Q72.The number of critical points of the function f(x) = (x −2)2/3(2x + 1) is (1) 1 (2) 2 (3) 0 (4) 3 6
What This Question Tests
The question involves finding the local maxima of a trigonometric function by determining its critical points and using the first derivative test for sign change in the given interval.
Concepts Tested
Formulas Used
f'(x) = 0 for critical points
Second derivative test or sign change of f'(x) for local extrema
📚 NCERT Sections This Tests
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9.17 — (A) Sin I¢C = 1.44/1.68 Which Gives I¢C = 59°. Total Internal Reflection
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9.17 (a) sin i¢c = 1.44/1.68 which gives i¢c = 59°. Total internal reflection takes place when i > 59° or when r < rmax = 31°. Now, (sin i /sin r max max ) = 1.68 , which gives imax ~ 60°. Thus, all incident rays of angles in the range 0 < i < 60° will suffer total internal reflections in the pipe. (If the length of the pipe is finite, which it is in practice, there will be a lower limit on i determined by the ratio of the diameter to the length of the pipe.) (b) If there is no outer coating, i¢c = sin–1(1/1.68) = 36.5°. Now, i = 90° will have r = 36.5° and i¢ = 53.5° which is greater than i¢c. Thus, all incident rays (in the range 53.5° < i < 90°) will suffer total internal reflections.
📋 Question Details
- Chapter
- Applications of Derivatives
- Topic
- Local maxima and minima
- Year
- 2024
- Shift
- 08 Apr Shift 1
- Q Number
- Q72
- Type
- MCQ
- NCERT Ref
- Class 12 Mathematics Ch 6: Applications of Derivatives
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