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Question
If Rolle’s theorem holds for the function f(x) = (x –2) log x, x ∈ [1, 2], show that the equation x log x = 2 – x is satisfied by at least one value of x in (1, 2).
Solution
The Rolle’s theorem holds for the function f(x) = (x – 2) log x, x ∈ [1, 2].
∴ there exists t least onee real number c ∈ (1, 2) such that f'(c) 0.
Now, f(x) = (x – 2) log x
∴ f'(x) = `d/dx[(x - 2) log x]`
= `(x - 2)."d"/"dx"(logx) + logx.d/dx(x - 2)`
= `(x - 2) xx (1)/x + (logx)(1 - 0)`
= `1 - (2)/x + logx`
∴ f'(c) = `1 - (2)/c + logc`
∴ f'(c) = 0 gives `1 - (2)/c + logc` = 0
∴ c – 2 + c log c = 0
∴ c log c = 2 – c, where c ∈ (1, 2)
∴ c satisfies the equation x log x = 2 – x, c ∈ (1, 2).
Hence, the equation log x = 2 – x is satisfied by at least one value of x i(1, 2).
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