A curve in the first quadrant passes through the point (1, 2), and the slope of its tangent at any point (x, y) of it is \(\displaystyle\frac{y}{x(x + 1)}\). If the curve also passes through the point (3, k), then k is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 3
The slope condition is the differential equation
\( \displaystyle \frac{dy}{dx} = \frac{y}{x(x + 1)} \)
Separating the variables:
\( \displaystyle \frac{dy}{y} = \frac{dx}{x(x + 1)} \)
By partial fractions:
\( \displaystyle \frac{1}{x(x + 1)} = \frac{1}{x} - \frac{1}{x + 1} \)
Integrating both sides (x and y are positive):
\(\log y = \log x - \log(x + 1) + \log c\)
\( \displaystyle y = \frac{cx}{x + 1} \)
Putting x = 1, y = 2:
\( \displaystyle 2 = \frac{c}{2} \)
\(c = 4\)
So the curve is \(y = \frac{4x}{x + 1}\). At x = 3:
\( \displaystyle k = \frac{4 \times 3}{3 + 1} = 3 \)