1

### JEE Main 2017 (Online) 9th April Morning Slot

The function f : N $\to$ N defined by f (x) = x $-$ 5 $\left[ {{x \over 5}} \right],$ Where N is the set of natural numbers and [x] denotes the greatest integer less than or equal to x, is :
A
one-one and onto
B
one-one but not onto.
C
onto but not one-one.
D
neither one-one nor onto.

## Explanation

f(1) = 1 - 5$\left[ {{1 \over 5}} \right]$ = 1

f(6) = 6 - 5$\left[ {{6 \over 5}} \right]$ = 1

So, this function is many to one.

f(10) = 10 - 5$\left[ {{10 \over 5}} \right]$ = 0 which is not present in the set of natural numbers.

So this function is neither one-one nor onto.
2

### JEE Main 2018 (Offline)

Two sets A and B are as under :

A = {($a$, b) $\in$ R $\times$ R : |$a$ - 5| < 1 and |b - 5| < 1};

B = {($a$, b) $\in$ R $\times$ R : 4($a$ - 6)2 + 9(b - 5)2 $\le$ 36 };

Then
A
neither A $\subset$ B nor B $\subset$ A
B
B $\subset$ A
C
A $\subset$ B
D
A $\cap$ B = $\phi$ ( an empty set )

## Explanation

Given,

$4{\left( {a - 6} \right)^2} + 9{\left( {b - 5} \right)^2} \le 36$

Let $a - 6 = x$ and $b - 5 = y$

$\therefore\,\,\,\,$ $4{x^2} + 9{y^2} \le 36$

$\Rightarrow \,\,\,\,{{{x^2}} \over 9} + {{{y^2}} \over 4} \le 1$

This is a equation of ellipse.

This ellipse will look like this, According to set A,

$\left| {a - 5} \right| < 1$

as $a - 6 = x$ then $a - 5 = x + 1$

$\therefore\,\,\,$ $\left| {x + 1} \right| < 1$

$\Rightarrow \,\,\, - 1 < x + 1 < 1$

$\Rightarrow \,\,\, - 2 < x < 0$

$\left| {b - 5} \right| < 1$

as $b - 5 = y$

$\therefore\,\,\,$ $\left| y \right| < 1$

$\Rightarrow \,\,\,\, - 1 < y < 1$

This will look like this, By combining both those graphs it will look like this, To check entire set A is inside of B or not put any paint of set A on the equation ${{{x^2}} \over 9} + {{{y^2}} \over 4} \le 1$ and if this inequality satisfies, then it is inside B.

By looking at the graph you can surely say (0, 1) or (0, $-$1) inside the graph. But to check point ($-2$, 1) or ($-$2, $-$1) is inside of the ellipse or not, put on the ${{{x^2}} \over 9} + {{{y^2}} \over 4} \le 1.$

Putting ($-$2, 1) on the inequality,

LHS $= {{\left( { - 2} \right){}^2} \over 9} + {{{1^2}} \over 4}$

$= {4 \over 9} + {1 \over 4}$

$= {{25} \over {36}} < 1$

$\therefore\,\,\,$ Inequality holds.

So, $\left( { - 2,1} \right)$ is inside the ellipse.

Similarly by checking we can see $\left( { - 2, - 1} \right)$ is also inside the ellipse.

Hence, we can say entire set A is inside of the set B.

$\therefore\,\,\,\,$ A $\subset$ B
3

### JEE Main 2018 (Online) 15th April Morning Slot

Consider the following two binary relations on the set A = {a, b, c} :
R1 = {(c, a), (b, b), (a, c), (c, c), (b, c), (a, a)} and
R2 = {(a, b), (b, a), (c, c), (c, a), (a, a), (b, b), (a, c)}.
Then :
A
both R1 and R2 are not symmetric.
B
R1 is not symmetric but it is transitive.
C
R2 is symmetric but it is not transitive.
D
both R1 and R2 are transitive.

## Explanation

Here both R1 and R2 are symmetric as for any (x, y) $\in$ R1, we have (y, x) $\in$ R1 and similarly for any (x, y) $\in$ R2, we have (y, x) $\in$ R2

In R1, (b, c) $\in$ R1, (c, a) $\in$ R1 but (b,a) $\notin$ R1

Similarly in R2, (b, a) $\in$ R2, (a, c) $\in$ R2 but (b, c) $\notin$ R2

$\therefore$ R1 and R2 are not transitive.
4

### JEE Main 2018 (Online) 15th April Evening Slot

Let f : A $\to$ B be a function defined as f(x) = ${{x - 1} \over {x - 2}},$ Where A = R $-$ {2} and B = R $-$ {1}. Then   f   is :
A
invertible and ${f^{ - 1}}(y) =$ ${{3y - 1} \over {y - 1}}$
B
invertible and ${f^{ - 1}}\left( y \right) = {{2y - 1} \over {y - 1}}$
C
invertible and ${f^{ - 1}}\left( y \right) = {{2y + 1} \over {y - 1}}$
D
not invertible

## Explanation

Assume,

y = f(x)

$\Rightarrow$ y = ${{x - 1} \over {x - 2}}$

$\Rightarrow$ yx - 2y = x - 1

$\Rightarrow$ (y - 1)x = 2y - 1

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

As on the given domain the function is invertible and its inverse can be computed as shown above.