JEE Mathematics 2023 Question with Solution

Given: The equation considered in the solution is

$$3\left(x^2 + \frac{1}{x^2}\right) - 2\left(x + \frac{1}{x}\right) + 5 = 0$$

Find: The number of real solutions.

Using

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

let

$$t = x + \frac{1}{x}$$

Then the equation becomes

$$3\left(t^2 - 2\right) - 2t + 5 = 0$$

so

$$3t^2 - 2t - 1 = 0$$

Now factorize:

$$3t^2 - 3t + t - 1 = 0$$ $$(t - 1)(3t + 1) = 0$$

Hence

$$t = 1 \quad \text{or} \quad t = -\frac{1}{3}$$

Now check

$$x + \frac{1}{x} = 1$$

and

$$x + \frac{1}{x} = -\frac{1}{3}$$

For real $x$, the value of $x + \frac{1}{x}$ must satisfy $t \ge 2$ or $t \le -2$. Neither $1$ nor $-\frac{1}{3}$ satisfies this, so there is no real solution.

Therefore, the number of real solutions is $0$. The solution explicitly states that the correct option is D, which disagrees with the listed options; among the given options, the defensible answer corresponding to $0$ is B.

Given: The worked solution uses the substitution $t = x + \frac{1}{x}$. Find: Whether any real $x$ exists.

From

$$t = x + \frac{1}{x}$$

we have the identity

$$x^2 + \frac{1}{x^2} = t^2 - 2$$

Substituting into the equation used in the solution gives

$$3(t^2 - 2) - 2t + 5 = 0$$ $$3t^2 - 6 - 2t + 5 = 0$$ $$3t^2 - 2t - 1 = 0$$

Factorizing,

$$(t - 1)(3t + 1) = 0$$

Thus,

$$t = 1, -\frac{1}{3}$$

But for real $x$,

$$t = x + \frac{1}{x}$$

must lie outside the interval $(-2,2)$. Since both obtained values lie inside that interval, neither produces a real value of $x$.

Therefore, there are no real solutions, so the correct value is $0$ and the matching listed option is B.