Question

Range of the function f(x) =
(3sinx+cosx - 2)/
(3sinx + 4cosx +10)

Pls help me fast
I will mark branliest​

Answer 1

Answer:

Solution : 

f(x)=(3sinx−4cosx−10)(3sinx+4cosx−10)f(x)=(3sinx-4cosx-10)(3sinx+4cosx-10)

f(x)=[(3sinx−10)−4cosx][(3sinx−10)+4cosx]f(x)=[(3sinx-10)-4cosx][(3sinx-10)+4cosx]

f(x)=(3sinx−10)2−16cos2xf(x)=(3sinx-10)2-16cos2x

f(x)=9sin2x+100−60sinx−16(1−sin2x)f(x)=9sin2x+100-60sinx-16(1-sin2x)

f(x)=25sin2x+84−60sinxf(x)=25sin2x+84-60sinx

f(x)=(5sinx−6)2+48f(x)=(5sinx-6)2+48

at sinx=1

f(x)=(5−6)2+48=49f(x)=(5-6)2+48=49

Minimum value off(x)−−−−√=49−−√=7f(x)=49=7.