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Algebra: Functions Test1
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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)= md2 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)= md2 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}{16x5}$  
$ \displaystyle \frac{1}{x}$ 
Question 3 Explanation:
f(x) = 2x +3, g(x) = (x 3)/2
fog(x)=2{(x 3)/2}+ 3 =x3+3 =x
gof(x)=(2x+33)/2=x
Therefore correct option is (b)
Alternatively,
it can be easily observed that g(x)=f^{1}(x)
Therefore, fog(x)=gof(x)
fog(x)=2{(x 3)/2}+ 3 =x3+3 =x
gof(x)=(2x+33)/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(x3) = (x 33)/2 =(x6)/2
Therefore 2x+3 =(x6)/2
Solving for x we will get x=4.
=>g(x3) = (x 33)/2 =(x6)/2
Therefore 2x+3 =(x6)/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)
x  
x^{2}  
$ \displaystyle \frac{5x+3}{4x1}$  
$ \displaystyle \frac{\left( x+3 \right)\,\left( 5x+3 \right)}{\left( 4x5 \right)\,\left( 4x1 \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) =x^{2}
fog(x)=x and gof(x)=0
Therefore,
(gofofogogof) (x) (fog)(x)
=(gofofog) (x) (fog)(x)
=(x) (gof) (x) =x^{2}
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