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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 thenWhat 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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