Inverse Trigonometric Functions

Case 1: 1 $\le \frac{1}{4x^2-1}\le 1$

4x2 – 1 $\ge 1or4x^2-1\le -1$

$x^2\ge \frac{2}{4}or{x}^2\le 0$

So x $\in (-\infty ,-\frac{1}{\sqrt[]{2}}]\cup [\frac{1}{\sqrt[]{2}},\infty )\cup \left\{0\right\}$

 $x^8-x^7-x^6+x^5+3x^4-4x^3-2x^2+4x-1=0$

$\Rightarrow x^7\left(x-1\right)-x^5\left(x-1\right)+3x^3\left(x-1\right)-x\left(x^2-1\right)+2x\left(1-x\right)+\left(x-1\right)=0$

$\Rightarrow \left(x-1\right)\left(x^2-1\right)\left(x^5+3x-1\right)=0\therefore x=\pm 1$ are roots of above equation and x5 + 3x – 1 is a monotonic term hence vanishes at exactly one value of x other then 1 or 1.

$\therefore$ 3 real roots.

Let and are the roots of $\left(p^2+q^2\right)x^2-2q\left(p+r\right)x+q^2+r^2=0$

$\therefore \alpha +\beta >0and\alpha \beta >0$ Also, it has a common root with x² + 2x – 8 = 0

$\therefore$ The common root between above two equations is 4.

$\Rightarrow 16\left(p^2+q^2\right)-8q\left(p+r\right)+q^2+r^2=0\Rightarrow \left(16p^2-8pq+q^2\right)+\left(16q^2-8qr+r^2\right)=0$

$\Rightarrow {\left(4p-q\right)}^2+{\left(4q-r\right)}^2=0\Rightarrow q=4pandr=16p\therefore \frac{q^2+r^2}{p^2}=\frac{16p^2+256p^2}{p^2}=272$

 $Let\frac{si{n}^{-1}x}{\alpha }=\frac{co{s}^{-1}x}{\beta }=k\Rightarrow si{n}^{-1}x+co{s}^{-1}x=k\left(\alpha +\beta \right)\Rightarrow \alpha +\beta =\frac{\pi }{2k}$

Now $\frac{2\pi \alpha }{\alpha +\beta }=\frac{2\pi \alpha }{\frac{\pi }{2k}}\Rightarrow 4k\alpha =4si{n}^{-1}x.Heresin\left(\frac{2\pi \alpha }{\alpha +\beta }\right)=sin\left(4si{n}^{-1}x\right)$

$\Rightarrow sin4\theta =4x\sqrt[]{1-x^2}\left(1-2x^2\right)$

Kindly go through the solution

Kindly go through the solution

Kindly go through the solution