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The slope of the tangent to the curve y = x2-x at the point, where the line y= 2 cuts the curve in the first quadrant, is
- a)2
- b)3
- c)-3
- d)-2
Correct answer is option 'B'. Can you explain this answer?
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The slope of the tangent to the curve y = x2-x at the point, where the...
Finding the Point of Intersection
To find the point of intersection of the line y= 2 and the curve y = x2-x, we can equate the two equations and solve for x.
2 = x2-x
Rearranging the equation, we get:
x2 - x - 2 = 0
This is a quadratic equation that can be solved using the quadratic formula:
x = (-b ± √(b2-4ac))/2a
Plugging in the values, we get:
x = (1 ± √(1+8))/2
x = (1 ± √9)/2
x = (1 ± 3)/2
So, x = -1 or x = 2. Since we are looking for the point of intersection in the first quadrant, we can take x = 2.
Therefore, the point of intersection is (2, 2).
Finding the Slope of the Tangent
To find the slope of the tangent to the curve y = x2-x at the point (2, 2), we need to take the derivative of the function and evaluate it at x = 2.
y = x2-x
dy/dx = 2x-1
At x = 2,
dy/dx = 2(2)-1 = 3
Therefore, the slope of the tangent to the curve y = x2-x at the point (2, 2) is 3.
Answer: Option B (3)
To find the point of intersection of the line y= 2 and the curve y = x2-x, we can equate the two equations and solve for x.
2 = x2-x
Rearranging the equation, we get:
x2 - x - 2 = 0
This is a quadratic equation that can be solved using the quadratic formula:
x = (-b ± √(b2-4ac))/2a
Plugging in the values, we get:
x = (1 ± √(1+8))/2
x = (1 ± √9)/2
x = (1 ± 3)/2
So, x = -1 or x = 2. Since we are looking for the point of intersection in the first quadrant, we can take x = 2.
Therefore, the point of intersection is (2, 2).
Finding the Slope of the Tangent
To find the slope of the tangent to the curve y = x2-x at the point (2, 2), we need to take the derivative of the function and evaluate it at x = 2.
y = x2-x
dy/dx = 2x-1
At x = 2,
dy/dx = 2(2)-1 = 3
Therefore, the slope of the tangent to the curve y = x2-x at the point (2, 2) is 3.
Answer: Option B (3)
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The slope of the tangent to the curve y = x2-x at the point, where the line y= 2 cuts the curve in the first quadrant, isa)2b)3c)-3d)-2Correct answer is option 'B'. Can you explain this answer?
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