TS EAMCET 4 May 2018 Shift 1 Solved Paper

Consider the equation.
f(x)=2x3+10x−13
Therefore,
f′(x)=6x2+10>0,∀x∈ℝ
So, the function f(x) is strictly increasing.
Thus,
f(1)=2+10−13
=−1<0
And,
f(2)=16+20−13
=23>0
So, the one and only one root e∈(1,2) where e is theeccentricity of the conic.
As 1<e<2
This implies that the conic is a hyperbola.