Match List I with List II.
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List – I |
List – II |
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|
(A) |
R = {(x, y) : x and y are student of the same school} |
(I) |
Symmetric |
|
(B) |
R = {(L1, L2) : L1 ⊥ L2, L1, L2 ∈ L, where L is a set of all lines} |
(II) |
One - one |
|
(C) |
A function f : R → R defined by f(x) = 2 - 3x is |
(III) |
Bijective |
|
(D) |
A function f : [0, 1] → R defined by f(x) = 1 + x2 is f : [0, 1] → R |
(IV) |
Equivalence |
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 2
A. R = {(x, y) : x, y ∈ same school}
so (x, x) ∈ R for every x ∈ A → IV
(x, y) ∈ R ⇒ (y, x) ∈ R
(x, y) ∈ R , (y, z) ∈ R ⇒ (x, x) ∈ R equivalence
B. R = {(L1, L2) : L1 ⊥ L2 for L1, L2 ∈ L
so (L1, L1) ∉ R as on line is perpendicular to itself
(L1, L2) ∈ R ⇒ (L2, L1) ∈ R (symmetric) → I
(L1, L2) ∈ R , (L2, L3) ∈ R
⇒ (L1, L3) ∉ R as L1 ⊥ L2 L2 ⊥ L3
⇒ L1 || L2
C. y = 2 - 3x → linear function
so $\frac{2-y}{3} = x$ ⇒ for every y ∈ R → III
there event atleast one x
y1 = y2 ⇒ 2 - 3x1 = 2 - 3x2
⇒ x1 = x2 ⇒ bijective
D. f : [0, 1] → R
y = 1 + x2
so $x = \sqrt{1-y}$
⇒ one, one → II
for y = 5 not defined but for every y, x is unique as x ∈ [0, 1]