If set A contains 5 elements and set B contains 6 elements, then the number of one-one onto mappings from A to B is:
- (a)720
- (b)120
- (c)0
- (d)none of the above.
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Relations and Functions - ISC Class 12 Mathematics Questions with Answers, Page 3
58 past-paper questions on Relations and Functions from ISC Class 12 Mathematics papers (2027-2017), newest first, in full. Questions 41-58 are on this page, 20 to a page. Tap "Show answer" under a question to see its answer.
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The function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \sin(3x + 2)$, $\forall x \in \mathbb{R}$ is:
- (a)One-One
- (b)Onto
- (c)Neither one-one nor onto
- (d)one-one but not onto.
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Consider the mapping $f: A \to B$ is defined by $f(x) = \frac{x-1}{x-2}$ such that $f(x)$ is one-one onto.
Based on the above information, answer the following questions by choosing the correct option.
(i)[2.0]
Domain of $f(x)$ is:
- (a)$\mathbb{R} - \{2\}$
- (b)$\mathbb{R}$
- (c)$\mathbb{R} - \{1, 2\}$
- (d)$\mathbb{R} - \{0\}$
(ii)[2.0]
Range of $f(x)$ is:
- (a)$\mathbb{R} - \{2\}$
- (b)$\mathbb{R}$
- (c)$\mathbb{R} - \{1\}$
- (d)$\mathbb{R} - \{0\}$
(iii)[2.0]
If $g(x) = 2f(x) - 1$, then $g(x)$ in terms of $x$ is:
- (a)$\frac{x+2}{x}$
- (b)$\frac{x+1}{x-2}$
- (c)$\frac{x-2}{x}$
- (d)$\frac{x}{x-2}$
(iv)[2.0]
A function $f(x)$ is said to be one-one if:
- (a)$f(x_1) = f(x_2) \implies x_1 = x_2$
- (b)$f(-x_1) = f(-x_2) \implies -x_1 = x_2$
- (c)$f(x_1) = f(x_2) \implies -x_1 = x_2$
- (d)$-f(x_1) = f(x_2) \implies x_1 = x_2$
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Let A be the set of all students of a boy’s school. Then the relation R in A is defined by: $R = \{(a,b) : a \text{ is sister of } b\}$ is
- (a)an equivalence relation
- (b)symmetric relation
- (c)an empty relation
- (d)a universal relation
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Let $R$ be the relation in the set $N$, given by $R = \{(x, y) : x = y + 3 \wedge y > 5\}$.
Choose the correct answer from the following:
- (a)$(7, 4) \in R$.
- (b)$(9, 6) \in R$.
- (c)$(4, 1) \in R$.
- (d)$(8, 5) \in R$.
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Write total number of functions from set A to set B, where $A = \{1, 2, 3\}$, $B = \{p, q, r, s\}$.
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If $A = \{1, 2, 3\}$, $B = \{x, y\}$, then the number of functions from $A$ to $B$ is:
- (a)3
- (b)6
- (c)8
- (d)12
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Show that the function $f : N \to N$, defined by $f(x) = 5x - 3, \forall x \in N$, is one-one function but not onto function.
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If the function $f : \mathbb{R} \to \mathbb{R}$ be defined as $f(x) = \frac{3x + 4}{5x - 7}, x \ne \frac{7}{5}$ and $g : \mathbb{R} \to \mathbb{R}$ be defined as $g(x) = \frac{7x + 4}{5x - 3}, x \ne \frac{3}{5}$, show that $(g \circ f)(x) = (f \circ g)(x)$.
Given: $f(x) = \frac{3x + 4}{5x - 7}, x \ne \frac{7}{5}$ and $g(x) = \frac{7x + 4}{5x - 3}, x \ne \frac{3}{5}$
To show: $(g \circ f)(x) = (f \circ g)(x)$
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Determine whether the binary operation $*$ on $\mathbb{R}$ defined by $a * b = |a - b|$ is commutative. Also, find the value of $(-3) * 2$.
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If $f : A \to A$ and $A = \mathbb{R} - \left\\{\frac{8}{5}\right\\}$, show that the function $f(x) = \frac{8x + 3}{5x - 8}$ is one – one onto. Hence, find $f^{-1}$.
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If $f : \mathbb{R} \to \mathbb{R}$, $f(x) = x^3$ and $g : \mathbb{R} \to \mathbb{R}$, $g(x) = 2x^2 + 1$, and $\mathbb{R}$ is the set of real numbers, then find $fog(x)$ and $gof(x)$.
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The binary operation $* : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ is defined as $a * b = 2a + b$.
Find $(2 * 3) * 4$.
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A binary operation $*$ defined on $\mathbb{Q} - \{1\}$ is given by $a * b = a + b - ab$. Find the identity element.
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If the function $f(x) = \sqrt{2x-3}$ is invertible then find its inverse. Hence prove that $(f \circ f^{-1})(x) = x$.
Given: $f(x) = \sqrt{2x-3}$ is invertible
To show: $(f \circ f^{-1})(x) = x$
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Let $R^+$ be the set of all positive real numbers and $f: R^+ \to [4, \infty): f(x) = x^2 + 4$. Show that inverse of $f$ exists and find $f^{-1}$.
Given: Let $R^+$ be the set of all positive real numbers and $f: R^+ \to [4, \infty): f(x) = x^2 + 4$
To show: Inverse of $f$ exists and find $f^{-1}$
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If $A$, $B$ and $C$ are the elements of Boolean algebra, simplify the expression $(A' + B')(A + C') + B'(B + C)$. Draw the simplified circuit.
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Prove that locus of $z$ is circle and find its centre and radius if $\frac{z-i}{z-1}$ is purely imaginary.
Given: $\frac{z-i}{z-1}$ is purely imaginary.
To show: Locus of $z$ is circle with centre $(1/2, 1/2)$ and radius $1/\sqrt{2}$.
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