Find the range of function f:[0,1]→R,f(x)=x3-x2+4x+2sin-1x.
[-(π+2),0]
[0,(4+π)]
[2,3]
[0,(2+π)]
Explanation for correct option.
Step:1 Simplify the function f(x).
f(x)=x3-x2+4x+2sin-1x⇒f(x)=x3+2sin-1x-x2+4x+4-4⇒f(x)=x3+2sin-1x-(x2-4x+4)+4⇒f(x)=x3+2sin-1x-(x-2)2+4⇒f(x)=x3+2sin-1x+4-(x-2)2
Here, the part x3+2sin-1x+4 of the function is increasing.
Step: Calculate the range of f(x)
Here, in
f(x),x∈(0,1)∴(x-2)∈(-2,-1)⇒(x-2)2∈(4,1)⇒(x-2)minatx=1;(x-2)maxatx=0.
f(x)maxatx=1∴f(x)x=1=4+1+π-1=4+πf(x)minatx=0;∴f(x)x=0=0+0+4-4=0
Range : [0,(4+π)].
Therefore, Option(B) is the correct answer.