what can we say about $f(x)= |\cos x| + |\sin x|$ If function $f:\mathbb R \to  [0,2]$ is defined by $f(x)= |\cos x| + |\sin x|$ then


$f$ is one one
$f$ is onto
$f$ is differentiable on $\mathbb R$
The minimum value of $f$ is $1$.


I countered option 3 by fact that $f (x)$ is not differentiable at $x=\pi$ so it is not differentiable on $\mathbb R$.

I countered option 4 by inequality that $|\cos x|+|\sin x|\ge|\cos x +\sin x | \ge \sqrt{2}$ so min value is $\sqrt{2}$ 

I have doubt in deciding option 1 and 2. And also what is correct option
thanks. Btw book says answer is option 4
Since $\cos$ and $\sin$ are periodic with the same period, so is $f$, so $f$ being one to one is out of the question. It should be easy to see that $1$ is false.
The second point is false because $f(x)$ cannot reach $0$ at any value of $x$.
Your answer for 3 is correct.
Your answer for $4$ is incorrect. Your inequality, $\cos x + \sin x \geq \sqrt 2$, is not true at all. For example, $\cos 0 + \sin 0 = 1 &lt; \sqrt 2.$
A function $f:a \to B$ is 1-1 iff $\forall a,b \in A$, $f(a)=f(b) \implies a=b$


But your function is periodic. You can easily provide a counter example by picking some value and then 'moving up' the number line by a full period and showing that $f(a)=f(b)$ but $a \not= b$ 


A function $f:A \to B$ is onto means that: $ \forall b \in B$ $ \exists  a \in A$ such that $f(a)=b$


There does not exist any $x \in \mathbb{R}$ such that $f(x)=o$. This proves 2 is false.


you're correct 
consider f(0): |cos(0)| + |sin(0)| = |1| + |0| = 1 &lt; $\sqrt2$