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The gradient of the curve y + px +qy = 0 at (1, 1) is 1/2. The values of p and q are
- a)(–1, 1)
- b)(2, –1)
- c)(1, 2)
- d)none of these
Correct answer is 'A'. Can you explain this answer?
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Verified Answer
The gradient of the curve y + px +qy = 0 at (1, 1) is 1/2. The values ...
+px+qy=0
y+qy=-px
(1+q)y=-px
y=-[p/(1+q)]x
This is a linear function, so has constant gradient at all points on the curve. Hence
-p/(1+q)=1/2
2p=-(1+q)
But there is an issue: you have stated that the curve's gradient is 1/2 at the point (1,1) but the curve does not cross through this point! Regardless of our choices for p and q satisfying the expressions above this paragraph, the equation of the curve will always simplify to y=0.5x, which crosses through the origin (0,0), as well as (1,0.5) and (2,1) - but not (-1,1).
For your curve to pass through (-1,1), we would need to add a constant term, like so:
y+px+qy=1/2
Most Upvoted Answer
The gradient of the curve y + px +qy = 0 at (1, 1) is 1/2. The values ...
y+px+qy=0
y+qy=-px
(1+q)y=-px
y=-[p/(1+q)]x
This is a linear function, so has constant gradient at all points on the curve. Hence
-p/(1+q)=1/2
2p=-(1+q)
But there is an issue: you have stated that the curve's gradient is 1/2 at the point (1,1) but the curve does not cross through this point! Regardless of our choices for p and q satisfying the expressions above this paragraph, the equation of the curve will always simplify to y=0.5x, which crosses through the origin (0,0), as well as (1,0.5) and (2,1) - but not (-1,1).
For your curve to pass through (-1,1), we would need to add a constant term, like so:
y+px+qy=1/2
y+qy=-px
(1+q)y=-px
y=-[p/(1+q)]x
This is a linear function, so has constant gradient at all points on the curve. Hence
-p/(1+q)=1/2
2p=-(1+q)
But there is an issue: you have stated that the curve's gradient is 1/2 at the point (1,1) but the curve does not cross through this point! Regardless of our choices for p and q satisfying the expressions above this paragraph, the equation of the curve will always simplify to y=0.5x, which crosses through the origin (0,0), as well as (1,0.5) and (2,1) - but not (-1,1).
For your curve to pass through (-1,1), we would need to add a constant term, like so:
y+px+qy=1/2
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Community Answer
The gradient of the curve y + px +qy = 0 at (1, 1) is 1/2. The values ...
1, 1/2
b)1/2, 1
c)-1/2, 1
d)1, -1/2
To find the gradient of the curve at a point (x, y), we can use implicit differentiation. Taking the derivative of the equation with respect to x, we get:
p + q(dy/dx) = 0
Since we are given that the gradient at (1, 1) is 1/2, we can substitute x = 1 and y = 1 into the equation above:
p + q(dy/dx) = 0
p + q(1/2) = 0
p + q/2 = 0
2p + q = 0
Now, we need to find the values of p and q that satisfy this equation. To do this, we can substitute p = -q/2 into the original equation:
y(px - qy) = 0
1(-q/2) - q(1) = 0
-q/2 - q = 0
-q/2 - 2q/2 = 0
-3q/2 = 0
q = 0
Substituting q = 0 back into the equation p + q/2 = 0, we get:
p + 0/2 = 0
p = 0
Therefore, the values of p and q are 0 and 0, respectively. None of the given answer choices match these values, so the correct answer is none of the above.
b)1/2, 1
c)-1/2, 1
d)1, -1/2
To find the gradient of the curve at a point (x, y), we can use implicit differentiation. Taking the derivative of the equation with respect to x, we get:
p + q(dy/dx) = 0
Since we are given that the gradient at (1, 1) is 1/2, we can substitute x = 1 and y = 1 into the equation above:
p + q(dy/dx) = 0
p + q(1/2) = 0
p + q/2 = 0
2p + q = 0
Now, we need to find the values of p and q that satisfy this equation. To do this, we can substitute p = -q/2 into the original equation:
y(px - qy) = 0
1(-q/2) - q(1) = 0
-q/2 - q = 0
-q/2 - 2q/2 = 0
-3q/2 = 0
q = 0
Substituting q = 0 back into the equation p + q/2 = 0, we get:
p + 0/2 = 0
p = 0
Therefore, the values of p and q are 0 and 0, respectively. None of the given answer choices match these values, so the correct answer is none of the above.
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