Hello student ,

f : A → B is surjection and f(x) = cos x.

Now B = f(A)

A = {0, pi/6 , pi/4 , pi/3 , pi/2}

f(0) = cos(0) = 1

f( pi/6 ) = cos( pi/6) = sqrt of 3 / 2

f( pi/4 ) = cos( pi/4) = sqrt of 1/2

f( pi/3) = cos( pi/3) = 1/2

f( pi/2) = cos( pi/2) = 0

Therefore B = f(A) = {1 , sqrt of 3 /2 , sqrt of 1 /2 , 1/2 , 0 }

Hi aspirant,

A surjective function is a function whose image is equal to its co-domain. Also, the range, co-domain and the image of a surjective function are all equal. So every element in A will have a value defined by f(x) in B.

So results are cos0=1, cos pi/6= √3/2, cos pi/4= 1/√2, Cos pi/3= 1/2 and cos pi/2= 0

Hope it helps!!