Q71.Let Sk = ∑kr=1 tan−1( 22r+1+32r+16r ), then k→∞Sk (1) tan−1( 23 ) (2) π2 (3) cot−1( 23 ) (4) tan−1(3)
What This Question Tests
The question tests the ability to manipulate the argument of an inverse tangent function into a form suitable for telescoping sum and then evaluate the limit of the sum as the number of terms approaches infinity.
Concepts Tested
Formulas Used
tan^-1(x) - tan^-1(y) = tan^-1((x-y)/(1+xy))
Limit of sum as k -> infinity
📚 NCERT Sections This Tests
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.
2.2 — A Regular Hexagon Of Side 10 Cm Has A Charge 5 Mc At Each Of Its
Physics Class 11 · Chapter 2
2.2 A regular hexagon of side 10 cm has a charge 5 mC at each of its vertices. Calculate the potential at the centre of the hexagon.
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
- Inverse Trigonometric Functions
- Topic
- Summation of inverse tangent series
- Year
- 2021
- Shift
- 16 Mar Shift 1
- Q Number
- Q71
- Type
- MCQ
- NCERT Ref
- Class 12 Mathematics Ch 2: Inverse Trigonometric Functions
More from this Chapter
Q90.The value of cot (cosec−1 53 + tan−1 23 ) is (1) 6 (2) 3 17 17 (3) 4 (4) 5 17 17
Q78.A value of tan−1 (sin (cos−1 (√2 ))) (1) π (2) π 4 2 (3) π (4) π 3 6
Q67.A value of x for which sin (cot−1(1 + x)) = cos (tan−1 x), is : (1) −12 (2) 1 (3) 0 (4) 1 2
Q67.The number of solutions of the equation, sin−1 x = 2 tan−1 x (in principal values) is : (1) 1 (2) 4 (3) 2 (4) 3