Let f-0-0- - be a differentiable function such that f1 - e and limt - x t2f2x-x2f2tt-x-0- If fx-1- then x is equal to

Lim_t→ xt^2f^2(x) x^2f^2(t)t x=0 Usingf L'Hospital rule ⇒ nbsp;lim_t → x 2t· f^2(x) x^2(2f(t))· f'(t)1=0 ⇒ 2xf^2(x) 2x^2f(x)f'(x)=0 ⇒ 2x f(x)(f(x) xf'(x))=0 ⇒ ...

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Standard XII

Mathematics

L'Hospital Rule to Remove Indeterminate Form

Let f:0,∞→0, ...

Question

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

e

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2 e

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\frac{1}{e}

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\frac{1}{2 e}

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Solution

The correct option is C $\frac{1}{e}$
$lim t \to x \frac{t^{2} f^{2} (x) - x^{2} f^{2} (t)}{t - x} = 0$
Usingf L'Hospital rule
$\Rightarrow lim t \to x \frac{2 t \cdot f^{2} (x) - x^{2} (2 f (t)) \cdot f^{'} (t)}{1} = 0$
$\Rightarrow 2 x f^{2} (x) - 2 x^{2} f (x) f^{'} (x) = 0$
$\Rightarrow 2 x f (x) (f (x) - x f^{'} (x)) = 0$
$\Rightarrow \frac{f^{'} (x)}{f (x)} = \frac{1}{x}$ $(∵ x \neq 0, f (x) \neq 0)$
$\Rightarrow ln f (x) = ln (x) + ln C$
$\Rightarrow f (x) = C x$
If $x = 1 \Rightarrow C = e$
$∴ f (x) = e x$
If $f (x) = 1 \Rightarrow x = \frac{1}{e}$

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Let f:(0,∞)→(0,∞) be a differentiable function such that f(1)=e and limt→xt2f2(x)−x2f2(t)t−x=0. If f(x)=1, then x is equal to

The correct option is C 1elimt→xt2f2(x)−x2f2(t)t−x=0 Usingf L'Hospital rule ⇒limt→x2t⋅f2(x)−x2(2f(t))⋅f′(t)1=0 ⇒2xf2(x)−2x2f(x)f′(x)=0 ⇒2xf(x)(f(x)−xf′(x))=0 ⇒f′(x)f(x)=1x (∵x≠0,f(x)≠0) ⇒lnf(x)=ln(x)+lnC ⇒f(x)=Cx If x=1⇒C=e ∴f(x)=ex If f(x)=1⇒x=1e

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