1
MCQ (Single Correct Answer)

JEE Main 2018 (Online) 16th April Morning Slot

Let N denote the set of all natural numbers. Define two binary relations on N as R = {(x, y) $$ \in $$ N $$ \times $$ N : 2x + y = 10} and R2 = {(x, y) $$ \in $$ N $$ \times $$ N : x + 2y = 10}. Then :
A
Range of R1 is {2, 4, 8).
B
Range of R2 is {1, 2, 3, 4}.
C
Both R1 and R2 are symmetric relations.
D
Both R1 and R2 are transitive relations.

Explanation

For R1; 2x + y = 10 and x, y $$ \in $$ N possible values for x and y are :

x = 1, y = 8    i.e.   (1, 8);

x = 2, y = 6    i.e    (2, 6);

x = 3, y = 4    i.e    (3, 4);

x = 4, y = 2    i.e    (4, 2)

$$\therefore\,\,\,$$ R1 = { (1, 8), (2, 6), (3, 4), (4, 2) }

$$\therefore\,\,\,$$ Range of R1 is {2, 4, 6, 8}

R1 is not symmetric.

R1 is not transitive also as

(3, 4), (4, 2) $$ \in $$ R , but (3, 2) $$ \notin $$ R1

For R2 : x + 2y = 10 and x, y $$ \in $$ N

Possible values of x, and y are :

x = 8, y= 1    i.e    (8, 1)

x = 6, y = 2    i.e    (6, 2)

x = 4, y = 3    i.e    (4, 3) and

x = 2, y = 4    i.e    (2, 4)

$$\therefore\,\,\,$$ R2 = {(8, 1) (6, 2) (4, 3) (2, 4)}

$$\therefore\,\,\,$$ Range of R2 = $$\left\{ {1,2,3,4} \right\}$$

R2 is not symmetric and transitive
2
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Morning Slot

For $$x \in R - \left\{ {0,1} \right\}$$, Let f1(x) = $$1\over x$$, f2 (x) = 1 – x

and f3 (x) = $$1 \over {1 - x}$$ be three given

functions. If a function, J(x) satisfies

(f2 o J o f1) (x) = f3 (x) then J(x) is equal to :
A
f1 (x)
B
$$1 \over x$$ f3 (x)
C
f2 (x)
D
f3 (x)

Explanation

Given,

f1(x) = $${1 \over x}$$

f2(x) = 1 $$-$$ x

f3(x) = $${1 \over {1 - x}}$$

(f2 $$ \cdot $$ J $$ \cdot $$ f1) (x) = f3(x)

$$ \Rightarrow $$   f2 {J(f1(x))} = f3(x)

$$ \Rightarrow $$   f2{J ($${1 \over x}$$)} = $${1 \over {1 - x}}$$

$$ \Rightarrow $$   1 $$-$$ J($${1 \over x}$$) = $${1 \over {1 - x}}$$

$$ \Rightarrow $$  J($${{1 \over x}}$$) = 1 $$-$$ $${{1 \over {1 - x}}}$$

$$ \Rightarrow $$   J ($${{1 \over { x}}}$$) = $${{ - x} \over {1 - x}}$$ = $${x \over {x - 1}}$$

Put x inplace of $${1 \over x}$$

$$ \therefore $$   J(x) = $${{{1 \over x}} \over {{1 \over x} - 1}}$$

= $${1 \over {1 - x}} = {f_3}\left( x \right)$$
3
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Evening Slot

Let A = {x$$ \in $$R : x is not a positive integer}.

Define a function $$f$$ : A $$ \to $$  R   as  $$f(x)$$ $${{2x} \over {x - 1}}$$,

then $$f$$ is :
A
not injective
B
neither injective nor surjective
C
surjective but not injective
D
injective but not surjective

Explanation

f(x) = $${{2x} \over {x - 1}}$$

f(x) = 2 + $${2 \over {x - 1}}$$

f'(x) = $$-$$ $${2 \over {{{\left( {x - 1} \right)}^2}}}$$ < 0 $$\forall $$ x $$ \in $$ R

Hence f(x) is strictly decreasing

So, f(x) is one-one

Range : Let y = $${{2x} \over {x - 1}}$$

xy $$-$$ y = 2x

$$ \Rightarrow $$  x(y $$-$$ 2) = y

$$ \Rightarrow $$  x = $${y \over {y - 2}}$$

given that x $$ \in $$ R : x is not a +ve integer

$$ \therefore $$  $${y \over {y - 2}} \ne $$ N    (N $$ \to $$ Natural number)

$$ \Rightarrow $$  y $$ \ne $$ Ny $$-$$ 2N

$$ \Rightarrow $$  y $$ \ne $$ $${{2N} \over {N - 1}}$$

So range $$ \notin $$ R (in to function)
4
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 10th January Evening Slot

Let N be the set of natural numbers and two functions f and g be defined as f, g : N $$ \to $$ N such that

f(n) = $$\left\{ {\matrix{ {{{n + 1} \over 2};} & {if\,\,n\,\,is\,\,odd} \cr {{n \over 2};} & {if\,\,n\,\,is\,\,even} \cr } \,\,} \right.$$;

      and g(n) = n $$-$$($$-$$ 1)n.

Then fog is -
A
neither one-one nor onto
B
onto but not one-one
C
both one-one and onto
D
one-one but not onto

Explanation

f(x) = $$\left\{ {\matrix{ {{{n + 1} \over 2};} & {if\,\,n\,\,is\,\,odd} \cr {{n \over 2};} & {if\,\,n\,\,is\,\,even} \cr } \,\,} \right.$$;

g(x) = n $$-$$ ($$-$$ 1)n $$\left\{ {\matrix{ {n + 1;\,\,\,\,n\,\,is\,\,odd} \cr {n - 1;\,\,\,\,n\,\,is\,\,even} \cr } } \right.$$

f(g(n)) = $$\left\{ {\matrix{ {{n \over 2};\,\,\,\,n\,\,is\,\,even} \cr {{{n + 1} \over 2};\,\,\,\,n\,\,is\,\,odd} \cr } } \right.$$

$$ \therefore $$  many one but onto

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