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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
    A
    2
    B
    6
    C
    8
    D
    -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
    Question 2
    If it is given  that a >b  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, ..., then the relation Ma [md (a) mn (a, b)]=mn [a,md (Ma (a, b))] does not hold if
    A
    a< 0, b < 0,
    B
    a> 0, b > 0
    C
    a> 0, b < 0, I a I < I b I
    D
    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 won’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)=
    A
    1
    B
    gof(x)
    C
    $ \displaystyle \frac{15x+9}{16x-5}$
    D
    $ \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)
    A
    - 3
    B
    1/4
    C
    -4
    D
    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 then What is value of (gofofogogof) (x) (fogofog)(x)
    A
    x
    B
    x2
    C
    $ \displaystyle \frac{5x+3}{4x-1}$
    D
    $ \displaystyle \frac{\left( x+3 \right)\,\left( 5x+3 \right)}{\left( 4x-5 \right)\,\left( 4x-1 \right)}$
    Question 5 Explanation: 
    g(x)=f-1(x),
    fog(x)=x and gof(x)=0
    Therefore,
    (gofofogogof) (x) (fog)(x)
    =(gofofog) (x) (fog)(x)
    =(x) (gof) (x) =x2
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