslash \left( \frac{A+B}{2},C\,2 \right) \right) \right)\\=\backslash \left( x\left( \frac{A+B}{2}\times C\,2 \right) \right)\\(c)X\left( \left( \backslash \left( \left( A,B \right),2 \right),C,3 \right) \right)\\=X\left( \left( \backslash \left( \frac{A+B}{2},2 \right),C,3 \right) \right)\\=X\left( \left( A+B,C,3 \right) \right)\\=X\left( \frac{A+B+C}{2},3 \right)=\frac{A+B+C}{6}\\(d)\,X\left( \backslash \left( \left( \backslash \left( \left( A,B \right),2 \right),C \right),2 \right),3 \right)\\=X\left( \backslash \left( \left( \backslash \left( \left( A,B \right),2 \right),C \right),2 \right),3 \right)\\=X\left( \backslash \left( \left( \backslash \left( \frac{A+B}{2},2 \right),C \right),2 \right),3 \right)\\=X\left( \backslash \left( \left( A+B,C \right),2 \right),3 \right)\\=X\left( \backslash \left( \frac{A+B+C}{2},2 \right),3 \right)\\=X\left( \backslash \left( A+B+C,3 \right) \right)=\frac{A+B+C}{3}\end{array}$

Question 4 |

x and yare non-zero real numbersf (x, y) = + (x +y)

^{0.5},if (x +y)^{0.5}is real otherwise = (x +y)^{2}g (x, y) = (x +y)^{2}if (x + y)^{0.5}is real, otherwise =- (x +y)For Â which of the following is f (x, y) necessarily greater than g (x, y)optionx and y are positive | |

x and y are negative | |

x and y are greater than - 1 | |

None of these |

Question 4 Explanation:

When both x and y are positive

f (x, y) = + (x +y)

When both x and y are negative

f(x, y) = (x +y)

IfÂ Â 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.

f (x, y) = + (x +y)

^{0.5}g (x, y) = (x +y)^{2}so g(x,y)>f(x,y)When both x and y are negative

f(x, y) = (x +y)

^{2}g (x, y) = -(x +y)IfÂ Â 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)^{0.5}is real otherwise = (x +y)^{2}g (x, y) = (x +y)^{2}if (x + y)^{0.5}is real, otherwise =- (x +y) Which of the following is necessarily false?f(x,y) >g(x,y) for 0 | |

f(x,y) > g (x,y) when x,y<-1 | |

f (x, y) > g (x, y) for x, y > 1 | |

None of these |

Question 5 Explanation:

In this case ,f (x, y) = + (x +y)

This is true cause (x+y)<1 for whichÂ (x +y)

In this case, f(x, y) = (x +y)

This is also true cause for (x+y)<-2Â (x +y)

In this case f (x, y) = + (x +y)

This is false because for(x+y)>2 (x +y)

So the answer is c

^{0.5Â }Â Â g (x, y) = (x +y)^{2}This is true cause (x+y)<1 for whichÂ (x +y)

^{0.5}>(x +y)^{2}In this case, f(x, y) = (x +y)

^{2Â Â Â Â Â }g (x, y) = -(x +y)This is also true cause for (x+y)<-2Â (x +y)

^{2}>-(x+y)In this case f (x, y) = + (x +y)

^{0.5Â }Â Â 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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