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## Algebra: Functions Test-1

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Question 1 |

*If md (x) = I x I, mn (x, y) = minimum of x and y and Ma (a, b, c, ....) = maximum of a, b, c, ...*Value of Ma [md (a),mn (md (b), a), mn (ab,md (ac))] where a = - 2, b = - 3, c = 4 is

2 | |

6 | |

8 | |

-2 |

Question 1 Explanation:

Ma [md (a),mn (md (b), a), mn (ab,md (ac))]

Segregating the individual terms inside the third bracket:

i) md(a) =|a|. In this case as a=-2 and md(a)=2

ii)md (ac)= md|-2 x 4|= 8 using a = - 2, c = 4

iii) mn((ab,md (ac)) = mn(6,8)=6

iv) mn (md (b), a) =mn (3,-2) =-2. Ma [md (a),mn (md (b), a), mn (ab,md (ac))] =[2,-2,6] =6

Segregating the individual terms inside the third bracket:

i) md(a) =|a|. In this case as a=-2 and md(a)=2

ii)md (ac)= md|-2 x 4|= 8 using a = - 2, c = 4

iii) mn((ab,md (ac)) = mn(6,8)=6

iv) mn (md (b), a) =mn (3,-2) =-2. Ma [md (a),mn (md (b), a), mn (ab,md (ac))] =[2,-2,6] =6

Question 2 |

*If it is givenÂ Â that a >b Â and if md (x) = I x I, mn (x, y) = minimum of x and y and Ma (a, b, c, ....) = maximum of a, b, c, ...,Â*then the relation Ma [md (a) mn (a, b)]=mn [a,md (Ma (a, b))] does not hold if

a< 0, b < 0, | |

a> 0, b > 0 | |

a> 0, b < 0, I a I < I b I | |

a> 0, b < 0, I a I > I b I |

Question 2 Explanation:

Segregating and simplifying:

Considering a>b,

ma [md (a), b)]=mn [a,md (a)]

Now, whether a is greater than zero or less than zero wonâ€™t affect md (a)=|a|.

Then, ma(|a|,b) will not be equal to mn [a,md (a)] if a<0,

Only one option has a<0.

Considering a>b,

ma [md (a), b)]=mn [a,md (a)]

Now, whether a is greater than zero or less than zero wonâ€™t affect md (a)=|a|.

Then, ma(|a|,b) will not be equal to mn [a,md (a)] if a<0,

Only one option has a<0.

Question 3 |

*f(x) = 2x +3 and g(x) = (x -3)/2 then,*

*fog*(x)=

1 | |

gof(x) | |

$ \displaystyle \frac{15x+9}{16x-5}$ | |

$ \displaystyle \frac{1}{x}$ |

Question 3 Explanation:

*f*(x) = 2x +3,

*g*(x) = (x -3)/2

*fog*(x)=2{(x -3)/2}+ 3 =x-3+3 =x

*gof*(x)=(2x+3-3)/2=x

Therefore correct option is (b)

**Alternatively**,

it can be easily observed that g(x)=f

^{-1}(x)

Therefore,

*fog*(x)=

*gof*(x)

Question 4 |

*f*(x) = 2x +3 and

*g*(x) = (x -3)/2 then For what value of x;

*f*(x) =

*g*(x - 3)

- 3 | |

1/4 | |

-4 | |

None of these |

Question 4 Explanation:

*g*(x) = (x -3)/2

*=>g*(x-3) = (x -3-3)/2 =(x-6)/2

Therefore 2x+3 =(x-6)/2

Solving for x we will get x=-4.

Question 5 |

*f*(x) = 2x +3 and

*g*(x) = (x -3)/2 thenWhat is value of (

*gofofogogof)*(x) (

*fogofog*)(x)

x | |

x ^{2} | |

$ \displaystyle \frac{5x+3}{4x-1}$ | |

$ \displaystyle \frac{\left( x+3 \right)\,\left( 5x+3 \right)}{\left( 4x-5 \right)\,\left( 4x-1 \right)}$ |

Question 5 Explanation:

g(x)=f

Therefore,

(

=(x) (

^{-1}(x),*fog*(x)=x and*gof*(x)=0Therefore,

(

*gofofogogof)*(x) (*fog*)(x)**=**(*gofofog)*(x) (*fog*)(x)=(x) (

*gof)*(x) =x^{2}
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