Modulus Transformations

Two operations. One inside the function, one outside. Completely different effects.

|f(x)| — the fold

Take the graph of f(x). Every part that dips below the x-axis gets reflected upward. The positive parts are untouched.

For y = |sin x|:

• The portion on [π, 2π] where sin x is negative gets folded up
• The result is always ≥ 0
• The period halves — from 2π to π — because you now get two humps per old cycle

Toggle |f(x)| in the explorer below and watch what happens to the negative arch.

For y = |cos x|: same principle. Period halves to π.

Why this matters for JEE: Questions about number of solutions to |sin x| = k become trivial once you see the graph. If 0 < k < 1, the horizontal line y = k cuts 2n arches in [0, nπ].

f(|x|) — the mirror

Take the right half of f(x), i.e., the graph for x ≥ 0. Mirror it to the left. The result is always even — symmetric about the y-axis.

For y = sin|x|:

• For x ≥ 0: behaves exactly like sin x
• For x < 0: mirrors the right half
• This is an odd function mirrored into an even one — note that sin|x| is even, not odd

Toggle f(|x|) in the explorer and note the sharp corner at x = 0 — that's where the mirror joins.

The critical difference

		f(x)		f(|x|)
What happens	negative parts fold up	left side mirrors right
Symmetry	not necessarily symmetric	always even (y-axis symmetry)
Period of |sin x|	π	—
Differentiable at x = 0?	depends on f	no (corner if f(0) ≠ 0 or f'(0⁺) ≠ 0)

Common exam questions using these

1. Sketch y = |sin 2x| for x ∈ [0, 2π]. How many zeroes?
   Period of sin 2x is π. Period of |sin 2x| is π/2. So on [0, 2π] there are 4 complete arches → 5 zeroes (including endpoints).

2. Is sin|x| differentiable at x = 0?
   Left derivative: −cos(0) = −1. Right derivative: cos(0) = +1. Not equal → not differentiable at 0.

3. Find the period of |sin x + cos x|.
   sin x + cos x = √2 sin(x + π/4), period 2π. The modulus halves it to π.

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Next: Reciprocal curves — drawing cosec x from sin x without memorising a new shape.