Continuity

Question

Given the piecewise function shown, with f(x)=ax+1 for x \le 2 and f(x)=bx-3 for x>2, find a relation making it continuous.

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Continuity at the junction point x = 2 requires the left-hand limit, the right-hand limit, and the function value to coincide, which is the defining condition for a piecewise function to join without a jump. The left piece gives f(2^-) = a(2) + 1 = 2a + 1, and since x = 2 is included on the left, this is also f(2). The right piece gives f(2^+) = b(2) - 3 = 2b - 3. Equating the one-sided limits yields 2a + 1 = 2b - 3, which is the continuity relation. Option 2a - 1 = 2b + 3 moves both constants to the wrong sides, reversing the signs of the intercepts and giving a false relation. Option a = b ignores the differing intercepts entirely. Option a + b = 0 has no basis in the junction condition. The governing JEE pattern is matching left and right limits at a breakpoint. Plausibility check: choosing a = 1 gives 3 = 2b - 3, so b = 3, and both pieces meet at the point (2,3) seen in the figure, confirming a seamless graph.

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