Modulus Inequalities PYQ | JEE Main 2025 | Fastest Method

 

❓ Question

Find the number of solutions of

$$|2^x-1|+|4-2^x|<3.$$

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✍️ Solution

Let

$$t=2^x \qquad (t>0).$$

Then the inequality becomes

$$|t-1|+|4-t|<3.$$

The critical points are

$$t=1,\;4$$

which correspond to

$$x=0,\;2.$$

Now solve case-wise.

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Case 1: $x<0$ ($0<t<1$)

$$|t-1|=1-t,\qquad |4-t|=4-t$$
$$(1-t)+(4-t)<3$$
$$5-2t<3$$
$$2t>2$$
$$t>1,$$

which contradicts $t<1$.

No solution.

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Case 2: $0\le x<2$ ($1\le t<4$)

$$|t-1|=t-1,\qquad |4-t|=4-t$$
$$(t-1)+(4-t)=3.$$

The inequality becomes

$$3<3,$$

which is impossible.

No solution.

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Case 3: $x\ge2$ ($t\ge4$)

$$|t-1|=t-1,\qquad |4-t|=t-4$$
$$(t-1)+(t-4)<3$$
$$2t-5<3$$
$$2t<8$$
$$t<4,$$

which contradicts $t\ge4$.

No solution.

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✅ Final Answer

$$\boxed{0}$$

There are no real solutions to the given inequality.