Let a smooth curve y=f(x) be such that the slope of the tangent at any point (x, y) on it is directly proportional to (-y/x). If the curve passes through the points (1,2) and (8,1), then |y(1/8)| is equal to

Solution

Maths sir · 4 steps

Separate the variables and the curve becomes y = Cx⁻ᵏ; the two points fix both constants.

1. Set up. Proportional means equal up to a constant, so dy/dx = -ky/x for some constant k. Put all the y terms on one side and all the x terms on the other: dy/y = -kdx/x.
2. Integrate. Integrating gives ln|y| = -k ln|x| + c. Take the exponential of both sides to get the curve.

   y = C x⁻ᵏ

3. Use both points. At (1,2), 2 = C × 1, so C = 2. At (8,1), 1 = 2 × 8⁻ᵏ, so 8ᵏ = 2 and k = 1/3. The curve is y = 2x^(-1/3).
4. So. y(1/8) = 2(1/8)^(-1/3) = 2 × 8^(1/3) = 2 × 2 = 4. So |y(1/8)| = 4, option (b).

The idea to keep. Whenever the slope is a constant times y/x, the curve is y = Cxⁿ. One point fixes C, a second point fixes n, and logs are only a step on the way.

Answer from the NTA final answer key. Solution by PrepWiser.