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Part BCSIR NET December 2025non-lipschitz-does-not-automatically-mean-non-unique-check-whether-the-sign-oscillates

Non lipschitz does not automatically mean non unique check whether the sign oscillates

Consider the following statements: (P) The initial value problem y′ = f(y), where f(y) = y·cos(1/y) for y ≠ 0 and f(0) = 0, y(0) = 0, has at most one solution. (Q) If z(x) is the solution of the initial value problem z′=(1−z4)100/(1+z2),z(0)=0z' = (1-z^{4})^{100}/(1+z^{2}), z(0) = 0, then for each α∈(0,2)\alpha \in (0,2), there exists xα∈Rx_\alpha \in \mathbb{R} such that z(xα)=1−αz(x_\alpha) = 1-\alpha. Then which of the following statements is true?

  1. A.Both (P) and (Q) are true.✓ Correct
  2. B.(P) is true, (Q) is FALSE.
  3. C.(P) is FALSE, (Q) is true.
  4. D.Both (P) and (Q) are FALSE.

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests converse assumed.

See pricing

50 are analysed free — try those first.

The trap it tests

Converse assumed

The theorem runs one way. You used it in the other.

Drill statements like this

Related counterexample: Every IVP has a unique solution

More on this topic

The chapter behind this: Existence, uniqueness and how far the solution extends — free to read

From Ordinary Differential Equations › Existence–uniqueness, Picard, Lipschitz

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

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