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Question 15 The graph drawn below depicts (A) ๐‘ฆ=ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1) ๐‘ฅ (B) ๐‘ฆ=ใ€–๐‘๐‘œ๐‘ ใ€—^(โˆ’1) ๐‘ฅ (C) ๐‘ฆ=ใ€–๐‘๐‘œ๐‘ ๐‘’๐‘ใ€—^(โˆ’1) ๐‘ฅ (D) ๐‘ฆ=ใ€–๐‘๐‘œ๐‘กใ€—^(โˆ’1) ๐‘ฅ We can solve this using options From graph For x = โˆ’1, y = ๐œ‹ For x = 0, y = ๐œ‹/2 For x = 1, y = 0 Checking all options for x = โˆ’1, 0, 1 For x = โˆ’1 Option (A) - y So, the correct answer is (B) Question 15 The graph drawn below depicts (A) y=sin^(โˆ’1) x (B) ๐‘ฆ=ใ€–๐‘๐‘œ๐‘ ใ€—^(โˆ’1) ๐‘ฅ (C) ๐‘ฆ=ใ€–๐‘๐‘œ๐‘ ๐‘’๐‘ใ€—^(โˆ’1) ๐‘ฅ (D) ๐‘ฆ=ใ€–๐‘๐‘œ๐‘กใ€—^(โˆ’1) ๐‘ฅ We can solve this using options From graph For x = โˆ’1, y = ๐œ‹ For x = 0, y = ๐œ‹/2 For x = 1, y = 0 If we do normal trigonometry, then this basically means, For ฮธ = ๐œ‹, then __ ฮธ = โˆ’1 For ฮธ = ๐œ‹/2, then __ ฮธ = 0 For ฮธ = 0, then __ ฮธ = 1 Since cos 0 = 1 And, cos ๐œ‹/2 = 0 โˆด This is graph of ๐’š=ใ€–๐’„๐’๐’”ใ€—^(โˆ’๐Ÿ) ๐’™ So, the correct answer is (B)

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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