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JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)

Let the area of the region \(x, y) : 2y leq x^2 + 3 , y + |x| leq 3 , y geq |x - 1|\ be A. Then 6A is equal to:

Solution & Explanation

### Related Formula textArea = int_a^b (y_textupper - y_textlower) \, dx ### Core Logic Plotting the boundary lines and tracking intersection points yields a composite geometric region bounding a central area.
Area Under Curves diagram for Q57 - JEE Main 2025 Morning
Area Under Curves diagram for Q57 - JEE Main 2025 Morning
### Step 1: Set up the integral pieces The bounded region A can be conceptualized as a total bounding box/rectangle minus specific external integrals: A = 4 - 2 int_0^1 left[ (3 - x) - left( fracx^2 + 32 right) right] \, dx ### Step 2: Evaluate the Integral A = 4 - 2 left[ 3x - fracx^22 - fracx^36 - frac32x right]_0^1 A = 4 - 2 left[ 3 - frac12 - frac16 - frac32 right] = 4 - 2 left[ frac56 right] = 4 - frac53 = frac73 ### Step 3: Calculate 6A 6A = 6 times frac73 = 14 ### Pattern Recognition When dealing with multiple absolute functions (|x|, |x-1|), check for coordinate mirror symmetry across vertical axes to slash total required calculus computations in half. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Application of Integrals

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