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

Reason (R ): sec -1 (-2)=-Ο€/4

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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.

[Class 12] Assertion (A): The domain of the function sec^-1 2x is (βˆ’βˆž, - CBSE Class 12 Sample Paper for 2023 Boards

part 2 - Question 19 [Assertion Reasoning] - CBSE Class 12 Sample Paper for 2023 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12
part 3 - Question 19 [Assertion Reasoning] - CBSE Class 12 Sample Paper for 2023 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12
part 4 - Question 19 [Assertion Reasoning] - CBSE Class 12 Sample Paper for 2023 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12
part 5 - Question 19 [Assertion Reasoning] - CBSE Class 12 Sample Paper for 2023 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12

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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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo