Q.
Let f:R→R defined by f(x)=2x3+2x2+300x+5sinx, then f is
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Relations and Functions - Part 2
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Solution:
f(x)=2x3+2x2+300x+5sinx f′(x)=6x2+4x+300+5cosx =2(3x2+2x+147)+(6+5cosx)
Since −1≤cosx≤1∀x∈R ∴6+5cosx>0
let g(x)=3x2+2x+147
Since a=3>0 and D=(2)2−4×3×147<0 ∴g(x)>0 f′(x)>0 ∴f(x) is an increasing function therefore said to be 1−1 function.
Now, f(∞)=−∞ and f(∞)=∞ ∴f(x) is continuous also. ∴ Range of f= codomain of f=R. ∴f is onto.