Let A be a 3 $$\times$$ 3 real matrix such that

$$A\left( {\matrix{ 1 \cr 1 \cr 0 \cr } } \right) = \left( {\matrix{ 1 \cr 1 \cr 0 \cr } } \right);A\left( {\matrix{ 1 \cr 0 \cr 1 \cr } } \right) = \left( {\matrix{ { - 1} \cr 0 \cr 1 \cr } } \right)$$ and $$A\left( {\matrix{ 0 \cr 0 \cr 1 \cr } } \right) = \left( {\matrix{ 1 \cr 1 \cr 2 \cr } } \right)$$.

If $$X = {({x_1},{x_2},{x_3})^T}$$ and I is an identity matrix of order 3, then the system $$(A - 2I)X = \left( {\matrix{ 4 \cr 1 \cr 1 \cr } } \right)$$ has :

Let f : R $$\to$$ R be defined as $$f(x) = {x^3} + x - 5$$. If g(x) is a function such that $$f(g(x)) = x,\forall 'x' \in R$$, then g'(63) is equal to ________________.

If $${1 \over {2\,.\,{3^{10}}}} + {1 \over {{2^2}\,.\,{3^9}}} + \,\,.....\,\, + \,\,{1 \over {{2^{10}}\,.\,3}} = {K \over {{2^{10}}\,.\,{3^{10}}}}$$, then the remainder when K is divided by 6 is :

Let f(x) be a polynomial function such that $$f(x) + f'(x) + f''(x) = {x^5} + 64$$. Then, the value of $$\mathop {\lim }\limits_{x \to 1} {{f(x)} \over {x - 1}}$$ is equal to: