Question

Let a and b are constants f(x)=a sin x+bx+3x^2. If f(-2)=5 then f(2)=
a)-5
b)5
c)0
d)19​

Answer 1

Answer:

$\mathrm{solve\:for\:a,\:f=a\sin \left(x\right)+bx+3x^2\quad :\quad a=\frac{f-bx-3x^2}{\sin \left(x\right)};\quad \:x\ne \:2\pi n,\:x\ne \:\pi +2\pi n}$

Step-by-step explanation:

$f=a\sin \left(x\right)+bx+3x^2$

$\mathrm{Switch\:sides}$

$a\sin \left(x\right)+bx+3x^2=f$

$\mathrm{Subtract\:}bx+3x^2\mathrm{\:from\:both\:sides}$

$a\sin \left(x\right)+bx+3x^2-\left(bx+3x^2\right)=f-\left(bx+3x^2\right)$

$a\sin \left(x\right)=f-bx-3x^2$

$\mathrm{Divide\:both\:sides\:by\:}\sin \left(x\right);\quad \:x\ne \:2\pi n,\:x\ne \:\pi +2\pi n$

$\frac{a\sin \left(x\right)}{\sin \left(x\right)}=\frac{f}{\sin \left(x\right)}-\frac{bx}{\sin \left(x\right)}-\frac{3x^2}{\sin \left(x\right)};\quad \:x\ne \:2\pi n,\:x\ne \:\pi +2\pi n$

$a=\frac{f-bx-3x^2}{\sin \left(x\right)};\quad \:x\ne \:2\pi n,\:x\ne \:\pi +2\pi n$

If f(-2)=5 then f(2)= -5