A uniform thin rod AB of length $$L$$ has linear mass density $$\mu(x) = a + \frac{bx}{L}$$, where $$x$$ is measured from A. If the CM of the rod lies at a distance of $$\left(\frac{7}{12}L\right)$$ from A, then $$a$$ and $$b$$ are related as:

$$X_{cm} = \frac{\int_{0}^{L} x \cdot dm}{\int_{0}^{L} dm}$$

$$M = \int_{0}^{L} \mu(x) dx = \int_{0}^{L} \left(a + \frac{bx}{L}\right) dx$$

$$M = \left[ ax + \frac{bx^2}{2L} \right]_{0}^{L} = aL + \frac{bL^2}{2L}$$

$$M = L \left( a + \frac{b}{2} \right)$$

$$\int_{0}^{L} x \cdot \mu(x) dx = \int_{0}^{L} x \left( a + \frac{bx}{L} \right) dx = \int_{0}^{L} \left( ax + \frac{bx^2}{L} \right) dx$$

$$ = \left[ \frac{ax^2}{2} + \frac{bx^3}{3L} \right]_{0}^{L} = \frac{aL^2}{2} + \frac{bL^3}{3L}$$ $$ = L^2 \left( \frac{a}{2} + \frac{b}{3} \right)$$

$$X_{cm} = \frac{L^2 \left( \frac{a}{2} + \frac{b}{3} \right)}{L \left( a + \frac{b}{2} \right)} = \frac{7}{12}L$$

$$\frac{\frac{a}{2} + \frac{b}{3}}{a + \frac{b}{2}} = \frac{7}{12}$$

$$\frac{3a + 2b}{3(2a + b)} = \frac{7}{12}$$ 

Solving the equation gives $$b = 2a$$