JEE MAIN - Mathematics (2020 - 4th September Evening Slot - No. 7)
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 :
$$\mathop {\lim }\limits_{t \to x} {{{t^2}{f^2}(x) - {x^2}{f^2}(t)} \over {t - x}} = 0$$. If f(x) = 1, then x is equal to :
$${1 \over e}$$
e
$${1 \over 2e}$$
2e
Explanation
$$\mathop {\lim }\limits_{t \to x} {{{t^2}{f^2}(x) - {x^2}{f^2}(t)} \over {t - x}} = 0$$
(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
$$ \Rightarrow x = {1 \over e}$$
(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
$$ \Rightarrow x = {1 \over e}$$
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