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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 Ist quadrant, is
- a)2
- b)3
- c)–3
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
The slope of the tangent to the curve y = x2 –x at the point, wh...
The point at which line y=2 cut the curve can be found out by
2=x^2-x
x^2-x-2=0
on solving we will get,
x=-1,2
since, the point is in first quadrant, So, x=2
Thus, point is (2,2).
Slope of the curve can be found out by differentiating the the equation of curve with respect to the x.
dy/dx= 2x-1
Now, put the value of x
dy/dx=2(2)-1=4-1=3
Therefore, the slope is 3.
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The slope of the tangent to the curve y = x2 –x at the point, wh...
Finding the Point of Intersection
To find the point of intersection between the curve y = x^2 and the line y = 2, we need to substitute y = 2 in the equation of the curve and solve for x.
2 = x^2
Taking the square root of both sides, we get:
x = ±√2
Since the line y = 2 cuts the curve in the first quadrant, we take the positive root:
x = √2
Finding the Slope of the Tangent
To find the slope of the tangent to the curve at the point (√2, 2), we need to take the derivative of the curve with respect to x and evaluate it at x = √2.
y = x^2
Taking the derivative with respect to x, we get:
dy/dx = 2x
Evaluating at x = √2, we get:
dy/dx = 2√2
Therefore, the slope of the tangent to the curve at the point (√2, 2) is 2√2.
Simplifying, we get:
2√2 = 2(1.414) = 2.828
Therefore, the correct answer is option B, 3 (rounded to the nearest integer).
To find the point of intersection between the curve y = x^2 and the line y = 2, we need to substitute y = 2 in the equation of the curve and solve for x.
2 = x^2
Taking the square root of both sides, we get:
x = ±√2
Since the line y = 2 cuts the curve in the first quadrant, we take the positive root:
x = √2
Finding the Slope of the Tangent
To find the slope of the tangent to the curve at the point (√2, 2), we need to take the derivative of the curve with respect to x and evaluate it at x = √2.
y = x^2
Taking the derivative with respect to x, we get:
dy/dx = 2x
Evaluating at x = √2, we get:
dy/dx = 2√2
Therefore, the slope of the tangent to the curve at the point (√2, 2) is 2√2.
Simplifying, we get:
2√2 = 2(1.414) = 2.828
Therefore, the correct answer is option B, 3 (rounded to the nearest integer).
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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 Ist quadrant, isa)2b)3c)–3d)none of theseCorrect answer is option 'B'. Can you explain this answer?
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