Question 8 - CBSE Class 12 Sample Paper for 2020 Boards

Last updated at Oct. 9, 2019 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

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Question 8 - CBSE Class 12 Sample Paper for 2020 Boards

∫ 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

∫ dx/√(9 - 25x ² )

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

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