Continuity Condition

A piecewise function is continuous at the junction point x = 1 if the left-hand limit equals the right-hand limit and equals the function value there.

For x<1, f(x)=15, so

lim x→1⁻ f(x) = 15

For x≥1, f(x)=cx, so

lim x→1⁺ f(x) = c⋅1 = c

and f(1)=c.

Setting them equal gives c = 15, making f continuous everywhere.

The continuous function f(x)=15 if x<1 cx if x≥1 is a classic piecewise continuity problem for IIT JAM math preparation. Candidates must find the integer value of c that ensures continuity at x=1, where the definition changes. This tests limit matching for left and right approaches, a core concept in calculus.

Step-by-Step Solution

To check continuity at x=1 for this continuous function f(x)=15 if x<1 cx if x≥1:

• Left limit: As x→1⁻, f(x)=15 → limit = 15.
• Right limit: As x→1⁺, f(x)=cx → limit = c.
• Function value: f(1)=c.

For continuity, 15 = c, so c = 15. Verify elsewhere: constant 15 for x<1 is continuous; linear cx for x≥1 is continuous.

No options provided, but common distractors include:

• c = 1 (ignores left constant)
• c = 16 (off-by-one error)
• c = 0 (trivial mismatch)

Only c = 15 matches limits perfectly.

Graph Insight

Plot y=15 horizontal line left of x=1, y=15x line from x=1 onward. At c=15, it connects seamlessly without jump. For c≠15, discontinuity (removable type) occurs.

IIT JAM Exam Tips

Practice similar continuous function f(x)=15 if x<1 cx if x≥1 questions from prior papers. Focus on epsilon-delta if advanced, but limits suffice here. Time-saver: equate one-sided limits directly.