To determine which assertion is TRUE , let us analyze each function and its given option step-by-step.
1. Analysis of Option A: The function ψ 1 ( x ) \psi_1(x) ψ 1 ( x ) ψ 1 ( x ) = e − x + x , x ≥ 0 \psi_1(x) = e^{-x} + x, \quad x \ge 0 ψ 1 ( x ) = e − x + x , x ≥ 0
Differentiating with respect to x x x ψ 1 ′ ( x ) = − e − x + 1 = 1 − e − x \psi_1'(x) = -e^{-x} + 1 = 1 - e^{-x} ψ 1 ′ ( x ) = − e − x + 1 = 1 − e − x
For any x > 0 x > 0 x > 0 e − x < 1 e^{-x} < 1 e − x < 1 ψ 1 ′ ( x ) > 0 for all x > 0 \psi_1'(x) > 0 \quad \text{for all } x > 0 ψ 1 ′ ( x ) > 0 for all x > 0
Thus, ψ 1 ( x ) \psi_1(x) ψ 1 ( x ) x > 0 x > 0 x > 0 ψ 1 ( 0 ) = e 0 + 0 = 1 \psi_1(0) = e^0 + 0 = 1 ψ 1 ( 0 ) = e 0 + 0 = 1 ψ 1 ( x ) > 1 for all x > 0 \psi_1(x) > 1 \quad \text{for all } x > 0 ψ 1 ( x ) > 1 for all x > 0
Therefore, Option A (ψ 1 ( x ) ≤ 1 \psi_1(x) \le 1 ψ 1 ( x ) ≤ 1 FALSE .
2. Analysis of Option B: The function ψ 2 ( x ) \psi_2(x) ψ 2 ( x ) ψ 2 ( x ) = x 2 − 2 x − 2 e − x + 2 , x ≥ 0 \psi_2(x) = x^2 - 2x - 2e^{-x} + 2, \quad x \ge 0 ψ 2 ( x ) = x 2 − 2 x − 2 e − x + 2 , x ≥ 0
Differentiating with respect to x x x ψ 2 ′ ( x ) = 2 x − 2 + 2 e − x = 2 ( e − x + x − 1 ) = 2 ( ψ 1 ( x ) − 1 ) \psi_2'(x) = 2x - 2 + 2e^{-x} = 2(e^{-x} + x - 1) = 2(\psi_1(x) - 1) ψ 2 ′ ( x ) = 2 x − 2 + 2 e − x = 2 ( e − x + x − 1 ) = 2 ( ψ 1 ( x ) − 1 )
Since ψ 1 ( x ) > 1 \psi_1(x) > 1 ψ 1 ( x ) > 1 x > 0 x > 0 x > 0 ψ 2 ′ ( x ) > 0 for all x > 0 \psi_2'(x) > 0 \quad \text{for all } x > 0 ψ 2 ′ ( x ) > 0 for all x > 0
Thus, ψ 2 ( x ) \psi_2(x) ψ 2 ( x ) x > 0 x > 0 x > 0 ψ 2 ( 0 ) = 0 2 − 2 ( 0 ) − 2 e 0 + 2 = 0 \psi_2(0) = 0^2 - 2(0) - 2e^0 + 2 = 0 ψ 2 ( 0 ) = 0 2 − 2 ( 0 ) − 2 e 0 + 2 = 0 ψ 2 ( x ) > 0 for all x > 0 \psi_2(x) > 0 \quad \text{for all } x > 0 ψ 2 ( x ) > 0 for all x > 0
Therefore, Option B (ψ 2 ( x ) ≤ 0 \psi_2(x) \le 0 ψ 2 ( x ) ≤ 0 FALSE .
3. Analysis of Option C: The function f ( x ) f(x) f ( x ) f ( x ) = ∫ − x x ( ∣ t ∣ − t 2 ) e − t 2 d t , x > 0 f(x) = \int_{-x}^{x} (|t| - t^2) e^{-t^2} dt, \quad x > 0 f ( x ) = ∫ − x x ( ∣ t ∣ − t 2 ) e − t 2 d t , x > 0
Since the integrand ( ∣ t ∣ − t 2 ) e − t 2 (|t| - t^2) e^{-t^2} ( ∣ t ∣ − t 2 ) e − t 2 t t t f ( x ) f(x) f ( x ) f ( x ) = 2 ∫ 0 x ( t − t 2 ) e − t 2 d t = 2 ∫ 0 x t e − t 2 d t − 2 ∫ 0 x t 2 e − t 2 d t f(x) = 2 \int_{0}^{x} (t - t^2) e^{-t^2} dt = 2 \int_{0}^{x} t e^{-t^2} dt - 2 \int_{0}^{x} t^2 e^{-t^2} dt f ( x ) = 2 ∫ 0 x ( t − t 2 ) e − t 2 d t = 2 ∫ 0 x t e − t 2 d t − 2 ∫ 0 x t 2 e − t 2 d t
Evaluating the first integral:
2 ∫ 0 x t e − t 2 d t = [ − e − t 2 ] 0 x = 1 − e − x 2 2 \int_{0}^{x} t e^{-t^2} dt = \left[ -e^{-t^2} \right]_0^x = 1 - e^{-x^2} 2 ∫ 0 x t e − t 2 d t = [ − e − t 2 ] 0 x = 1 − e − x 2
So,
f ( x ) = 1 − e − x 2 − 2 ∫ 0 x t 2 e − t 2 d t f(x) = 1 - e^{-x^2} - 2 \int_{0}^{x} t^2 e^{-t^2} dt f ( x ) = 1 − e − x 2 − 2 ∫ 0 x t 2 e − t 2 d t
For any y > 0 y > 0 y > 0 e − y > 1 − y e^{-y} > 1 - y e − y > 1 − y y = t 2 y = t^2 y = t 2 e − t 2 > 1 − t 2 ⟹ t 2 e − t 2 > t 2 − t 4 e^{-t^2} > 1 - t^2 \implies t^2 e^{-t^2} > t^2 - t^4 e − t 2 > 1 − t 2 ⟹ t 2 e − t 2 > t 2 − t 4
Integrating from 0 0 0 x x x ∫ 0 x t 2 e − t 2 d t > ∫ 0 x ( t 2 − t 4 ) d t = x 3 3 − x 5 5 \int_{0}^{x} t^2 e^{-t^2} dt > \int_{0}^{x} (t^2 - t^4) dt = \frac{x^3}{3} - \frac{x^5}{5} ∫ 0 x t 2 e − t 2 d t > ∫ 0 x ( t 2 − t 4 ) d t = 3 x 3 − 5 x 5
Multiplying by − 2 -2 − 2 − 2 ∫ 0 x t 2 e − t 2 d t < − 2 3 x 3 + 2 5 x 5 -2 \int_{0}^{x} t^2 e^{-t^2} dt < -\frac{2}{3}x^3 + \frac{2}{5}x^5 − 2 ∫ 0 x t 2 e − t 2 d t < − 3 2 x 3 + 5 2 x 5
Substituting this back into f ( x ) f(x) f ( x ) f ( x ) < 1 − e − x 2 − 2 3 x 3 + 2 5 x 5 f(x) < 1 - e^{-x^2} - \frac{2}{3}x^3 + \frac{2}{5}x^5 f ( x ) < 1 − e − x 2 − 3 2 x 3 + 5 2 x 5
Therefore, Option C (f ( x ) ≥ 1 − e − x 2 − 2 3 x 3 + 2 5 x 5 f(x) \ge 1 - e^{-x^2} - \frac{2}{3}x^3 + \frac{2}{5}x^5 f ( x ) ≥ 1 − e − x 2 − 3 2 x 3 + 5 2 x 5 FALSE .
4. Analysis of Option D: The function g ( x ) g(x) g ( x ) g ( x ) = ∫ 0 x 2 t e − t d t , x > 0 g(x) = \int_{0}^{x^2} \sqrt{t} e^{-t} dt, \quad x > 0 g ( x ) = ∫ 0 x 2 t e − t d t , x > 0
Using the substitution t = u 2 ⟹ d t = 2 u d u t = u^2 \implies dt = 2u \, du t = u 2 ⟹ d t = 2 u d u g ( x ) = ∫ 0 x u e − u 2 ( 2 u d u ) = 2 ∫ 0 x u 2 e − u 2 d u g(x) = \int_{0}^{x} u e^{-u^2} (2u \, du) = 2 \int_{0}^{x} u^2 e^{-u^2} du g ( x ) = ∫ 0 x u e − u 2 ( 2 u d u ) = 2 ∫ 0 x u 2 e − u 2 d u
Consider the Taylor series expansion bound for e − y e^{-y} e − y y > 0 y > 0 y > 0 h ( y ) = 1 − y + y 2 2 − e − y h(y) = 1 - y + \frac{y^2}{2} - e^{-y} h ( y ) = 1 − y + 2 y 2 − e − y
Note that h ( 0 ) = 0 h(0) = 0 h ( 0 ) = 0 h ′ ( y ) = − 1 + y + e − y = ψ 1 ( y ) − 1 > 0 for all y > 0 h'(y) = -1 + y + e^{-y} = \psi_1(y) - 1 > 0 \quad \text{for all } y > 0 h ′ ( y ) = − 1 + y + e − y = ψ 1 ( y ) − 1 > 0 for all y > 0
Since h ′ ( y ) > 0 h'(y) > 0 h ′ ( y ) > 0 y > 0 y > 0 y > 0 h ( y ) > 0 h(y) > 0 h ( y ) > 0 y > 0 y > 0 y > 0 e − y < 1 − y + y 2 2 e^{-y} < 1 - y + \frac{y^2}{2} e − y < 1 − y + 2 y 2
Substituting y = u 2 y = u^2 y = u 2 e − u 2 < 1 − u 2 + u 4 2 e^{-u^2} < 1 - u^2 + \frac{u^4}{2} e − u 2 < 1 − u 2 + 2 u 4
Thus,
u 2 e − u 2 < u 2 − u 4 + u 6 2 u^2 e^{-u^2} < u^2 - u^4 + \frac{u^6}{2} u 2 e − u 2 < u 2 − u 4 + 2 u 6
Integrating from 0 0 0 x x x ∫ 0 x u 2 e − u 2 d u < ∫ 0 x ( u 2 − u 4 + u 6 2 ) d u = x 3 3 − x 5 5 + x 7 14 \int_{0}^{x} u^2 e^{-u^2} du < \int_{0}^{x} \left(u^2 - u^4 + \frac{u^6}{2}\right) du = \frac{x^3}{3} - \frac{x^5}{5} + \frac{x^7}{14} ∫ 0 x u 2 e − u 2 d u < ∫ 0 x ( u 2 − u 4 + 2 u 6 ) d u = 3 x 3 − 5 x 5 + 14 x 7
Multiplying both sides by 2 2 2 g ( x ) = 2 ∫ 0 x u 2 e − u 2 d u < 2 3 x 3 − 2 5 x 5 + 1 7 x 7 g(x) = 2 \int_{0}^{x} u^2 e^{-u^2} du < \frac{2}{3}x^3 - \frac{2}{5}x^5 + \frac{1}{7}x^7 g ( x ) = 2 ∫ 0 x u 2 e − u 2 d u < 3 2 x 3 − 5 2 x 5 + 7 1 x 7
Hence, g ( x ) ≤ 2 3 x 3 − 2 5 x 5 + 1 7 x 7 g(x) \le \frac{2}{3}x^3 - \frac{2}{5}x^5 + \frac{1}{7}x^7 g ( x ) ≤ 3 2 x 3 − 5 2 x 5 + 7 1 x 7 x ∈ ( 0 , 1 2 ) x \in \left(0, \frac{1}{2}\right) x ∈ ( 0 , 2 1 )
Therefore, Option D is TRUE .
Conclusion: The correct statement is D .