Mathematics Times – July 2019

maximum value is

9

2

when x 2.

7.Sol: Method 1:

x xy y x xy y x xy y^2        ^23 ^22  2 2^22 

 

2

   3.3 2x y 9

The maximum value is 9, and this can be

attained when x y  0.

(^22)  (^22) 
1
3
x xy y x xy y    
^22 
2
2
3
  x xy y
 
1 2 2
.3 1
3 3
   x y
The minimum value of x xy y^2  ^2 is 1 when
x y  1.
The answer is 1 9 10 .
Method 2:
Let x xy y k^2   ^2 (1)
x xy y^2   ^23 (2)
   ^22
3
2 1 :
2
k
x y

   (3)
2 1 :2 3   xy k (4)
     
2 3
3 4 : 3
2
k
x y k

    
9
0 9
2
k
k

    (5)
     
2 3
3 4 : 3
2
k
x y k

    
3 3
0 1
2
k
k

    (6)
The maximum value is 9, and the minimum
value is 1.
The answer is 1 9 10 .
8.Sol:Since,
2 6x x y x x y^2      ^2 0,2  (^3) ^20
Solve for x: 0  x 3.
x y x x x y x x^2       ^2 2 2 6^22  ^28 
(^)     0 x 4 16^2.
Since 0  x 3 ,the greatest value for x is 3.
Therefore, the greatest value of x y x^2  ^22 is
16 1 15  when x 3.
9.Sol: Method 1
Suppose that k x y 3 4 is a possible value.
Substituting yk x^3  4 into
x y x y^2    ^2 14 6 6, we get
   
2 2
16 x k x x k x      3 224 24 3 96,
which simplifies to
25 2 3 76x k x k k^2        ^2 24 96 0 (1)
if the line 3 4x y k  intersects the given
circle, the discriminant of (1) must be
nonnegative.
Thus we get
   
(^22)
3 76 25k    k k24 96 0, which
simplifies to k k   (^73)  7 0.
Hence    7 k 73.
The largest possible Value can have is 73 when
47
5
y and
47
5
y.
10.Sol: The domain is
4 13 0x 
13
4
 x
Let 4 13 , 0x t t  .
Then  
1 7
4 13 4 13
2 2
y x    x
 
1 2 7 1 2
1 3
2 2 2
     t t t
Since  
7.Sol: Method 1:
8.Sol:
9.Sol: Method 1
10.Sol:
1 2 1
0, 1
2 2
t t  .
1 7
3
2 2
y  

. The range of y x x   2 3 4 13is

y R ,

7

2

y.