Question

How many real solution does the equation x?+ 14x5 +16x3 + 30x-560 = 0 have? [A (A) 1 (B)3 (C) 5 (D)7

JEE/Engineering Exams

Maths

Solution

4.0 (1 ratings)

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