Find the maximum value of the function f(2)=3+4x2-x4

If ya meant find the max value of
f(x)=3+4x²-x⁴

take derivitive
f'(x)=8x-4x³
find where it equals 0
0=8x-4x³
0=4x(2-x²)
0=4x((√2)-x)((√2)+x)
set each to zero

0=4x
0=x

0=(√2)-x
x=√2

0=(√2)+x
-√2=x

ok, test the points around them to find if they are max or min

f'(-2)>0
f(-1)<0
f(1)>0
f(2)<0
so it goes from +, -√2, -, 0, +, √2, -
max is where it goes from + to -
that is at -√2 and √2

find f(√2) and f(-√2) to find which one is bigger
trick question, they are the same

max occurs at f(√2) and f(-√2)
the result is
f(√2)=f(-√2)=7
the max value is 7

In the first triangle 80+20+p=180 (since sum of all angles of a triangle is 180)so p= 80also in second triangle 55+45+q=180 (since sum of all angles of a triangle is 180)so q=80so we can conclude that p=80 degrees q=80 degrees so p=q=80 degreeshence option b is correct