If $f$ is an even function defined on the interval $(-5, 5)$, then four real values of $x$ satisfying the equation $f(x) = f\left( {\frac{{x + 1}}{{x + 2}}} \right)$ are
$\frac{{ - 3 - \sqrt 5 }}{2},\frac{{ - 3 + \sqrt 5 }}{2},\frac{{3 - \sqrt 5 }}{2},\frac{{3 + \sqrt 5 }}{2}$
$\frac{{ - 5 + \sqrt 3 }}{2},\frac{{ - 3 + \sqrt 5 }}{2},\frac{{3 + \sqrt 5 }}{2},\frac{{3 - \sqrt 5 }}{2}$
$\frac{{3 - \sqrt 5 }}{2},\frac{{3 + \sqrt 5 }}{2},\frac{{ - 3 - \sqrt 5 }}{2},\frac{{5 + \sqrt 3 }}{2}$
$ - 3 - \sqrt 5 , - 3 + \sqrt 5 ,3 - \sqrt 5 ,3 + \sqrt 5$
Consider the function $\mathrm{f}:\left[\frac{1}{2}, 1\right] \rightarrow \mathrm{R}$ defined by $f(x)=4 \sqrt{2} x^3-3 \sqrt{2} x-1$. Consider the statements
$(I)$ The curve $y=f(x)$ intersects the $x$-axis exactly at one point
$(II)$ The curve $y=f(x)$ intersects the $x$-axis at $\mathrm{x}=\cos \frac{\pi}{12}$
Then
If domain of the function $\log _e\left(\frac{6 x^2+5 x+1}{2 x-1}\right)+\cos ^{-1}\left(\frac{2 x^2-3 x+4}{3 x-5}\right)$ is $(\alpha, \beta) \cup(\gamma, \delta]$, then $18\left(\alpha^2+\beta^2+\gamma^2+\delta^2\right)$ is equal to $....$.
Let $R _{1}$ and $R _{2}$ be two relations defined as follows :
$R _{1}=\left\{( a , b ) \in R ^{2}: a ^{2}+ b ^{2} \in Q \right\}$ and $R _{2}=\left\{( a , b ) \in R ^{2}: a ^{2}+ b ^{2} \notin Q \right\}$
where $Q$ is the set of all rational numbers. Then
The domain of definition of the function $f (x) = {\log _{\left[ {x + \frac{1}{x}} \right]}}|{x^2} - x - 6|+ ^{16-x}C_{2x-1} + ^{20-3x}P_{2x-5}$ is
Where $[x]$ denotes greatest integer function.
A real valued function $f(x)$ satisfies the function equation $f(x - y) = f(x)f(y) - f(a - x)f(a + y)$ where a is a given constant and $f(0) = 1$, $f(2a - x)$ is equal to