JEE Advanced 2018 · Paper 1, Mathematics question 10

Sum the series (all converge for $|x|<\tfrac12$):
$$\sum_{i\ge1}x^{i+1}-x\sum_{i\ge1}\left(\frac x2\right)^i=\frac{x^2}{1-x}-\frac{x^2}{2-x}=\frac{x^2}{(1-x)(2-x)},$$
$$\sum_{i\ge1}\left(-\frac x2\right)^i-\sum_{i\ge1}(-x)^i=-\frac x{2+x}+\frac x{1+x}=\frac x{(1+x)(2+x)}.$$

Simplify. Since $\dfrac\pi2-\cos^{-1}u=\sin^{-1}u$, the equation is $\sin^{-1}A=\sin^{-1}B$, i.e. $A=B$:
$$\frac{x^2}{(1-x)(2-x)}=\frac x{(1+x)(2+x)}.$$

Roots.

• $x=0$ is one root.
• Otherwise $x(1+x)(2+x)=(1-x)(2-x)$, i.e. $g(x)=x^3+2x^2+5x-2=0$. Here $g'=3x^2+4x+5>0$, $g(0)=-2$ and $g\left(\tfrac12\right)=\tfrac98>0$. So $g$ is increasing: it is negative on $\left(-\tfrac12,0\right]$ and has exactly one root in $\left(0,\tfrac12\right)$.

Total: 2 solutions.

Answer: 2