A.  (3,0) 
B.  (1,0) 
C.  (0.5,0.5) 
D.  (0.5,0) 
Option: D Explanation :
If (0, 0) is substituted in the equation, we get 0, which is less than 2. So, any point on the same side as that of the origin, should yield a value less than 2, when substituted in the equation.
Hence the result
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A.  F: R^{3} → R^{2} defined by f (x, y, z) = (x, z) 
B.  F: R^{3} → R^{3} defined by f (x, y, z) = (x, y  1, z) 
C. 
F: R^{2} → R^{2} defined by F (x,y) = (2 x, y x)

D.  F: R^{2} → R^{2} defined by f (x,y) = (y, x) 
Option: B Explanation :
A function F(x , y ) x (a, b) is linear if F (ax, ay) = a F (x,y) and F (a+x, b+y) = F (a, b)+ F (x,y).
The functions given in options (a), (c), (d) satisfy these two conditions but option (b) doesn't
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The entire graph of the function f(x) = x2 + kx  x + 9 is strictly above the xaxis if and only if
A.   3 < k < 5 
B.   3 < k < 2 
C.   3 < k < 7 
D.   5 < k < 7 
Option: D Explanation :
y = x^{2 }+ (k 1) x + 9 = (x + (k 1)/2)^{2}  ((k 1)/2? + 9
For the entire graph to lie above the xaxis. y should be greater than 0, for all x.
So, 9  ((k 1) / 2) ^{2 } > 0 i.e., (k  7) (k+ 5) < 0 or  5 < k < 7
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A. 
x=4,y=4

B. 
x=3,y=3

C.  x=8,y=7.25 
D.  x=3,y=4 
Option: C Explanation :
x_{(view port) }= (XV_{ max}  XV _{min}) (XW  XW _{min}) / (XW _{max } XW _{min}) + XV_{min}
y_{(viewport) }= (YV_{max}  YV_{min})( YW  YW_{min}) / (YW_{max} YW_{min} ) + YV_{min}
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A. 
(0, 1, 1)

B. 
(k, 0, 1 k) for any real k

C.  (1,2, 7) 
D.  there exists no solution 
Option: D Explanation :
The determinant given above = (2,1,6) Click on Discuss to view users comments. 