2 Marks - Page 2 - Maths STD 12 Science Questions - Vidyadip

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2 Marks

Question 512 Marks

Let A = {a, b, c} and the relation R be defined on A as follows: R = {(a, a), (b, c), (a, b)}. Then, write minimum number of ordered pairs to be added in R to make it reflexive and transitive.

Answer

We have,
A = {a, b, c} and R = {(a, a), (b, c), (a, b)}.
R can be a reflexive relation only when elements (b, b) and (c, c) are added to it.
R can be a transitive relation only when the element (a, c) is added to it.
So, the minimum number of ordered pairs to be added in R is 3.

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Question 522 Marks

Let R₀ denote the set of all non-zero real numbers and let A = R₀ × R₀. If '*' is a binary operation on adefined by,
(a, b) * (c, d) = (ac, bd) for all (a, b), (c, d) ∈ A
Find the invertible element in A.

Answer

Let (m, n) be the inverse of $(\text{a, b})\forall(\text{a, b})\in\text{A}$. Then,
(a, b) * (m, n) = (1, 1)
Implies that (am, bn) = (1, 1)
Implies that am = 1 & bn = 1
Implies that $\text{m}=\frac{1}{\text{a}}\text{ and }\text{n}=\frac{1}{\text{b}}$
Thus, $\Big(\frac{1}{\text{a}},\frac{1}{\text{b}}\Big)$ is the inverse of $(\text{a, b})\forall(\text{a, b})\in\text{A}$.

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Question 532 Marks

If f : C → C is defined by f(x) = (x - 2)³, write f^-1(-1).

Answer

Let f^-1(1) = x ......(1)
⇒ f(x) = -1
⇒ (x - 2)³ = -1
$\Rightarrow\ \text{x}-2=-1\ \text{or }-\omega\text{ or }-\omega^2$
as the roots of $(-1)^\frac{1}{3}$ are $-1,-\omega\text{ and }-\omega^2,$ where $\omega=\frac{1\pm\text{i}\sqrt{3}}{2}$
$\Rightarrow\ \text{x}=-1+2\text{ or }2-\omega\text{ or }2-\omega^2=1,2-\omega,2-\omega$
$\Rightarrow\ \text{f}^{-1}(-1)=1,2-\omega,2-\omega^2$ [from 1]

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Question 542 Marks

Let * be the binary operation on N defined by a * b = H.C.F. of a and b. Is * commutative? Is * associative? Does there exist identity for this binary operation on N?

Answer

a * b = H.C.F. of a and b.

a * b = H.C.F. of a and b = H.C.F. of b and a = b * a

Therefore, operation * is commutative.

(a * b) * c = (H.C.F. of a and b) * c = H.C.F. of (H.C.F. of a and b) and c

= H.C.F. of a, b and c = a * (b * c)

Therefore, the operation is associative.

$1*\text{a}=\text{a}*1\neq\text{a}.$

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Question 552 Marks

Let f: X → Y be an invertible function. Show that f has unique inverse. (Hint: suppose g₁ and g₂ are two inverses of f. Then for all y ∈ Y, fog₁(y) = 1_Y(y) = fog₂(y). Use one-one ness of f).

Answer

Given: f: X → Y be an invertible function.
Thus f is 1 – 1 and onto and therefore f⁻¹ exists.
Let g₁ and g₂ be two inverses of f . Then for all $\text{y}\in\text{Y},$
fog₁(y) = I_y(y) = fog₂(y) $\therefore$ fog₁(y) = fog₂(y)
⇒ f[g₁(y)] = f[g_2(y)] ⇒ g₁(y) = g₂(y)
$\therefore$ The inverse is unique and hence f has a unique inverse.

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Question 562 Marks

If A = {1, 2, 3, 4} define relations on A which have properties of being:
Reflexive, symmetric and transitive.

Answer

The relation on A having properties of being symmetric, reflexive and transitive is,
R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (2, 1)}
The relation R is an equivalence relation on A.

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Question 572 Marks

Let C denote the set of all complex numbers. A function f : C → C is defined by f(x) = x³. Write f^-1(1).

Answer

f : R → R defined by f(x) = x³
$\therefore$ f^-1(x³) = x
$\Rightarrow\ \text{f}^{-1}(1)=\{1,\omega,\omega^2\}$ $[\because\ \sqrt[3]{1}=\{1,\omega,\omega^2\}]$

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Question 582 Marks

Find gof and fog when f : R → R and g : R → R are defined by:
f(x) = 2x + x² and g(x) = x³

Answer

Given: f : R → R and g : R → R
Therefore, gof : R → R and fog : R → R
f(x) = 2x + x² and g(x) = x³
gof(x) = g(f(x)) = g(2x + x²)
gof(x) = g(2x + x²)³
fog(x) = f(g(x)) = f(x³)
$\therefore$ fog(x) = 2x³ + x⁶

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Question 592 Marks

Consider f: R₊→ [4, ∞) given by f(x) = x² + 4. Show that f is invertible with the inverse f^–1 of f given by $f^{-1}(\text{y})=\sqrt{\text{y}-4},$ where R₊ is the set of all non-negative real numbers.

Answer

Consider $f:\text{R}_{+}\rightarrow[4,\infty]$ and f(x) = x² + 4.
Let $\text{x}_1,\text{x}_2\in\text{R}\rightarrow[4,\infty],\text{ then }f(\text{x}_1)=\text{x}_{1}^{2}+4\text{ and }f(\text{x}_2)=\text{x}_{2}^{2}+4$
$\Rightarrow\ \text{x}_{1}^{2}+4=\text{x}_{2}^{2}+4\Rightarrow\text{x}_1=\text{x}_2\ \ \ \ \ \therefore f\text{ is one-one.}$
Now $\text{y}=\text{x}^2+4\Rightarrow\text{x}=\sqrt{\text{y}-4}$ as x > 0
$\therefore\ \ f\left(\sqrt{\text{y}-4}\right)=\left(\sqrt{\text{y}-4}\right)^2+4=\text{y}\ \ f(\text{x})=\text{y}\ \ \ \ \therefore\ f\text{ is onto.}$
Therefore, f(x) is invertible and $f^{-1}(\text{y})=\sqrt{\text{y}-4}.$

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Question 602 Marks

If the mappings f and g are given by f = {(1, 2), (3, 5), (4, 1)} and g = {(2, 3), (5, 1), (1, 3)}, write fog.

Answer

We are given that, f = {(1, 2), (3, 5), (4, 1)} and g = {(2, 3), (5, 1), (1, 3)}

Now, the domain of g is {2, 5, 1}

We know that, fog(x) = f{g(x)}

$\therefore$ fog(2) = f{g(2)} = f(3) = 5

$\therefore$ fog(5) = f{g(5)} = f(1) = 2

$\therefore$ fog(1) = f{g(1)} = f(3) = 5

Therefore, fog = {(2, 5), (5, 2), (1, 5)}

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Question 612 Marks

Write the identity element for the binary operation * defined on the set R of all real numbers by the rule $\text{a}\times\text{b}=\frac{3\text{ab}}{7}\ \forall\text{ a, b}\in\text{R}$.

Answer

We have,
$\text{a}\times\text{b}=\frac{3\text{ab}}{7}$
Let e be the identity element with respect to *. Then
a * e = a
$\Rightarrow\frac{3\text{ae}}{7}=\text{a}\ \Rightarrow\text{e}=\frac{7}{3}$

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Question 622 Marks

Let * be a binary operation on the set Q of rational numbers as follows:
a * b = a – b

Answer

a * b = a - b = -(b - a) = -b * a
$\therefore$ operation is not commutative.
(a * b) * c = (a - b) * c = (a - b) - c = a - b - c
And a * b (b * c) = a * (b - c) = a - (b - c) = a - b + c
Here, $(\text{a}*\text{b})*\text{c}\neq\text{a}*(\text{b}*\text{c})$
$\therefore$ operation * is not associative.

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Question 632 Marks

For each binary operation * defined below, determine whether * is commutative or associative.
On Q, define $\text{a} * \text{b} =\frac{\text{ab}}{2}$

Answer

For commutativity: $\text{a}*\text{b}=\frac{\text{ab}}{2}\ \text{and}\ \text{b}*\text{a}=\frac{\text{ba}}{2}=\frac{\text{ab}}{2}=\text{a}*\text{b}$
For associativity: $\text{a}*(\text{b}*\text{c})=\text{a}*\Big(\frac{\text{bc}}{2}\Big)=\frac{\text{abc/2}}{2}=\frac{\text{abc}}{4}$
Also, $(\text{a}*\text{b})*\text{c}=\Big(\frac{\text{ab}}{2}\Big)*\text{c}=\frac{\text{abc}/2}{2}=\frac{\text{abc}}{4}$
$\therefore$ a * (b * c) = (a * b) * c
Therefore, the operation * is commutative and associative.

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Question 642 Marks

Let S = {a, b, c}. Find the total number of binary operations on S.

Answer

Number of binary operations on a set with n elements is n².
Here, S = {a, b, c}
Number of elements in S = 3
Number of binary operations on a set with 3 elements is $3^{3^{2}}=3^9$

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Question 652 Marks

Find which of the binary operations are commutative and which are associative.
Show that none of the operations given above has identity.

Answer

Let the identity be I.

$\text{a}*\text{I}=\text{a - I}\neq\text{a}$
$\text{a}*\text{I}=\text{a}^2-\text{I}^2\neq\text{a}$
$\text{a}*\text{I}=\text{a + aI}\neq\text{a}$
$\text{a}*\text{I}=(\text{a - I})^2\neq\text{a}$
$\text{a}*\text{I}=\frac{\text{aI}}{4}\neq\text{a}$
$\text{a}*\text{I}=\text{aI}^2\neq\text{a}$

Therefore, none of the operations given above has identity.

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Question 662 Marks

Let * be a binary operation on the set Q of rational numbers as follows:
a * b = a²+ b²

Answer

a * b = a² + b² = b² + a² = b * a
$\therefore$ operation is commutative.
(a * b) * c = (a² + b²) * c = (a² + b²) + c² = a² + b² + c²
And a * b (b * c) = a * (b² + c²) = a² + (b² + c²)²
Here, $(\text{a}*\text{b})*\text{c}\neq\text{a}*(\text{b}*\text{c})$
$\therefore$ operation * is not associative.

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Question 672 Marks

Which one of the following graphs represents a function?

Answer

Figure (a) represents a function f : R → R
Whereas fig (b) does not represent a function.

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Question 682 Marks

Determine whether the following operations define a binary operation on the given set or not:

$'\odot'$ on N defined by $\text{a}\odot\text{b}=\text{a}^{\text{b}}+\text{b}^{\text{a}}$ for all $\text{a, b}\in\text{N.}$

Answer

Let $\text{a, b}\in\text{N.}$ Then,
$\text{a}^{\text{b}},\text{b}^{\text{a}}\in\text{N}$
$\Rightarrow\ \text{a}^{\text{b}}+\text{b}^{\text{a}}\in\text{N}$ $\big[\because$ Addition is binary operation on N$\big]$
$\Rightarrow\ \text{a}\odot\text{b}\in\text{N}$
Thus, $\odot$ is a binary operation on N.

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Question 692 Marks

Give an example of a relation which is,
Transitive but neither reflexive nor symmetric.

Answer

Let R be the relation on A such that
R = {(1, 2), (2, 3), (1, 3)}
The relation R on A is transitive, but neither symmetric nor reflexive.

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Question 702 Marks

Let 'o' be a binary operation on the set Q₀ of all non-zero rational numbers defined by $\text{a}\ ^*\ \text{b}=\frac{\text{ab}}{2}$ for all $\text{a},\text{b}\in\text{Q}_0.$
Find the invertible elements of Q₀.

Answer

We have,
$\text{a }^*\text{ b}=\frac{\text{ab}}{2}$ for all $\text{a},\text{b}\in\text{Q}_0$
Let $\text{b}\in\text{Q}_0$ be the inverse of $\text{a}\in\text{Q}_0$ with respect to *, then,
a * b = b * a = e for all $\text{a}\in\text{Q}_0$
$\Rightarrow\frac{\text{ab}}{2}=\text{e}\Rightarrow\frac{\text{ab}}{2}=2$
$\Rightarrow\text{b}=\frac{4}{\text{a}}$
Thus, $\text{b}=\frac{4}{\text{a}}$ is the inverse of a with respect to *.

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Question 712 Marks

Find the identity element in the set I⁺ of all positive integers defined by a * b = a + b for all a, b ∈ I⁺.

Answer

Let e be the identity element in I⁺ with respect to * such that
a * e = a = e * a, $\forall\ \text{a}\in\text{I}^{+}$
a * e = a and e * a = a, $\forall\ \text{a}\in\text{I}^{+}$
a + e = a and e + a = a, $\forall\ \text{a}\in\text{I}^{+}$
e = 0, $\forall\ \text{a}\in\text{I}^{+}$
Thus, 0 is the identity element in I⁺ with respect to *.

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Question 722 Marks

Let C be the set of complex numbers. Prove that the mapping f : C → R given by f(z) = |z|, ∀ z ∈ C, is neither one-one nor onto.

Answer

The mapping f : C → R

Given, f(z) = |z|, ∀ z ∈ C

f(1) = |1| = 1

f(-1) = |-1| = 1

f(1) = f(-1)

$\text{But}\ 1\neq-1$

So, f(z) is not one-one. Also, f(z) is not onto as there is no pre-image for any negative element of R under the mapping f(z).

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Question 732 Marks

Find the number of all onto functions from the set A = {1, 2, 3, ..., n} to itself.

Answer

We know that every onto function from A to itself is one-one.
Therefore, the number of one-one functions = number of bijections =n!

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Question 742 Marks

Let '*' be a binary operation on N defined by a * b = 1.c.m. (a, b) for all $\text{a, b}\in\text{N}$
Find 2 * 4, 3 * 5, 1 * 6.

Answer

a * b = 1.c.m. (a, b)
2 * 4 = 1.c.m. (2, 4)
= 4
3 * 5 = 1.c.m. (3, 5)
= 15
1 * 6 = 1.c.m. (1, 6)
= 6

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Question 752 Marks

Let 'o' be a binary operation on the set Q₀ of all non-zero rational numbers defined by $\text{a}\ ^*\ \text{b}=\frac{\text{ab}}{2}$ for all $\text{a},\text{b}\in\text{Q}_0.$
Show that 'o' is both commutative and associate.

Answer

We have,

$\text{a }^*\text{ b}=\frac{\text{ab}}{2}$ for all $\text{a},\text{b}\in\text{Q}_0$

Commutativity:

Let $\text{a},\text{b}\in\text{Q}_0,$ then

$\Rightarrow\text{a }^*\text{ b}=\frac{\text{ab}}{2}=\frac{\text{ba}}{2}=\text{a }^*\text{ b}$

$\Rightarrow\text{a }^*\text{ b}=\text{b }^*\text{ a}$

Thus, * is commutative on Q₀.

Associativity:

Let $\text{a},\text{b},\text{c}\in\text{Q}_0,$ then

$\Rightarrow(\text{a }^*\text{ b})\ ^*\ \text{c}=\frac{\text{ab}}{2}\ ....(1)$

and, $\text{a }^*\ (\text{b }^*\text{ c})=\text{a }^*\ \frac{\text{bc}}{2}=\frac{\text{abc}}{4}\ ....(2)$

From (1) & (2)

$(\text{a }^*\text{ b})\ ^*\ \text{c}=\text{a }^*\ (\text{b }^*\text{ c})$

⇒ * is accosiative on Q₀.

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Question 762 Marks

If f : A → A, g : A → A are two bijections, then prove that:
fog is an injection.

Answer

Given: A → A, g : A → A are two bijections.
Then, fog : A → A
Injectivity of fog: Let x and y be two elements of the domain (A), such that
(fog)(x) = (fog)(y)
⇒ f(g(x)) = f(g(y))
⇒ g(x) = g(y) (As, f is one-one)
⇒ x = y (As, g is one-one)
So, fog is an injection.

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Question 772 Marks

Let S = {a, b, c} and T = {1, 2, 3}. Find F^–1 of the following functions F from S to T, if it exists:
F = {(a, 3), (b, 2), (c, 1)}

Answer

S = {a, b, c}, T = {1, 2, 3}
F: S → T is defined as:
F ={(a, 3), (b, 2), (c, 1)}
⇒ F(a) = 3, F(b) = 2, F(c) = 1
Therefore, F^-1: T → S is given by
F^-1 = {(3, a), (2, b), (1, c)}.

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Question 782 Marks

Write the total number of binary operations on a set consisting of two elements.

Answer

Number of binary operations on a set with n elements $=\text{n}^{\text{n}^2}$
Here, Number of binary operations on a set with 2 elements $=2^{2^2}$
= 2⁴
=16

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Question 792 Marks

Which of the following functions from A to B are one-one and onto?
f₁ = {(1, 3), (2, 5), (3, 7)}; A = {1, 2, 3}, B = {3, 5, 7}

Answer

f₁ = {(1, 3), (2, 5), (3, 7)}
A = {1, 2, 3}, B = {3, 5, 7}
We can earily observe that in f₁ every element of A has different image from B.
$\therefore$ f₁ in not one-one.
Also, each element of B is the image of some element of A.
$\therefore$ f₁ in not on to.

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Question 802 Marks

If f : R → R be defined by f(x) = x⁴, write f^-1(1).

Answer

Let f^-1(1) = x ......(1)
⇒ f(x) = 1
⇒ x⁴ = 1
⇒ x⁴ - 1 = 0
⇒ (x² - 1)(x² + 1) = 0 [Using identity: a² - b² = (a - b)(a + b)]
⇒ (x - 1)(x + 1)(x² + 1) = 0 [Using identity: a² - b² = (a - b)(a + b)]
$\Rightarrow\ \text{x}=\pm1\ [\text{as x}\in\text{R}]$
⇒ f^-1(1) = {-1, 1} [from (1)]

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Question 812 Marks

If A = {2, 3, 4}, B = {1, 3, 7} and R = {(x, y): x ∈ A, y ∈ B and x < y} is a relation from A to B, then write R^-1.

Answer

Since R = {(x, y): x ∈ A, y ∈ B and x < y}
R = {(2, 3), (2, 7), (3, 7), (4, 7)}
Hence, R^-1 = {(3, 2), (7, 2), (7, 3), (7, 4)}

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Question 822 Marks

If f : A → B is an injection, such that range of f = {a}, determine the number of elements in A.

Answer

Range of f = {a}
Therefore, the number of images of f = 1
Since, is an injection, there will be exactly one image for each element of f.
Therefore, number of element in A = 1.

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Question 832 Marks

If the binary operation * on the set Z of integers is defined by a * b = a + 3b², find the value of 2 * 4.

Answer

Given: a * b = a + 3b²
Here,
2 * 4 = 2 + 3(4)²
= 2 + 3(16)
= 2 + 48
= 50

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Question 842 Marks

Prove that the operation * on the set $\text{M}=\Bigg\{\begin{bmatrix}\text{a} & 0 \\0 & \text{b} \end{bmatrix};\text{ a, b}\in\text{R}-\{0\}\Bigg\}$ defined by A * B = AB is a binary operation.

Answer

Given that * is an operation that is valid on the set $\text{M}=\Bigg\{\left(\begin{array}{c}\text{a}&0\\ 0&\text{b}\end{array}\right):\text{b}\in \text{R}-\big\{0\big\}\Bigg\}$ and it is defined as given: A * B = AB.
According to the problem it is given that on applying the operation * fore two given numbers in the set 'M' it gives a number in the set 'M' as a result of the operation.
$\Rightarrow \text{A}*\text{B}\in \text{M}...(1)$
Let us take $\text{A}=\left(\begin{array}{c}\text{a}&0\\ 0&\text{b}\end{array}\right)\text{ and }\text{B}=\left(\begin{array}{c}\text{c}&0\\ 0&\text{d}\end{array}\right)$ here $\text{a}\in \text{R},\ \text{c}\in \text{R},\ \text{d}\in \text{R}$ then,
$\Rightarrow \text{AB}=\left(\begin{array}{c}\text{a}&0\\ 0&\text{b}\end{array}\right)\times\left(\begin{array}{c}\text{c}&0\\ 0&\text{d}\end{array}\right)$
$\Rightarrow \text{AB}=\begin{pmatrix}((\text{a}\times\text{c})+(0\times 0))&((\text{a}\times0)+(0\times \text{d}))\$(0\times\text{c})+(\text{b}\times 0))&((0\times0)+(\text{b}\times\text{d})) \end{pmatrix}$
$\Rightarrow \text{Ab}=\begin{pmatrix}(\text{ac}+0)&(0+0)\$0+0)&(0+\text{bd}) \end{pmatrix}$
$\Rightarrow \text{AB}=\begin{pmatrix} \text{ac}&0\\0&\text{bd}\end{pmatrix}$
Since $\text{b}\in \text{R}$ and $\text{c}\in \text{R}$ then $\text{ac}\in \text{R}$
And also $\text{b}\in \text{R}$ and $\text{d}\in \text{R}$ then $\text{bd}\in \text{R}$
$\Rightarrow \text{AB}\in \text{R}$

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Question 852 Marks

Let A = {3, 5, 7}, B = {2, 6, 10} and R be a relation from A to B defined by R = {(x, y): x and y are relatively prime}. Then, write R and R^-1.

Answer

R = {(x, y): x and y are relatively prime}
Then,
R = {(3, 2), (5, 2), (7, 2), (3, 10), (7, 10), (5, 6), (7, 6)}
So, R^-1 = {(2, 3), (2, 5), (2, 7), (10, 3), (10, 7), (6, 5), (6, 7)}

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Question 862 Marks

Write the domain of the real function $\text{f(x)}=\frac{1}{\sqrt{|\text{x}|-\text{x}}}.$

Answer

Case-1: When x > 0
|x| = x
$\Rightarrow\ \frac{1}{\sqrt{|\text{x}|-\text{x}}}=\frac{1}{\sqrt{\text{x}-\text{x}}}=\frac{1}{0}=\infty$
Case-2: When x < 0
|x| = -x
$\Rightarrow\ \frac{1}{\sqrt{|\text{x}|-\text{x}}}=\frac{1}{\sqrt{-\text{x}-\text{x}}}=\frac{1}{\sqrt{-2\text{x}}}$ (exists because when x < 0, -2x > 0)
⇒ f(x) is defined when x < 0
So, domain $=(-\infty,0)$

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Question 872 Marks

Define an equivalence relation.

Answer

A relation R on a set A is said to be equivalence relation on a if R is:
Reflexive, Symmetric and Transitive.
R = {(x, y): x = y} on the set of real numbers is an equivalence relation.

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Question 882 Marks

f: Z → Z given by f(x) = x²

Answer

f: Z → Z is given by,
f(x) = x²
It is seen that for f(= 1) = f(1) = 1, but $-1\neq1.$
$\therefore$ f is not injective.
Now, $-2\in\text{Z}.$ But, there does not exist any element $\text{x}\in\text{Z}$ such that f(x) = x² = -2.
$\therefore$ f is not surjective.
Hence, function f is neither injective but not surjective.

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Question 892 Marks

If A = {a, b, c, d} and the function f = {(a, b), (b, d), (c, a), (d, c)}, write f^-1.

Answer

We are given that, f = {(a, b), (b, d), (c, a), (d, c)}
An inverse relation is the set of ordered pairs obtained by interchanging the first and second elements of each pair in the original relation.
$\therefore$ f^-1 = {(b, a), (d, b), (a, c), (c, d)}

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Question 902 Marks

Let A = {0, 1, 2, 3} and R be a relation on A defined as R = {(0, 0), (0, 1), (0, 3), (1, 0), (1, 1), (2, 2), (3, 0), (3, 3)}. Is R reflexive? symmetric? transitive?

Answer

We have,

R = {(0, 0), (0, 1), (0, 3), (1, 0), (1, 1), (2, 2), (3, 0), (3, 3)}

As, $(\text{a, a})\in\text{R}\ \forall\ \text{a}\in\text{A}$

So, R is a reflexive relation.

Also, $(\text{a, b})\in\text{R}$ and $(\text{b, a})\in\text{R}$

So, R is a symmetric as well

And, $(0,1)\in\text{R}$ but $(1,2)\notin\text{R}$ and $(2,3)\notin\text{R}$

So, R is not a transitive relation.

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Question 912 Marks

Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one.

Answer

It is given that a = {1, 2, 3}, B = {4, 5, 6, 7}
f: A → B is defined as f = {(1, 4), (2, 5), (3, 6)}.
$\therefore$ f(1) = 4, f(2) = 5, f(3) = 6
It is seen that the images of distinct elements of A under f are distinct.
Hence, function f is one-one.

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Question 922 Marks

Let * be a binary operation, on the set of all non-zero real numbers, given by
$\text{a}\times\text{b}=\frac{\text{ab}}{5}\ \forall\text{ a, b}\in\text{R}-\{0\}$
Write the value of x given by 2 * (x * 5) = 10.

Answer

Given: 2 * (x * 5) = 10
Here,
$2\times\Big(\frac{5\text{x}}{5}\Big)=10$
Implies that 2 * x = 10
Implies that $\frac{2\text{x}}{5}=10$
Implies that $\text{x}=\frac{10\times5}{2}$
Implies that x = 25

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Question 932 Marks

Show that the relation R in the set R of real numbers, defined as R = {(a, b) : a ≤ b²} is neither reflexive nor symmetric nor transitive.

Answer

$\text{R}=\big\{(\text{a},\text{b}):\text{a}\leq\text{b}^2\big\},$ Relation R is defined as the set of real numbers.

(i)	Whether $(\text{a},\text{a})\in\text{R},\ \text{then}\ \text{a}\leq\text{a}^2$ which is false.	$\therefore$	R is not reflexive.
(ii)	Whether (a,b) = (b, a) , then $\text{a}\leq\text{b}^2\ \text{and}\ \text{b}\leq\text{a}^2,$ it is false	$\therefore$	R is not symmetric.
(iii)	$\text{Now }\text{a}\leq\text{b}^2\ \text{and}\ \text{b}\leq\text{c}^2\Rightarrow\text{a}\leq\text{c}^4,$ which is false	$\therefore$	R is not transitive.

Therefore, R is neither reflexive, nor symmetric and nor transitive.

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Question 942 Marks

Write the total number of one-one functions from set A = {1, 2, 3, 4} to set B = {a, b, c}.

Answer

A has 4 elements and B has 3 elements.
Also, one-one function is only possible from A to B if nA ≤ nB.
But, here nA > nB
So, the number of one-one functions from A to B is 0.

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Question 952 Marks

Consider f: {1, 2, 3} → {a, b, c} given by f(1) = a, f(2) = b and f(3) = c. Find f^–1 and show that (f^–1)^–1 = f.

Answer

f = {(1, a), (2, b), (3, c)}, then it is clear that f is 1 – 1 and onto and therefore f^-1 exists.
Also, f^-1 = {(1, a), (b, 2), (c, 3)} and (f^-1)^-1 = {(1, a), (2, b), (3, c)} = f
Hence, (f^-1)^-1 = f.

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Question 962 Marks

If f : R → R is defined by f(x) = 3x + 2, find f(f(x)).

Answer

f(f(x)) = f(3x + 2)
= 3(3x + 2) + 2
= 9x + 6 + 2
= 9x + 8

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Question 972 Marks

Let A = {1, 2, 3, 4} and B = {a, b} be two sets. Write the total number of onto functions from A to B.

Answer

When two sets A and B have m and n elements respectively, then the number of onto functions from A to B is,

$\begin{cases}\sum_{\text{r}=1}^\text{n}(-1)^\text{r}\text{ nC}_\text{r}\text{r}^\text{m},&\text{if m}\geq\text{n}\\0,&\text{if m}<\text{n}\end{cases}$

Here, number of elements in A = 4 = m

Number of elements in B = 2 = n

So, m > n

Number of onto functions

$=\sum_{\text{r}=1}^2(-1)^\text{r}2\text{C}_\text{r}\text{r}^4$

$= (-1)^12\text{C}_11^4 + (-1)^22\text{C}_22^4$

$= -2 + 16$

$= 14$

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Question 982 Marks

If functions f : A → B and g : B → A satisfy gof = I_A, then show that f is oneone and g is onto.

Answer

Given that, f : A → B and g : B → A satisfy gof = I_A,

$\because$ gof = I_A

⇒ gof{f(x₁)} = gof{f(x₂)}

⇒ g(x₁) = g(x₂) $[\because\text{gof}=\text{I}_\text{A}]$

$\therefore$ x₁ = x₂

Hence, f is one-one and g is onto.

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Question 992 Marks

Let A = {a, b, c, d} and f : A → A be given by f = {(a, b), (b, d), (c, a), (d, c)}. Write f ^-1.

Answer

We have,
A = {a, b, c, d} and f : A → A be given by
f = {(a, b), (b, d), (c, a), (d, c)}
(Since, the elements of a function when interchanged gives inverse function. Therefore, f^-1 = {(b, a), (d, b), (a, c), (c, d)})

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Question 1002 Marks

Define a transitive relation.

Answer

A relation R on a set A is said to be transitive if
$(\text{a, b})\in\text{R}$ and $(\text{b, c})\in\text{R}$
$\Rightarrow\ (\text{a, c})\in\text{R}$ for all $\text{a, b, c}\in\text{R}$
i.e., aRb and bRc
⇒ aRc for all $\text{a, b, c}\in\text{R}$

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2 Marks - Page 2 - Maths STD 12 Science Questions - Vidyadip