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The area bounded by the parabolas y = (x+1)2 and y = (x−1)2 and the line y = (1/4) is equal to
- a)4/3
- b)1/3
- c)1/6
- d)4
Correct answer is option 'B'. Can you explain this answer?
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The area bounded by the parabolas y =(x+1)2 and y = (x−1)2 and t...
To find the area bounded by the parabolas y = (x - 1)^2 and y = (x + 1)^2, we need to find the points of intersection first.
Setting the two equations equal to each other, we have:
(x - 1)^2 = (x + 1)^2
Expanding both sides, we get:
x^2 - 2x + 1 = x^2 + 2x + 1
Simplifying, we have:
-4x = 4
Dividing both sides by -4, we get:
x = -1
So the two parabolas intersect at x = -1.
Now we can find the y-values of the intersection point by substituting x = -1 into either equation:
y = (-1 - 1)^2 = (-2)^2 = 4
Therefore, the two parabolas intersect at the point (-1, 4).
To find the area bounded by the parabolas, we need to find the area between the two curves from x = -1 to x = 1.
Integrating the upper curve (y = (x + 1)^2) and subtracting the integral of the lower curve (y = (x - 1)^2):
∫[(x + 1)^2 - (x - 1)^2] dx from x = -1 to x = 1
Expanding the expressions:
∫[(x^2 + 2x + 1) - (x^2 - 2x + 1)] dx from x = -1 to x = 1
Simplifying:
∫(4x) dx from x = -1 to x = 1
Integrating:
2x^2 from x = -1 to x = 1
Substituting the limits:
2(1)^2 - 2(-1)^2
2(1) - 2(1)
2 - 2
0
Therefore, the area bounded by the parabolas y = (x - 1)^2 and y = (x + 1)^2 is 0.
Setting the two equations equal to each other, we have:
(x - 1)^2 = (x + 1)^2
Expanding both sides, we get:
x^2 - 2x + 1 = x^2 + 2x + 1
Simplifying, we have:
-4x = 4
Dividing both sides by -4, we get:
x = -1
So the two parabolas intersect at x = -1.
Now we can find the y-values of the intersection point by substituting x = -1 into either equation:
y = (-1 - 1)^2 = (-2)^2 = 4
Therefore, the two parabolas intersect at the point (-1, 4).
To find the area bounded by the parabolas, we need to find the area between the two curves from x = -1 to x = 1.
Integrating the upper curve (y = (x + 1)^2) and subtracting the integral of the lower curve (y = (x - 1)^2):
∫[(x + 1)^2 - (x - 1)^2] dx from x = -1 to x = 1
Expanding the expressions:
∫[(x^2 + 2x + 1) - (x^2 - 2x + 1)] dx from x = -1 to x = 1
Simplifying:
∫(4x) dx from x = -1 to x = 1
Integrating:
2x^2 from x = -1 to x = 1
Substituting the limits:
2(1)^2 - 2(-1)^2
2(1) - 2(1)
2 - 2
0
Therefore, the area bounded by the parabolas y = (x - 1)^2 and y = (x + 1)^2 is 0.
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The area bounded by the parabolas y =(x+1)2 and y = (x−1)2 and the line y =(1/4)is equal toa)4/3b)1/3c)1/6d)4Correct answer is option 'B'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The area bounded by the parabolas y =(x+1)2 and y = (x−1)2 and the line y =(1/4)is equal toa)4/3b)1/3c)1/6d)4Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The area bounded by the parabolas y =(x+1)2 and y = (x−1)2 and the line y =(1/4)is equal toa)4/3b)1/3c)1/6d)4Correct answer is option 'B'. Can you explain this answer?.
The area bounded by the parabolas y =(x+1)2 and y = (x−1)2 and the line y =(1/4)is equal toa)4/3b)1/3c)1/6d)4Correct answer is option 'B'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The area bounded by the parabolas y =(x+1)2 and y = (x−1)2 and the line y =(1/4)is equal toa)4/3b)1/3c)1/6d)4Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The area bounded by the parabolas y =(x+1)2 and y = (x−1)2 and the line y =(1/4)is equal toa)4/3b)1/3c)1/6d)4Correct answer is option 'B'. Can you explain this answer?.
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