JEE Mathematics 2023 Question with Solution

Given: The parabola has focus $\left( -\frac{1}{2}, 0 \right)$ and directrix $y = -\frac{1}{2}$.

Find: How many elements the set $S$ contains.

From the extracted solution, the parabola is written as

$$\left(x+\frac{1}{2}\right)^2 = \left(y+\frac{1}{4}\right)$$

so

$$y = x^2 + x$$

Thus $f(x) = x^2 + x$.

Now the given condition becomes

$$\tan^{-1}\left(\sqrt{x(x+1)} + \sin^{-1}\sqrt{x^2+x+1}\right) = \frac{\pi}{2}$$

the solution then uses the domain conditions

$$0 \le x^2 + x + 1 \le 1$$

which gives

$$x^2 + x \le 0$$

Also, for $\sqrt{f(x)}$ to be defined,

$$x^2 + x \ge 0$$

Combining both,

$$x^2 + x = 0$$

Therefore,

$$x = 0, -1$$

Hence, the set $S$ contains exactly two elements.

Note: The solution concludes with two elements, but the solution's also marks the correct option as C, which conflicts with the listed options. Based on the worked solution, the correct statement is Exactly two elements.

Given: Focus $\left( -\frac{1}{2}, 0 \right)$ and directrix $y = -\frac{1}{2}$.

Find: The number of elements in $S$.

The page shows:

$$\left(x+\frac{1}{2}\right)^2 = \left(y+\frac{1}{4}\right)$$

Hence,

$$y = x^2 + x$$

Now substitute in the set condition. The page rewrites the condition and imposes:

$$0 \le x^2 + x + 1 \le 1$$

So,

$$x^2 + x \le 0$$

Also,

$$x^2 + x \ge 0$$

Therefore,

$$x^2 + x = 0$$

Factoring,

$$x(x+1)=0$$

Thus,

$$x=0,-1$$

So $S$ contains $2$ elements, corresponding to option A among the listed options.