Let f:R→R be defined by f(x) = kx, if x≤1; 2x+3, if x>1 be continuous. Then find the possible value of k.

f(x)=kx if x≤1=2x+3 if x>1So, L.H.L.=R.H.L.kx=2x+3k=2x+3For x=1, k=2(1)+3=5Hence, option '1' is incorrect. Let's re-examine the solution. For continuity at x=1, the left-hand limit (LHL) must equal the right-hand limit (RHL) and the function value at x=1. LHL = lim (x→1⁻) f(x) = lim (x→1⁻) kx = kRHL = lim (x→1⁺) f(x) = lim (x→1⁺) (2x+3) = 2(1)+3 = 5f(1) = kFor continuity, LHL = RHL = f(1), so k = 5. Therefore, the possible value of k is 5. However, 5 is not among the given options. Let's reconsider the problem statement. If the question is asking for the value of k that makes the function continuous at x=1, then we must have:lim (x→1⁻) f(x) = lim (x→1⁺) f(x) = f(1)lim (x→1⁻) kx = lim (x→1⁺) (2x+3) = k k = 5The correct answer is k=5. Since 5 is not an option, there might be an error in the provided options.