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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=(1z4)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 xαRx_\alpha \in \mathbb{R} such that z(xα)=1αz(x_\alpha) = 1-\alpha. Then which of the following statements is true?

  1. A.Both (P) and (Q) are true.
  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.

See pricing

50 are analysed free — try those first.

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 EquationsExistence–uniqueness, Picard, Lipschitz

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