Question

Let f be a positive derivable function which satisfies the relation
f ²(xy) = f ²(x) + f ²(y) + 3xy – 3x – 3y $\forall$ x, y > 0 and f '(1) = $\frac{1}{\sqrt{3}}$

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Linked Question 1

The value of $\underset{x \to \infty}{Lim} \frac{\sqrt{27 x + 1}}{f (x)}$ is equal to

$9 \sqrt{3}$

$3 \sqrt{3}$

2 f (y) f '(y) = 3 – $\frac{1}{y}$
$\int 2 f (y) f' (y) dy = \int (3 - \frac{1}{y}) dy + C$
(f(y))² = 3y – ln y + c
       y = 1, 3 = 3 – 0 + c ⇒ c = 0
$∴ f (y) = \sqrt{3 y - \ln y}$
$\Rightarrow f (x) = \sqrt{3 x - \ln x}$
(i) $\underset{x \to 0}{Lim} \frac{\sqrt{27 x + 1}}{\sqrt{3 x - \ln x}} = \underset{x \to 0}{Lim} \frac{\sqrt{x} \sqrt{27 + \frac{1}{x}}}{\sqrt{x} \sqrt{3 - \frac{\ln x}{x}}} = 3$
(ii) $\int \frac{\ln x - 1}{x^{\frac{3}{2}} f (x)} dx = \int \frac{\ln x - 1}{x^{\frac{3}{2}} \sqrt{3 x - \ln x}} dx = \int \frac{\ln x - 1}{x^{2} \sqrt{3 - \frac{\ln x}{x}}} dx$
Putting 3 – $\frac{\ln x}{x}$ = t²
$- (\frac{x \cdot \frac{1}{x} - l n x}{x^{2}}) dx$ = 2t dt   ⇒ $\frac{\ln x - 1}{x^{2}}$ dx = 2t dt
$∴ \int \frac{2 t dt}{t}$ dt   ⇒ 2t + c ⇒ 2 $\sqrt{3 - \frac{l n x}{x}} + c \Rightarrow 2 \sqrt{\frac{3 x - l n x}{x}} + c \Rightarrow \frac{2 f (x)}{\sqrt{x}} + c$

Linked Question 2

$\int \frac{\ln x - 1}{x^{\frac{3}{2}} f (x)} dx$ equals

[Note : C is the constant of integration.]

$\frac{2 \sqrt{f (x)}}{x^{\frac{3}{2}}} + C$

$2 \sqrt{\frac{f (x)}{x}} + C$

$\sqrt{\frac{f (x)}{x}} + C$

$\frac{2 f (x)}{\sqrt{x}} + C$