### Every continuous function $f : [0,1] \to [0,1]$ has a fixed point.

Result: Let $f : [0,1] \to [0,1]$ be a continuous function then there exists $c \in [0,1]$ such that f(c) = c.

Proof:
If $f(0) = 0$ or $f(1) = 1$ then nothing to prove. We assume $f(0) \ne 0$ and $f(1) \ne 1$.
Consider $g(x) = f(x) - x$, then $g$ is a continuous function from $[0,1]$ to $[0,1]$.
Since $f$ is a function from $[0,1]$ to $[0,1$ we have $g(0) = f(0) - 0 > 0$ and $g(1) = f(1) - 1 < 0$. Now by intermediate value theorem there exists $x$ in $[0,1]$ such that $g(x) = 0 = f(x) - x$ and the proof is done.

### NBHM 2020 PART A Question 4 Solution $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$
Evaluate : $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$ Solution : \int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx = \int_{-\infty}^{\inft...