To solve the given problem, we start by solving the given first-order differential equation.
The given differential equation for x ∈ ( 0 , ∞ ) x \in (0, \infty) x ∈ ( 0 , ∞ ) f ′ ( x ) = 2 − f ( x ) x f'(x) = 2 - \frac{f(x)}{x} f ′ ( x ) = 2 − x f ( x )
Rearranging the terms, we get:
f ′ ( x ) + 1 x f ( x ) = 2 f'(x) + \frac{1}{x} f(x) = 2 f ′ ( x ) + x 1 f ( x ) = 2
This is a standard first-order linear differential equation of the form d y d x + P ( x ) y = Q ( x ) \frac{dy}{dx} + P(x) y = Q(x) d x d y + P ( x ) y = Q ( x ) P ( x ) = 1 x P(x) = \frac{1}{x} P ( x ) = x 1 Q ( x ) = 2 Q(x) = 2 Q ( x ) = 2
The integrating factor (I.F. \text{I.F.} I.F. I.F. = e ∫ 1 x d x = e ln x = x ( since x > 0 ) \text{I.F.} = e^{\int \frac{1}{x} \, dx} = e^{\ln x} = x \quad (\text{since } x > 0) I.F. = e ∫ x 1 d x = e l n x = x ( since x > 0 )
Multiplying both sides of the differential equation by the integrating factor x x x x f ′ ( x ) + f ( x ) = 2 x x f'(x) + f(x) = 2x x f ′ ( x ) + f ( x ) = 2 x d d x [ x f ( x ) ] = 2 x \frac{d}{dx} \left[ x f(x) \right] = 2x d x d [ x f ( x ) ] = 2 x
Integrating both sides with respect to x x x x f ( x ) = ∫ 2 x d x x f(x) = \int 2x \, dx x f ( x ) = ∫ 2 x d x x f ( x ) = x 2 + C x f(x) = x^2 + C x f ( x ) = x 2 + C
where C C C f ( x ) = x + C x f(x) = x + \frac{C}{x} f ( x ) = x + x C
We are given that f ( 1 ) ≠ 1 f(1) \neq 1 f ( 1 ) = 1 f ( x ) f(x) f ( x ) x = 1 x = 1 x = 1 f ( 1 ) = 1 + C f(1) = 1 + C f ( 1 ) = 1 + C
Since f ( 1 ) ≠ 1 f(1) \neq 1 f ( 1 ) = 1 C ≠ 0 C \neq 0 C = 0
Now, differentiating f ( x ) f(x) f ( x ) x x x f ′ ( x ) = 1 − C x 2 f'(x) = 1 - \frac{C}{x^2} f ′ ( x ) = 1 − x 2 C
Now, let us evaluate each option:
Option (A): We need to find lim x → 0 + f ′ ( 1 x ) \lim_{x \rightarrow 0^+} f'\left(\frac{1}{x}\right) lim x → 0 + f ′ ( x 1 )
Substituting x ↦ 1 x x \mapsto \frac{1}{x} x ↦ x 1 f ′ ( x ) f'(x) f ′ ( x ) f ′ ( 1 x ) = 1 − C ( 1 x ) 2 = 1 − C x 2 f'\left(\frac{1}{x}\right) = 1 - \frac{C}{\left(\frac{1}{x}\right)^2} = 1 - C x^2 f ′ ( x 1 ) = 1 − ( x 1 ) 2 C = 1 − C x 2
Taking the limit as x → 0 + x \to 0^+ x → 0 + lim x → 0 + f ′ ( 1 x ) = lim x → 0 + ( 1 − C x 2 ) = 1 − 0 = 1 \lim_{x \rightarrow 0^+} f'\left(\frac{1}{x}\right) = \lim_{x \rightarrow 0^+} (1 - C x^2) = 1 - 0 = 1 lim x → 0 + f ′ ( x 1 ) = lim x → 0 + ( 1 − C x 2 ) = 1 − 0 = 1
Thus, Option (A) is correct.
Option (B): We need to find lim x → 0 + x f ( 1 x ) \lim_{x \rightarrow 0^+} x f\left(\frac{1}{x}\right) lim x → 0 + x f ( x 1 )
Substituting x ↦ 1 x x \mapsto \frac{1}{x} x ↦ x 1 f ( x ) f(x) f ( x ) f ( 1 x ) = 1 x + C x f\left(\frac{1}{x}\right) = \frac{1}{x} + C x f ( x 1 ) = x 1 + C x
Multiplying by x x x x f ( 1 x ) = x ( 1 x + C x ) = 1 + C x 2 x f\left(\frac{1}{x}\right) = x \left(\frac{1}{x} + C x\right) = 1 + C x^2 x f ( x 1 ) = x ( x 1 + C x ) = 1 + C x 2
Taking the limit as x → 0 + x \to 0^+ x → 0 + lim x → 0 + x f ( 1 x ) = lim x → 0 + ( 1 + C x 2 ) = 1 ≠ 2 \lim_{x \rightarrow 0^+} x f\left(\frac{1}{x}\right) = \lim_{x \rightarrow 0^+} (1 + C x^2) = 1 \neq 2 lim x → 0 + x f ( x 1 ) = lim x → 0 + ( 1 + C x 2 ) = 1 = 2
Thus, Option (B) is incorrect.
Option (C): We need to evaluate lim x → 0 + x 2 f ′ ( x ) \lim_{x \rightarrow 0^+} x^2 f'(x) lim x → 0 + x 2 f ′ ( x )
Using f ′ ( x ) = 1 − C x 2 f'(x) = 1 - \frac{C}{x^2} f ′ ( x ) = 1 − x 2 C x 2 f ′ ( x ) = x 2 ( 1 − C x 2 ) = x 2 − C x^2 f'(x) = x^2 \left(1 - \frac{C}{x^2}\right) = x^2 - C x 2 f ′ ( x ) = x 2 ( 1 − x 2 C ) = x 2 − C
Taking the limit as x → 0 + x \to 0^+ x → 0 + lim x → 0 + x 2 f ′ ( x ) = lim x → 0 + ( x 2 − C ) = − C \lim_{x \rightarrow 0^+} x^2 f'(x) = \lim_{x \rightarrow 0^+} (x^2 - C) = -C lim x → 0 + x 2 f ′ ( x ) = lim x → 0 + ( x 2 − C ) = − C
Since C ≠ 0 C \neq 0 C = 0 − C ≠ 0 -C \neq 0 − C = 0
Thus, Option (C) is incorrect.
Option (D): We need to test if ∣ f ( x ) ∣ ≤ 2 |f(x)| \leq 2 ∣ f ( x ) ∣ ≤ 2 x ∈ ( 0 , 2 ) x \in (0, 2) x ∈ ( 0 , 2 )
Since f ( x ) = x + C x f(x) = x + \frac{C}{x} f ( x ) = x + x C C ≠ 0 C \neq 0 C = 0 lim x → 0 + ∣ f ( x ) ∣ = lim x → 0 + ∣ x + C x ∣ = ∞ \lim_{x \rightarrow 0^+} |f(x)| = \lim_{x \rightarrow 0^+} \left| x + \frac{C}{x} \right| = \infty lim x → 0 + ∣ f ( x ) ∣ = lim x → 0 + x + x C = ∞
Since f ( x ) f(x) f ( x ) x → 0 + x \to 0^+ x → 0 + ∣ f ( x ) ∣ ≤ 2 |f(x)| \leq 2 ∣ f ( x ) ∣ ≤ 2 x ∈ ( 0 , 2 ) x \in (0, 2) x ∈ ( 0 , 2 )
Thus, Option (D) is incorrect.
Conclusion: The only correct option is (A) .