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Inverse D-operator (part 1)

Authored by TP emath

Mathematics

University - Professional Development

Used 3+ times

Inverse D-operator (part 1)
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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Which inverse D-operator formula is used to compute 1D2(5)\frac{1}{D^2}\left(5\right)  ?

1F(D)(k)=1F(0)⋅k\frac{1}{F\left(D\right)}\left(k\right)=\frac{1}{F\left(0\right)}\cdot k  

1D[f(x)]=∫f(x) dx\frac{1}{D}\left[f\left(x\right)\right]=\int f\left(x\right)\ dx  

1F(D)[ekx]=1F(k)⋅ekx\frac{1}{F\left(D\right)}\left[e^{kx}\right]=\frac{1}{F\left(k\right)}\cdot e^{kx}  

1F(D)(k)=1F(k)⋅k\frac{1}{F\left(D\right)}\left(k\right)=\frac{1}{F\left(k\right)}\cdot k  

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Which inverse D-operator formula is used to compute 1D2−1(5)\frac{1}{D^2-1}\left(5\right)  ?

1F(D)(k)=1F(0)⋅k\frac{1}{F\left(D\right)}\left(k\right)=\frac{1}{F\left(0\right)}\cdot k  

1D[f(x)]=∫f(x) dx\frac{1}{D}\left[f\left(x\right)\right]=\int f\left(x\right)\ dx  

1F(D)[ekx]=1F(k)⋅ekx\frac{1}{F\left(D\right)}\left[e^{kx}\right]=\frac{1}{F\left(k\right)}\cdot e^{kx}  

1F(D)(k)=1F(k)⋅k\frac{1}{F\left(D\right)}\left(k\right)=\frac{1}{F\left(k\right)}\cdot k  

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What does it mean by the formula 1F(D)(k)=1F(0)⋅k\frac{1}{F\left(D\right)}\left(k\right)=\frac{1}{F\left(0\right)}\cdot k  ?

replace all D by 0 

replace all k by 0

 replace all D by k

multiply F by 0 

4.

MULTIPLE SELECT QUESTION

45 sec • 1 pt

Which inverse D-operator formula is used to compute 1D2−1(e2x)\frac{1}{D^2-1}\left(e^{2x}\right)  ?

1F(D)(k)=1F(0)⋅k\frac{1}{F\left(D\right)}\left(k\right)=\frac{1}{F\left(0\right)}\cdot k  

1F(D)[ekxV(x)]=ekx⋅1F(D+k)[V(x)]\frac{1}{F\left(D\right)}\left[e^{kx}V\left(x\right)\right]=e^{kx}\cdot\frac{1}{F\left(D+k\right)}\left[V\left(x\right)\right]  

1F(D)[ekx]=1F(k)⋅ekx\frac{1}{F\left(D\right)}\left[e^{kx}\right]=\frac{1}{F\left(k\right)}\cdot e^{kx}  

1F(D)(k)=1F(k)⋅k\frac{1}{F\left(D\right)}\left(k\right)=\frac{1}{F\left(k\right)}\cdot k  

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What does it mean by the formula 1F(D)[ekx]=1F(k)⋅ekx\frac{1}{F\left(D\right)}\left[e^{kx}\right]=\frac{1}{F\left(k\right)}\cdot e^{kx}  ?

replace all D by 0 

replace all k by 0

 replace all D by k

multiply all D by k 

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Which inverse D-operator formula is used to compute 1D2−1(ex)\frac{1}{D^2-1}\left(e^x\right)  ?

1F(D)(k)=1F(0)⋅k\frac{1}{F\left(D\right)}\left(k\right)=\frac{1}{F\left(0\right)}\cdot k  

1F(D)[ekxV(x)]=ekx⋅1F(D+k)[V(x)]\frac{1}{F\left(D\right)}\left[e^{kx}V\left(x\right)\right]=e^{kx}\cdot\frac{1}{F\left(D+k\right)}\left[V\left(x\right)\right]  

1F(D)[ekx]=1F(k)⋅ekx\frac{1}{F\left(D\right)}\left[e^{kx}\right]=\frac{1}{F\left(k\right)}\cdot e^{kx}  

1F(D)(k)=1F(k)⋅k\frac{1}{F\left(D\right)}\left(k\right)=\frac{1}{F\left(k\right)}\cdot k  

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What does it mean by the formula 1F(D)[ekx V(x)]=ekx⋅1F(D+k)[V(x)]\frac{1}{F\left(D\right)}\left[e^{kx}\ V\left(x\right)\right]=e^{kx}\cdot\frac{1}{F\left(D+k\right)}\left[V\left(x\right)\right]  ?

put ekxe^{kx}  outside the operator and replace all D by (D+k)

 replace all D by (D+k)

 put ekxe^{kx}  outside the operator and add F(D)F\left(D\right) by k

replace F(D)F\left(D\right)  by k 

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