For, 1+6x – 2×3 and, e-x + 2xex – x2ex

For, 1+6x – 2×3 and, e-x + 2xex – x2ex

The Correct Answer And Eplanation is:

To solve this problem, we are asked to find a linear differential operator that annihilates the given functions. An annihilator is a differential operator L(D)L(D) such that L(D)[f(x)]=0L(D)[f(x)] = 0, where DD is the derivative operator ddx\frac{d}{dx}.

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🔷 Function 1: f1(x)=1+6x−2x3f_1(x) = 1 + 6x – 2x^3

This is a polynomial of degree 3. In general:

• A polynomial of degree nn is annihilated by the operator Dn+1D^{n+1}, since the (n+1)th(n+1)^{\text{th}} derivative of a polynomial of degree nn is zero.

Here:

• The degree is 3 ⇒ Use D4D^4

✅ Annihilator for f1(x)f_1(x) is:
D4\boxed{D^4}

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🔷 Function 2: f2(x)=e−x+2xex−x2exf_2(x) = e^{-x} + 2x e^x – x^2 e^x

This is a linear combination of exponential functions and polynomials multiplied by exponentials. Let’s break it down:

Terms:

1. e−xe^{-x}
2. 2xex2x e^x
3. −x2ex-x^2 e^x

We treat each term:

• e−xe^{-x} ⇒ annihilated by (D+1)(D + 1)
• xneaxx^n e^{ax} ⇒ annihilated by (D−a)n+1(D – a)^{n+1}

So:

• 2xex2x e^x is xexx e^x: (D−1)2(D – 1)^2 annihilates it
• −x2ex-x^2 e^x: (D−1)3(D – 1)^3 annihilates it

But since we want an operator that annihilates all three terms, we take the least common multiple (LCM) of the annihilators.

Annihilators:

• e−xe^{-x} → D+1D + 1
• x2exx^2 e^x → (D−1)3(D – 1)^3

✅ So, the annihilator is:
(D+1)(D−1)3\boxed{(D + 1)(D – 1)^3}

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✅ Final Answers:

1. For 1+6x−2×31 + 6x – 2x^3: