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For the function to be continuous at x=2, the left-hand limit (LHL) must be equal to the right-hand limit (RHL) at x=2.
LHL at x=2: lim (x->2-) f(x) = lim (x->2-) kx2 = k(2)2 = 4k
RHL at x=2: lim (x->2+) f(x) = lim (x->2+) (3x+5) = 3(2) + 5 = 6 + 5 = 11
For continuity, LHL = RHL, so 4k = 11
Solving for k: k = 11/4
Correct Answer: \frac{11}{4}
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