Animated Solution for Mathematics - Functions: Let $f(x)=|x-1|$. Then

My dear student, welcome to a journey of mathematical discovery. In the high-stakes arena of JEE Advanced, we are often conditioned to look for patterns, to seek the elegance of symmetry, and to trust the standard algebraic identities we have memorized.

But today, we are going to learn the most important lesson in a mathematician's toolkit: the power of healthy skepticism.

We are presented with the function $f(x)=|x-1|$. At first glance, it looks harmless. It is an absolute value function, a staple of our curriculum.

But let us pause and visualize it. If you were to sketch this on your graph paper, you would not draw a V-shape centered at the origin $(0,0)$. Instead, you would shift that vertex to the right, landing firmly at $(1,0)$.

That single unit of horizontal shift is not just a geometric detail; it is a fundamental transformation that shatters the standard algebraic properties we often take for granted.

Testing Option (b)

Let us begin by addressing the most common trap: the assumption of linearity. Many students see an expression and instinctively want to distribute it. Option (b) suggests that $f(x+y)=f(x)+f(y)$.

This is the hallmark of a linear function, like $f(x)=mx$. But is our function linear? Let us put it to the test.

Imagine we choose two values, $x=2$ and $y=5$. If the property held, the function would distribute across the sum.

Let us calculate the left-hand side (LHS):

Now, let us look at the right-hand side (RHS): $f(2)+f(5)$. We calculate $f(2)=|2-1|=1$ and $f(5)=|5-1|=4$. Adding these gives us $1+4=5$.

We are left with a stark reality: $6 eq 5$. The function does not distribute. By simply picking two numbers, we have dismantled the illusion of linearity.

Testing Option (a)

Next, let us examine Option (a), which claims that $f(x^2)=(f(x){)}^2$. This is a bold claim. It suggests that squaring the input is equivalent to squaring the output.

Let us test this with $x=2$. For the LHS, we calculate $f(2^2)$, which is $f(4)$. Substituting into our function, we get:

Now, for the RHS, we calculate $(f(2){)}^2$. We know $f(2)=|2-1|=1$. Squaring this gives us $1^2=1$.

Again, we see that $3 eq 1$. The property fails. This teaches us that the absolute value function, especially when shifted, does not preserve the power of the input in the way we might hope.

Testing Option (c)

Finally, we arrive at Option (c): $f(|x|)=|f(x)|$. This one feels intuitive, does it not? Absolute values are all about symmetry. But let us be rigorous.

Let us test a negative value, $x=-5$, to see if the function handles the negative sign as we expect. For the LHS, we calculate $f(|-5|)$, which simplifies to $f(5)$. Substituting this into our function, we get:

Now, for the RHS, we calculate $|f(-5)|$. First, we find $f(-5)=|-5-1|=|-6|=6$. Then, we take the absolute value of that result: $|6|=6$.

Once more, we find that $4 eq 6$. The symmetry is broken. The shift of the vertex to $x=1$ means that the function is no longer symmetric about the y-axis, and thus, the standard properties of absolute values are fundamentally altered.

We have systematically tested each option. We have used counter-examples to disprove the claims of linearity, power preservation, and symmetry.

In the context of the JEE, this is a vital skill. When you encounter a question asking you to verify properties, do not waste time trying to prove them for all $x$. Instead, hunt for the counter-example.

Since options (a), (b), and (c) have all been proven false, we are left with the only remaining logical conclusion: (d) None of these.

My dear student, remember this: mathematics is not just about memorizing formulas; it is about testing the boundaries of those formulas. The shift in $f(x)=|x-1|$ is a reminder that even the smallest change in a function's definition can have profound consequences on its algebraic behavior. Stay curious, stay skeptical, and keep calculating.