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Exercise 8.1 · Q1
Q.

State whether yy is a function of xx in the following two cases. Justify your answer.

(i)

xxyy
−3-3−6-6
−2-2−1-1
1100
1155
2200

(ii)

xxyy
−3-344
−2-244
−1-144
2244
3344
Andaman Nicobar CbseNCERTSubjective· 2mImportance★★★★★est
5% · 1/20 Questions
✓ Free question

Test each table against the definition of a function: every input xx must map to exactly one output yy.

A relation ff from set XX to set YY is a function if every x∈Xx \in X has one and only one image y∈Yy \in Y. If some xx is paired with two (or more) different yy-values, the relation is not a function.

  1. Case (i). Read the table:

    xx−3-3−2-2111122
    yy−6-6−1-1005500

    The input x=1x=1 appears twice, once paired with y=0y=0 and once with y=5y=5. Since x=1x=1 does not have a unique image, the vertical-line test fails at x=1x=1.

x=1  ⟹  y=0 and y=5(two images for one input)x=1 \implies y=0 \text{ and } y=5 \quad (\text{two images for one input})

Hence case (i) is not a function.

  1. Case (ii). Read the table:

    xx−3-3−2-2−1-12233
    yy4444444444

    Every value of xx (−3,−2,−1,2,3-3,-2,-1,2,3) is distinct, and each is paired with exactly one yy-value (here all equal to 44). Repetition of the output y=4y=4 is allowed — a function only forbids repetition of the input.

Every x has a unique image y=4\text{Every } x \text{ has a unique image } y=4

Hence case (ii) is a function — it is the constant function f(x)=4f(x)=4.

  1. Self-check. In (i) the two pairs (1,0)(1,0) and (1,5)(1,5) share the same first coordinate but different second coordinates — this alone disqualifies a relation from being a function, regardless of the other rows. In (ii) all five xx-values are pairwise distinct, so uniqueness of image is automatic.
✓Final answer

  1. Not a function — x=1x=1 maps to both 00 and 55.
  2. Is a function — every xx has the single image y=4y=4 (the constant function f(x)=4f(x)=4).

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