Q72.If for p ≠q ≠0 , then function f(x) = 7√p(729+x)−3 is continuous at x = 0 , then 3√729+qx−9 (1) 7pqf(0) −1 = 0 (2) 63qf(0) −p2 = 0 (3) 21qf(0) −p2 = 0 (4) 7pq f(0) −9 = 0
What This Question Tests
The problem requires evaluating a limit of an indeterminate form (0/0) for a function to be continuous at x=0, which can be done using L'Hopital's rule or algebraic manipulation with binomial approximation, and then relating it to the given options.
Concepts Tested
Formulas Used
lim (x->0) ( (a+x)^n - a^n ) / x = n*a^(n-1)
Condition for continuity: lim(x->c) f(x) = f(c)
📚 NCERT Sections This Tests
1.3 — Define The Following Terms:
Chemistry Class 11 · Chapter 1
1.3 Define the following terms: (i) Mole fraction (ii) Molality (iii) Molarity (iv) Mass percentage.
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.
14.2 — Which Of The Statements Given In Exercise 14.1 Is True For P-Type
Physics Class 12 · Chapter 14
14.2 Which of the statements given in Exercise 14.1 is true for p-type semiconductos.
📋 Question Details
- Chapter
- Limits & Continuity
- Topic
- Continuity of a function
- Year
- 2022
- Shift
- 27 Jul Shift 2
- Q Number
- Q72
- Type
- MCQ
- NCERT Ref
- Class 12 Mathematics Ch 5: Continuity and Differentiability
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