Q65.If the function f(x) = {k1(xk2−π)2cos x,−1, xx ≤π> π to: (1) ( 21 , 1) (2) (1, 0) (3) ( 21 , −1) (4) (1, 1) + c, where c is a constant of integration, then g(0) is
What This Question Tests
This question requires ensuring continuity and differentiability of a piecewise function at the point where the definition changes, often involving L'Hôpital's Rule for evaluating limits of derivatives.
Concepts Tested
Formulas Used
lim f(x) as x→a
f'(x) at x=a
L'Hôpital's Rule
📚 NCERT Sections This Tests
1.18 — A Point Charge Of 2.0 Mc Is At The Centre Of A Cubic Gaussian
Physics Class 11 · Chapter 1
1.18 A point charge of 2.0 mC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?
2.1 — Two Charges 5 × 10–8 C And –3 × 10–8 C Are Located 16 Cm Apart. At
Physics Class 11 · Chapter 2
2.1 Two charges 5 × 10–8 C and –3 × 10–8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.
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.
📋 Question Details
- Chapter
- Limits & Continuity
- Topic
- Differentiability of piecewise functions
- Year
- 2020
- Shift
- 05 Sep Shift 1
- Q Number
- Q65
- Type
- MCQ
- NCERT Ref
- Class 12 Mathematics Ch 5: Continuity and Differentiability
More from this Chapter
Q97.The function f : R ∼{0} →R given by f(x) = x1 − e2x−12 can be made continuous at x = 0 by defining f(0) as (1) 2 (2) −1 (3) 0 (4) 1
Q92.Let f(x) = −1) sin ( {(x0, if x = 1 JEE Main 2008 JEE Main Previous Year Paper (1) f is neither differentiable at x = 0 nor at x = 1 (2) f is differentiable at x = 0 and at x = 1 (3) f is differentiable at x = 0 but not at x = 1 (4) f is differentiable at x = 1 but not at x = 0
Q70.Let f : R →R be a positive increasing function with limx→∞ f(3x)f(x) = 1. Then limx→∞ f(2x)f(x) (1) 2 (2) 3 3 2 (3) 3 (4) 1
Q71.Consider the following statements P : Suman is brilliant Q : Suman is rich R : Suman is honest The negation of the statement "Suman is brilliant and dishonest if and only if Suman is rich" can be expressed as (1) ∼(Q ↔(P∧∼R)) (2) ∼Q ↔∼P ∧R (3) ∼(P∧∼R) ↔Q (4) ∼P ∧(Q ↔∼R)