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Algebra: Functions Test-2

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Question 1
f(x) = 2x +3 and g(x) = (x -3)/2 then What is the value of fo (fog) o (gof) (x)
A
x
B
x2
C
2x+3
D
$ \displaystyle \frac{x+3}{4x-5}$
Question 1 Explanation: 
fog(x)=gof (x)=x
fo (fog) o (gof) (x) = fo (fog) (x) =f (x)=2x+3.
Therefore (c) is the correct answer.
Question 2
If le (x, y) = least of (x, y), mo (x) = I X I,me (x, y) = maximum of (x, y).
Find the value of me (a+ mo (Ie (a, b)); mo (a+ me (mo (a)mo (b))), at a = - 2 and b =-3
A
1
B
0
C
5
D
3
Question 2 Explanation: 
Segregating and Simplifying:
i)mo (a+ me (mo (a)mo (b)) =mo (a+ me (mo (a),3)
=mo (a+ me (2,3))
=mo (a+ 3)
= |-2+3|
=1
ii) a+ mo (Ie (a, b))
=a+mo(-3)
=-2+3
=1
The maximum value of both the terms is 1.
Question 3
If le (x, y) = least of (x, y), mo (x) = I X I,me (x, y) = maximum of (x, y) Which of the following must always be correct for a, b >0
A
mo (le (a, b)) > (me (mo (a),mo (b)))
B
mo (le (a, b))> (me (mo (a),mo (b))
C
mo (le (a, b))< (Ie(mo (a)), mo (b))
D
mo (le (a, b)) = le (mo (a),mo (b))
Question 3 Explanation: 
First observation since both a and b are greater than 0,
mo(a)=a and mo(b)=b.
Therefore,
(me (mo (a),mo (b))) = me (a,b)
le (mo (a),mo (b))=le (a,b)
Now since both a and b are positive,
we can conclude that mo (le (a, b)) = le (a, b)
Question 4
If le (x, y) = least of (x, y), mo (x) = I X I,me (x, y) = maximum of (x, y) For what values of a is me (a2 - 3a, a - 3) <0?
A
1
B
0
C
a<0 anda<3
D
a<0 ora<3
Question 4 Explanation: 
Capture ty
Case I. a < 0, a3 - 3a > a - 3 = a (a - 3) < 0 or 0 < a < 3 which is not true.
Case II. 0 < a < 3, a (a - 3) < 0 or 0 < a < 3 which is true.
Case III. a = 3, me (0, 0) < 0 not true.
Case IV. a> 3, a (a - 3) < 0 or 0 < a < 3 not true
The above graph clearly shows that the maximum value of the function occurs when a lies between 0 and 3. Therefore, option b is correct. The above graph clearly shows that the maximum value of the function occurs when a lies between 0 and 3.
Question 5
If le (x, y) = least of (x, y), mo (x) = I X I,me (x, y) = maximum of (x, y) For what values of a le (a2 - 3a, a - 3) < 0
A
1
B
0
C
a
D
a
Question 5 Explanation: 
Capture tu
Again in case I, a < 0; a - 3 < 0 or a < 3
(from last Question) can be true
In case II, 0 < a < 3; a - 3 < 0 or a < 3 can be true
In case III, a = 3, le (0, 0) = 0 < 0, not true
In case IV, a> 3, a - 3 < 0 or a < 3 not true
The graph clearly shows that the minimum value of the function lies between 0 and 3. Therefore the answer will be option (b).
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