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JEE Mathematics 2025 Question with Solution

Consider the sets A={(x,y)∈R×R:x2+y2=25}A = \{(x, y) \in \mathbb{R} \times \mathbb{R} : x^2 + y^2 = 25\}, B={(x,y)∈R×R:x2+9y2=144}B = \{(x, y) \in \mathbb{R} \times \mathbb{R} : x^2 + 9y^2 = 144\}, C={(x,y)∈Z×Z:x2+y2≤4}C = \{(x, y) \in \mathbb{Z} \times \mathbb{Z} : x^2 + y^2 \leq 4\}, and D=A∩BD = A \cap B. The total number of one-one functions from the set DD to the set CC is:

  • A

    1512015120

  • B

    1932019320

  • C

    1716017160

  • D

    1829018290

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: A={(x,y)∈R2:x2+y2=25}A = \{(x,y) \in \mathbb{R}^2 : x^2 + y^2 = 25\}, B={(x,y)∈R2:x2+9y2=144}B = \{(x,y) \in \mathbb{R}^2 : x^2 + 9y^2 = 144\}, C={(x,y)∈Z2:x2+y2≤4}C = \{(x,y) \in \mathbb{Z}^2 : x^2 + y^2 \le 4\}, and D=A∩BD = A \cap B.

Find: The total number of one-one functions from DD to CC.

First, find the number of elements in DD by solving the intersection of AA and BB:

x2+y2=25x2+9y2=144\begin{aligned} x^2 + y^2 &= 25 \\ x^2 + 9y^2 &= 144 \end{aligned}

Subtracting the first equation from the second,

8y2=1198y^2 = 119

so,

y2=1198y^2 = \frac{119}{8}

Then,

x2=25−1198=818x^2 = 25 - \frac{119}{8} = \frac{81}{8}

Counting Elements and Injective Functions

Hence the intersection points are

(±818,±1198)\left(\pm \sqrt{\frac{81}{8}}, \pm \sqrt{\frac{119}{8}}\right)

All four sign combinations are possible, so

∣D∣=4|D| = 4

Now count the integer points in C={(x,y)∈Z2:x2+y2≤4}C = \{(x,y) \in \mathbb{Z}^2 : x^2 + y^2 \le 4\}.

Possible values are:

x2+y2=0⇒(0,0)giving 1 pointx2+y2=1⇒(±1,0),(0,±1)giving 4 pointsx2+y2=2⇒(±1,±1)giving 4 pointsx2+y2=4⇒(±2,0),(0,±2)giving 4 points\begin{aligned} x^2 + y^2 = 0 &\Rightarrow (0,0) \quad \text{giving } 1 \text{ point} \\ x^2 + y^2 = 1 &\Rightarrow (\pm 1,0), (0,\pm 1) \quad \text{giving } 4 \text{ points} \\ x^2 + y^2 = 2 &\Rightarrow (\pm 1, \pm 1) \quad \text{giving } 4 \text{ points} \\ x^2 + y^2 = 4 &\Rightarrow (\pm 2,0), (0,\pm 2) \quad \text{giving } 4 \text{ points} \end{aligned}

Therefore,

∣C∣=1+4+4+4=13|C| = 1 + 4 + 4 + 4 = 13

The number of one-one functions from a set of size mm to a set of size nn is

nPm=n!(n−m)!{}^nP_m = \frac{n!}{(n-m)!}

So the number of one-one functions from DD to CC is

13P4=13!9!=13⋅12⋅11⋅10=17160{}^{13}P_4 = \frac{13!}{9!} = 13 \cdot 12 \cdot 11 \cdot 10 = 17160

Therefore, the correct option is C.

Common mistakes

  • Counting the points of CC incorrectly by missing lattice points such as (±1,±1)(\pm 1, \pm 1). This is wrong because all integer pairs satisfying x2+y2≤4x^2 + y^2 \le 4 must be included. List cases by values 0,1,2,40, 1, 2, 4 systematically.

  • Assuming the intersection DD has only 22 points instead of 44. This is wrong because both coordinates can independently take positive and negative values once x2x^2 and y2y^2 are fixed. Count all sign combinations.

  • Using 13413^4 instead of permutations for one-one functions. This is wrong because injective mappings cannot assign the same image to two different elements of DD. Use 13P4=13⋅12⋅11⋅10{}^{13}P_4 = 13 \cdot 12 \cdot 11 \cdot 10 instead.

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