CBSE Class 12 Maths Exemplar Chapter 1: Relations and Functions NCERT Solutions

The given relation is R = \{(a, a), (b, c), (a, b)\} on the set A = \{a, b, c\}.

To make the relation R reflexive, it must contain pairs of the form (x, x) for every element x in set A. Currently, R contains (a, a). Therefore, we need to add (b, b) and (c, c) to R to ensure reflexivity.

To make the relation R transitive, if (x, y) \in R and (y, z) \in R, then (x, z) must also be in R.

In the given relation R, we have (a, b) \in R and (b, c) \in R. For R to be transitive, the ordered pair (a, c) must be added.

So, the minimum number of ordered pairs to be added to R to make it reflexive and transitive are (b, b), (c, c), and (a, c). The total minimum number of ordered pairs to be added is 3.

The given function is $f(x)\; =\; \backslash sqrt\{25\; -\; x^2\}$.

For the function $f(x)$ to be real-valued, the expression under the square root must be non-negative. Therefore, we must have:

$$25\; -\; x^2\; \backslash ge\; 0$$

Rearranging the inequality, we get:

$$x^2\; \backslash le\; 25$$

Taking the square root of both sides, we find the range of x:

$$-5\; \backslash le\; x\; \backslash le\; +5$$

The domain D of the function is the set of all possible values of x. Thus, the domain is the closed interval $[-5,\; 5]$.

$D\; =\; [-5,\; 5]$

Given the functions $f(x)\; =\; 2x\; +\; 1$ and $g(x)\; =\; x^2\; -\; 2$, both defined for all real numbers $x\; \backslash in\; R$.

The composition of functions $gof$ is defined as $gof(x)\; =\; g(f(x))$.

First, substitute the expression for $f(x)$ into $g(x)$:

$gof(x)\; =\; g(f(x))\; =\; g(2x\; +\; 1)$

Now, apply the function $g$ to the expression $(2x\; +\; 1)$. The function $g(x)$ squares its input and subtracts 2. So:

$g(2x\; +\; 1)\; =\; (2x\; +\; 1)^2\; -\; 2$

Expand the squared term:

$(2x\; +\; 1)^2\; =\; (2x)^2\; +\; 2(2x)(1)\; +\; 1^2\; =\; 4x^2\; +\; 4x\; +\; 1$

Substitute this back into the expression for $gof(x)$:

$gof(x)\; =\; (4x^2\; +\; 4x\; +\; 1)\; -\; 2$

Simplify the expression:

$gof(x)\; =\; 4x^2\; +\; 4x\; -\; 1$

Therefore, the composition $gof$ is $4x^2\; +\; 4x\; -\; 1$.

Given the function $f(x)\; =\; 2x\; -\; 3$, defined for all real numbers $x\; \backslash in\; R$.

To find the inverse function $f^\{-1\}(x)$, we first set $y\; =\; f(x)$:

$$y\; =\; 2x\; -\; 3$$

Next, we swap $x$ and $y$ to represent the inverse relationship:

$$x\; =\; 2y\; -\; 3$$

Now, we solve this equation for $y$ to find the expression for $f^\{-1\}(x)$:

$$x\; +\; 3\; =\; 2y$$

$$y\; =\; \backslash frac\{x+3\}\{2\}$$

Therefore, the inverse function is $f^\{-1\}(x)\; =\; \backslash frac\{x+3\}\{2\}$.

Given the set $A\; =\; \backslash \{a,\; b,\; c,\; d\backslash \}$ and the function $f$ defined as a set of ordered pairs:

$f\; =\; \backslash \{(a,\; b),\; (b,\; d),\; (c,\; a),\; (d,\; c)\backslash \}$

The inverse function, denoted by $f^\{-1\}$, is obtained by reversing the order of each ordered pair in the definition of $f$. If $(x,\; y)\; \backslash in\; f$, then $(y,\; x)\; \backslash in\; f^\{-1\}$.

Applying this rule to each pair in $f$:

• For $(a,\; b)\; \backslash in\; f$, we have $(b,\; a)\; \backslash in\; f^\{-1\}$.
• For $(b,\; d)\; \backslash in\; f$, we have $(d,\; b)\; \backslash in\; f^\{-1\}$.
• For $(c,\; a)\; \backslash in\; f$, we have $(a,\; c)\; \backslash in\; f^\{-1\}$.
• For $(d,\; c)\; \backslash in\; f$, we have $(c,\; d)\; \backslash in\; f^\{-1\}$.

Therefore, the inverse function is:

$f^\{-1\}\; =\; \backslash \{(b,\; a),\; (d,\; b),\; (a,\; c),\; (c,\; d)\backslash \}$

Given the function $f(x)\; =\; x^2\; -\; 3x\; +\; 2$.

We need to find the composition $f\backslash \{f(x)\backslash \}$, which means we substitute $f(x)$ into itself.

$f\backslash \{f(x)\backslash \}\; =\; f(x^2\; -\; 3x\; +\; 2)$

Now, we replace every instance of $x$ in the definition of $f(x)$ with the expression $(x^2\; -\; 3x\; +\; 2)$:

$f\backslash \{f(x)\backslash \}\; =\; (x^2\; -\; 3x\; +\; 2)^2\; -\; 3(x^2\; -\; 3x\; +\; 2)\; +\; 2$

First, expand the squared term $(x^2\; -\; 3x\; +\; 2)^2$. Using the formula $(a+b+c)^2\; =\; a^2+b^2+c^2+2ab+2bc+2ca$, where $a=x^2$, $b=-3x$, $c=2$:

$(x^2\; -\; 3x\; +\; 2)^2\; =\; (x^2)^2\; +\; (-3x)^2\; +\; (2)^2\; +\; 2(x^2)(-3x)\; +\; 2(-3x)(2)\; +\; 2(x^2)(2)$

$=\; x^4\; +\; 9x^2\; +\; 4\; -\; 6x^3\; -\; 12x\; +\; 4x^2$

$=\; x^4\; -\; 6x^3\; +\; (9x^2\; +\; 4x^2)\; -\; 12x\; +\; 4$

$=\; x^4\; -\; 6x^3\; +\; 13x^2\; -\; 12x\; +\; 4$

Next, distribute the $-3$ in the second term:

$-3(x^2\; -\; 3x\; +\; 2)\; =\; -3x^2\; +\; 9x\; -\; 6$

Now, substitute these expanded terms back into the expression for $f\backslash \{f(x)\backslash \}$:

$f\backslash \{f(x)\backslash \}\; =\; (x^4\; -\; 6x^3\; +\; 13x^2\; -\; 12x\; +\; 4)\; +\; (-3x^2\; +\; 9x\; -\; 6)\; +\; 2$

Combine like terms:

$f\backslash \{f(x)\backslash \}\; =\; x^4\; -\; 6x^3\; +\; (13x^2\; -\; 3x^2)\; +\; (-12x\; +\; 9x)\; +\; (4\; -\; 6\; +\; 2)$

$f\backslash \{f(x)\backslash \}\; =\; x^4\; -\; 6x^3\; +\; 10x^2\; -\; 3x\; +\; 0$

So, $f\backslash \{f(x)\backslash \}\; =\; x^4\; -\; 6x^3\; +\; 10x^2\; -\; 3x$.

The given set of ordered pairs is $g\; =\; \backslash \{(1,\; 1),\; (2,\; 3),\; (3,\; 5),\; (4,\; 7)\backslash \}$.

To determine if $g$ is a function, we check if each element in the domain (the first element of each pair) maps to exactly one element in the codomain (the second element of each pair). In this set, the domain elements are 1, 2, 3, and 4. Each of these elements appears only once as the first element of an ordered pair. Therefore, $g$ is a function.

Now, we are given that the function $g$ can be described by the linear equation $g(x)\; =\; \backslash alpha\; x\; +\; \backslash beta$. We can use the given ordered pairs to form equations and solve for $\backslash alpha$ and $\backslash beta$.

Using the pair $(1,\; 1)$, we have $g(1)\; =\; 1$:

$\backslash alpha(1)\; +\; \backslash beta\; =\; 1\; \backslash implies\; \backslash alpha\; +\; \backslash beta\; =\; 1\; \backslash quad\; \backslash dots(i)$

Using the pair $(2,\; 3)$, we have $g(2)\; =\; 3$:

$\backslash alpha(2)\; +\; \backslash beta\; =\; 3\; \backslash implies\; 2\backslash alpha\; +\; \backslash beta\; =\; 3\; \backslash quad\; \backslash dots(ii)$

We now have a system of two linear equations with two variables, $\backslash alpha$ and $\backslash beta$.

Subtract equation (i) from equation (ii):

$(2\backslash alpha\; +\; \backslash beta)\; -\; (\backslash alpha\; +\; \backslash beta)\; =\; 3\; -\; 1$

$2\backslash alpha\; +\; \backslash beta\; -\; \backslash alpha\; -\; \backslash beta\; =\; 2$

$\backslash alpha\; =\; 2$

Substitute the value of $\backslash alpha\; =\; 2$ back into equation (i):

$2\; +\; \backslash beta\; =\; 1$

$\backslash beta\; =\; 1\; -\; 2$

$\backslash beta\; =\; -1$

Thus, the values assigned to $\backslash alpha$ and $\backslash beta$ are $\backslash alpha\; =\; 2$ and $\backslash beta\; =\; -1$. The function is $g(x)\; =\; 2x\; -\; 1$.

We can verify this with the other pairs: $g(3)\; =\; 2(3)\; -\; 1\; =\; 6\; -\; 1\; =\; 5$ (correct) and $g(4)\; =\; 2(4)\; -\; 1\; =\; 8\; -\; 1\; =\; 7$ (correct).