Let a function ƒ : [0, 5] $$ \to $$ R be continuous, ƒ(1) = 3 and F be defined as :

$$F(x) = \int\limits_1^x {{t^2}g(t)dt} $$ , where $$g(t) = \int\limits_1^t {f(u)du} $$

Then for the function F, the point x = 1 is :

In the expansion of $${\left( {{x \over {\cos \theta }} + {1 \over {x\sin \theta }}} \right)^{16}}$$, if $${\ell _1}$$ is the least value of the term independent of x when $${\pi \over 8} \le \theta \le {\pi \over 4}$$ and $${\ell _2}$$ is the least value of the term independent of x when $${\pi \over {16}} \le \theta \le {\pi \over 8}$$, then the ratio $${\ell _2}$$ : $${\ell _1}$$ is equal to :

Let a – 2b + c = 1.

If $$f(x)=\left| {\matrix{ {x + a} & {x + 2} & {x + 1} \cr {x + b} & {x + 3} & {x + 2} \cr {x + c} & {x + 4} & {x + 3} \cr } } \right|$$, then:

The following system of linear equations
7x + 6y – 2z = 0
3x + 4y + 2z = 0
x – 2y – 6z = 0, has