12th Grade > Mathematics
RELATIONS AND FUNCTIONS II MCQs
Relations And Functions
Total Questions : 89
| Page 2 of 9 pages
Answer: Option D. -> (x−72)13
:
D
We have, f : R → R, g: R → R defined by f(x) = 2x - 3 and g (x) = x3 + 5
It can be checked that f(x) and g(x) are bijective functions
∴ fo g is also bijective and (fog) = f(g(x)) = f (x3+5)=2(x3+5)−3=2x3+7
(fog)(x)=y⇒2x3+7=y⇒x=(y−72)13
∴(fog)−1(x)=(x−72)13,xϵR
∴ The correct answer is (d).
:
D
We have, f : R → R, g: R → R defined by f(x) = 2x - 3 and g (x) = x3 + 5
It can be checked that f(x) and g(x) are bijective functions
∴ fo g is also bijective and (fog) = f(g(x)) = f (x3+5)=2(x3+5)−3=2x3+7
(fog)(x)=y⇒2x3+7=y⇒x=(y−72)13
∴(fog)−1(x)=(x−72)13,xϵR
∴ The correct answer is (d).
Answer: Option B. -> R−{−1}
:
B
y=2+x2−x⇒2y−yx=2+x⇒x(y+1)=2y−2⇒x=2y−2y+1⇒f−1(x)=2x−2x+1
∴ Range of f= Domain of f−1=R−{−1}
:
B
y=2+x2−x⇒2y−yx=2+x⇒x(y+1)=2y−2⇒x=2y−2y+1⇒f−1(x)=2x−2x+1
∴ Range of f= Domain of f−1=R−{−1}
Answer: Option D. -> [14,∞)
:
D
tan−1(x2+x+a)≥0⇒x2+x+a≥0
⇒D≤0⇒1−4a≤0⇒a≥14 ; D is discriminant of quadratic equation.
⇒aϵ[14,∞)
:
D
tan−1(x2+x+a)≥0⇒x2+x+a≥0
⇒D≤0⇒1−4a≤0⇒a≥14 ; D is discriminant of quadratic equation.
⇒aϵ[14,∞)
Answer: Option D. -> an equivalence relation
:
D
(a, b) R (a, b) because a + b = b + a. So, r is reflexive.
(a, b)R (c, d) ⇒ a+d = b+c ⇒ c+b = d+a
⇒ (c,d) R (a,b)
So, R is symmetric.
(a, b) R (c, d) and (c, d) R (e, f)
⇒ a + d = b + c, c + f = d + e
Adding, a + d + c + f = b + c + d +e
⇒ a + f = b + e
⇒ (a, b) R (e, f).
∴ R is transitive.
:
D
(a, b) R (a, b) because a + b = b + a. So, r is reflexive.
(a, b)R (c, d) ⇒ a+d = b+c ⇒ c+b = d+a
⇒ (c,d) R (a,b)
So, R is symmetric.
(a, b) R (c, d) and (c, d) R (e, f)
⇒ a + d = b + c, c + f = d + e
Adding, a + d + c + f = b + c + d +e
⇒ a + f = b + e
⇒ (a, b) R (e, f).
∴ R is transitive.
Answer: Option D. -> A subset of P × Q
:
D
A relation from P to Q is a subset of P × Q.
:
D
A relation from P to Q is a subset of P × Q.
Answer: Option B. -> −74
:
B
2f(x)−3f(1x)=x2……(i)Replacingxby1x2f(1x)−3f(x)=1x2……(ii)
Solving (i) and (ii) we get
−5f(x)=2x2+3x2f(x)=−15(2x2+3x2)∴f(2)=−15(8+34)=−74
:
B
2f(x)−3f(1x)=x2……(i)Replacingxby1x2f(1x)−3f(x)=1x2……(ii)
Solving (i) and (ii) we get
−5f(x)=2x2+3x2f(x)=−15(2x2+3x2)∴f(2)=−15(8+34)=−74
Answer: Option C. -> {(1, 2), (5, 1), (3, 1)}
:
C
( x,y) ∈ R⇔ (y,x) ∈R−1, ∴ R−1={(3,1),(5,1),(1,2)}.
:
C
( x,y) ∈ R⇔ (y,x) ∈R−1, ∴ R−1={(3,1),(5,1),(1,2)}.
Answer: Option A. -> 29
:
A
n ( A× A) = n(A)n(A) = 32 = 9
So, the total number of subsets of A× A is 29
and a subset of A× A is a relation over the set A .
:
A
n ( A× A) = n(A)n(A) = 32 = 9
So, the total number of subsets of A× A is 29
and a subset of A× A is a relation over the set A .
Answer: Option B. -> {(4, 1), (5, 2), (6, 3),.....}
:
B
R={(a,b):a,b∈N,a−b=3}={((n+3),n):n∈N}
={(4,1),(5,2),(6,3), .......} .
:
B
R={(a,b):a,b∈N,a−b=3}={((n+3),n):n∈N}
={(4,1),(5,2),(6,3), .......} .
Answer: Option C. -> 13(x2−4x+3)
:
C
f(x)+2f(1−x)=x2+1 .........(i)
Replacing x by 1 – x
f(1−x)+2f(x)=(1−x2)+1 .......(ii)
multiplying (ii) by 2 and subtracting it from (1), we get
−3f(x)=x2−2(1−x)2−1
3f(x)=2(1−x)2+1−x2=(1−x)(2−2x+1+x)=(1−x)(3−x)=x2−4x+3
f(x)=13(x2−4x+3)
:
C
f(x)+2f(1−x)=x2+1 .........(i)
Replacing x by 1 – x
f(1−x)+2f(x)=(1−x2)+1 .......(ii)
multiplying (ii) by 2 and subtracting it from (1), we get
−3f(x)=x2−2(1−x)2−1
3f(x)=2(1−x)2+1−x2=(1−x)(2−2x+1+x)=(1−x)(3−x)=x2−4x+3
f(x)=13(x2−4x+3)
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