The correct option is B x9−f(x)
A)
As range of f(x) is (0,1),
Let g(x)=x9−f(x)
∴g(x)=0⇒x=(f(x))19
x will lie (0,1).
B)
Let g(x)=x−π2−x∫0f(t)cost dt
g(0)=−π2∫0f(t)cost dt ⇒g(0)<0
Now,
g(1)=1−π2−1∫0f(t)cost dt
The minimum value of g(1) takes place when π2−1∫0f(t)cost dt is maximum
Max.⎛⎜⎝π2−1∫0f(t)cost dt⎞⎟⎠=π2−1
Min.(g(1))=2−π2>0
∴g(1)>0
As g(0)⋅g(1)<0 there is atleast one value of x∈(0,1) where g(x) will be zero.
C)
Let g(x)=ex−x∫0f(t)sint dt
g′(x)=ex−f(x)sinx
When x∈(0,1) then ex∈(1,e) and (f(x)sinx)∈(0,1)
∴g′(x)>0 ∀ x∈(0,1)
So g(x) monotonically increasing function.
Now, g(0)=1
So g(x)>1 ∀ x∈(0,1)
D)
It will always remain positive in (0,1)
Hence option A and B are correct.