JEE Advanced · 2018 · 20 May 2018 · Paper 1 · Inverse Trigonometric Functions

Simplify the sums (for $|x|<1$, $|x/2|<1$):

${\sum }_{i=1}^{\infty }{x}^{i+1}=\frac{x^2}{1-x}$, ${\sum }_{i=1}^{\infty }(x/2{)}^i=\frac{x/2}{1-x/2}=\frac{x}{2-x}$.

LHS argument: $\frac{x^2}{1-x}-x\cdot \frac{x}{2-x}=\frac{x^2(2-x)-x^2(1-x)}{(1-x)(2-x)}=\frac{x^2}{(1-x)(2-x)}$.

Similarly for RHS: $\frac{-x/2}{1+x/2}-\frac{-x}{1+x}=\frac{x}{(1+x)(2+x)}$.

Using ${\sin }^{-1}(a)=\frac{\pi }{2}-{\cos }^{-1}(b)$ means ${\sin }^{-1}(a)={\sin }^{-1}(b)$, so $a=b$:

$\frac{x^2}{(1-x)(2-x)}=\frac{x}{(1+x)(2+x)}$.

This gives $x=0$ or $x^3+2x^2+5x-2=0$.

Let $f(x)=x^3+2x^2+5x-2$. Then $f^{\prime }(x)=3x^2+4x+5>0$ (discriminant negative), so $f$ is strictly increasing.

$f(0)=-2<0$, $f(1/2)=1/8+1/2+5/2-2>0$.

So there is exactly one root in $(0,1/2)$.

Total: 2 solutions (including $x=0$).