Question

We assign certain score to a function depending on type of function it is. If $f(x)$ is one-one then score given to $f(x)$ is 1, if many-one then score given to $f(x)$ is 2. Similarly, if $f(x)$ is onto, into, odd, even, neither even nor odd, periodic and aperiodic then score given to $f(x)$ is 3, 4, 5, 6, 7, 8 and 9 respectively. Then net score of $f: R \rightarrow R$, $f(x)=\frac{(2^{\sin x}+1)^2}{2^{\sin x}}$ is __.

Answer 1

20

Answer 2

The given function is $f: R \rightarrow R$, $f(x)=\frac{(2^{\sin x}+1)^2}{2^{\sin x}}$.

First, simplify the expression for $f(x)$: $f(x) = \frac{(2^{\sin x})^2 + 2 \cdot 2^{\sin x} \cdot 1 + 1^2}{2^{\sin x}}$ $f(x) = \frac{2^{2\sin x} + 2 \cdot 2^{\sin x} + 1}{2^{\sin x}}$ $f(x) = \frac{2^{2\sin x}}{2^{\sin x}} + \frac{2 \cdot 2^{\sin x}}{2^{\sin x}} + \frac{1}{2^{\sin x}}$ $f(x) = 2^{\sin x} + 2 + 2^{-\sin x}$ $f(x) = 2^{\sin x} + 2^{- \sin x} + 2$

Now, let's analyze the properties of $f(x)$:

1. One-one or Many-one:

A function is one-one if distinct inputs map to distinct outputs. It is many-one if at least two distinct inputs map to the same output. We know that $\sin x$ is a periodic function, meaning $\sin x$ takes the same value for multiple $x$ values (e.g., $\sin 0 = 0$ and $\sin \pi = 0$). Let's check $f(0)$ and $f(\pi)$: $f(0) = 2^{\sin 0} + 2^{-\sin 0} + 2 = 2^0 + 2^0 + 2 = 1+1+2 = 4$. $f(\pi) = 2^{\sin \pi} + 2^{-\sin \pi} + 2 = 2^0 + 2^0 + 2 = 1+1+2 = 4$. Since $0 \neq \pi$ but $f(0) = f(\pi)$, the function is many-one. Score for many-one: 2.

2. Onto or Into:

A function $f: A \rightarrow B$ is onto if its range is equal to its codomain ($B$). It is into if its range is a proper subset of its codomain ($B$). The codomain is given as $R$. Let's find the range of $f(x)$. Let $u = \sin x$. Since $x \in R$, $u \in [-1, 1]$. So, $f(x) = g(u) = 2^u + 2^{-u} + 2$. Consider the function $h(u) = 2^u + 2^{-u}$. To find its range for $u \in [-1, 1]$, we can analyze its derivative or check values at critical points and endpoints. $h'(u) = 2^u \ln 2 - 2^{-u} \ln 2 = \ln 2 (2^u - 2^{-u})$. Setting $h'(u) = 0 \implies 2^u = 2^{-u} \implies u = -u \implies 2u = 0 \implies u=0$. At $u=0$, $h(0) = 2^0 + 2^0 = 1+1=2$. This is a minimum value. At the endpoints of the interval $[-1, 1]$: For $u=1$, $h(1) = 2^1 + 2^{-1} = 2 + 1/2 = 5/2$. For $u=-1$, $h(-1) = 2^{-1} + 2^{-(-1)} = 1/2 + 2 = 5/2$. So, the range of $h(u)$ for $u \in [-1, 1]$ is $[2, 5/2]$. Therefore, the range of $f(x) = h(\sin x) + 2$ is $[2+2, 5/2+2] = [4, 9/2]$. Since the range $[4, 9/2]$ is a proper subset of the codomain $R$, the function is into. Score for into: 4.

3. Odd, Even, or Neither even nor odd:

A function $f(x)$ is even if $f(-x) = f(x)$. It is odd if $f(-x) = -f(x)$. Let's check $f(-x)$: $f(-x) = 2^{\sin(-x)} + 2^{-\sin(-x)} + 2$ Since $\sin(-x) = -\sin x$: $f(-x) = 2^{-\sin x} + 2^{-(-\sin x)} + 2$ $f(-x) = 2^{-\sin x} + 2^{\sin x} + 2$ This is identical to $f(x)$. So, $f(-x) = f(x)$. The function is even. Score for even: 6.

4. Periodic or Aperiodic:

A function $f(x)$ is periodic if there exists a positive real number $T$ such that $f(x+T) = f(x)$ for all $x$ in the domain. We know that $\sin x$ is a periodic function with period $2\pi$. Let's check $f(x+2\pi)$: $f(x+2\pi) = 2^{\sin(x+2\pi)} + 2^{-\sin(x+2\pi)} + 2$ Since $\sin(x+2\pi) = \sin x$: $f(x+2\pi) = 2^{\sin x} + 2^{-\sin x} + 2 = f(x)$. So, the function is periodic with period $2\pi$. Score for periodic: 8.

Net Score Calculation:

Sum the scores for each property: Net Score = (Score for Many-one) + (Score for Into) + (Score for Even) + (Score for Periodic) Net Score = $2 + 4 + 6 + 8 = 20$.