If the function f (x)=x2+ax is even, then a =
F (x) is an even function,
Then f (-x)= f (x),
Specifically, f (-1)= f (1)
I.e.1-a=1+a
Solve a=0
F (x) is an even function,
Then f (-x)= f (x),
Specifically, f (-1)=f (1)
I.e.1-a=1+a
Solve a=0
If f (x), g (x) is an odd function, g (x) is even and f (x)+g (x)=1/(x2-x+1), find f (x) If f (x), g (x) defined on R is an odd function g (x) is even and f (x)+g (x)=1/(x2-x+1), find f (x) If f (x), g (x) is defined on R as an odd function g (x) is even and f (x)+g (x)=1/(x2-x+1), find f (x)
Because f (x)+g (x)=1/(x2-x+1)(1)
So f (-x)+g (-x)=1/(x2+x+1)
Also f (-x)=-f (x), g (-x)=g (x)
So -f (x)+g (x)=1/(x2+x+1)(2)
(1) Formula -(2),
2F (x)=1/(x2-x+1)-1/(x2+x+1)
F (x)= x/[(x2+1)2-x2]
Under what conditions are the vectors a+b and a-b parallel? Under what conditions is the vector a+b perpendicular to a-b?
A=xb
When vector a is parallel to vector b,
X1y2=x2y1
Under what conditions are the vectors (a+b) and (a-b) parallel?
One is 0, or both are equal
It is proved that (a+b) and (a-b) are parallel vectors:(a+b)×(a-b)=0
-Ab =0 so one of them is 0
You don't have to prove they' re equal, do you?
One of them is 0, or both are equal
It is proved that (a+b) and (a-b) are parallel vectors:(a+b)×(a-b)=0
-Ab =0 so one of them is 0
You don't have to prove they' re equal, do you?
Given vector (1.2) vector b (-3,2), when K is what value,1k vector a is perpendicular to vector a-vector b? 2K vector a+ vector b is parallel to vector a-3 vector b? Are they the same or opposite when they are parallel?
First, vector a-vector b=(4,0) k-vector a=(k,2k) because vector a (1,2), vector b (-3,2), vector a-vector a=(4,0) k-vector a=(k,2k), vector a+vector b=(k-3,2k+2) a-3 vector b=(1,2)-3(-3,2)=(10,8) K-vector a+vector b...
The first question is that because vector a (1,2), vector b (-3,2), vector a-vector b=(4,0) k vector a=(k,2k) is perpendicular to vector a-vector b so that 4k+0=0, i.e. k=0, the second question K vector a+vector b=(k-3,2k+2) vector a-3 vector b=(1,2)-3(-3,2)=(10,8) K vector a+vector b...