Q74.The integral ∫20 ||x −1| −x|dx is equal to
What This Question Tests
This question requires evaluating a limit of an indeterminate form (0/0) using L'Hôpital's Rule or algebraic manipulation involving summation formulas.
Concepts Tested
Formulas Used
L'Hôpital's Rule for 0/0 form
Sum of arithmetic progression
📚 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.
12.4 — The Ground State Energy Of Hydrogen Atom Is –13.6 Ev. What Are The
Physics Class 12 · Chapter 12
12.4 The ground state energy of hydrogen atom is –13.6 eV. What are the kinetic and potential energies of the electron in this state?
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.
📋 Question Details
- Chapter
- Calculus
- Topic
- Limits and Continuity
- Year
- 2020
- Shift
- 02 Sep Shift 1
- Q Number
- Q74
- Type
- Numerical
- NCERT Ref
- Class 12 Mathematics Ch 5: Continuity and Differentiability (Limits are a prerequisite)
More from this Chapter
Q64.If the tangent to the curve y = x + sin y at a point (a, b) is parallel to the line joining (0, 23 ) and ( 21 , 2) , then (1) b = a (2) |b −a| = 1 (3) |a + b| = 1 (4) b = π2 + a JEE Main 2020 (02 Sep Shift 1) JEE Main Previous Year Paper
Q65.If p(x) be a polynomial of degree three that has a local maximum value 8 at x = 1 and a local minimum value 4 at x = 2 then p(0) is equal to (1) 6 (2) −12 (3) 24 (4) 12
Q66.Let P(h, k) be a point on the curve y = x2 + 7x + 2 , nearest to the line, y = 3x −3 . Then the equation of the normal to the curve at P is (1) x + 3y + 26 = 0 (2) x + 3y −62 = 0 (3) x −3y −11 = 0 (4) x −3y + 22 = 0
Q67.Area (in sq. units) of the region outside |x|2 + |y|3 = 1 and inside the ellipse x24 + y29 = 1 is (1) 6(π −2) (2) 3(π −2) (3) 3(4 −π) (4) 6(4 −π) x, y > 0, y(0) = 1. If y(π) = a