Let a curve y=f(x), x in(0, infty) pass throu | vedclass

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(C) The equation of the tangent at $R(b, f(b))$ is given by:$y - f(b) = f'(b)(x - b)$Since this tangent passes through $S(0, c)$,we have:$c - f(b) = f

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$5$

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(C) The equation of the tangent at $R(b, f(b))$ is given by:$y - f(b) = f'(b)(x - b)$Since this tangent passes through $S(0, c)$,we have:$c - f(b) = f'(b)(0 - b)$$c - f(b) = -b f'(b)$Given $bc = 3$,we have $c = \frac{3}{b}$. Substituting this into the equation:$\frac{3}{b} - f(b) = -b f'(b)$$b f'(b) - f(b) = -\frac{3}{b}$Dividing both sides by $b^2$:$\frac{b f'(b) - f(b)}{b^2} = -\frac{3}{b^3}$This is the derivative of $\frac{f(b)}{b}$ with respect to $b$:$\frac{d}{db} \left( \frac{f(b)}{b} \right) = -\frac{3}{b^3}$Integrating both sides with respect to $b$:$\frac{f(b)}{b} = \int -3b^{-3} db = \frac{3}{2b^2} + \lambda$$f(b) = \frac{3}{2b} + \lambda b$Since the curve passes through $P(1, 3/2)$:$\frac{3}{2} = \frac{3}{2(1)} + \lambda(1) \Rightarrow \lambda = 0$Thus,$f(x) = \frac{3}{2x}$.Since the curve passes through $Q(a, 1/2)$:$f(a) = \frac{3}{2a} = \frac{1}{2} \Rightarrow a = 3$So,$Q$ is $(3, 1/2)$.The distance $PQ$ is given by:$PQ^2 = (3 - 1)^2 + (1/2 - 3/2)^2 = 2^2 + (-1)^2 = 4 + 1 = 5$.

Let a curve $y=f(x), x \in(0, \infty)$ pass through the points $P\left(1, \frac{3}{2}\right)$ and $Q\left(a, \frac{1}{2}\right)$. If the tangent at any point $R(b, f(b))$ to the given curve cuts the $y$-axis at the point $S(0, c)$ such that $bc=3$,then $(PQ)^2$ is equal to $.........$.

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