| A. | (3,0) |
| B. | (1,0) |
| C. | (0.5,0.5) |
| D. | (0.5,0) |
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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: R3 → R2 defined by f (x, y, z) = (x, z) |
| B. | F: R3 → R3 defined by f (x, y, z) = (x, y - 1, z) |
| C. |
F: R2 → R2 defined by F (x,y) = (2 x, y -x)
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| D. | F: R2 → R2 defined by f (x,y) = (y, x) |
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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 x-axis if and only if
| A. | - 3 < k < 5 |
| B. | - 3 < k < 2 |
| C. | - 3 < k < 7 |
| D. | - 5 < k < 7 |
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Option: D Explanation :
y = x2 + (k- 1) x + 9 = (x + (k -1)/2)2 - ((k- 1)/2? + 9
For the entire graph to lie above the x-axis. 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
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| B. |
x=3,y=3
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| C. | x=8,y=7.25 |
| D. | x=3,y=4 |
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Option: C Explanation :
x(view port) = (XV max - XV min) (XW - XW min) / (XW max - XW min) + XVmin
y(viewport) = (YVmax - YVmin)( YW - YWmin) / (YWmax- YWmin ) + YVmin
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| A. |
(0, 1, 1)
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| B. |
(k, 0, 1 -k) for any real k
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| C. | (-1,2, -7) |
| D. | there exists no solution |
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Option: D Explanation :
The determinant given above = (2,1,6) Click on Discuss to view users comments. |
