"Let $f:\left( {0,\infty } \right) \to \left( {0,\infty } \right)$ be a differentiable function such that f(1) = e and $$\mathop {\lim }\limits_{t \to x} {{{t^2}{f^2}(x) - {x^2}{f^2}(t)} \over {t - x}} = 0$$. If f(x) = 1, then x is equal to :"

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Accepted Answer

"$$\mathop {\lim }\limits_{t \to x} {{{t^2}{f^2}(x) - {x^2}{f^2}(t)} \over {t - x}} = 0$$

(Using L'Hospital's Rule)

$$ \Rightarrow \mathop {\lim }\limits_{t \to x} {{2t{f^2}(x) - 2{x^2}f(t).f'(t)} \over 1} = 0$$

$\Rightarrow$ 2xf²(x) - 2x².f(x).f'(x) = 0

$\Rightarrow$ 2xf(x){f(x) - xf'(x)} = 0

[x $\ne$ 0, f(x) $\ne$ 0 as given
function $f:\left( {0,\infty } \right) \to \left( {0,\infty } \right)$ only takes positive value as input and output]

$\Rightarrow f(x) = xf'(x)$

$\Rightarrow {{f'(x)} \over {f(x)}} = {1 \over x}$

Integrating w.r.t x, we get

$\Rightarrow ln\,f(x) = ln\,x + ln\,C$

$\Rightarrow f(x) = Cx$

$\because$ f(1) = e

$\Rightarrow C = e;\,so\,f(x) = ex$

When f(x) = 1 = ex

$\Rightarrow x = {1 \over e}$"

Let $f:\left( {0,\infty } \right) \to \left( {0,\infty } \right)$ be a differentiable function such that f(1) = e and
$$\mathop {\lim }\limits_{t \to x} {{{t^2}{f^2}(x) - {x^2}{f^2}(t)} \over {t - x}} = 0$$. If f(x) = 1, then x is equal to :

Solution

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Let $f:\left( {0,\infty } \right) \to \left( {0,\infty } \right)$ be a differentiable function such that f(1) = e and $$\mathop {\lim }\limits_{t \to x} {{{t^2}{f^2}(x) - {x^2}{f^2}(t)} \over {t - x}} = 0$$. If f(x) = 1, then x is equal to :

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More Limits, Continuity and Differentiability questions

Drill 25 more like these. Every day. Free.

Let $f:\left( {0,\infty } \right) \to \left( {0,\infty } \right)$ be a differentiable function such that f(1) = e and
$$\mathop {\lim }\limits_{t \to x} {{{t^2}{f^2}(x) - {x^2}{f^2}(t)} \over {t - x}} = 0$$. If f(x) = 1, then x is equal to :