If `f(x)=ax^(4)-bx^(2)+x+5 and f(-3)=2, then f(3) is`

if f(x)=ax^(7)+bx^(3)+cx-5,f(-7)=7 then f(7) is

Let f(x)=ax^(4)+bx^(2)+3x+7 and f(-4)=2286 and f=(4)=N . The number of ways in which N can be resolved as a product of two divisors which are relatively prime:

If f(x)=ax^2+bx+2 , where f(-1)=7 and f(1)=5, then: f(a,b)-=

If f(x)=ax^2-3x+c where f(2)=3 and f(3) =10, then: (a,c)-=

If: f(x)=bx^2+cx+d and: f(x+1)-f(x)=8x+3, then:

If f (x) =ax ^(2) + bx + 2 and f (1) =3, f (4) =42, then a and b respectively are

If f(x)=ax^(2)+bx+c and f(-1) ge -4 , f(1) le 0 and f(3) ge 5 , then the least value of a is

f(x)=x^(4)+ax^(3)+bx^(2)+cx+d and f(1)=10,f(2)=20,f(3)=30 then find the value of f(12)+f(-8)