Let $\alpha$ and $\beta$ be the roots of the equation $5 x^{2}+6 x-2=0 .$ If $S_{n}=\alpha^{n}+\beta^{n}, n=1,2,3 \ldots$ then :
$5 \mathrm{S}_{6}+6 \mathrm{S}_{5}=2 \mathrm{S}_{4}$
$5 \mathrm{S}_{6}+6 \mathrm{S}_{5}+2 \mathrm{S}_{4}=0$
$6 \mathrm{S}_{6}+5 \mathrm{S}_{5}+2 \mathrm{S}_{4}=0$
$6 \mathrm{S}_{6}+5 \mathrm{S}_{5}=2 \mathrm{S}_{4}$
All the points $(x, y)$ in the plane satisfying the equation $x^2+2 x \sin (x y)+1=0$ lie on
Let $a, b, c, d$ be numbers in the set $\{1,2,3,4,5,6\}$ such that the curves $y=2 x^3+a x+b$ and $y=2 x^3+c x+d$ have no point in common. The maximum possible value of $(a-c)^2+b-d$ is
Let $\alpha, \beta(\alpha>\beta)$ be the roots of the quadratic equation $x ^{2}- x -4=0$. If $P _{ a }=\alpha^{ n }-\beta^{ n }, n \in N$, then $\frac{ P _{15} P _{16}- P _{14} P _{16}- P _{15}^{2}+ P _{14} P _{15}}{ P _{13} P _{14}}$ is equal to$......$
Let $r$ be a real number and $n \in N$ be such that the polynomial $2 x^2+2 x+1$ divides the polynomial $(x+1)^n-r$. Then, $(n, r)$ can be
If the quadratic equation ${x^2} + \left( {2 - \tan \theta } \right)x - \left( {1 + \tan \theta } \right) = 0$ has $2$ integral roots, then sum of all possible values of $\theta $ in interval $(0, 2\pi )$ is $k\pi $, then $k$ equals