Let f: R arrow R defined by f ( x )= cos -1(- - x ) where x is fractional part function. Then which of the following is/are correct?

Question

Let f: R arrow R defined by f ( x )= cos -1(- - x ) where x is fractional part function. Then which of the following is/are correct? (A) f is many one but not even function. (B) Range of f contains two prime numbers. (C) f is aperiodic. (D) Graph of f does not lie below x-axis

Tardigrade Admin · Accepted Answer

We have f(x)= cos -1(- -x ) D f = all real As 0 ≤ - x <1 ∀ x ∈ R ⇒ -1<- - x ≤ 0 So R f =[(π/2), π) Clearly, f is neither even nor odd. But f ( x +1)= f ( x ) ⇒ f is periodic with period 1.

Q. Let $f : R \to R$ defined by $f (x) = cos^{- 1} (- {- x})$ where ${x}$ is fractional part function. Then which of the following is/are correct?

f is many one but not even function.

Range of f contains two prime numbers.

f is aperiodic.

Graph of f does not lie below x-axis

We have $f (x) = cos^{- 1} (- {- x})$
$D_{f} =$ all real
As $0 \leq {- x} < 1\forall x \in R$
$\Rightarrow - 1 < - {- x} \leq 0$
So $R_{f} = [\frac{π}{2}, π)$
Clearly, $f$ is neither even nor odd.
But $f (x + 1) = f (x) \Rightarrow f$ is periodic with period 1.

Q. Let f:R→R defined by f(x)=cos−1(−{−x}) where {x} is fractional part function. Then which of the following is/are correct?