If $\begin{vmatrix} x+1 & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda^2 \end{vmatrix} = \frac{9}{8}(103x+81)$,then $\lambda$ and $\frac{\lambda}{3}$ are the roots of the equation:

Let $a = \lim_{x \to 1} \left( \frac{x}{\ln x} - \frac{1}{x \ln x} \right)$,$b = \lim_{x \to 0} \frac{x^3 - 16x}{4x + x^2}$,$c = \lim_{x \to 0} \frac{\ln(1 + \sin x)}{x}$,and $d = \lim_{x \to -1} \frac{(x + 1)^3}{3(\sin(x + 1) - (x + 1))}$. Then the matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is:

Suppose $a_1, a_2, \dots$ are real numbers,with $a_1 \neq 0$. If $a_1, a_2, a_3, \dots$ are in $A.P.$,then:

If $A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$ and $\det(A^n - I) = 1 - \lambda^n$ for $n \in N$,then $\lambda$ is:

The determinant $\left| \begin{array}{ccc} a^2 & a^2 - (b - c)^2 & bc \\ b^2 & b^2 - (c - a)^2 & ca \\ c^2 & c^2 - (a - b)^2 & ab \end{array} \right|$ is divisible by :

Let the numbers $2, b, c$ be in an $A.P.$ and $A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & b & c \\ 4 & b^2 & c^2 \end{bmatrix}$. If $\det(A) \in [2, 16]$,then $c$ lies in the interval: