Check sibling questions

Assertion (A): The domain of the function sec -1 2x is (-∞,1/2]∪[1/2,∞ )

Reason (R ): sec -1 (-2)=-π/4

 

In the following questions, a statement of assertion (A) is followed by a statement of Reason (R). Choose the correct answer out of the following choices.

(a) Both A and R are true and R is the correct explanation of A.

(b) Both A and R are true but R is not the correct explanation of A.

(c) A is true but R is false.

(d) A is false but R is true.

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Transcript

Question 19 Assertion (A): The domain of the function 〖𝑠𝑒𝑐〗^(βˆ’1) 2π‘₯ 𝑖𝑠 (βˆ’βˆž,1/2]βˆͺ[1/2,∞ ) Reason (R): 〖𝑠𝑒𝑐〗^(βˆ’1) (βˆ’2)=βˆ’πœ‹/4 In the following questions, a statement of assertion (A) is followed by a statement of Reason (R). Choose the correct answer out of the following choices. (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true. Checking Assertion Assertion (A): The domain of function 〖𝑠𝑒𝑐〗^(βˆ’1) 2π‘₯ 𝑖𝑠 (βˆ’βˆž,1/2]βˆͺ[1/2,∞ ) We know that Domain of 〖𝑠𝑒𝑐〗^(βˆ’1) ΞΈ 𝑖𝑠 (βˆ’βˆž,1]βˆͺ[1,∞ ) Thus, βˆ’βˆž<θ≀1 and 1≀ ΞΈ<∞ Putting ΞΈ = 2x βˆ’βˆž<2π‘₯≀1 and 1≀2π‘₯<∞ Dividing both sides by 2 βˆ’βˆž/2<2π‘₯/2≀1/2 and 1/2≀2π‘₯/2<∞/2 βˆ’βˆž<π’™β‰€πŸ/𝟐 and 𝟏/πŸβ‰€π’™<∞ Thus, Domain of 〖𝑠𝑒𝑐〗^(βˆ’1) 2π‘₯ 𝑖𝑠 (βˆ’βˆž,𝟏/𝟐]βˆͺ[𝟏/𝟐,∞ ) Thus, Assertion is true Checking Reason Reason (R): 〖𝑠𝑒𝑐〗^(βˆ’1) (βˆ’2)=βˆ’πœ‹/4 Let y = secβˆ’1 (βˆ’2) y = πœ‹βˆ’ secβˆ’1 (2) y = πœ‹ βˆ’ πœ‹/3 y = πŸπ…/πŸ‘ Since Range of secβˆ’1 is [0, Ο€] – {πœ‹/2} Hence, Principal Value is πŸπ…/πŸ‘ Thus, Reasoning is false So, Assertion is true Reasoning is false So, the correct answer is (c)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.