Ex 7.2, 38 - Class 12
Last updated at Dec. 8, 2016 by Teachoo
Transcript
Ex 7.2, 38 10𝑥9+ 10𝑥 log𝑒10𝑥10+ 10𝑥 dx equals (A) 10𝑥 – 𝑥10 + 𝐶 (B) 10𝑥+ 𝑥10+𝐶 (C) 10𝑥 – 𝑥10−1 + 𝐶 (D) log 10𝑥+𝑥10 + 𝐶 Step 1: Let 𝑥10+ 10𝑥= 𝑡 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 10𝑥10−1+ 10𝑥 𝑙𝑜𝑔10= 𝑑𝑡𝑑𝑥 10𝑥9+ 10𝑥 𝑙𝑜𝑔10= 𝑑𝑡𝑑𝑥 𝑑𝑥= 𝑑𝑡 10𝑥9 + 10𝑥 𝑙𝑜𝑔10 Step 2: Integrating the function 10𝑥 910𝑥 𝑙𝑜𝑔10 𝑥10+ 10𝑥 . 𝑑𝑥 Putting 𝑥10+ 10𝑥=𝑡 & 𝑑𝑥= 𝑑𝑡 10𝑥9 + 10𝑥 𝑙𝑜𝑔10 = 10𝑥 910𝑥 𝑙𝑜𝑔10𝑡 . 𝑑𝑡 10𝑥9+ 10𝑥 𝑙𝑜𝑔10 = 1𝑡.𝑑𝑡 = log 𝑡+𝐶 = log 10𝑥+ 𝑥10+𝐶 ∴ Option D is correct.
Ex 7.2
Ex 7.2, 1
Ex 7.2, 2
Ex 7.2, 3
Ex 7.2, 4
Ex 7.2, 5
Ex 7.2, 6
Ex 7.2, 7
Ex 7.2, 8
Ex 7.2, 9
Ex 7.2, 10
Ex 7.2, 11
Ex 7.2, 12
Ex 7.2, 13
Ex 7.2, 14
Ex 7.2, 15
Ex 7.2, 16
Ex 7.2, 17
Ex 7.2, 18
Ex 7.2, 19
Ex 7.2, 20 Important
Ex 7.2, 21
Ex 7.2, 22
Ex 7.2, 23
Ex 7.2, 24
Ex 7.2, 25
Ex 7.2, 26 Important
Ex 7.2, 27
Ex 7.2, 28
Ex 7.2, 29
Ex 7.2, 30
Ex 7.2, 31
Ex 7.2, 32
Ex 7.2, 33
Ex 7.2, 34
Ex 7.2, 35 Important
Ex 7.2, 36 Important
Ex 7.2, 37
Ex 7.2, 38 You are here
Ex 7.2, 39
About the Author
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.