Q84.Let f : [−2, 3] →[0, ∞) be a continuous function such that f(1 −x) = f(x) for all x ∈[−2, 3]. If R1 is the numerical value of the area of the region bounded by y = f(x), x = −2, x = 3 and the axis of x and R2 = ∫3−2 xf(x)dx, then : (1) 3R1 = 2R2 (2) 2R1 = 3R2 (3) R1 = R2 (4) R1 = 2R2
What This Question Tests
This question combines the concept of area under a curve with properties of definite integrals, particularly those related to symmetric intervals and even/odd functions, modified by a given functional property f(1-x) = f(x).
Concepts Tested
Formulas Used
R1 = ∫₋₂³ f(x)dx
R2 = ∫₋₂³ xf(x)dx
∫₋ₐᵃ xf(x)dx = 0 if f is even
∫₋ₐᵃ f(x)dx = 2∫₀ᵃ f(x)dx if f is even
📚 NCERT Sections This Tests
9.15 — Apply Mirror Equation And The Condition:
Physics Class 12 · Chapter 9
9.15 Apply mirror equation and the condition: (a) f < 0 (concave mirror); u < 0 (object on left) (b) f > 0; u < 0 (c) f > 0 (convex mirror) and u < 0 (d) f < 0 (concave mirror); f < u < 0 to deduce the desired result.
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📋 Question Details
- Chapter
- Definite Integration & Area
- Topic
- Properties of definite integrals and area under curve
- Year
- 2013
- Shift
- 25 Apr Online
- Q Number
- Q84
- Type
- MCQ
- NCERT Ref
- Class 12 Mathematics Ch 7: Integrals
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