Very very simple, try it

All the values of any function $f$ of the form $f(x)=a.sinx+b.cosx$ lies in the interval $\left[ -\sqrt{a^2+b^2}, \sqrt{a^2+b^2} \right]$

Therefore $min.\{ 3.sinx+4.cosx \}=-\sqrt{3^2+4^2}=-5$

$\quad \quad \quad \quad max. \{ 3.sinx+4.cosx \}=\sqrt{3^2+4^2}=5$

$\implies min. \{ f(x) \}= -5+5 =0$ , $\quad \& \quad max. \{ f(x) \}=5+5=10$

Hence $min.\{f(x)\}+max.\{f(x)\}=0+10=\boxed{10}$

Entre cosseno e seno, o cosseno tem o coeficiente maior que o seno. Então se adotar o menor e o maior valor possível para o cosseno, o seno tem valor 0.

cosx = -1

3 0 + 4 (-1) + 5 = 1

cosx = 1

3 0 + 4 1 + 5 = 9

Somando os dois valores: 9 + 1 = 10

The maximum value of $sins$ and $cosx$ is $1$ and the minimum value is $-1$ so I just substituted the values to get the answer. By this method you can solve this question in under 20 seconds.

i used a formula i learnt in trig .....

for any expression in the form "a.sinx + b.cosx + c", its values lie between

root(a^2 + b^2) - c^2 <= a.sinx + b.cosx + c <= root(a^2 + b^2) + c^2

substitute a,b,c ..... hence the result

Just graph it.

max and min value of asinx + bcosx+ c are (a+b)^1/2 +c and - (a+b)^1/2 +c their sum is 2c= 10

-√(3^2+4^2)<=3.sinx+4.cosx<=+√(3^2+4^2) Adding 5 => 0<=f(x)<=10 Minimum value=0 Maximum value =10 0+10=10