Single Choice

If sin $$^{-1}(x-\dfrac{x^2}{2}+\dfrac{x^3}{4}+..............\infty)+$$ cos $$^{-1}(x^2-\dfrac{x^4}{2}+\dfrac{x^6}{4}-...........\infty)$$$$=\dfrac {\pi}{2}$$ and $$0 < x < \sqrt{2}$$ then x =

A$$\dfrac{1}{2}$$
B$$1$$
Correct Answer
C$$-\dfrac{1}{2}$$
D$$-1$$

Solution

Given $$\text{sin}^{-1}(x-\dfrac{x^2}{2}+\dfrac{x^3}{4}+..............\infty)+$$ $$\text{cos}^{-1}(x^2-\dfrac{x^4}{2}+\dfrac{x^6}{4}-...........\infty)$$=\dfrac {\pi}{2}$$
As we know that
$$\text{sin}^{-1} x+\text{cos}^{-1} x=\dfrac{\pi}{2}$$
So $$x-\dfrac{x^2}{2}+\dfrac{x^3}{4}+..............\infty=$$ $$x^2-\dfrac{x^4}{2}+\dfrac{x^6}{4}-...........\infty$$
As we know that
Sum of infinite $$GP$$ i.e $$a+a r+a r^2+\cdots\infty=\dfrac{a}{1-r}$$
$$\implies \dfrac{x}{1+\dfrac{x}{2}}=\dfrac{x^2}{1+\dfrac{x^2}{2}}$$
$$\implies \dfrac{1}{x}+\dfrac{1}{2}=\dfrac{1}{x^2}+\dfrac{1}{2}$$
$$\implies x=x^2\implies x=1$$

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