Maths

Let  $A=\left(\begin{array}{cc}\hfill 2\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 2\hfill \end{array}\right).$ If $B=I{-}^5\text{}{C}_1\left(adjA\right){+}^5\text{}{C}_2{\left(adjA\right)}^2-..{.}^5\text{}{C}_5{\left(adjA\right)}^5,$  then the sum of all elements of the matrix B is

Let  $A=\left(\begin{array}{cc}\hfill 2\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 2\hfill \end{array}\right).$ If $B=I{-}^5\text{?}{C}_1\left(adjA\right){+}^5\text{?}{C}_2{\left(adjA\right)}^2-..{.}^5\text{?}{C}_5{\left(adjA\right)}^5,$  then the sum of all elements of the matrix B is

Let arg(z) represent the principal argument of the complex number z. Then $\left|z\right|=3$ and art(z – 1) – arg(z + 1) = $\frac{\pi }{4}$  intersect

Let a be root of the equation 1 + x² + x⁴ = 0. Then the value of a¹⁰¹¹ + a²⁰²² - a³⁰³³ is equal to:

Let the lines y + 2x = $\sqrt[]{11}+7\sqrt[]{7}$ and $2y+x=2\sqrt[]{11}+6\sqrt[]{7}$ be normal to a circle C : ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2.$ If the line $\sqrt[]{11}y-3x=\frac{5\sqrt[]{77}}{3}+11$ is tangent to the circle C, then the value of $\sqrt[]{11}y-3x=\frac{5\sqrt[]{77}}{3}+11$  is equal to…………….

The number of elements in the set $\left\{z=a+ib\in C:a,b\in Zand1<\left|z-3+2i\right|<4\right\}$  is……….

The number of elements in the set $\left\{z=a+ib\in C:a,b\in Zand1<\left|z-3+2i\right|<4\right\}$  is……….

The number of positive integers k such that the constant term in the binomial expansion of ${\left(2x^3+\frac{3}{x^k}\right)}^{12},x\ne 0is{2}^8.\mathcal{l},where\mathcal{l}$  is an odd integer, is……….

Let A = $\left\{1,a_1,a_2....a_{18},77\right\}$  be a set of integers with 1 < a₁ < a₂ < … < a₁₈ < 77. Let the set A + A = $\left\{x+y:x,y\in A\right\}$  contains exactly 39 elements. Then, the value of a₁ + a₂ +…. + a₁₈ is equal to…………

Let $\mathcal{l}$  be a line which is normal to the curve y = 2x² + x + 2 at a point P on the curve. If the point Q(6, 4) lies on the line $\mathcal{l}$  and O is origin, then the area of the triangle OPQ is equal to………