Given:

The given statement is “A function could have a property that f(−x)=f(x) for all x”.

Calculation:

Consider the function f(x)=x2.

Now, check whether the above mentioned function satisfies the property f(−x)=f(x) or not.

f(−x)=(−x)2=x2=f(x)

Therefore, the statement is true.

Given:

The given identity is “cos(a+b)=cosa+cosb, for all a and b in [0,2π]”.

Calculation:

Take a=π2 and b=π2, and substitute it, in both the parts of the identity as shown below.

cos(a+b)=cos(π2+π2)=cos(π)=−1

Similarly, calculate the right hand side of the identity as follows.

cosa+cosb=cos(π2)+cos(π2)=0+0=0

That is, cos(π2+π2)≠cos(π2)+cos(π2) implies cos(a+b)≠cosa+cosb.

Therefore, the statement is false.

Given:

The given statement is “If f is a linear function of the form f(x)=mx+b, then f(u+v)=f(u)+f(v), for all u and v.”

Calculation:

Take u=1 and v=1 ,and compute the following.

f(1+1)=f(1)+f(1)        (1)

The right hand side of equation (1) is given by,

f(1+1)=f(2)=2m+b

And, the left hand side of equation (1) is given by,

f(1)+f(1)=(m+b)+(m+b)=2m+2b

Thus, f(1+1)≠f(1)+f(1) that implies f(u+v)≠f(u)+f(v).

Therefore, the statement is false.

Given:

The given statement is “The function f(x)=1−x has the property that f(f(x))=x”.

Calculation:

Consider the function f(x)=1−x.

Calculate f(f(x))=x as shown below.

f(f(x))=f(1−x)=1−(1−x)=1−1+x=x

Therefore, the statement is true.

Given:

The given statement is “The set {x:|x+3|  >4} can be drawn on the number line without lifting pencil”.

Calculation:

The given set {x:|x+3|  >4} can be written as the union of the disjoint intervals (−∞,−7) and (1,∞).

Therefore, the set {x:|x+3|  >4} cannot be drawn on the number line without lifting the pencil.

Therefore, the statement is false.

Given:

The given statement is “The equation, log10(xy)=(log10x)(log10y).”

Calculation:

Take, x=10 and y=10 ,and compute both the sides of the above mentioned equation as shown below.

log10(10×10)=log10100=2

Similarly, calculate the right hand side of the equation as shown below.

(log1010)(log1010)=1×1=1

Then, log10(10×10)≠(log1010)(log1010) implies log10(xy)≠(log10x)(log10y).

Therefore, the statement is false.

Given:

The given statement is “The equation, sin−1(sin(2π))=0.”

Calculation:

Consider sin−1(sin(2π))=0.

Then, compute the following.

sin−1(sin(2π))=sin−1(0)=0

Thus, sin−1(sin(2π))=0.

Therefore, the statement is true.