3 Marks - 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) = x² - 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) = 1² + 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) = 1² - 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) = 2² - 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 x₁, x₂ $\in$ 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(x₁) = f(x₂)
$ \Rightarrow \frac{{{x_1} - 2}}{{{x_1} - 3}} = \frac{{{x_2} - 2}}{{{x_2} - 3}}$
$\Rightarrow$ (x₁ - 2)(x₂ - 3) = (x₂ - 2)(x₁ - 3)
$\Rightarrow$ x₁x₂ - 3x₁ - 2x₂ +6 = x₁x₂ - 2x₁ - 3x₂ + 6
$\Rightarrow$ -3x₁ - 2x₂ = -2x₁ - 3x₂
$\Rightarrow$ x₁ = x₂ $\therefore$ 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 $\leq$ b³} is reflexive, symmetric or transitive.

Answer

For (a, a), a < a³ which is false. $\therefore$ R is not reflexive.
For (a, b), a < b³ and (b, a), b > a³ which is false. $\therefore$R is not symmetric.
For a < b² b < c³. Now b < c$$³ implies b³ < c⁹

Thus, we get a < c⁹, therefore (a,c) does not belong to $\mathbb {R}$ 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 $\leq$ b²} is neither reﬂexive nor symmetric nor transitive.

Answer

R = {(a, b): a $\leq$ b²}
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 < 4².
But, 4 is not less than 1².
$\therefore$ (4, 1) $\notin$ R
$\therefore$ R is not symmetric.
Now,
(3,2), (2, 1.5) $\in$ R
(as 3 < 2² = 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 = {(L₁, L₂) : L₁ || L₂}. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4.

Answer

L₁||L₁ i.e (L₁, L₁) $\in$ R Hence reflexive
Let $(L_1,L_2)\in R$, then
L₁||L₂ which implies L₂ ||L1$$
$\Rightarrow$ (L₂, L$_1$) $\in$ R Hence symmetric
We know the
L₁||L₂ and L₂||L₃
Then L₁|| L₃
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 = {(P₁, P₂) : P₁ and P₂ 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 = {(P₁, P₂): P₁ and P₂ have same number of sides}

Consider the element (P₁, P₁), it shows P₁ and P₁ have same number of sides. Therefore, R is reflexive.
If (P₁, P₂) $\in$ R then P₁ and P₂ have same no.of pages .Then, P₂ and P₁ have same no. of pages so (P₂, P₁) $\in$ R
$\therefore$ R is symmetric.
If (P₁, P₂) $\in$ R and (P₂, P₃) $\in$ R then P₁ and P₂ have same no. of pages and P₂ and P₃ have same no. of pages .This implies P₁ and P₃ have same no. of pages and hence (P₁, P₃) $\in$ 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 = {(T₁, T₂) : T₁ is similar to T₂}, is equivalence relation. Consider three right angle triangles T₁ with sides 3, 4, 5, T₂ with sides 5, 12, 13 and T₃ with sides 6, 8, 10. Which triangles among T₁, T₂ and T₃ are related?

Answer

Part I: R = {(T₁, T₂): T₁ is similar to T₂} and T₁, T₂ are triangles.
We know that each triangle similar to itself and thus (T₁, T₁)$\in$ R $\therefore$ R is reflexive.
Also if two triangles are similar, then T₁ $\cong$ T₂ $\Rightarrow$ T₁ $\cong$ T₂ $\therefore$ R is symmetric.
Again, if T₁ $\cong$ T₂ and T₂$\cong$ T₃ $\Rightarrow$ then T₁$\cong$ T₃ $\therefore$ R is transitive.
Therefore, R is an equivalent relation.
Part II: It is given that T₁, T₂ and T₃ are right angled triangles.
$\Rightarrow$ T₁ with sides 3, 4, 5, T₂ with sides 5, 12, 13 and T₃ 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, T₁ and T₃ 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(x₁, y₁) and Q(x₂, y₂) 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) $\Rightarrow$ x² + y² = r² 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 - Maths STD 12 Science Questions - Vidyadip