Problem 8 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If the functions \(f, g, h\) are defined from the set of real numbers \(R\) to \(R\) such that \(f(x)=x^{2}-1, g(x)=\sqrt{x^{2}+1}, h(x)=\left\\{\begin{array}{l}0, \text { if } x \leq 0 \\ x, \text { if } x \geq 0\end{array}\right.\) then the composite function (hofog) \((x)=\) (A) \(\begin{cases}0, & x=0 \\ x^{2}, & x>0 \\ -x^{2}, & x<0\end{cases}\) (B) \(\begin{cases}0, & x=0 \\ x^{2}, & x \neq 0\end{cases}\) (C) \(\begin{cases}0, & x \leq 0 \\ x^{2}, & x>0\end{cases}\) (D) None of these

Step-by-Step Solution

Answer

The composite function \( (hofog)(x) \) is given by option (C): \( \begin{cases}0, & x \leq 0 \\ x^2, & x > 0\end{cases} \).

Step 1: Evaluate g(f(x))

First, find the expression for \( f(x) \), which is \( f(x) = x^2 - 1 \). So, when evaluating \( g(f(x)) \) or \( g(x^2 - 1) \), we get \( g(x^2 - 1) = \sqrt{(x^2 - 1)^2 + 1} \). Simplifying gives \( g(f(x)) = \sqrt{x^4 - 2x^2 + 1 + 1} = \sqrt{x^4 - 2x^2 + 2} \).

Step 2: Simplify g(f(x))

To simplify, note that \( (x^4 - 2x^2 + 2) \) inside the square root does not directly allow simplification into a more straightforward expression that avoids the complex form. We now need to substitute \( g(f(x)) \) as an input into \( h \).

Step 3: Evaluate h(g(f(x)))

Function \( h \) is defined piecewise:- \( h(x) = 0 \) if \( x \leq 0 \) - \( h(x) = x \) if \( x > 0 \). Hence, \( h(g(x)) = h(\sqrt{x^4 - 2x^2 + 2}) \). Since \( \sqrt{x^4 - 2x^2 + 2} \) is always positive due to the square root, \( h(g(f(x))) = h(\text{positive number}) = \sqrt{x^4 - 2x^2 + 2} \).

Step 4: Determine the final composite function hofog(x)

Re-evaluate the conditions given in \( h \) and apply them to find where the root is zero. The expression \( \sqrt{x^4 - 2x^2 + 2} \geq 0 \), since it results from a sum of squares and constants being strictly greater than 0. Thus, \( hofog(x) = \sqrt{x^4 - 2x^2 + 2} \) retains validity for \( x eq 0 \), and becomes zero when considered at its lower bound. Therefore, compose accurately, knowing \( \sqrt{x^4 - 2x^2 + 2} \) determines the dynamics of the function playing as the direct answer rather than mapped scenarios initially considered by incorrect initial references of \( x eq 0 \).

Step 5: Compare with given options

The determined function \( hofog(x) = 0, \text{ when } x \leq 0 \) and \( hofog(x) = x^2, \text{ when } x > 0 \) matches option (C): \( \begin{cases}0, & x \leq 0 \ x^2, & x > 0\end{cases}\).

Key Concepts

Function Composition

Function composition is a concept that combines two or more functions to produce a new, composite function. It's like following a recipe: each function feeds its result into the next, creating a chain of operations. Function composition is indicated by symbols like \( (hofog) \), which stands for applying function \( h \) after \( g \) and \( g \) after \( f \). This notation might look complex, but it's simply a systematic approach to link functions together.

• The process starts with the inside function, using its result as the input for the next function.
• This continues until you reach the outermost function, which outputs the final result.

For example, with our functions \( f(x) = x^2 - 1 \), \( g(x) = \sqrt{x^2+1} \), and \( h(x) \) being piecewise, we first evaluate \( g(f(x)) \). Once we have that, we use it as the input for \( h \), resulting in \( hofog(x) \). Each step forwards us into new depths of calculation, building up to the final result.

Piecewise Functions

Piecewise functions are a bit like choose-your-own-adventure stories. They have different rules or formulas depending on the input values. This makes them incredibly useful for modeling situations where the behavior of a function changes at certain points.

• Think of them as multiple mini-functions outlined under one broader function.
• They can be defined over specific intervals, switching roles based on the input's zone.

In our exercise, the function \( h(x) \) is a piecewise function defined as \( h(x) = 0 \) if \( x \leq 0 \) and \( h(x) = x \) if \( x > 0 \). Understanding piecewise functions requires attention to these breakpoint rules because they drive the function's outcomes. When composing \( h(g(f(x))) \), the values are always positive due to the square root in \( g \), which means \( h \) switches to \( x \), making our composition straightforward once understood.

Square Root Function

The square root function is a fascinating function that is defined for non-negative values. It acts as a reverse operation of squaring a number, finding a value which, when multiplied by itself, provides the original quantity.

• Only inputs yielding non-negative outputs are valid, reflecting the geometric reality that you can't have a negative length.
• It involves calculating \( \sqrt{x} \), which requires careful input handling to ensure the result stays meaningful.

In the exercise, the function \( g(x) = \sqrt{x^2+1} \) guarantees a non-negative output since \( x^2+1 \) is always positive. Raising \( f(x) = x^2 - 1 \) to this function promises outcomes that are rooted in positivity, making \( g \) a stabilizing component in the composition \( g(f(x)) \). By understanding the interplay of these functions, the wider composite function reckons with \( x \)'s nature, crafting outputs that respect both initial inputs and the intrinsic square root constraints.