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प्रश्न
Show that the function given by `f(x) = (log x)/x` has maximum at x = e.
उत्तर
We have f (x) = `log x/x, x > 0`
Differentiating w.r.t. x , we get
⇒ `f' (x) = (x(1/x) - (log x)*1)/x^2`
`= (1 - log x)/x^2`
For maximum / minimum, f (x) = 0
⇒ `(1 - log x)/x^2 = 0`
⇒ log x = 1 .....(∵x2 ≠ 0)
⇒ x = e
Again differentiating w.r.t x, we get
`f'' (x) = (x^2 (-1/x) - (1 - log x) 2x)/x^4`
`= (-x - 2x + 2x log x)/x^4`
`= (x (2 log x - 3))/x^4`
`= (2 log x - 3)/x^3`
Also, f'' (e) = `(2 log e - 3)/e^3`
`= (2.1 - 3)/e^3` ....(∵ loge e = 1)
`= -1/e^3 < 0`
⇒ f (x) has a maximum value at x = e.
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