The range of `f(x)=(x+1)(x+2)(x+3)(x+4)+5` for `x in [-6,6]` is [4, 5045] (b) [0, 5045] `[-20 , 5045]` (d) none of these

Options

A. [4, 5045]
B. [0, 5045]
C. [-20, 5045]
D. None of these

Answer

A

Explanation

Let `g(x)=(x+1)(x+2)(x+3)(x+4)`
The rough of `g(x)` is given as follows,

` :. g(x)=(x+1)(x+2)(x+3)(x+4)`
`=(x+1)(x+4)(x+2)(x+3)`
`=(x^(2)+5x+4)(x^(2)+5x+6)`
`=t(t+2)=(t+1)^(2)-1 " where " t=x^(2)+5x`
Now, `g_(min)= -1,` for which ` x^(2)+5x= -1 ` has real roots in `[-6,6].`
Then, the range of `f(x)` is `[4,5045].`