Relations And Functions Ii(11th And 12th > Mathematics ) Questions and answers for exam Preparation

11th And 12th > Mathematics

RELATIONS AND FUNCTIONS II MCQs

Total Questions : 30 | Page 1 of 3 pages

Inverse exists for a function which is

Injective
Surjective
Bijective
Many-one

Discuss Question

Answer: Option C. -> Bijective
:
C
We have seen that inverse exists only when function is one-one and onto, i.e. Bijective.

Question 2.

Let X be a family of sets and R be a relation on X defined by 'A is disjoint from B'. Then R is

Reflexive
Symmetric
Anti-symmetric
Transitive

Discuss Question

Answer: Option B. -> Symmetric
:
B

Clearly, the relation is symmetric but it is neither reflexive nor transitive.

Question 3.

Let R be a relation on the set N of natural numbers defined by nRm

⇔ n is a factor of m (i.e. n(m). Then R is

Reflexive and symmetric
Transitive and symmetric
Equivalence
Reflexive, transitive but not symmetric

Discuss Question

Answer: Option D. -> Reflexive, transitive but not symmetric
:
D

Since n | n for all n $i n$ N,

therefore R is reflexive.

Since 2 | 6 but 6 | 2, therefore R is not symmetric.

Let n R m and m R p $\Rightarrow$ n|m and m|p $\Rightarrow$ n|p $\Rightarrow$ nRp

So. R is transitive.

Question 4.

If f (x) +f (x+a) + f(x+2a) +....+ f (x+na) = constant; $\forall x$ $ϵ R$ and $a > 0$ and f(x) is periodic,then period of f(x), is

(n+1) a
$e^{n + 1} a$
na
$e^{n a}$

Discuss Question

Answer: Option A. -> (n+1) a
:
A

f (x) + f (x + a) + f (x + 2 a) + . . . . + f (x + n a) = k

replacing

x = x + a

f (x + a) + f (x + 2 a) + . . . . + f (x + (n + 1) a) = k

On subtracting second equation from first one -

f (x) - f (x + (n + 1) a) = 0 f (x) = f (x + (n + 1) a ∴ T = (n + 1) a

Question 5.

The function $f (x) = c o s \sqrt{x}$ is

Periodic with period $2 \sqrt{π}$
Periodic with period $\sqrt{π}$
Periodic with period $4 π^{2}$
Not a periodic function

Discuss Question

Answer: Option D. -> Not a periodic function
:
D
Try drawing

c o s \sqrt{x}

graph. It’s not periodic.

Question 6.

$T h e i n v e r s e o f f (x) = {(5 - {(x - 8)}^{5})}^{\frac{1}{3}} i s$

$5 - (x - 8)^{5}$
$8 + (5 - x^{3})^{\frac{1}{5}}$
$8 - (5 - x^{3})^{\frac{1}{5}}$
$(5 - (x - 8)^{\frac{1}{5}})^{3}$

Discuss Question

Answer: Option B. ->

8 + (5 - x^{3})^{\frac{1}{5}}

:
B

$y = f (x) = {(5 - {(x - 8)}^{5})}^{\frac{1}{3}}$
then $y^{3} = 5 - (x - 8)^{5} \Rightarrow (x - 8)^{5} = 5 - y^{3}$
$\Rightarrow x = 8 + (5 - y^{3})^{\frac{1}{5}}$
Let, $z = g (x) = 8 + (5 - x^{3})^{\frac{1}{5}}$
To check, $f (g (x)) = [5 - {(x - 8)}^{5}]^{\frac{1}{3}}$
$= {(5 - {[{(5 - x^{3})}^{\frac{1}{5}}]}^{5})}^{\frac{1}{3}} = (5 - 5 + x^{3})^{\frac{1}{3}} = x$
similarly , we can show that $g (f (x)) = x$
Hence, $g (x) = 8 + (5 - x^{3})^{\frac{1}{5}}$ is the inverse of $f (x)$