Q82.The maximum area (in sq. units) of a rectangle having its base on the x− axis and its other two vertices on the parabola, y = 12 −x2 such that the rectangle lies inside the parabola, is : (1) 20√2 (2) 32 (3) 36 (4) 18√3
What This Question Tests
This question requires setting up an area function for the rectangle based on the parabola's equation, then finding its maximum value by using the first derivative test (setting the derivative to zero).
Concepts Tested
Formulas Used
Area = length * width
d/dx(f(x))=0 for extremum
📚 NCERT Sections This Tests
9.5 — A Small Bulb Is Placed At The Bottom Of A Tank Containing Water To A
Physics Class 12 · Chapter 9
9.5 A small bulb is placed at the bottom of a tank containing water to a depth of 80cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)
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.
9.18 — For Fixed Distance S Between Object And Screen, The Lens Equation
Physics Class 12 · Chapter 9
9.18 For fixed distance s between object and screen, the lens equation does not give a real solution for u or v if f is greater than s/4. Therefore, fmax = 0.75 m.
📋 Question Details
- Chapter
- Applications of Derivatives
- Topic
- Maxima and Minima
- Year
- 2019
- Shift
- 12 Jan Shift 1
- Q Number
- Q82
- Type
- MCQ
- NCERT Ref
- Class 12 Mathematics Ch 6: Applications of Derivatives
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