Let 𝜙 and 𝜓 be two linearly independent solutions of the ordinary differential equation

𝑦′′ + (2 − cos 𝑥) 𝑦 = 0, 𝑥 ∈ ℝ .

Let 𝛼, 𝛽 ∈ ℝ be such that 𝛼 < 𝛽, 𝜙(𝛼) = 𝜙(𝛽) = 0 and 𝜙(𝑥) ≠ 0 for all 𝑥 ∈ (𝛼, 𝛽).

Consider the following statements:

𝑃: 𝜙′ (𝛼)𝜙′ (𝛽) > 0.

𝑄: 𝜙(𝑥)𝜓(𝑥) ≠ 0 for all 𝑥 ∈ (𝛼, 𝛽).

Then 

Given -

Let 𝜙 and 𝜓 be two linearly independent solutions of the ordinary differential equation

𝑦′′ + (2 − cos 𝑥) 𝑦 = 0, 𝑥 ∈ ℝ .

Let 𝛼, 𝛽 ∈ ℝ be such that 𝛼 < 𝛽, 𝜙(𝛼) = 𝜙(𝛽) = 0 and 𝜙(𝑥) ≠ 0 for all 𝑥 ∈ (𝛼, 𝛽).

Concept -

(i) If function f(x) is increasing then f'(x) > 0

(ii) If function f(x) is decreasing then f'(x) < 0

(iii) Sturn Separation Theorem - If y₁(x) & y₂(x) are two continuous linearly independent solutions to homogeneous differential equation y'' + Py' + Qy =   0

having α & β being successive roots of y₁(x), then there exist exactly one root of y₂ in (α , β ), i.e.  there exist x ∈  (α , β ) such that y₂(x) = 0

Explanation -

For statement (P) -

Let 𝛼, 𝛽 ∈ ℝ be such that 𝛼 < 𝛽, 𝜙(𝛼) = 𝜙(𝛽) = 0 and 𝜙(𝑥) ≠ 0 for all 𝑥 ∈ (𝛼, 𝛽).

So now we have two path I and II which is shown below 

If we take the path (I) then \(\phi'(\alpha)>0 \ \ and \ \ \phi'(\beta)<0 \implies \phi'(\alpha)\phi'(\beta)<0\)

Hence Statement P is false.

If we take the path (II) then \(\phi'(\alpha)<0 \ \ and \ \ \phi'(\beta)>0 \implies \phi'(\alpha)\phi'(\beta)<0\)

Hence Statement P is false.

For statement (Q) -

Now Use the Sturn Separation Theorem -

If y1(x) & y2(x) are two continuous linearly independent solutions to homogeneous differential equation y'' + Py' + Qy =   0

having α & β being successive roots of y1(x), then there exist exactly one root of y2 in (α , β ), i.e.  there exist x ∈  (α , β ) such that y2(x) = 0

Now here we have 𝜙(𝛼) = 𝜙(𝛽) = 0 then there exist x ∈  (α , β ) such that 𝜓(𝑥) = 0

So there exist x ∈  (α , β ) such that 𝜙(𝑥)𝜓(𝑥) =  0

Hence Statement Q is false.

So option (3) is correct.