Let $f:\mathbf{R}-\{0\}\to (-\infty ,1)$ be a polynomial of degree 2 , satisfying $f(x)f\left(\frac{1}{x}\right)=f(x)+f\left(\frac{1}{x}\right)$. If $f(\mathrm{K})=-2\mathrm{K}$, then the sum of squares of all possible values of K is :

The product of all the rational roots of the equation ${\left(x^2-9x+11\right)}^2-(x-4)(x-5)=3$, is equal to

The number of real solution(s) of the equation $x^2+3x+2=\min \{|x-3|,|x+2|\}$ is :

The sum, of the squares of all the roots of the equation $x^2+|2x-3|-4=0$, is

The sum of the squares of the roots of $|x-2{|}^2+|x-2|-2=0$ and the squares of the roots of $x^2-2|x-3|-5=0$, is

Let ${\mathrm{P}}_{\mathrm{n}}={\alpha }^{\mathrm{n}}+{\beta }^{\mathrm{n}},\mathrm{n}\in \mathrm{N}$. If ${\mathrm{P}}_{10}=123,{\mathrm{P}}_9=76,{\mathrm{P}}_8=47$ and ${\mathrm{P}}_1=1$, then the quadratic equation having roots $\frac{1}{\alpha }$ and $\frac{1}{\beta }$ is :

The number of real roots of the equation $x |x - 2| + 3|x - 3| + 1 = 0$ is :

Let the set of all values of $p\in \mathbb{R}$, for which both the roots of the equation $x^2-(p+2)x+(2p+9)=0$ are negative real numbers, be the interval $(\alpha ,\beta ]$. Then $\beta -2\alpha$ is equal to

Let $\alpha$ and $\beta$ be the roots of $x^2+\sqrt{3}x-16=0$, and $\gamma$ and $\delta$ be the roots of $x^2+3x-1=0$. If $P_n=$ ${\alpha }^n+{\beta }^n$ and $Q_n={\gamma }^n+{\hat{o}}^n$, then $\frac{P_{25}+\sqrt{3}{P}_{24}}{2P_{23}}+\frac{Q_{25}-Q_{23}}{Q_{24}}$ is equal to

Consider the equation $x^2+4x-n=0$, where $n\in [20,100]$ is a natural number. Then the number of all distinct values of $n$, for which the given equation has integral roots, is equal to

Let the equation $x(x+2)(12-k)=2$ have equal roots. Then the distance of the point $\left(k,\frac{k}{2}\right)$ from the line $3 x+4 y+5=0$ is

Let the foci of a hyperbola be $(1,14)$ and $(1,-12)$. If it passes through the point $(1,6)$, then the length of its latus-rectum is :

Let one focus of the hyperbola $\mathrm{H}:\frac{x^2}{{\mathrm{a}}^2}-\frac{y^2}{{\mathrm{b}}^2}=1$ be at $(\sqrt{10},0)$ and the corresponding directrix be $x=\frac{9}{\sqrt{10}}$. If $e$ and $l$ respectively are the eccentricity and the length of the latus rectum of H , then $9\left(e^2+l\right)$ is equal to :

Let e1 and e2 be the eccentricities of the ellipse $\frac{x^2}{b^2}+\frac{y^2}{25}=1$ and the hyperbola $\frac{x^2}{16}-\frac{y^2}{b^2}=1$, respectively. If b < 5 and e1e2 = 1, then the eccentricity of the ellipse having its axes along the coordinate axes and passing through all four foci (two of the ellipse and two of the hyperbola) is :

Let the sum of the focal distances of the point $\mathrm{P}(4,3)$ on the hyperbola $\mathrm{H}:\frac{x^2}{{\mathrm{a}}^2}-\frac{y^2}{{\mathrm{b}}^2}=1$ be $8\sqrt{\frac{5}{3}}$. If for H , the length of the latus rectum is $l$ and the product of the focal distances of the point P is m , then $9l^2+6\mathrm{m}$ is equal to :

Let the area of a $△PQR$ with vertices $P(5,4), Q(-2,4)$ and $R(a, b)$ be 35 square units. If its orthocenter and centroid are $O\left(2,\frac{14}{5}\right)$ and $C(c, d)$ respectively, then $c+2 d$ is equal to

Let ${\mathrm{L}}_1:\frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2}$ and ${\mathrm{L}}_2:\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z}{1}$ be two lines. Let $L_3$ be a line passing through the point $(\alpha ,\beta ,\gamma )$ and be perpendicular to both $L_1$ and $L_2$. If $L_3$ intersects ${\mathrm{L}}_1$, then $|5\alpha -11\beta -8\gamma |$ equals :

Let a straight line $L$ pass through the point $P(2, -1, 3)$ and be perpendicular to the lines $\frac{x-1}{2}=\frac{y+1}{1}=\frac{z-3}{-2}$ and $\frac{x-3}{1}=\frac{y-2}{3}=\frac{z+2}{4}$. If the line $L$ intersects the $yz$-plane at the point $Q$, then the distance between the points $P$ and $Q$ is:

Let P be the foot of the perpendicular from the point $(1,2,2)$ on the line $\mathrm{L}:\frac{x-1}{1}=\frac{y+1}{-1}=\frac{z-2}{2}$. Let the line $\vec{r}=(-\hat{i}+\hat{j}-2\hat{k})+\lambda (\hat{i}-\hat{j}+\hat{k}),\lambda \in \mathbf{R}$, intersect the line L at Q . Then $2(\mathrm{PQ}{)}^2$ is equal to :

Let ${\mathrm{L}}_1:\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and ${\mathrm{L}}_2:\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ be two lines. Then which of the following points lies on the line of the shortest distance between ${\mathrm{L}}_1$ and ${\mathrm{L}}_2$ ?