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JEE Main 2023 · 29 Jan · Shift 1

JEE Main 2023 Mathematics question on Limits Continuity and Differentiability

Let x=2x = 2 be a root of the equation x2+px+q=0x^2 + px + q = 0 and
f(x)={1−cos⁡(x2−4px+q2+8q+16)(x−2p)4,x≠2p0,x=2pf(x) = \begin{cases} \dfrac{1 - \cos(x^2 - 4px + q^2 + 8q + 16)}{(x - 2p)^4}, & x \neq 2p \\ 0, & x = 2p \end{cases}
Then lim⁡x→2p−[f(x)]\lim_{x \to 2p^-} [f(x)],
where [⋅][\cdot] denotes greatest integer function, is
  1. A.
    −1-1
  2. B.
    11
  3. C.
    22
  4. D.
    00
Show answer and solution

Correct option: D

Solution

Since x = 2 is a root of x² + px + q = 0, we have 4 + 2p + q = 0, so q = -2p - 4.

Now find the argument of cosine: x² - 4px + q² + 8q + 16.

At x = 2p: (2p)² - 4p(2p) + q² + 8q + 16 = 4p² - 8p² + q² + 8q + 16 = -4p² + (q+4)².

Since q = -2p - 4, we get q + 4 = -2p, so (q+4)² = 4p².

Thus the argument = -4p² + 4p² = 0 at x = 2p, confirming 0/0 form.

Let t = x - 2p → 0⁻. The numerator argument becomes:
x² - 4px + q² + 8q + 16 = (x - 2p)² + (q+4)² - 4p² + ...

More carefully: x² - 4px + q² + 8q + 16 = (x-2p)² - 4p² + (q+4)² = t² - 4p² + 4p² = t².

So f(x) = (1 - cos(t²)) / t⁴.

Using 1 - cos(u) ≈ u²/2 for small u, with u = t²:
f(x) ≈ (t⁴/2) / t⁴

[JEE Main 2023 · 29 Jan · Shift 1]

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