3 Marks Question - MATHS STD 12 Science Questions - Vidyadip

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Question 13 Marks

Let $A = \{-1, 0, 1, 2\}, B = \{-4, -2, 0, 2\}$ and $f, g : A \rightarrow B$ be the functions defined by $f(x) = x^2 - x, x \in A$ and $g(x) = 2\left| {x - \frac{1}{2}} \right| - 1,x \in A.$ Are $f$ and $g$ equal? Justify your answer. $($Hint: One may note that two functions $f : A \rightarrow B$ and $g : A \rightarrow B$ such that$ f(a) = g(a) \forall$ a $\in A,$ are called equal functions$).$

Answer

When $x = -1$ then $f(x) = 1^2 + 1 = 2$ and $g(x) $
$= 2\left| { - 1 - \frac{1}{2}} \right| - 1 = 2$
At $x = 0, f(0) = 0$ and $g(0) $
$= 2\left| { - \frac{1}{2}} \right| - 1 $
$= 2 \times \frac{1}{2} - 1 = 0$
At $x = 1, f(1) = 1^2 - 1 = 0$ and $g(1) $
$= 2\left| {1 - \frac{1}{2}} \right| - 1 $
$= 2 \times \frac{1}{2} - 1 = 0$
At $x = 2, f(2) $
$= 2^2 - 2 = 2$ and $g(2) $
$= 2\left| {2 - \frac{1}{2}} \right| - 1 = 3 - 1 = 2$
Thus for each a $\in A, f(a) = g(a)$
Therefore, $f$ and $g$ are equal function.

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Question 23 Marks

If f: N $\to$ N is defined by f(n) = $\left\{ \begin{array} { l } { \frac { n + 1 } { 2 } , \text { if } n \text { is odd } } \\ { \frac { n } { 2 } , \text { if } n \text { is even } } \end{array} \right.$for all n $ \in$ N. State whether the function f is bijective. Justify your answer.

Answer

The given function is $f : N$ $ \to$ $N$ such that
$ f ( n ) = \left\{ \begin{array} { l l } { \frac { n + 1 } { 2 } , } & { \text { if } n \text { is odd } } \\ { \frac { n } { 2 } } & { \text { if } n \text { is even } } \end{array} \right.$
One-one function
Here,$f(1) =$ $\frac { 1 + 1 } { 2 } = \frac { 2 } { 2 } $$= 1$
and $f(2) =$ $\frac 22$$= 1$
$\therefore$ $f(n)$ is not a one-one function because at two distinct values of domain (N), f(n) has same image.

Onto function
If n is an odd natural number, then $ 2n- 1$ is also an odd natural number.
Now, $f(2n - 1) =$ $\frac { 2 n - 1 + 1 } { 2 }$$= n$
Again, if n is an even natural number, then $2n$ is also an even natural number.
$f(2n) =$ $\frac{2n}{2}$ $= n$

From Equations (i) and (ii), we observe that for each n (whether even or odd) there exists its pre-image in N,
i.e, Range = co-domain
$\therefore $ f is onto
f(x) is not one-one but onto
So, it is not a bijective function.

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Question 33 Marks

Let $A = R - \{3\}$ and $B = R - \{1\}.$ Consider the function $f : A \rightarrow B$ defined by $f(x) = \left( {\frac{{x - 2}}{{x - 3}}} \right)$ Is f one-one and onto? Justify your answer.

Answer

$A = R - \{3\}$ and $B = R - \{1\}$ and $f(x) = \frac{{x - 2}}{{x - 3}}$
Let $x_1, x_2 \in A,$ then $f({x_1}) = \frac{{{x_1} - 2}}{{{x_1} - 3}}$ and $f({x_2}) = \frac{{{x_2} - 2}}{{{x_2} - 3}}$
Now, for $f(x_1) = f(x_2)$
$ \Rightarrow \frac{{{x_1} - 2}}{{{x_1} - 3}} = \frac{{{x_2} - 2}}{{{x_2} - 3}}$
$\Rightarrow (x_1 - 2)(x_2 - 3) = (x_2 - 2)(x_1 - 3)$
$\Rightarrow x_1x_2 - 3x_1 - 2x_2 +6 = x_1x_2 - 2x_1 - 3x_2 + 6$
$\Rightarrow -3x_1 - 2x_2 = -2x_1 - 3x_2$
$\Rightarrow x_1 = x_2$
$\therefore f$ is one-one function.
Now, $y = \frac{{x - 2}}{{x - 3}}$
$\Rightarrow y(x - 3) = x - 2$
$\Rightarrow xy - 3y = x - 2$
$\Rightarrow x(y - 1) = 3y - 2$
$\Rightarrow x = \frac{{3y - 2}}{{y - 1}}$
$\therefore f\left( {\frac{{3y - 2}}{{y - 1}}} \right) = \frac{{\frac{{3y - 2}}{{y - 1}} - 2}}{{\frac{{3y - 2}}{{y - 1}} - 3}}$
$ = \frac{{3y - 2 - 2y + 2}}{{3y - 2 - 3y + 3}} = y$
$\Rightarrow f(x) = y$
Therefore, $f$ is an onto function.

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Question 43 Marks

Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a, b) : |a - b| is even}, is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}.

Answer

A = {1, 2, 3, 4, 5} and R = {(a, b) : |a - b| is even}, then R = {(1, 3), (1, 5), (3, 5), (2, 4)}

For (a, a), |a - a| = 0 which is even. $\therefore$ R is reflexive.
If |a - b| is even, then |b - a| is also even. $\therefore$ R is symmetric.
Now, if |a - b| and |b - c| is even then |a - b + b - c| is even
$\Rightarrow$ |a - c| is also even. $\therefore$ R is transitive.
Therefore, R is an equivalence relation.
Elements of {1, 3, 5} are related to each other.
Since |1 - 3| = 2, |3 - 5| = 2, |1 - 5| = 4 all are even numbers
$\Rightarrow$ Elements of {1, 3, 5} are related to each other.
Similarly elements of (2, 4) are related to each other.
Since |2 - 4| = 2 an even number, then no element of the set {1, 3, 5} is related to any element of (2, 4).
Hence no element of {1, 3, 5} is related to any element of {2, 4}.

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Question 53 Marks

Show that the relation R in the set A of all the books in a library of a college, given by R = {(x, y) : x and y have same number of pages} is an equivalence relation.

Answer

Books x and x have same number of pages $\Rightarrow$ (x, x) $\in$ R $\therefore$R is reflexive.
If (x, y) $\in$ R ,then x and y have same no. of pages
$\Rightarrow$ y and x have same no. of pages
$\Rightarrow$ (y, x) $\in$ R $\therefore$ R is symmetric.
Now if (x, y) $\in$ R, (y, z) $\in$ R. Then
x and y have same no. of pages and y and z have same no. of pages. This implies x and z have same no. of pages.
$\Rightarrow$ (x, z) $\in$ R $\therefore$ R is transitive.
Since R is reflexive, symmetric and transistive, therefore, R is an equivalence relation.

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Question 63 Marks

Check whether the relation $R$ in $R$ defined by $R = \{(a, b) : a \leq b^3\}$ is reflexive, symmetric or transitive.

Answer

For $(a, a), a < a^3$ which is false. $\therefore R$ is not reflexive.
For $(a, b), a < b^3$ and $(b, a), b > a^3$ which is false. $\therefore R$ is not symmetric.
For $a < b^2 b < c^3.$ Now $b < c^3$ implies $b^3 < c^9$

Thus, we get $a < c^9,$ therefore $(a,c)$ does not belong to $R$ and hence $R$ is not transitive.
Therefore, $R$ is neither reflexive, nor symmetric and nor transitive.

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Question 73 Marks

Show that the relation R in R defined as R = {(a, b) : a $\leq$ b}, is reflexive and transitive but not symmetric.

Answer

a $\leq$ a which is true, so (a, a) $\in$ R, $\therefore$ R is reflexive.
a $\leq$ b but b $\leq$ a which is false. $\therefore$ R is not symmetric.
a $\leq$ b and b $\leq$ c $\Rightarrow$ a $\leq$ c which is true. $\therefore$ R is transitive.

Therefore, R is reflexive and transitive but not symmetric.

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Question 83 Marks

Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b ): b = a +1} is reflexive, symmetric or transitive.

Answer

Let A = {1, 2, 3, 4, 5, 6}
Relation R is defined on set A as:
R = {(a, b): b = a + 1}
Therefore, R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}
We find (a, a) $\notin$ R, where a $\in$ A.
For instance (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) $\notin$ R
Therefore, R is not reflexive.
It can be observed that (1, 2) $\in$ R, but (2, 1) $\notin$ R.
Therefore, R is not symmetric.
Now, (1, 2), (2, 3) $\in$ R
But, (1, 3) $\notin$ R
Therefore, R is not transitive
Hence, R is neither reflexive, nor symmetric, nor transitive.

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Question 93 Marks

Show that the relation $R$ in the set $R$ of real numbers, defined as $R = \{(a, b): a \leq b^2\}$ is neither reﬂexive nor symmetric nor transitive.

Answer

$R = \{(a, b): a \leq b^2\}$
It can be observed that
$(\frac{1}{2}, \frac{1}{2}) \in R,$ since $\frac{1}{2} > (\frac{1}{2})^2 = \frac{1}{4}$
$\therefore R$ is not reflexive.
Now, $(1, 4) \in R$ as $1 < 4^2.$
But, $4$ is not less than $1^2.$
$\therefore (4, 1) \notin R$
$\therefore R$ is not symmetric.
Now,
$(3,2), (2, 1.5) \in R$
$($as $3 < 2^2 = 4$ and $2< (1.5)^2 = 2.25)$
But, $3 > (1.5)^2 = 2.25$
$\therefore (3, 1.5) \notin R$
$\therefore R$ is not transitive.
Hence, $R$ is neither reflexive, nor symmetric, nor transitive.

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Question 103 Marks

Let $L$ be the set of all lines in xy plane and $R$ be the relation in $L$ define as $R = {(L_1, L_2) : L_1 || L_2}$. Show that $R$ is an equivalence relation. Find the set of all lines related to the line $y = 2x + 4.$

Answer

$L_1\|L_1 i.e (L_1, L_1) \in R$ Hence reflexive
Let $(L_1,L_2)\in R$ , then
$L_1\|L_2$ which implies $L_2 \|L1 $
$ \Rightarrow (L_2, L _1 ) \in R$ Hence symmetric
We know the
$L_1\|L_2 and L_2\|L_3$
Then $L_{1 }\| L_3$
Therefore, $(L_1,L_2)\in R$ and $(L_2,L_3)\in R $ implies $(L_1,L_3)\in R$
Hence Transitive
Hence, $R$ is an equivalence relation.
Any line parallel to $y = 2x + 4$ is of the form $y = 2x + K$, where $k$ is a real number.
Therefore, set of all lines parallel to $y = 2x + 4$ is $y : y = 2x + k, k$ is a real number

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Question 113 Marks

Determine whether the below relation is reflexive, symmetric and transitive:
Relation R in the set Z of all integers defined as
R = {(x, y) : x – y is an integer}

Answer

It is given that Relation R in the set Z of all integers is defined as
R = {(x, y) : x – y is an integer}
Now, for every x $\in$ Z, (x, x) $\in$ R, as x - x = 0 is an integer.
$\Rightarrow$ R is reflexive.
Next, for every x, y $\in$ Z if (x, y) $\in$ R, then x - y is an integer.
$\Rightarrow$ -(x - y) is also an integer.
$\Rightarrow$ (y - x) is an integer.
$\Rightarrow$ (y - x) $\in$ R
$\Rightarrow$ R is symmetric.
Further, Take (x, y) ,(y, z) $\in$ R where x, y, z $\in$ R,
$\Rightarrow$ (x - y) and (y - z) are integers.
$\Rightarrow$ (x - z) = (x - y) + (y - x) is an integer.
$\Rightarrow$ (x, z) $\in$ R
$\Rightarrow$ R is transitive.
Therefore, R is reflexive, symmetric and transitive.

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Question 123 Marks

Show that the relation $R$ defined in the set A of all polygons as $R = (P_1, P_2) : P_1$ and $P_2$ have same number of sides, is an equivalence relation. What is the set of all elements in A related to the right angle triangle $T$ with sides $3, 4,$ and $5?$

Answer

Part $I: R = \{(P_1, P_2): P_1$ and $P_2$ have same number of sides$\}$

Consider the element $(P_1, P_1),$ it shows $P_1$ and $P_1$ have same number of sides. Therefore, $R$ is reflexive.
If $(P_1, P_2) \in R$ then $P_1$ and $P_2$ have same no.of pages .Then, $P_2$ and $P_1$ have same no. of pages so $(P_2, P_1) \in R$
$\therefore R$ is symmetric.
If $(P_1, P_2) \in R$ and $(P_2, P_3) \in R$ then $P_1$ and $P_2$ have same no. of pages and $P_2$ and $P_3$ have same no. of pages .This implies $P_1$ and $P_3$ have same no. of pages and hence $(P_1, P_3) \in R ,$ therefore, $R$ is transitive.
Therefore, $R$ is an equivalent relation.

Part $II:$ Since the relation considers only the number of sides,
therefore, all the triangles are similar to the given triangle.

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Question 133 Marks

Determine whether the below relation is reflexive, symmetric and transitive:
Relation R in the set A = {1, 2, 3, 4, 5, 6} as
R = {(x, y) : y is divisible by x}

Answer

It is given that relation R on the set A = {1, 2, 3, 4, 5, 6} as
R = {(x, y) : y is divisible by x}
We know that any number 'x' is divisible by itself.
$\Rightarrow$ (x, x) $\in$ R $\forall$ $x\in A$
$\Rightarrow$ R is reflexive.
Now, (2, 4) $\in$ R but (4, 2) $\notin$ R.
$\Rightarrow$ R is not symmetric.
Let (x,y), (y,z) $\in$ R.
$\Rightarrow$y is divisible by x and z is divisible by y.
$\Rightarrow$ z is divisible by x.
$\Rightarrow$ (x,z) $\in$ R
$\Rightarrow$ R is transitive.
Therefore, R is reflexive and transitive but not symmetric.

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Question 143 Marks

Show that the relation $R$ defined in the set $A$ of all triangles as $R = \{(T_1, T_2) : T_1$ is similar to $T_2\},$ is equivalence relation. Consider three right angle triangles $T_1$ with sides $3, 4, 5, T_2$ with sides $5, 12, 13$ and $T_3$ with sides $6, 8, 10.$ Which triangles among $T_1, T_2$ and $T_3$ are related?

Answer

Part $I: R = \{(T_1, T_2): T_1$ is similar to $T_2\}$ and $T_1, T_2$ are triangles.
We know that each triangle similar to itself and thus $(T_1, T_1)\in R$
$\therefore R$ is reflexive.
Also if two triangles are similar, then $T_1 \cong T_2$
$\Rightarrow T_1 \cong T_2$
$\therefore R$ is symmetric.
Again, if $T_1 \cong T_2$ and $T_2\cong T_3$
$\Rightarrow$ then $T_1\cong T_3$
$\therefore R$ is transitive.
Therefore, $R$ is an equivalent relation.
Part $II:$ It is given that $T_1, T_2$ and $T_3$ are right angled triangles.
$\Rightarrow T_1$ with sides $3, 4, 5, T_2$ with sides $5, 12, 13$ and $T_3$ with sides $6, 8, 10$
Since, two triangles are similar if corresponding sides are proportional.
Therefore, $\frac{3}{6} = \frac{4}{8} = \frac{5}{{10}} = \frac{1}{2}$
Therefore, $T_1$ and $T_3$ are related.

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Question 153 Marks

Determine whether the below relation is reflexive, symmetric and transitive:
Relation R in the set N of natural numbers is defined as
R = {(x, y) : y = x + 5 and x < 4}

Answer

It is given that Relation R in the set N of natural numbers defined as
R = {(x, y) : y = x + 5 and x < 4}
Clearly,
R = {(1, 6), (2, 7), (3, 8)}
Reflexive
A relation is said to be reflexive if (x, x) $\in$ R, where x is from domain. we can see that (1,1) $\notin$ R.
$\Rightarrow$ R is not reflexive.
Symmetric
A relation is said to be symmetric if (y, x) $\in$ R whenever (x, y) $\in$ R.
Here, (1,6) $\in$ R, but (6,1) $\notin$ R
$\Rightarrow$ R is not symmetric.
Transitive
Now, since there is no pair in R such that (x, y) and (y, z) $\in$ R, then (x, z) cannot belong to R.
$\therefore$ R is not transitive.
Therefore, R is neither reflexive, nor symmetric, nor transitive.

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Question 163 Marks

Show that the relation $R$ in the set $A$ of points in a plane given by $R = \{(P, Q) :$ distance of the point $P$ from the origin is same as the distance of the point $Q$ from the origin$\}$, is an equivalence relation. Further, show that the set of all points related to a point $P \ne (0, 0)$ is the circle passing through $P$ with origin as centre.

Answer

Part $I: R = \{(P, Q):$ distance of the point $P$ from the origin is the same as the distance of the point $Q$ from the origin$\}$
Let $P(x_1, y_1)$ and $Q(x_2, y_2)$ and $O (0, 0).$
$\therefore OP = OQ$
$\Rightarrow \sqrt {x_1^2 + y_1^2} = \sqrt {x_2^2 + y_2^2}$
$\Rightarrow x_1^2 + y_1^2 = x_2^2 + y_2^2$
Now, For $(P, P), OP = OP$
$\therefore R$ is reflexive.
Also $OP = OQ$ and $OQ = OP$
$\Rightarrow (P, Q) = (Q, P) \in R$
$\therefore R$ is symmetric.
Also $OP = OQ$ and $OQ = OR$
$\Rightarrow OP = OQ$
$\therefore R$ is transitive.
Therefore, $R$ is an equivalent relation.
Part $II:$ As $x_1^2 + y_1^2 = x_2^2 + y_2^2 = {r^2} ($let$)$
$\Rightarrow x^2 + y^2 = r^2$ which represents a circle with centre $(0, 0)$ and radius $r.$

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Question 173 Marks

Determine whether the below relations is reflexive, symmetric and transitive:
Relation R in the set A = {1, 2, 3, ..., 13, 14} defined as R = {(x, y) : 3x – y = 0}.

Answer

The Relation R on the set A = {1, 2, 3, ...., 13, 14}, is defined as
R = {(x, y) : 3x - y = 0}
Then, R = {(1, 3), (2, 6), (3, 9), (4, 12)}
Reflexive
A relation is said to be reflexive if (x, x) $\in$ R, for every x in the domain.
Clearly, R is not reflexive as (1,1), (2,2) …… (14,14) $\notin$ R
Symmetric
A relation is said to be symmetric if (y, x) $\in$ R whenever (x, y) $\in$ R.
But here, R is not symmetric as (1,3) $\in$ R, but (3,1) $\notin$ R
Transitive
A relation is said to be transitive if (x, z) $\in$ R whenever (x, y) $\in$ R and (y, z) $\in$ R
But here, R is not transitive as (1,3), (3,9) $\in$ R, but (1,9) $\notin$ R
Therefore, R is neither reflexive, nor symmetric, nor transitive.

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3 Marks Question - MATHS STD 12 Science Questions - Vidyadip