Let a curve y = f ( x ) , x ( 0 ,) pass through the points P ( 1 , 3 2 ) and Q ( a , 1 2 ) . 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 b c = 3 , then ( P Q ) 2

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Solution

Q.
$Let a curve y = f (x), x \in (0, \infty) pass through the points P (1, \frac{3}{2}) and Q (a, \frac{1}{2}) .$

$If the tangent at any point R (b, f (b)) to the given curve cuts the y-axis at the point S (0, c)$ $such that b c = 3, then {(P Q)}^{2} is equal to$ ______. [2023]

Ans.
(5)

$Equation of tangent at R (b, f (b)) is$

$y - f (b) = f' (b) (x - b)$

$Now, it passes through S (0, c)$

$∴$ $c - f (b) = f' (b) (0 - b)$

$\Rightarrow \frac{3}{b} - f (b) = - b f' (b) \Rightarrow b f' (b) - f (b) = - \frac{3}{b}$

$\Rightarrow \frac{b f' (b) - f (b)}{b^{2}} = - \frac{3}{b^{3}}$ $\Rightarrow$ $d (\frac{f (b)}{b}) = - \frac{3}{b^{3}}$

$\Rightarrow \frac{f (b)}{b} = \frac{3}{2 b^{2}} + M$

$Now, the curve passes through P (1, \frac{3}{2}) .$

$∴$ $\frac{3}{2} = \frac{3}{2} + M So, f (b) = \frac{3}{2 b}$

$Also, it passes through Q (a, \frac{1}{2})$

$\Rightarrow \frac{1}{2} = \frac{3}{2 a} \Rightarrow a = 3 \Rightarrow Q \neq (3, \frac{1}{2})$

$∴$ ${(P Q)}^{2} = 2^{2} + 1^{2} = 5$

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