Q.

Consider the sets A={(x,y)∈R×R : x2+y2=25}, B={(x,y)∈R×R : x2+9y2=144}, C={(x,y)∈Z×Z : x2+y2≤4} and D=A∩B. The total number of one-one functions from the set D to the set C is:          [2025]

1 18290  
2 15120  
3 17160  
4 19320  

Ans.

(3)

We Have, A : x2+y2=25           ... (i)

B : x2144+y216=1          ... (ii)

C : x2+y2≤4          ... (iii)

Solving (i) and (ii), we get

x2+9(25–x2)=144 ⇒ –8x2=–81 ⇒ x=±922

From (i), 818+y2=25 ⇒ y2=25 –818 ⇒ y=±11922

As, D=A∩B

={(922,11922),(922,–11922),(–922,11922),(–922,–11922)}

⇒ n(D)=4

Also, C={(x,y)∈Z×Z : x2+y2≤4}

={(0, 2), (0, –2), (2, 0), (–2, 0), (1, 1), (–1, –1), (–1, 1), (1, –1), (0, 1), (0, –1), (1, 0), (–1, 0), (0, 0)}.

⇒n(C)=13

∴   Total number of one-one function from D to C = 13P4 = 17160.