Find the principal value of cot - 1 ( 3 )

Given that, cot - 1 ( 3 ) Let cot - 1 ( 3 ) = θ We know that, the range of principal value branch of y = cot - 1 x is ( 0 , π ) Now, 3 = cotθ ⇒ cot π 6 = cotθ ⇒ θ = π 6 ∈ 0 , π The obtained value is in the range of 0 , π . Therefore, the principal value of cot - 1 ( 3 ) = π 6 ∈ 0 , π .

Find the domain of f x = sin - 1 x 2 + 1 2 x .

Given f x = sin - 1 x 2 + 1 2 x We know that domain of sin - 1 x is - 1 , 1 Now, - 1 ≤ x 2 + 1 2 x ≤ 1 Consider, - 1 ≤ x 2 + 1 2 x ⇒ - 1 + 1 ≤ x 2 + 1 2 x + 1 ⇒ x 2 + 1 + 2 x 2 x ≥ 0 ⇒ x + 1 2 2 x ≥ 0 x + 1 2 is non-negative for all values of x . In order to get x + 1 2 2 x ≥ 0 we must have x > 0 ∴ x + 1 2 2 x ≥ 0 ⇒ x = - 1 and x > 0 Consider, x 2 + 1 2 x ≤ 1 ⇒ x 2 + 1 2 x - 1 ≤ 1 - 1 ⇒ x 2 + 1 - 2 x 2 x ≤ 0 ⇒ x - 1 2 2 x ≤ 0 x - 1 2 is non-negative for all values of x . In order to get x - 1 2 2 x ≤ 0 we must have x < 0 ∴ x - 1 2 2 x ≤ 0 ⇒ x = 1 and x < 0 . The only values of x satisfying inequalities are - 1 and 1 ∴ Domain of f x is - 1 , 1 .

E M B I B E

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 2 - Inverse Trigonometric Functions>Exercise

Embibe Experts Solutions for Chapter: Inverse Trigonometric Functions, Exercise 1: Exercise

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Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Inverse Trigonometric Functions, Exercise 1: Exercise

Attempt the practice questions on Chapter 2: Inverse Trigonometric Functions, Exercise 1: Exercise with hints and solutions to strengthen your understanding. Mathematics Crash Course (Based on Revised Syllabus-2023) solutions are prepared by Experienced Embibe Experts.