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## Algebra: Functions Test-6

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Question 1 |

x and y are non-zero real numbers

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

g (x, y) = (x +y)

If f(x, y) =g (x, y) then

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)If f(x, y) =g (x, y) then

x=y | |

x+y= 1 | |

x+y=-2 | |

Both b and c |

Question 1 Explanation:

Going through all the options

The value of x and y could be anything so cant be sure about the functions

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

In this case both are 1

So f(x, y) =g (x, y)

The value of x and y could be anything so cant be sure about the functions

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

^{0.5 }g (x, y) = (x +y)^{2}In this case both are 1

So f(x, y) =g (x, y)

Question 2 |

y=ax+b | |

y=a+bx+cx ^{2} | |

y=e ^{ax+b} | |

None of these |

Question 2 Explanation:

(a) Can be eliminated as looking at the data we can conclude that there is no linear relation between x and y

(b) Using the various values of x and y we can create the following equations

4=a+b+c -----------(1)

6=a+2b+4c---------(2)

14=a+3b+9c--------(3)

22=a+4b+16c

32=a+5b+25c

44=a+6b+36c

We will find the values of a,b and c using the first 3 equations and will check if the values satisfy the other 3 equations

Using eq 1 and 2 we get

2=b+3c

Using 2 and 3 we get

8=b+5c

Solving we get c=3,b=-7 and a=8

The values of a, b and c satisfy all the equations to a large limit.

(c) As the exponential function grows more quickly for small positive values as compared to large values,

while quite the opposite is happening here. So can’t be a match

(b) Using the various values of x and y we can create the following equations

4=a+b+c -----------(1)

6=a+2b+4c---------(2)

14=a+3b+9c--------(3)

22=a+4b+16c

32=a+5b+25c

44=a+6b+36c

We will find the values of a,b and c using the first 3 equations and will check if the values satisfy the other 3 equations

Using eq 1 and 2 we get

2=b+3c

Using 2 and 3 we get

8=b+5c

Solving we get c=3,b=-7 and a=8

The values of a, b and c satisfy all the equations to a large limit.

(c) As the exponential function grows more quickly for small positive values as compared to large values,

while quite the opposite is happening here. So can’t be a match

Question 3 |

If f(0, y) = y + 1, and f(x + I, y) =f (x, f (x, y)). Then, what is the value of f(1,2) ?

1 | |

2 | |

3 | |

4 |

Question 3 Explanation:

F(1,2)=f(0,3)
F(0,3)=4
So,f(1,2)=4

Question 4 |

Functions m and M are defined as follows:m (a, b, c) = min (a + b, c, a)M(a, b, c) = max (a + b, c, a)If a = - 2, b = - 3 and c = 2 what is the maximum between
$ \displaystyle \left[ \frac{m\text{ }\left( a,\text{ }b,\text{ }c \right)+M\text{ }\left( a,\text{ }b,\text{ }c \right)}{2} \right]and\left[ \frac{m\text{ }\left( a,\text{ }b,\text{ }c \right)\text{ }M\text{ }\left( a,\text{ }b,\text{ }c \right)}{2} \right]$

3/2 | |

7/2 | |

-3/2 | |

-7/2 |

Question 4 Explanation:

[m (a, b, c)+M (a, b, c)]/2=-3/2
[m (a, b, c) –M (a, b, c)]/2=-7/2
Max of the 2 is -3/2 which is the answer

Question 5 |

Functions m and M are defined as follows:
m (a, b, c) = min (a + b, c, a)
M(a, b, c) = max (a + b, c, a)If a b, and c are negative, then what gives the minimum of a and b

m (a,b,c) | |

–M (-a, a,-b) | |

m (a+b, b,c) | |

None of these |

Question 5 Explanation:

As all a,b and c are negative

(a) m (a,b,c)=min(a+b,c,a)

(b) –M (-a, a,-b)=-max(0,-b,-a)

(c) m (a+b, b,c)=min(a+2b,c,a+b)

(a) m (a,b,c)=min(a+b,c,a)

(b) –M (-a, a,-b)=-max(0,-b,-a)

(c) m (a+b, b,c)=min(a+2b,c,a+b)

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