Let be differentiable in and f(1) = 1. Then the value of ea, such that f(a) = 0, is equal to __________. [2024]

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Solution

Q.
Let $f (x) = \sqrt{\lim_{r \to x} {\frac{2 r^{2} [(f {(r))}^{2} - f (x) f (r)]}{r^{2} - x^{2}} - r^{3} e^{\frac{f (r)}{r}}}}$ be differentiable in $(- \infty, 0) \cup (0, \infty)$ and f(1) = 1. Then the value of ea, such that f(a) = 0, is equal to __________. [2024]

Ans.
(2)

$f (x) = \sqrt{\lim_{r \to x} {\frac{2 r^{2} ((f {(r))}^{2} - f (x) f (r))}{r^{2} - x^{2}} - r^{3} e^{f (r) / r}}}$

Using L'Hospital rule, we have

$f (x) = \sqrt{\lim_{r \to x} {\frac{2 r^{2} [2 f (r) \cdot f' (r) - f (x) \cdot f' (r)] + [f {(r)}^{2} - f (x) \cdot f (r)] \cdot 4 r}{2 r} - r^{3} e^{\frac{f (r)}{r}}}}$

$= \sqrt{\frac{2 x^{2} [2 f (x) \cdot f' (x) - f (x) \cdot f' (x)] + 0}{2 x} - x^{3} e^{\frac{f (x)}{x}}}$

$∴ f (x) = \sqrt{x f (x) \cdot f' (x) - x^{3} \cdot e^{\frac{f (x)}{x}}}$

Let y = f(x)

$y^{2} = x y \cdot \frac{d y}{d x} - x^{3} e^{\frac{y}{x}} \Rightarrow \frac{d y}{d x} = \frac{y^{2} + x^{3} e^{y / x}}{x y}$

Put $y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$

$\Rightarrow \frac{v^{2} x^{2} + x^{3} e^{\frac{v x}{x}}}{x \cdot v x} = v + x \frac{d v}{d x} \Rightarrow \frac{v^{2} + x e^{v}}{v} - v = x \frac{d v}{d x}$

$\Rightarrow \frac{e^{v}}{v} = \frac{d v}{d x} \Rightarrow d x = v e^{- v} d v$

$\Rightarrow \int d x = \int v \cdot e^{- v} d v \Rightarrow x + c = - e^{- v} (v + 1)$

$\Rightarrow x + c = - e^{- y / x} (\frac{y}{x} + 1)$ ... (i)

$∵ y (1) = 1$

$\Rightarrow 1 + c = - e^{- 1} (2)$

$\Rightarrow c = - 1 - \frac{2}{e}$ ... (ii)

Now, f(a) = 0

$∴ a + c = - e^{- 0 / a} (\frac{0}{a} + 1)$ [Using (i)]

a + c = –1

$a = - 1 - c = - 1 + 1 + \frac{2}{e} = \frac{2}{e}$ [Using (ii)]

$∴$ ea = 2.

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