The number of points, where the curve f(x)=e⁸ˣ-e⁶ˣ-3 e⁴ˣ-e²ˣ+1, x ∈ ℝ cuts x-axis, is equal to _________.

Solution

Maths sir · 5 steps

Put t = e²ˣ and you get a reciprocal quartic. Divide by t² and it becomes a quadratic.

1. Substitute. Let t = e²ˣ, so t > 0 for every real x, and each positive t gives exactly one x. Then f(x) = 0 becomes t⁴ - t³ - 3t² - t + 1 = 0.
2. Divide by t squared. The coefficients read the same from both ends, so divide by t²: (t² + 1/t²) - (t + 1/t) - 3 = 0. Put y = t + 1/t, which means t² + 1/t² = y² - 2.

   y² - y - 5 = 0 ⇒ y = (1 ± √21)/2

3. Keep valid y. For t > 0, t + 1/t ≥ 2. Since √21 ≈ 4.58, y = (1 + √21)/2 ≈ 2.79 is allowed, and y ≈ -1.79 is rejected.
4. Back to t. t + 1/t = y gives t² - yt + 1 = 0. Its discriminant y² - 4 > 0, the product of the roots is 1 and the sum is positive, so both roots are distinct and positive.
5. So. Two distinct positive values of t give two distinct real x. Both are simple roots, so the curve crosses the x-axis there. The curve cuts the x-axis at 2 points.

The common slip: 1. Stopping at the single valid value of t + 1/t gives 1. Each value of y greater than 2 gives two positive values of t, so the count is 2.

The idea to keep. When coefficients are symmetric, divide by the middle power and set y = t + 1/t. Remember y ≥ 2 for t > 0, and each valid y > 2 gives two positive t.

Answer from the NTA final answer key. Solution by PrepWiser.