See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

f(x + y + z) = f(x) f(y) f(z) Putting y = z = –1 f(x – 2) = f(x) f(–1) f(–1) Putting x = 2 f(0) = 4(f(–1))2  f(0) is positive Now, putting x = 0 = y = z f(0) = f(0)3  f(0) = 1 Again, putting y = 2 and z = 0 f(x + 2) = f(x) f(2) f(0)  f(x + 2) = 4f(x)  f(x + 2) = 4f(x)  f(2) = 12 For More Material Join: @JEEAdvanced_2024