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
 A: *

*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, $\cos 0 + \sin 0 = 1 < \sqrt 2.$

A: *

*false, because $f(0)=f(\pi)=1$

*false. if exists $x\in\mathbb{R}$ such that $|\cos(x)|+|\sin(x)|=0 $, then
$\cos(x)=\sin(x)=0 \Rightarrow \cos^2(x)+\sin^2(x)=0$. Absurd!


*false.


$f'(0) = \displaystyle\lim_{t\to0}\frac{f(t)-f(0)}{t} = \displaystyle\lim_{t\to0}\frac{|\cos(t)|+|\sin(t)|-1}{t}$
But
$\displaystyle\lim_{t\to0^+}\frac{|\cos(t)|+|\sin(t)|-1}{t}= \displaystyle\lim_{t\to0^+}\frac{\cos(t)+\sin(t)-1}{t} \stackrel{L'h}{=} \displaystyle\lim_{t\to0^+}(-\sin(t)+\cos(t)) = 1$
$\displaystyle\lim_{t\to0^-}\frac{|\cos(t)|+|\sin(t)|-1}{t}= \displaystyle\lim_{t\to0^-}\frac{\cos(t)-\sin(t)-1}{t} \stackrel{L'h}{=} \displaystyle\lim_{t\to0^-}(-\sin(t)-\cos(t)) = -1$
So, the limit does not exists.


*true.


$f(0) = |\cos(0)|+|\sin(0)| = 1$ and
$f(x)^2 = \cos^2(x)+2|\sin(x)||\cos(x)|+\sin^2(x) = 1+|\sin(2x)| \ge 1$
Then $f(x)\ge 1, \forall x\in\mathbb{R}$ (remember that $f(x)\ge 0$)
A: *

*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 < $\sqrt2$
