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Question 8 - CBSE Class 12 Sample Paper for 2020 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at Dec. 16, 2024 by Teachoo

∫ dx/√(9 - 25x 2 )

(a) sin -1 (5x/3)+ c                                

(b) 1/5 sin -1 (5x/3) + c

(c) 1/6 log ((3 + 5x)/(3 -5x)) + c          

(d) 1/30 log ((3 + 5x)/(3 -5x)) + c

Note : - This is same as Ex 7.4, 4 of NCERT – Chapter 7 Class 12

Check the answer here https://www.teachoo.com/5001/719/Ex-7.4--4---Integrate-1---root-9---25-x2---Class-12-NCERT/category/Ex-7.4/

 

Transcript

Question 8 ∫ 𝑑𝑥/√(9 − 25𝑥^2 ) (a) sin-1 (5𝑥/3)+ c (b) 1/5 sin-1 (5𝑥/3) + c (c) 1/6 log ((3 + 5𝑥)/(3 −5𝑥)) + c (d) 1/30 log ((3 + 5𝑥)/(3 −5𝑥)) + c ∫1▒1/√(9 − 25𝑥^2 ) 𝑑𝑥 =∫1▒1/√(25(9/25 − 𝑥^2 ) ) 𝑑𝑥 =∫1▒1/(√25 √(9/25 − 𝑥^2 )) 𝑑𝑥 =1/5 ∫1▒1/√(9/25 − 𝑥^2 ) 𝑑𝑥 =1/5 ∫1▒1/√((3/5)^2 − 𝑥^2 ) 𝑑𝑥 It is of the form ∫1▒𝑑𝑥/√(𝑎^2 − 𝑥^2 ) =〖𝑠𝑖𝑛〗^(−1)⁡〖𝑥/𝑎〗 +𝑐 ∴ Replacing 𝑎 by 3/5 , we get =1/5 [sin^(−1)⁡〖𝑥/(3/5)〗 +𝐶1] =𝟏/𝟓 〖𝒔𝒊𝒏〗^(−𝟏)⁡〖𝟓𝒙/𝟑〗 +𝑪 So, (b) is the correct answer
  1. Class 12
  2. Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

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Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

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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