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Question
If f '(x) = `1/"x" + "x"` and f(1) = `5/2`, then f(x) = log x + `"x"^2/2` + ______
Solution
If f '(x) = `1/"x" + "x"` and f(1) = `5/2`, then f(x) = log x + `"x"^2/2` + 2
Explanation:
f(x) = ∫f '(x) dx
`= int (1/"x" + "x")` dx
f(x) = log |x| + `"x"^2/2 + "c"` ...(i)
f(1) = `5/2`
f(1) = log 1 + `1^2/2` + c
∴ `5/2 = 0 + 1/2 + "c"` ...(∵ log 1 = 0)
∴ c = `5/2 - 1/2`
∴ = `4/2` = 2
∴ c = 2
∴ f(x) = log |x| + `"x"^2/2` + 2
Notes
The answer in the textbook is incorrect.
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