Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

Ex 7.1

Ex 7.1,1

Ex 7.1, 2

Ex 7.1, 3

Ex 7.1, 4

Ex 7.1, 5

Ex 7.1, 6

Ex 7.1, 7

Ex 7.1, 8

Ex 7.1, 9

Ex 7.1, 10 Important

Ex 7.1, 11

Ex 7.1, 12

Ex 7.1, 13 Important

Ex 7.1, 14

Ex 7.1, 15

Ex 7.1, 16 Important

Ex 7.1, 17

Ex 7.1, 18 Important

Ex 7.1, 19

Ex 7.1, 20

Ex 7.1, 21 (MCQ)

Ex 7.1, 22 (MCQ) Important You are here

Last updated at May 29, 2023 by Teachoo

Ex 7.1, 22 If 𝑑/𝑑𝑥 f(x) = 4x3 − 3/𝑥4 such that f(2) = 0, then f(x) is x4 + 1/𝑥3 − 129/8 (B) x3 + 1/𝑥4 + 129/8 (C) x4 + 1/𝑥3 + 129/8 (D) x3 + 1/𝑥4 − 129/8 Given 𝑑/𝑑𝑥 f(x) = 4x3 − 3/𝑥4 Integrating both sides ∫1▒〖𝑑/𝑑𝑥 𝑓(𝑥) 〗=∫1▒(4𝑥^3− 3/𝑥^4 )𝑑𝑥 ∫1▒𝑑/𝑑𝑥 𝑓(𝑥)=4∫1▒〖𝑥^3 𝑑𝑥〗−3∫1▒〖1/𝑥^4 𝑑𝑥〗 𝑓(𝑥)=4∫1▒〖𝑥^3 𝑑𝑥〗−3∫1▒〖𝑥^(−4) 𝑑𝑥〗 𝑓(𝑥)=4 𝑥^(3 + 1)/(3 + 1)−3 𝑥^(−4 + 1)/(−4 + 1)+𝐶 𝑓(𝑥)=4 𝑥^4/4 − 3 𝑥^(−3)/(−3)+𝐶 𝑓(𝑥)=𝑥^4+𝑥^(−3)+𝐶 𝑓(𝑥)=𝑥^4+ 1/𝑥^3 +𝐶 Given 𝑓(2)=0 Putting 𝑥=2 in (1) 𝑓(2)=(2)^4+ 1/(2)^3 +𝐶 0=16+ 1/8 +𝐶 0= (128 + 1)/8 +𝐶 0= 129/8 +𝐶 𝐶=(−129)/8 Putting 𝐶=(−129)/8 in (1) 𝑓(𝑥)=𝑥^4+ 1/𝑥^3 +𝐶 ⇒ 𝒇(𝒙)=𝒙^𝟒+ 𝟏/𝒙^𝟑 −𝟏𝟐𝟗/𝟖 ∴ Option (A) is correct.