Let f be a differentiable function in the interval (0, ) such that f(1) =1 and for each x 0. Then 2f(2) + 3f(3) is equal to __________. [2024]

Question

Let f be a differentiable function in the interval (0, $\infty$ ) such that f(1) =1 and $\lim_{t \to x} \frac{t^{2} f (x) - x^{2} f (t)}{t - x} = 1$ for each x > 0. Then 2f(2) + 3f(3) is equal to __________. [2024]

Smart Achievers · Accepted Answer

(24)

We Have $\underset{t \to x}{l t} \frac{t^{2} f (x) - x^{2} f (t)}{t - x} = 1 (\frac{0}{0} form)$

$\Rightarrow \underset{t \to x}{l t} \frac{2 t f (x) - x^{2} f' (t)}{1} = 1 [L' Hospital Rule]$

$\Rightarrow 2 x f (x) - x^{2} f' (x) = 1 \Rightarrow f' (x) - \frac{2}{x} f (x) = \frac{- 1}{x^{2}}$

$I F = e^{\int \frac{- 2}{x} d x} = e^{- 2 \ln x} = \frac{1}{x^{2}}$

$∴ f (x) \frac{1}{x^{2}} = \int \frac{- 1}{x^{2}} \cdot \frac{1}{x^{2}} d x + C \Rightarrow f (x) \frac{1}{x^{2}} = \frac{1}{3 x^{3}} + C$

Now, f(1) = 1

$\Rightarrow \frac{2}{3} = C \Rightarrow f (x) = \frac{1}{3 x} + \frac{2 x^{2}}{3}$

Now, $2 f (2) + 3 f (3) = 2 [\frac{1}{6} + \frac{8}{3}] + 3 [\frac{1}{9} + \frac{18}{3}]$

$= \frac{1}{3} + \frac{16}{3} + \frac{1}{3} + 18 = 6 + 18 = 24$ .

Solution

Q.
Let f be a differentiable function in the interval (0, $\infty$ ) such that f(1) =1 and $\lim_{t \to x} \frac{t^{2} f (x) - x^{2} f (t)}{t - x} = 1$ for each x > 0. Then 2f(2) + 3f(3) is equal to __________. [2024]

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Solution

Q. Let f be a differentiable function in the interval (0, ∞) such that f(1) =1 and limt→xt2f(x)–x2f(t)t–x=1 for each x > 0. Then 2f(2) + 3f(3) is equal to __________. [2024]

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Q.
Let f be a differentiable function in the interval (0, $\infty$ ) such that f(1) =1 and $\lim_{t \to x} \frac{t^{2} f (x) - x^{2} f (t)}{t - x} = 1$ for each x > 0. Then 2f(2) + 3f(3) is equal to __________. [2024]