$$ \text { Let } f(x)=a x^{2}+b x+c \text { be such that } f(1)=3, f(-2)=\lambda \text { and } $$ $$f(3)=4$$. If $$f(0)+f(1)+f(-2)+f(3)=14$$, then $$\lambda$$ is equal to :

Let $$\alpha, \beta$$ and $$\gamma$$ be three positive real numbers. Let $$f(x)=\alpha x^{5}+\beta x^{3}+\gamma x, x \in \mathbf{R}$$ and $$g: \mathbf{R} \rightarrow \mathbf{R}$$ be such that $$g(f(x))=x$$ for all $$x \in \mathbf{R}$$. If $$\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \ldots, \mathrm{a}_{\mathrm{n}}$$ be in arithmetic progression with mean zero, then the value of $$f\left(g\left(\frac{1}{\mathrm{n}} \sum\limits_{i=1}^{\mathrm{n}} f\left(\mathrm{a}_{i}\right)\right)\right)$$ is equal to :

Let $$f, g: \mathbb{N}-\{1\} \rightarrow \mathbb{N}$$ be functions defined by $$f(a)=\alpha$$, where $$\alpha$$ is the maximum of the powers of those primes $$p$$ such that $$p^{\alpha}$$ divides $$a$$, and $$g(a)=a+1$$, for all $$a \in \mathbb{N}-\{1\}$$. Then, the function $$f+g$$ is

The number of bijective functions $$f:\{1,3,5,7, \ldots, 99\} \rightarrow\{2,4,6,8, \ldots .100\}$$, such that $$f(3) \geq f(9) \geq f(15) \geq f(21) \geq \ldots . . f(99)$$, is ____________.