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Application of Integrals
NCERT Exemplar Class 12
CBSE
Grade 12
Question:
The area of region bounded by $x^2 = 4y$ and line $x = 4y - 2$ is: (a) $9/8 \text{ sq units}$ (b) $9/4 \text{ sq units}$ (c) $9/2 \text{ sq units}$ (d) $3/4 \text{ sq units}$
Step-by-Step Solution
Intersection points x = -1, x = 2. \int_{-1}^2 ((x+2)/4 - x^2/4) dx = 9/8. [1.0 Mark]
--- 🎯 Official CBSE Marking Scheme: Selecting $9/8 \text{ sq units}$: 1.0 Mark
Correct Answer:$9/8 \text{ sq units}$
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