CBSE Class 12 Mathematics 2026 Main Paper 65/3/3 · Q335 Marksinjectivity_surjectivity

CBSE Class 12 Mathematics 2026 Main Paper 65/3/3 · Question 33 (Relations And Functions)

The printed problem specifies f:[0,∞)→A⊂ℕ, f(x)=4x²+12x+15. Examine whether it defines a function with that codomain. If the codomain may instead be real, determine the injectivity, range and whether 7 has a preimage.

✅ Step-by-Step Solution

Final Answer: No such A⊂ℕ exists. As a real-valued function the rule is injective, has range [15,∞), and never takes the value 7.

At x=1/4 the output is 73/4, which is not a natural number. Thus the given map to A⊂ℕ is invalid.
As a real-valued rule its derivative 8x+12 is positive for x≥0, proving injectivity. The rule is continuous, starts at 15, and grows without bound.
The real range is [15,∞), which makes the real-valued rule onto that interval. As 7 is below the range, it has no preimage.

📋 CBSE Stepwise Marking Scheme

[1 mark] At x=1/4 the output is 73/4, which is not a natural number. Thus the given map to A⊂ℕ is invalid.
[3 marks] As a real-valued rule its derivative 8x+12 is positive for x≥0, proving injectivity. The rule is continuous, starts at 15, and grows without bound.
[1 mark] The real range is [15,∞), which makes the real-valued rule onto that interval. As 7 is below the range, it has no preimage.