JEE Mathematics 2023 Question with Solution

Given: The equation is $e^{4x} + 8e^{3x} + 13e^{2x} - 8e^x + 1 = 0$.

Find: The number of real solutions and their nature.

Let $e^x = t$. Since $e^x > 0$ for all real $x$, we have $t > 0$.

Then the equation becomes

$$t^4 + 8t^3 + 13t^2 - 8t + 1 = 0$$

Dividing by $t^2$,

$$t^2 + 8t + 13 - \frac{8}{t} + \frac{1}{t^2} = 0$$

Rearrange as

$$\left(t - \frac{1}{t}\right)^2 + 8\left(t - \frac{1}{t}\right) + 15 = 0$$

Let

$$z = t - \frac{1}{t}$$

Then

$$z^2 + 8z + 15 = 0$$

So,

$$(z+3)(z+5)=0$$

Hence,

$$z = -3 \quad \text{or} \quad z = -5$$

Thus,

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

or

$$t - \frac{1}{t} = -5$$

These give

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

or

$$t^2 + 5t - 1 = 0$$

So,

$$t = \frac{-3 \pm \sqrt{13}}{2}$$

or

$$t = \frac{-5 \pm \sqrt{29}}{2}$$

Since $t=e^x>0$, only the positive roots are admissible:

$$t = \frac{\sqrt{13}-3}{2} \quad \text{or} \quad t = \frac{\sqrt{29}-5}{2}$$

Therefore,

$$x = \ln\left(\frac{\sqrt{13}-3}{2}\right) \quad \text{or} \quad x = \ln\left(\frac{\sqrt{29}-5}{2}\right)$$

Both admissible values of $t$ lie between $0$ and $1$, so both corresponding values of $x$ are negative.

Therefore, the equation has two solutions and both are negative. The correct option is B.

The raw option text on the page labels this statement as option A, but the solution working explicitly marks the correct option as B. Following the solution authority, the answer is taken as B.

Given: $e^{4x} + 8e^{3x} + 13e^{2x} - 8e^x + 1 = 0$

Find: How many real values of $x$ satisfy the equation.

1. Put $e^x=t$ with $t>0$.
2. Then

$$t^4 + 8t^3 + 13t^2 - 8t + 1 = 0$$

3. Divide by $t^2$:

$$t^2 + 8t + 13 - \frac{8}{t} + \frac{1}{t^2} = 0$$

4. Recognize

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

So the equation becomes

$$\left(t-\frac{1}{t}\right)^2 + 8\left(t-\frac{1}{t}\right) + 15 = 0$$

5. Let $z=t-\frac{1}{t}$. Then

$$z^2+8z+15=0$$

6. Factor:

$$(z+3)(z+5)=0$$

Thus $z=-3$ or $z=-5$. 7. For $z=-3$,

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

Multiplying by $t$,

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

8. For $z=-5$,

$$t-\frac{1}{t}=-5$$

Multiplying by $t$,

$$t^2+5t-1=0$$

9. Each quadratic gives one positive and one negative root, but only the positive roots are allowed because $t=e^x>0$.
10. Hence there are exactly two valid values of $t$, and both satisfy $$0

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