Consider the function f:[1/2, 1] → ℝ defined by f(x)=4 √2 x³-3 √2 x-1. Consider the statements
(I) The curve y=f(x) intersects the x-axis exactly at one point.
(II) The curve y=f(x) intersects the x-axis at x=cos π/12.
Then

Solution

Maths sir · 4 steps

See 4x³ - 3x and think cos 3θ. Put x = cos θ and the cubic becomes a one-line equation.

1. Spot the identity. Write f(x) = √2 (4x³ - 3x) - 1. Setting f(x) = 0 gives 4x³ - 3x = 1/√2, and 4 cos³ θ - 3 cos θ = cos 3θ.
2. Substitute. Put x = cos θ. Since x ∈ [1/2, 1], we can take θ ∈ [0, π/3], so 3θ ∈ [0, π].

   cos 3θ = 1/√2, 3θ ∈ [0, π] ⇒ 3θ = π/4 ⇒ θ = π/12

3. Check uniqueness. On [0, π] cosine takes each value only once, so π/4 is the only solution. You can confirm this another way: f'(x) = 3√2 (4x² - 1) ≥ 0 on the interval, so f is increasing. It goes from f(1/2) = -√2 - 1 < 0 to f(1) = √2 - 1 > 0, so it crosses the x-axis exactly once.
4. So. The curve meets the x-axis at exactly one point, x = cos π/12. Both (I) and (II) are correct, which is option (a).

The idea to keep. Whenever a cubic looks like 4x³ - 3x on a range inside [-1, 1], put x = cos θ. Then use the allowed range of 3θ to count the roots.

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