If g(x) = x² + x - 1 and (gof)(x) = 4x² - 10x + 5, then f(5/4) is equal to.
(A) 1/2
(B) -1/2
(C) -1/

Q: Wait, why do we reject the general form of $f(x)$? Can't we just find $f(x)$ first?

Great question! While finding $f(x)$ is mathematically possible, it is often a trap in competitive exams. In JEE Advanced, time is your most valuable currency. If the question asks for a specific value, calculating the general function is like building a whole factory just to produce one screw. It is inefficient and prone to algebraic errors. Always look for the 'direct path'—the path of least resistance—before diving into complex derivations.

Q: Is it always valid to substitute $x$ before finding the function?

Absolutely! If you are given an identity like $(g \circ f)(x) = \dots$, this holds true for all $x$ in the domain. Therefore, you are mathematically permitted to plug in any specific value of $x$ that interests you. It is a powerful tool to simplify complex functional equations into simple algebraic ones. Think of it as 'sampling' the function at a specific point rather than trying to map the entire landscape.

Q: How do I know if the quadratic will factorize?

In many competitive exams, if you end up with a quadratic equation where you need to find a specific value, it is often designed to be a perfect square or easily factorable. If you get a messy discriminant or irrational roots, pause and re-check your arithmetic—you might have missed a sign or a coefficient. The beauty of these problems often lies in their hidden symmetry!

Question

If $g (x) = x^2 + x - 1$ and $(gof) (x) = 4x^2 - 10x + 5$, then $f \left(\frac{5}{4}\right)$ is equal to.

Options:

(A) $\frac{1}{2}$

(B) $-\frac{1}{2}$

(C) $-\frac{1}{3}$

(D) $\frac{1}{3}$

Sigma Percentile · Accepted Answer

### The Art of the Shortcut: Mastering Composite Functions

Welcome, fellow traveler on the JEE journey! Today, we are going to dismantle a problem that, at first glance, might seem like a daunting task of algebraic manipulation. We are given two functions: an outer function $g(x) = x^2 + x - 1$ and a composite function $(g \circ f)(x) = 4x^2 - 10x + 5$. Our mission is to find the value of $f\left(\frac{5}{4}ight)$.

Many students, upon seeing this, immediately reach for their pens to find the general form of $f(x)$. They try to set up a quadratic equation for $f(x)$ and solve for it. While that is a valid path, it is often the long, winding road.

In the high-stakes environment of the JEE, we want the express train. Let's explore why we don't need to know the 'who' or 'what' of $f(x)$ to find its value at a specific point.

### The Philosophy of the Machine

Imagine a factory pipeline. The function $f(x)$ is the first machine; it takes an input $x$ and produces an output $f(x)$. The function $g(x)$ is the second machine; it takes the output of the first machine and processes it further.

The composition $(g \circ f)(x)$ is simply the final product of this two-stage process. We are told that when we feed $x = \frac{5}{4}$ into this pipeline, the final output is $4\left(\frac{5}{4}ight)^2 - 10\left(\frac{5}{4}ight) + 5$.

We don't need to know how the first machine works internally to know what it outputs when we feed it $\frac{5}{4}$. We only care about the value $f\left(\frac{5}{4}ight)$. Let's call this value $y$ for a moment. Our goal is to find $y$.

### The Strategic Pivot

By the definition of composition, we know that $(g \circ f)(x) = g(f(x))$. If we substitute $f(x)$ into the expression for $g(x)$, we get:

$$g(f(x)) = (f(x))^2 + f(x) - 1$$

Now, we have two ways to describe the same output. We have the 'theoretical' definition, which is $(f(x))^2 + f(x) - 1$, and we have the 'given' expression, which is $4x^2 - 10x + 5$. Since they represent the same thing, we equate them:

$$(f(x))^2 + f(x) - 1 = 4x^2 - 10x + 5$$

This is our master key. It connects the unknown function to the known variable $x$. Now, instead of solving for $f(x)$, we simply substitute $x = \frac{5}{4}$ into this equation. This transforms a functional equation into a simple algebraic one.

### The Calculation: Precision is Key

Let's evaluate the right-hand side (RHS) with care:

$$	ext{RHS} = 4\left(\frac{5}{4}ight)^2 - 10\left(\frac{5}{4}ight) + 5$$

Calculating the terms:

1. The first term: $4 	imes \frac{25}{16} = \frac{25}{4}$
2. The second term: $10 	imes \frac{5}{4} = \frac{50}{4}$
3. The constant term: $5 = \frac{20}{4}$

Combining these with a common denominator of $4$, we get:

$$	ext{RHS} = \frac{25 - 50 + 20}{4} = -\frac{5}{4}$$

Now, our equation looks like this:

$$\left(f\left(\frac{5}{4}ight)ight)^2 + f\left(\frac{5}{4}ight) - 1 = -\frac{5}{4}$$

### The Elegant Conclusion

We are almost there. Let's bring everything to one side to form a standard quadratic equation in terms of our unknown $y = f\left(\frac{5}{4}ight)$:

$$\left(f\left(\frac{5}{4}ight)ight)^2 + f\left(\frac{5}{4}ight) - 1 + \frac{5}{4} = 0$$

Since $-1 + \frac{5}{4} = \frac{1}{4}$, we have:

$$\left(f\left(\frac{5}{4}ight)ight)^2 + f\left(\frac{5}{4}ight) + \frac{1}{4} = 0$$

Look closely at this expression. It is the classic expansion of a perfect square: $a^2 + 2ab + b^2 = (a+b)^2$. Here, $a = f\left(\frac{5}{4}ight)$ and $b = \frac{1}{2}$. The equation simplifies perfectly to:

$$\left(f\left(\frac{5}{4}ight) + \frac{1}{2}ight)^2 = 0$$

Taking the square root of both sides, we find that $f\left(\frac{5}{4}ight) + \frac{1}{2} = 0$. This leads us directly to our final answer:

$$\mathbf{f\left(\frac{5}{4}ight) = -\frac{1}{2}}$$

And there you have it! We didn't need to find the general function or struggle with complex algebra. We simply used the properties of composition and a strategic substitution to reveal the answer. This is the essence of JEE problem-solving: identifying the most efficient path and trusting the math to guide you home.

Animated Solution for Mathematics - Functions: If $g (x) = x^{2} + x - 1$ and $(g o f) (x) = 4 x^{2} - 10 x + 5$ , then $f (\frac{5}{4})$ is equal to.

Select Answer:

Visualized Solution

Analyze the Given Functions

Define the Composition $g (f (x))$

Equate the Expressions

Substitute $x = \frac{5}{4}$

Evaluate the RHS Terms

Simplify the RHS

Form the Quadratic Equation

Recognize the Perfect Square

Solve for $f (\frac{5}{4})$

The Sigma Insight: Composite Functions

The Art of the Shortcut

The Philosophy of the Machine

The Strategic Pivot

The Calculation

The Elegant Conclusion

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Animated Solution for Mathematics - Functions: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If g(x)=x2+x−1 and (gof)(x)=4x2−10x+5, then f(45​) is equal to.