The key to this problem is to express the function value with the formula in brackets. Option a, consider using simultaneous expressions of cos2x = 1-2 sinx 2 and sin2x 2+cos2x 2 = 1, but the expressions on r are not unique and need to be segmented. The mapping from sinx to x in b is not injective (one-to-one correspondence), so it cannot represent the whole R. C makes z = x 2+1,but it cannot match the form of (x+ 1) 2. In D, z = x 2+2x, z+ 1 = (x+ 1) 2, f (z) = root sign (z+ 1) is what you want, so D is the correct answer, and the corresponding f(x)= root sign (.