Question

The number of real solutions of the equation, $${\sin }^{-1}\left(\sum _{i=1}^{\mathrm{\infty }}{x}^{i+1}-x\sum _{i=1}^{\mathrm{\infty }}{\left({\displaystyle \frac{x}{2}}\right)}^i\right)={\displaystyle \frac{\pi }{2}}-{\cos }^{-1}\left(\sum _{i=1}^{\mathrm{\infty }}{(-{\displaystyle \frac{x}{2}})}^i-\sum _{i=1}^{\mathrm{\infty }}{(-x)}^i\right)$$ lying in the interval $(-{\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{2}})$ is,

Answer 1

Hint: To solve this question first we will reduce the given equation into simplest form by using inverse trigonometric identity $${\sin }^{-1}x={\displaystyle \frac{\pi }{2}}-{\cos }^{-1}x$$, then we will expand the expansion and using sum of infinite G.P we will obtain the algebraic expression and then we will see how many root we obtain for that algebraic equation.

Complete step-by-step answer:
Now, we have $${\sin }^{-1}\left(\sum _{i=1}^{\mathrm{\infty }}{x}^{i+1}-x\sum _{i=1}^{\mathrm{\infty }}{\left({\displaystyle \frac{x}{2}}\right)}^i\right)={\displaystyle \frac{\pi }{2}}-{\cos }^{-1}\left(\sum _{i=1}^{\mathrm{\infty }}{(-{\displaystyle \frac{x}{2}})}^i-\sum _{i=1}^{\mathrm{\infty }}{(-x)}^i\right)$$
Let, $f(x)=\sum _{i=1}^{\mathrm{\infty }}{x}^{i+1}-x\sum _{i=1}^{\mathrm{\infty }}{\left({\displaystyle \frac{x}{2}}\right)}^i$ and $g(x)=\sum _{i=1}^{\mathrm{\infty }}{(-{\displaystyle \frac{x}{2}})}^i-\sum _{i=1}^{\mathrm{\infty }}{(-x)}^i$
So, we can write $${\sin }^{-1}\left(\sum _{i=1}^{\mathrm{\infty }}{x}^{i+1}-x\sum _{i=1}^{\mathrm{\infty }}{\left({\displaystyle \frac{x}{2}}\right)}^i\right)={\displaystyle \frac{\pi }{2}}-{\cos }^{-1}\left(\sum _{i=1}^{\mathrm{\infty }}{(-{\displaystyle \frac{x}{2}})}^i-\sum _{i=1}^{\mathrm{\infty }}{(-x)}^i\right)$$ as,
$${\sin }^{-1}f(x)={\displaystyle \frac{\pi }{2}}-{\cos }^{-1}g(x)$$
We know that, $${\sin }^{-1}x={\displaystyle \frac{\pi }{2}}-{\cos }^{-1}x$$,
So, re – writing $${\sin }^{-1}f(x)={\displaystyle \frac{\pi }{2}}-{\cos }^{-1}g(x)$$ as,
$${\sin }^{-1}f(x)={\sin }^{-1}g(x)$$
Now, we know that $${\sin }^{-1}x$$ is one - one function, so it is only true for $${\sin }^{-1}f(x)={\sin }^{-1}g(x)$$,
 if $f(x)=g(x)$ , that is
$\sum _{i=1}^{\mathrm{\infty }}{x}^{i+1}-x\sum _{i=1}^{\mathrm{\infty }}{\left({\displaystyle \frac{x}{2}}\right)}^i=\sum _{i=1}^{\mathrm{\infty }}{(-{\displaystyle \frac{x}{2}})}^i-\sum _{i=1}^{\mathrm{\infty }}{(-x)}^i$
Now, let us take $f(x)=\sum _{i=1}^{\mathrm{\infty }}{x}^{i+1}-x\sum _{i=1}^{\mathrm{\infty }}{\left({\displaystyle \frac{x}{2}}\right)}^i$
We can write f(x) as, $f(x)=\sum _{i=1}^{\mathrm{\infty }}{x}^i\cdot x-x\sum _{i=1}^{\mathrm{\infty }}{\left({\displaystyle \frac{x}{2}}\right)}^i$
Or, $f(x)=x\sum _{i=1}^{\mathrm{\infty }}{x}^i-x\sum _{i=1}^{\mathrm{\infty }}{\left({\displaystyle \frac{x}{2}}\right)}^i$
Now, $\sum _{i=1}^{\mathrm{\infty }}{x}^i$is G.P whose first term is x and common ratio is x,
So, $$x\sum _{i=1}^{\mathrm{\infty }}{x}^i={\displaystyle \frac{x^2}{1-x}}$$, as Sum of infinite G.P is ${\displaystyle \frac{a}{1-r}}$, whose first term is a and common ratio is r.
Similarly, $\sum _{i=1}^{\mathrm{\infty }}{\left({\displaystyle \frac{x}{2}}\right)}^i$is G.P whose first term is ${\displaystyle \frac{x}{2}}$ and common ratio is ${\displaystyle \frac{x}{2}}$,
So, $x\sum _{i=1}^{\mathrm{\infty }}{\left({\displaystyle \frac{x}{2}}\right)}^i={\displaystyle \frac{{\displaystyle \frac{x^2}{2}}}{1-\left({\displaystyle \frac{x}{2}}\right)}}={\displaystyle \frac{x^2}{2-x}}$.
So, $f(x)=x\sum _{i=1}^{\mathrm{\infty }}{x}^i-x\sum _{i=1}^{\mathrm{\infty }}{\left({\displaystyle \frac{x}{2}}\right)}^i={\displaystyle \frac{x^2}{1-x}}-{\displaystyle \frac{x^2}{2-x}}$
Now, in $f(x)=\sum _{i=1}^{\mathrm{\infty }}{x}^{i+1}-x\sum _{i=1}^{\mathrm{\infty }}{\left({\displaystyle \frac{x}{2}}\right)}^i$ and $g(x)=\sum _{i=1}^{\mathrm{\infty }}{(-{\displaystyle \frac{x}{2}})}^i-\sum _{i=1}^{\mathrm{\infty }}{(-x)}^i$, there is only difference in sign,
So, $g(x)=\sum _{i=1}^{\mathrm{\infty }}{(-{\displaystyle \frac{x}{2}})}^i-\sum _{i=1}^{\mathrm{\infty }}{(-x)}^i={\displaystyle \frac{-{\displaystyle \frac{x}{2}}}{1-(-{\displaystyle \frac{x}{2}})}}+{\displaystyle \frac{x}{1-(-x)}}$
Or on simplifying, we get
$g(x)=\sum _{i=1}^{\mathrm{\infty }}{(-{\displaystyle \frac{x}{2}})}^i-\sum _{i=1}^{\mathrm{\infty }}{(-x)}^i={\displaystyle \frac{x}{1+x}}-{\displaystyle \frac{x}{2+x}}$
As, we have $f(x)=g(x)$
So, ${\displaystyle \frac{x^2}{1-x}}-{\displaystyle \frac{x^2}{2-x}}={\displaystyle \frac{x}{1+x}}-{\displaystyle \frac{x}{2+x}}$
Now, we have to solve this algebraic expression,
Taking all expression on left side,
${\displaystyle \frac{x^2}{1-x}}-{\displaystyle \frac{x^2}{2-x}}-{\displaystyle \frac{x}{1+x}}+{\displaystyle \frac{x}{2+x}}=0$
Taking, x common, we get
$x\cdot ({\displaystyle \frac{x}{1-x}}-{\displaystyle \frac{x}{2-x}}-{\displaystyle \frac{1}{1+x}}+{\displaystyle \frac{1}{2+x}})=0$
So, x = 0 and $({\displaystyle \frac{x}{1-x}}-{\displaystyle \frac{x}{2-x}}-{\displaystyle \frac{1}{1+x}}+{\displaystyle \frac{1}{2+x}})=0$
Now, $$\begin{array}{rl}& {\displaystyle \frac{x}{1-x}}-{\displaystyle \frac{x}{2-x}}-{\displaystyle \frac{1}{1+x}}+{\displaystyle \frac{1}{2+x}}=0\\ & {\displaystyle \frac{x}{1-x}}-{\displaystyle \frac{x}{2-x}}={\displaystyle \frac{1}{1+x}}-{\displaystyle \frac{1}{2+x}}\end{array}$$
Taking L.C.M, we get
${\displaystyle \frac{2x-x^2-x+x^2}{(x-2)(x-1)}}={\displaystyle \frac{-1-x+2+x}{(x-2)(x-1)}}$
On solving we get
${\displaystyle \frac{x}{(x-2)(x-1)}}={\displaystyle \frac{1}{(x-2)(x-1)}}$
On cross – multiplying, we get
$x(x-2)(x-1)=(x-2)(x-1)$
On opening brackets, we get
$x^3+2x^2+5x-2=0$
Let, $f(x)=x^3+2x^2+5x-2$
Now, at $x=-{\displaystyle \frac{1}{2}}$ , we get
$f(-{\displaystyle \frac{1}{2}})={(-{\displaystyle \frac{1}{2}})}^3+2{(-{\displaystyle \frac{1}{2}})}^2+5(-{\displaystyle \frac{1}{2}})-2=-{\displaystyle \frac{1}{8}}+{\displaystyle \frac{1}{2}}-{\displaystyle \frac{5}{2}}-2=-{\displaystyle \frac{33}{8}}$
at $x={\displaystyle \frac{1}{2}}$ , we get
$f({\displaystyle \frac{1}{2}})={\left({\displaystyle \frac{1}{2}}\right)}^3+2{\left({\displaystyle \frac{1}{2}}\right)}^2+5\left({\displaystyle \frac{1}{2}}\right)-2={\displaystyle \frac{1}{8}}+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{5}{2}}-2={\displaystyle \frac{9}{8}}$
Now, $f(-{\displaystyle \frac{1}{2}})\cdot f({\displaystyle \frac{1}{2}})<0$, which means there exists one root in interval and one is x = 0.
So, we have two solutions for equation $${\sin }^{-1}\left(\sum _{i=1}^{\mathrm{\infty }}{x}^{i+1}-x\sum _{i=1}^{\mathrm{\infty }}{\left({\displaystyle \frac{x}{2}}\right)}^i\right)={\displaystyle \frac{\pi }{2}}-{\cos }^{-1}\left(\sum _{i=1}^{\mathrm{\infty }}{(-{\displaystyle \frac{x}{2}})}^i-\sum _{i=1}^{\mathrm{\infty }}{(-x)}^i\right)$$

Note: To, solve such question one must know the inverse trigonometric function properties such as $${\sin }^{-1}x={\displaystyle \frac{\pi }{2}}-{\cos }^{-1}x$$ and also how to open summation and what is sum of infinite geometric expansion which is equals to ${\displaystyle \frac{a}{1-r}}$, whose first term is a and common ratio is r.