Q73. equals limx→0 x2 (1) −π (2) 1 (3) −1 (4) π
What This Question Tests
The problem requires evaluating a limit involving trigonometric functions. It can be solved by using L'Hopital's rule or by approximating `sin(pi cos^2 x)` using `sin(pi-theta) = sin(theta)` and small angle approximations for sine.
Concepts Tested
Formulas Used
lim (x->0) sinx/x = 1
lim (x->0) sin(f(x))/f(x) = 1
cos^2 x = (1+cos 2x)/2
📚 NCERT Sections This Tests
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14.2 Which of the statements given in Exercise 14.1 is true for p-type semiconductos.
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📋 Question Details
- Chapter
- Limits & Continuity
- Topic
- Evaluation of limits using L'Hopital's rule or series expansion
- Year
- 2012
- Shift
- 26 May Online
- Q Number
- Q73
- Type
- MCQ
- NCERT Ref
- Class 12 Mathematics Ch 5: Continuity and Differentiability (Limits section is generally prerequisites)
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