Ex 7.1, 22 - Chapter 7 Class 12 Integrals

Transcript

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﷯ (B) x3 + 1﷮𝑥4﷯ − 129﷮8﷯ ﷮﷮ 𝑑﷮𝑑𝑥﷯𝑓 𝑥﷯﷯= ﷮﷮ 4 𝑥﷮3﷯− 3﷮ 𝑥﷮4﷯﷯﷯𝑑𝑥﷯ ﷮﷮ 𝑑﷮𝑑𝑥﷯﷯𝑓 𝑥﷯=4 ﷮﷮ 𝑥﷮3﷯ 𝑑𝑥﷯−3 ﷮﷮ 1﷮ 𝑥﷮4﷯﷯ 𝑑𝑥﷯ 𝑓 𝑥﷯=4 ﷮﷮ 𝑥﷮3﷯𝑑𝑥﷯−3 ﷮﷮ 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.