With math.h functions, what is the output of this code?\n\n#include<stdio.h>\n#include<math.h>\nint main()\n{\n float n = 1.54;\n printf("%f, %f\n", ceil(n), floor(n));\n return 0;\n}\n\nAssume default %f precision.

Introduction / Context:
ceil returns the least integer value greater than or equal to the argument; floor returns the greatest integer value less than or equal to the argument. Both return double.

Given Data / Assumptions:

• n = 1.54.
• ceil(1.54) = 2.0, floor(1.54) = 1.0.
• printf uses "%f" for both values, printing six fractional digits.

Concept / Approach:
Apply the definitions directly and print with fixed default precision.

Step-by-Step Solution:
Compute ceil(1.54) → 2.0.Compute floor(1.54) → 1.0.printf("%f, %f") prints "2.000000, 1.000000".

Verification / Alternative check:
Using integer casts would truncate toward zero, which also yields 1 for positive 1.54, but cast is not the same as floor for negative values.

Why Other Options Are Wrong:
Other numeric pairs do not match the mathematical definition of ceil and floor for 1.54.

Common Pitfalls:
Confusing truncation with floor for negative inputs; forgetting both functions return double.

Final Answer:
2.000000, 1.000000