Q83.The maximum area of a right angled triangle with hypotenuse h is : (1) h2 (2) h2 2√2 2 (3) h2 (4) h2 √2 4 = A(x)ecot−1 x + C , then A(x) is equal to :
What This Question Tests
This question tests the ability to formulate an optimization problem by expressing the area of a right-angled triangle in terms of its hypotenuse and one leg, then using calculus to find the maximum possible area.
Concepts Tested
Formulas Used
Area = (1/2) * base * height
a² + b² = h²
d/dx (Area_function) = 0
📚 NCERT Sections This Tests
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Physics Class 12 · Chapter 9
9.23 (a) At what distance should the lens be held from the card sheet in Exercise 9.22 in order to view the squares distinctly with the maximum possible magnifying power? (b) What is the magnification in this case? (c) Is the magnification equal to the magnifying power in this case? Explain.
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.)
12.5 — A Hydrogen Atom Initially In The Ground Level Absorbs A Photon,
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📋 Question Details
- Chapter
- Applications of Derivatives
- Topic
- Maxima and Minima (Optimization)
- Year
- 2013
- Shift
- 22 Apr Online
- Q Number
- Q83
- Type
- MCQ
- NCERT Ref
- Class 12 Mathematics Ch 6: Application of Derivatives
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