Question

( f(x)=x+14 x^{5}+16 x^{3}+30 x-560=0 )
( f(n)=7 x^{6}+70 x^{4}+48 x^{2}+30>0 )
it is shictly increasing
[
begin{array}{l}text { when } x rightarrow infty ; f(n) rightarrow infty x rightarrow-infty ; f(n) rightarrow-inftyend{array}
]
it cuts ( x ) anis at exactey
one point
( therefore ) it has only one real root

# Q.2 How many real solution does the equation x7+14x5 +16x3 +30x - 560 = 0 have? [AIEEE-2008] (A) 1 (B)3 (C) 5 (D) 7

Solution