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Exercise 1.5 · Q3

Q.The relation RR defined on a set A={0,−1,1,2}A=\{0,-1,1,2\} by xRyxRy if ∣x2+y2∣≤2|x^2+y^2|\le2, then which one of the following is true?

(1) R={(0,0),(0,−1),(0,1),(−1,0),(−1,1),(1,2),(1,0)}R=\{(0,0),(0,-1),(0,1),(-1,0),(-1,1),(1,2),(1,0)\}
(2) R−1={(0,0),(0,−1),(0,1),(−1,0),(1,0)}R^{-1}=\{(0,0),(0,-1),(0,1),(-1,0),(1,0)\}
(3) Domain of RR is {0,−1,1,2}\{0,-1,1,2\}
(4) Range of RR is {0,−1,1}\{0,-1,1\}
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Step 1. Since squares are never negative, ∣x2+y2∣=x2+y2|x^2+y^2|=x^2+y^2, so the condition is simply x2+y2≤2x^2+y^2\le2.

Step 2. Test every x∈{0,−1,1,2}x\in\{0,-1,1,2\}: for x=0x=0: need y2≤2⇒y∈{0,−1,1}y^2\le2\Rightarrow y\in\{0,-1,1\} (not 22, since 4>24>2). For x=−1x=-1: need 1+y2≤2⇒y2≤1⇒y∈{0,−1,1}1+y^2\le2\Rightarrow y^2\le1\Rightarrow y\in\{0,-1,1\}. For x=1x=1: same as x=−1x=-1, y∈{0,−1,1}y\in\{0,-1,1\}. For x=2x=2: need 4+y2≤24+y^2\le2, impossible for any yy.

Step 3. So R={(0,0),(0,−1),(0,1),(−1,0),(−1,−1),(−1,1),(1,0),(1,−1),(1,1)}R=\{(0,0),(0,-1),(0,1),(-1,0),(-1,-1),(-1,1),(1,0),(1,-1),(1,1)\} -- 9 pairs.

Step 4 (check options). (1) lists (1,2)(1,2), impossible since 1+4=5>21+4=5>2 -- wrong. (2) R−1R^{-1} has the same 9 pairs as RR itself here (since the condition is symmetric in x,yx,y), but option (2) lists only 5 -- incomplete, wrong. (3) claims domain includes 22, but x=2x=2 never appears (no valid yy) -- wrong. (4) Range == set of second coordinates appearing ={0,−1,1}=\{0,-1,1\} (never 22, by the same reasoning as the domain) -- this MATCHES exactly.

✓Final answer

Option (4): Range of RR is {0,−1,1}\{0,-1,1\}.

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