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

Reason (R ): sec -1 (-2)=-Ļ€/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 (āˆ’āˆž, - 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

Ā 

Ā 

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Transcript

Question 19 Assertion (A): The domain of the function ć€–š‘ š‘’š‘ć€—^(āˆ’1) 2š‘„ š‘–š‘  (āˆ’āˆž,1/2]∪[1/2,āˆž ) Reason (R): ć€–š‘ š‘’š‘ć€—^(āˆ’1) (āˆ’2)=āˆ’š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 ć€–š‘ š‘’š‘ć€—^(āˆ’1) 2š‘„ š‘–š‘  (āˆ’āˆž,1/2]∪[1/2,āˆž ) We know that Domain of ć€–š‘ š‘’š‘ć€—^(āˆ’1) Īø š‘–š‘  (āˆ’āˆž,1]∪[1,āˆž ) Thus, āˆ’āˆž<θ≤1 and 1≤ Īø<āˆž Putting Īø = 2x āˆ’āˆž<2š‘„ā‰¤1 and 1≤2š‘„<āˆž Dividing both sides by 2 āˆ’āˆž/2<2š‘„/2≤1/2 and 1/2≤2š‘„/2<āˆž/2 āˆ’āˆž<š’™ā‰¤šŸ/šŸ and šŸ/šŸā‰¤š’™<āˆž Thus, Domain of ć€–š‘ š‘’š‘ć€—^(āˆ’1) 2š‘„ š‘–š‘  (āˆ’āˆž,šŸ/šŸ]∪[šŸ/šŸ,āˆž ) Thus, Assertion is true Checking Reason Reason (R): ć€–š‘ š‘’š‘ć€—^(āˆ’1) (āˆ’2)=āˆ’šœ‹/4 Let y = secāˆ’1 (āˆ’2) y = šœ‹āˆ’ secāˆ’1 (2) y = šœ‹ āˆ’ šœ‹/3 y = šŸš…/šŸ‘ Since Range of secāˆ’1 is [0, Ļ€] – {šœ‹/2} Hence, Principal Value is šŸš…/šŸ‘ Thus, Reasoning is false So, Assertion is true Reasoning is false So, the correct answer is (c)

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