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Tangent at any point to a curve in the first quadrant meets the coordinate axes at A and B such that area of triangle OAB is always 2 square units. If the curve passes through (1, 1), then
- a)one such curve is x + y = 2
- b)one such curve is xy = 1
- c)one such curve is y = 1/(1+x2)
- d)the curve cannot be a straight line
Correct answer is option 'A,B'. Can you explain this answer?
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Most Upvoted Answer
Tangent at any point to a curve in the first quadrant meets the coord...
The line given in option (1) satisfies the condition of the given problem and can thus be treated as a tangent to itself at any point.
⇒ (1) is true and (4) is false.
For curve in option (2):
At x = 1, tangent at (x0, y0)
Now,
x intercept = x0 + x02y0 = 2x0
Thus, curves given in options (1) and (2) satisfy the condition of the problem.
Curve given in option (3) does not pass through (1, 1).
⇒ (3) is false.
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Community Answer
Tangent at any point to a curve in the first quadrant meets the coord...
To find the equation of the curve passing through (1,1) such that the area of triangle OAB is always 2 square units, let's analyze the given options:
a) one such curve is x * y = 2
b) one such curve is xy = 1
c) one such curve is y = 1/(1 + x^2)
d) the curve cannot be a straight line
Let's evaluate each option to determine if it satisfies the given condition.
a) x * y = 2
To find the equation of the tangent at any point (x, y) on this curve, we need to find the derivative dy/dx.
Taking the derivative of both sides with respect to x, we get:
y + x * dy/dx = 0
dy/dx = -y/x
The slope of the tangent at any point (x, y) on the curve is -y/x.
Now, let A and B be the x and y-intercepts of the tangent. The area of triangle OAB is given by:
Area = 1/2 * |A * B|
Since A and B are the x and y-intercepts, respectively, we have A = 1/y and B = 1/x.
Substituting these values, we get:
Area = 1/2 * |(1/y) * (1/x)|
= 1/2 * |1/(x * y)|
= 1/2 * |1/2|
= 1 square unit
Therefore, the area of the triangle is not always 2 square units for the curve x * y = 2. Hence, option 'a' is incorrect.
b) xy = 1
Similarly, taking the derivative dy/dx, we get:
y + x * dy/dx = 0
dy/dx = -y/x
The slope of the tangent at any point (x, y) on the curve is -y/x.
Using the same approach as above, we find that the area of triangle OAB is always 2 square units for this curve. Therefore, option 'b' is correct.
c) y = 1/(1 + x^2)
Taking the derivative dy/dx, we get:
dy/dx = -2x/(1 + x^2)^2
The slope of the tangent at any point (x, y) on the curve is -2x/(1 + x^2)^2.
Using the same approach as above, we find that the area of triangle OAB is not always 2 square units for this curve. Therefore, option 'c' is incorrect.
d) the curve cannot be a straight line
This statement is true because the area of triangle OAB will always be zero if the curve is a straight line passing through the origin (0,0). Therefore, option 'd' is correct.
In conclusion, the correct options are 'A' and 'D'. The curve can be represented by the equation xy = 1 or it cannot be a straight line.
a) one such curve is x * y = 2
b) one such curve is xy = 1
c) one such curve is y = 1/(1 + x^2)
d) the curve cannot be a straight line
Let's evaluate each option to determine if it satisfies the given condition.
a) x * y = 2
To find the equation of the tangent at any point (x, y) on this curve, we need to find the derivative dy/dx.
Taking the derivative of both sides with respect to x, we get:
y + x * dy/dx = 0
dy/dx = -y/x
The slope of the tangent at any point (x, y) on the curve is -y/x.
Now, let A and B be the x and y-intercepts of the tangent. The area of triangle OAB is given by:
Area = 1/2 * |A * B|
Since A and B are the x and y-intercepts, respectively, we have A = 1/y and B = 1/x.
Substituting these values, we get:
Area = 1/2 * |(1/y) * (1/x)|
= 1/2 * |1/(x * y)|
= 1/2 * |1/2|
= 1 square unit
Therefore, the area of the triangle is not always 2 square units for the curve x * y = 2. Hence, option 'a' is incorrect.
b) xy = 1
Similarly, taking the derivative dy/dx, we get:
y + x * dy/dx = 0
dy/dx = -y/x
The slope of the tangent at any point (x, y) on the curve is -y/x.
Using the same approach as above, we find that the area of triangle OAB is always 2 square units for this curve. Therefore, option 'b' is correct.
c) y = 1/(1 + x^2)
Taking the derivative dy/dx, we get:
dy/dx = -2x/(1 + x^2)^2
The slope of the tangent at any point (x, y) on the curve is -2x/(1 + x^2)^2.
Using the same approach as above, we find that the area of triangle OAB is not always 2 square units for this curve. Therefore, option 'c' is incorrect.
d) the curve cannot be a straight line
This statement is true because the area of triangle OAB will always be zero if the curve is a straight line passing through the origin (0,0). Therefore, option 'd' is correct.
In conclusion, the correct options are 'A' and 'D'. The curve can be represented by the equation xy = 1 or it cannot be a straight line.
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Tangent at any point to a curve in the first quadrant meets the coordinate axes at A and B such that area of triangle OAB is always 2 square units. If the curve passes through (1, 1), thena)one such curve is x + y = 2b)one such curve is xy = 1c)one such curve is y = 1/(1+x2)d)the curve cannot be a straight lineCorrect answer is option 'A,B'. Can you explain this answer?
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Tangent at any point to a curve in the first quadrant meets the coordinate axes at A and B such that area of triangle OAB is always 2 square units. If the curve passes through (1, 1), thena)one such curve is x + y = 2b)one such curve is xy = 1c)one such curve is y = 1/(1+x2)d)the curve cannot be a straight lineCorrect answer is option 'A,B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Tangent at any point to a curve in the first quadrant meets the coordinate axes at A and B such that area of triangle OAB is always 2 square units. If the curve passes through (1, 1), thena)one such curve is x + y = 2b)one such curve is xy = 1c)one such curve is y = 1/(1+x2)d)the curve cannot be a straight lineCorrect answer is option 'A,B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Tangent at any point to a curve in the first quadrant meets the coordinate axes at A and B such that area of triangle OAB is always 2 square units. If the curve passes through (1, 1), thena)one such curve is x + y = 2b)one such curve is xy = 1c)one such curve is y = 1/(1+x2)d)the curve cannot be a straight lineCorrect answer is option 'A,B'. Can you explain this answer?.
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