JEE Mathematics 2023 Question with Solution

Given: $x = 2$ is a root of $x^2 + px + q = 0$.

Find: $\lim_{x \to 2p} [f(x)]$.

From the root condition,

$$2^2 + 2p + q = 0$$

so

$$4 + 2p + q = 0$$

and hence

$$q = -2p - 4.$$

the solution states that after substitution, the expression inside the cosine simplifies to

$$x^2 - 4px + 4p^2 = (x - 2p)^2.$$

Therefore,

$$f(x) = \frac{1 - \cos\left((x - 2p)^2\right)}{(x - 2p)^4}.$$

Now use the standard limit

$$\lim_{t \to 0} \frac{1 - \cos t}{t^2} = \frac{1}{2}.$$

Let

$$t = (x - 2p)^2.$$

Then as $x \to 2p$, we have $t \to 0$, and

$$f(x) = \frac{1 - \cos t}{t^2} \to \frac{1}{2}.$$

So near $x = 2p$,

$$[f(x)] = \left[\frac{1}{2}\right] = 0.$$

Hence,

$$\lim_{x \to 2p} [f(x)] = 0.$$

Therefore, the correct option is C.

The solution contains a mismatch in the displayed function during working, but it explicitly concludes that the correct option is C, and the limit of the greatest integer value is $0$.

Given: $x = 2$ is a root of $x^2 + px + q = 0$.

Find: $\lim_{x \to 2p} [f(x)]$.

First obtain

$$q = -2p - 4.$$

The worked solution simplifies the cosine argument to

$$(x - 2p)^2.$$

So the expression becomes

$$f(x) = \frac{1 - \cos\left((x - 2p)^2\right)}{(x - 2p)^4}.$$

Write

$$u = (x - 2p)^2.$$

Then

$$f(x) = \frac{1 - \cos u}{u^2}.$$

As $u \to 0$,

$$\frac{1 - \cos u}{u^2} \to \frac{1}{2}.$$

Therefore,

$$[f(x)] \to \left[\frac{1}{2}\right] = 0.$$

So the correct option is C.