Kerala State Board New Syllabus Plus Two Maths Chapter Wise Previous Questions and Answers Chapter 1 Relations and Functions.

## Kerala Plus Two Maths Chapter Wise Previous Questions Chapter 1 Relations and Functions

### Plus Two Maths Relations and Functions 3 Marks Important Questions

Question 1.

Consider the set A = {1, 2, 3, 4, 5}, and B = {1, 4, 9, 16, 25} and a function ƒ : A → B defined by f(1) = 1, f(2) = 4, f(3) = 9, f(4) = 16 and f(5) = 25

(i) Show that f is one-to-one

(ii) Show that f is onto.

(iii) Does ƒ^{-1} exists? Explain (May – 2013)

Answer:

(i) ƒ = {(1,1), (2,4), (3,9), (4,16), (5,25)}

Every element in A is mapped to different elements in B. Therefore one-to-one.

(ii) R (ƒ) = {1, 4, 9, 16, 25} = B. Therefore onto.

(iii) Since f is one-to-one and onto function, ƒ^{-1} exists.

ƒ^{-1} = {(1,1), (4,2), (9,3), (16,4), (25,5)}

Question 2.

a) When a relation R on a set A is said to be reflexive

b) Show that ƒ : [-1, 1] → R given by \(f(x)=\frac{x}{x+2}\) is one-one (May – 2015)

Answer:

Question 3.

a) The function ƒ :N → N given by ƒ(x) = 2x

i) one-one and onto

ii) one-one and not onto

iii) not one-one and not onto

iv) onto but not one-one

b) Find goƒ(x), if ƒ(x) = 8x^{3} and g(x) = x^{1/2}

c) Let * be an operation such that a*b= LCM of a and b defined on the set A = {1,2,3,4,5}. Is * a binary operation? Justify your answer. (March-2016)

Answer:

a) ii) one-one and not onto

*b) Answered in previous years questions No. 2(ii) (6 Mark question)*

c) LCM of 2 and 3 is 6 ∉ A, therefore not a binary operation.

### Plus Two Maths Relations and Functions 4 Marks Important Questions

Question 1.

(i) ƒ : {1,2,3,4} → {5} defined by ƒ = {(1,5), (2,5), (3,5), (4,5)} Does the function is invertible?

(ii)

(iii) Let A = Nx N, N-set of natural numbers and * 1be a binary operation on A defined by (a,b) * (c,d) = (ac—bd,ud +bc). Show that* is commutative on A. (March -2011)

Answer:

(i) Inverse does not exists because fis not one-one.

Hence cummutative.

Question 2.

Let N be the set of Natural numbers. Consider the function ƒ: N → N defined by ƒ(x) = x + l, x ∈ N

(i) Prove that f is not onto

(ii) \(If g(x)=\left\{\begin{array}{ll}x-1, & x>1 \\ 1, & x=1\end{array}\right. then find g o f\)

(iii) Check whether goƒ is an onto function. (May 2011)

Answer:

(iii) Since f is not onto goƒ is also not onto.

Question 3.

(i) Give a relation on a set A = {1,2,3,4} which is reflexive , symmetric and not transitive.

(ii) Show that ƒ : [-1,1] → R given by \(f(x)=\frac{x}{x+2}\) is one-one.

(iii) Let ‘*’ be a binary operation on Q^{+} defined by a*b = \(a * b=\frac{a b}{6}\) ’.Find the inverse of 9 with respect to ’ * ’. (March -2012)

Answer:

(i) Given A = {1,2,3,4}

R = {(1,1)(2,2),(3,3),(4,4),(1,2),(2,1),(1,3),(3,1)}

Question 4.

(i) *:R x R → Ris given by a * b = 3a2 – b

Find the value of 2 * 3. Is ‘*’ commutative? Justify your answer.

(ii) ƒ :R → R is defined by ƒ(x) = x^{2} – 3x + 2 Find ƒoƒ (x) and ƒoƒ. (May 2012)

Answer:

Question 5.

(i) Consider ƒ : R → R given by ƒ(x) = 5x + 2

(a) Show that f is one-one.

(b) Is f invertible? Justify your answer.

(ii) Let * be a binary operation on N defined by a * b = HCF of a and b

(a) Is * commutative?

(b) Is * associative? (March-2013)

Answer:

(i) (a) Let x_{1}, x_{2}, ∈ R

ƒ(x_{1}) = ƒ(x_{2}) ⇒ 5x_{1} + 2 = 5x_{2} + 2

⇒ 5x_{2} = 5x_{2} ⇒ x_{1} = x_{2}

Therefore fis one-one.

(b) Yes.

Let y e range of ƒ

⇒ ƒ(x) = y ⇒ 5x + 2 = y

\(\Rightarrow x=\frac{y-2}{5} \in R\)

Therefore corresponding to every y ∈ R there existsa real number \(\frac{y-2}{5}\) Therefore f is onto.

Hence bijective, so invertible.

(ii) (a) Yes.

a * b = HCF (a,b) = HCF (b,a) = b * a

Hence commutative.

(b) Yes.

a * (b * c) = a* HF(b,c) = HCF(a,b,c)

(a*b) * c =HCF(a,b) * c HCF(a,b,c)

a * (b * c) = (a * b) * c

Hence associative.

Question 6.

(a) Let f: R → R be given by ƒ (x) = \(\frac{2 x+1}{3}\) find ƒoƒ and show that f is invertible.

(b) Find the identity element of the binary operation * on N defined by a * b = ab^{2}. (May 2014)

Answer:

Therefore f is onto.

Hence f is bijective and invertible.

(b) let ‘e’ be the identity element, then

Since e is not unique, this operation has no identity element.

Question 7.

a) What is the minimum number of pairs to form a non-zero reflexive relation on a set of n elements?

b) On the set R of real numbers, S is a relation defined as S = {(x,y)/X∈R, y ∈ R, x + y = xy}. Find a ∈ R such that ‘a’ is never the first element of an ordered pair in S. Also find b ∈ R such that ‘b’ is never the second element of an ordered pair in S.

c) Consider the function \(f(x)=\frac{3 x+4}{x-2}, x \neq 2\) Find a function on a suitable domain such that goƒ(x) = x = ƒog(x). (March 2015)

Answer:

Question 8.

(i) If ƒ: R → R and g: R → R defined by ƒ(x) = x^{2} and g(x) = x + 1, then goƒ (x) is

(a) (x + 1)^{2}

(b) x^{3} + l

(c) x^{2} + l

(d) x + l

(ii) Consider the function ƒ: N → N, given by ƒ(x) = x^{3}. Show that the function ‘ƒ’ is injective but not surjective.

(iii) The given table shows an operation on A = {p,q}

* | p | P |

P | P | P |

p | P | p |

(a) Is * a binary operation?

(b) * commutative? Give reason. (May 2016)

Answer:

(i) (C) x^{2} + 1

(ii) ƒ : N → N , given by ƒ(x) = x^{3}

for x,y ∈ N ⇒ ƒ(x) = ƒ(y)

x^{3} = y^{3} ⇒ x = y

There fore f is injective.

Now 2 ∈ N, but there does not exists any element x in domain N such that ƒ(x) = x^{3} = 2 their fore f is not surjective.

(iii) (a) Yes

(b) No, because p*q = q; q*p = p

⇒ p*q ≠ q*p

Question 9.

(i) Let R be a relation defined on A{1,2,3} by R = {(13),(3,1),(2,2)} is

(a) Reflexive

(b) Symmetric

(C) Transitive

(d) Reflexive but not transitive.

(ii) Find fog and gof if ƒ(x) = |x+1| and g(x) = 2x – 1

(iii) Let * be a binary operation defined on N x N by (a,b) * (c,d.) = (a + c, b + d)

Find the identity element for * if it exists. (March – 2017)

Answer:

(i) (b) Symmetric

(ii) ƒog(x) = |g(x) + 1| = |2x – 1 + 1| = |2x|

goƒ(x) = 2 ƒ(x) – 1 = 2 |x + 1| – 1

(iii) Let e =(e_{1}, e_{2}) be the identity element of the operation in ? N x N then, (a,b)*(e_{1}, e_{2}) = (a + e_{1}, b + e_{2}) ≠ (a,b) Since, e_{1} ≠ 0, e_{2} ≠ 0

Therefore identity element does not existš.

Question 10.

(i) If R = {(x,y) : x, y ∈ Z, x – y ∈ Z}, then the relation R is

(a) Reflexive but not transitive

(b) Reflexive but not symmetric

(C) Symmetric but not transitive

(d) An equivalence relation.

(ii) Let * be a binary operation on the set Q of rational numbers by a*b = 2a + b. Find 2 * (3 * 4) and (2 * 3) * 4.

(iii) Let ƒ : R → R, g : R → R be two one-one funçtions. Check whether gof is one-one or not. (May- 2017)

Answer:

(i) (d) An equivalance relation.

(ii) 2* (3 * 4) = 2 * 10 = 14

(2 * 3)* 4 = 7 * 4 = 18

(iii) ƒ : R → R, g : R → R

Let x_{1}, x_{2}, ∈ R

goƒ(x_{1}) = g(ƒ(x_{1})) = g(ƒ(x_{2})) = g(ƒ(x_{2}))

⇒ x_{1} = x_{2}

### Plus Two Maths Relations and Functions 6 Marks Important Questions

Question 1.

(i) (a) A function ƒ : X → Y is onto if range of ƒ = ………….

(b) Let ƒ : {1, 3, 4} {3, 4, 5} and

g: {3, 4, 5} → {6, 8, 10} be functions defined by

ƒ (1) = 3, ƒ (3) = 4, ƒ (4) = 5;

g (3) = 6, g(4) = 8, g(5) = 8 ,then (goƒ) (3) = …………..

(ii) Let Q be the set of Rational numbers and ‘*’ be the binary operation on Q defined by \(a * b=\frac{a b}{4}\) for all a,b in Q

(a) What is the identity element of ‘ * ’on Q?

(b) Find the inverse element of * ’ on Q.

(c) Show that a * (b * c) = (a * b) * c, ∀a,b,c ∈ Q.

Answer:

Question 2.

(i) Let R be the relation on the set N of natural numbers given by

R = {(a,b): a – b > 2, b>3}

Choose the correct answer

(a) (4, 1) ∈ R

(b) (5, 8) ∈ R

(c) (8, 7) ∈ R

(d) (10, 6) ∈ R

(ii) If ƒ(x) = 8x^{3} and g(x) = x^{1/3}, findg(ƒ(x)) and ƒ(g(x))

(iii) Let * be a binary operation on the set Q of rational numbers defined by a*b = \(\frac{a}{b}\). Check whether * is commutative and associative? (March – 2014, May – 2015, March – 2016)

Answre:

Question 3.

Let \(f(x)=\frac{x-1}{x-3}, x \neq 3\) and \(g(x)=\frac{x-3}{x-1}, x \neq 1\) be two functions defined on R.

(i) Find ƒog(x), x ≠ 0

(ii) Find ƒ^{-1 }(x) and g^{-1} (x), x ≠ 1

(iii) Find (goƒ)^{-1} (x) (May-2010)

Answer: