32. Show that $f: \mathbb{R} \to \mathbb{R}$ defined as $f(x) = \frac{x}{\sqrt{1+x^2}}$ is one-one but not onto. (function, domain, codomain, one-to-one, onto)

Step 1: Check for One-one (Injectivity)

To prove that the function $f(x)$ is one-one, we assume that $f(x_1) = f(x_2)$ for any two elements $x_1, x_2$ in the domain $\mathbb{R}$ and then show that this implies $x_1 = x_2$.

Step 2: Set up the equation for one-one

Substitute the function definition into the assumption $f(x_1) = f(x_2)$. This forms an equation that we will solve for $x_1$ and $x_2$.

Step 3: Square both sides

To eliminate the square roots, we square both sides of the equation. This operation simplifies the expression, removing the square root from the denominator.

Step 4: Rearrange the equation

Cross-multiply the terms to remove the denominators. This step helps in bringing all terms to a single line, making further algebraic manipulation easier.

Step 5: Expand and simplify

Expand the products on both sides and observe that the term $x_1^2 x_2^2$ cancels out from both sides, leading to $x_1^2 = x_2^2$.

Step 6: Analyze $x_1^2 = x_2^2$

From $x_1^2 = x_2^2$, we deduce that $x_1$ must be either equal to $x_2$ or $x_1$ must be equal to $-x_2$. We need to ensure that only $x_1 = x_2$ is possible for the function to be one-one.

Step 7: Check the sign of $f(x)$

Notice that $f(x)$ has the same sign as $x$. This means that $f(x_1) = f(x_2)$ cannot hold if $x_1$ and $x_2$ have opposite signs (unless both are zero), as one would produce a positive value and the other a negative value.

Step 8: Conclusion for one-one

Since $f(x_1) = f(x_2)$ implies that $f(x_1)$ and $f(x_2)$ must have the same sign, $x_1$ and $x_2$ must have the same sign. Therefore, from $x_1 = \pm x_2$, the only valid option is $x_1 = x_2$. This confirms that the function is one-one (injective).

Step 9: Check for Onto (Surjectivity)

To check if the function is onto, we need to determine if for every value $y$ in the codomain $\mathbb{R}$, there exists an $x$ in the domain $\mathbb{R}$ such that $f(x) = y$. We'll try to express $x$ in terms of $y$ to find the range of the function.

Step 10: Square both sides and solve for $x^2$

Square both sides of the equation $y = f(x)$ to eliminate the square root and then algebraically rearrange the equation to isolate $x^2$. This will define the relationship between $x$ and $y$ that must hold.

Step 11: Solve for $x$

Take the square root of both sides to find $x$ in terms of $y$. For $x$ to be a real number, the expression under the square root must be non-negative, and the denominator must not be zero.

Step 12: Determine the range restrictions

For $x$ to be defined, the term $1-y^2$ in the denominator must be strictly positive. This implies that $y^2$ must be less than $1$, which means $y$ must be between $-1$ and $1$ (exclusive). If $y = \pm 1$, the denominator becomes zero, making $x$ undefined. Also, consider the case where $y=0$, which gives $x=0$ and is valid.

Step 13: Consider the sign of $x$ and $y$

From the original function, $f(x) = \frac{x}{\sqrt{1+x^2}}$, we know that $f(x)$ has the same sign as $x$. Therefore, $y$ must have the same sign as $x$. If $y > 0$, we choose the positive square root for $x$. If $y < 0$, we choose the negative square root for $x$. This ensures consistency with the original function definition of $f(x)$.

Step 14: Final range determination

Combining the conditions, the range of the function $f(x)$ is the open interval $(-1, 1)$. This means that for any $y$ outside this interval (i.e., $y \le -1$ or $y \ge 1$), there is no real $x$ such that $f(x) = y$. Since the codomain is $\mathbb{R}$ and the range is $(-1, 1)$, the range is not equal to the codomain.

Step 15: Conclusion for Onto

Since the range of the function, which is the interval $(-1, 1)$, is not equal to the codomain, which is $\mathbb{R}$, the function is not onto (surjective). For example, there is no $x$ for which $f(x)=2$ because $2$ is not in the range $(-1{,}1)$.

More from this chapter

• Show that an onto function f:1,2,3 →1,2,3 is always one-one.
• Let A=-2,-1,0,1,2,3. Define a relation R on A by xRy if and only if y=max(x,1). If l is the number of elements in R, and m and n are the minimum numbers of orde
• 32. Show that f: R -> R defined as f(x) = x / sqrt(1 + x²) is one-one but not onto. (function, one-one, onto, real numbers, domain, codomain, range)
• Let A be the set of all students of a boys school. Show that the relation R in A given by R=(a, b): a is sister of b is the empty relation and R^()=(a, b): the
• Consider the identity function I_N: N → N defined as I_N(x)=x x N. Show that although I_N is onto but I_N+I_N: N → N defined as (I_N+I_N)(x)=I_N(x)+I_N(x)=x+x=2
• Prove that the relation R in the set A = 1, 2, 3, 4, 5 given by R = (a, b): |a - b| is even is an equivalence relation.
• Let f(x) = log_e x and g(x) = (x^4 - 2x³ + 3x² - 2x + 2)/(2x² - 2x + 1). Then the domain of f ° g is
• Let the domain of the function f(x) = cos^(-1) ( (4x + 5)/(3x - 7)) be [α, β] and the domain of g(x) = log_2(2 - 6log_7(2x + 5)) be [, ]. Then |7(α + β) + 4( +)

All Relations and Functions solutions →