3 Marks Question

Question 13 Marks

Let f, g: $R \to R$ be defined, respectively by f(x) = x + 1, g(x) = 2x – 3. Find f + g, f – g and $\frac{f}{g}$.

Answer

Here f (x) = x + 1 and g (x) = 2x – 3
Now (f + g) (x) =f (x) + g(x) = x + 1 + 2x - 3 = 3x - 2
(f - g) (x) = f(x) - g(x) = x + 1- (2x - 3) = x + 1 - 2x + 3 = -x + 4
$\frac{{(f)}}{{(g)}}(x) = \frac{{f(x)}}{{g(x)}} = \frac{{x + 1}}{{2x - 3}},x \ne \frac{3}{2}$

View full question & answer→

Question 23 Marks

The relation f is defined by $f ( x ) = \left\{ \begin{array} { l } { x ^ { 2 } , 0 \leq x \leq 3 } \\ { 3 x , 3 \leq x \leq 10 } \end{array} \right.$and the relation g is defined by $g ( x ) = \left\{ \begin{array} { l } { x ^ { 2 } , 0 \leq x \leq 2 } \\ { 3 x , 2 \leq x \leq 10 } \end{array} \right..$ Show that f is a function and g is not a function.

Answer

$f ( x ) = \left\{ \begin{array} { l } { x ^ { 2 } , 0 \leq x \leq 3 } \\ { 3 x , 3 \leq x \leq 10 } \end{array} \right.$
At $x = 3, f(x) = x^2$
$\therefore$ $f(3) = 3^2 = 9$
Also, at $x = 3, f(x) = 3x$
$\Rightarrow$ f(3) = $ 3 \times 3 = 9$
Since, f is defined at x = 3. Hence, f is a function.
Now, $g ( x ) = \left\{ \begin{array} { l } { x ^ { 2 } , 0 \leq x \leq 2 } \\ { 3 x , 2 \leq x \leq 10 } \end{array} \right.$
At $x = 2, g(x) = x^2$
$\Rightarrow$$ g(2) = 2^2 = 4$
Also, at x = 2, g(x) = 3x $\Rightarrow$$ g (2) = 3 \times 2 = 6$
At x = 2, relation g has two values.
$\therefore$ The relation g is not a function.

View full question & answer→

Question 33 Marks

Let $A = \{1, 2, 3, 4, 6\}.$ Let $R$ be the relation on $A$ defined by $\{(a, b): a, b \in A, b$ is exactly divisible by $a\}.$

Write $R$ in roster form
Find the domain of $R$
Find the range of $R.$

Answer

Here $A = \{1, 2, 3, 4, 6\}$
We have to form a set of ordered pairs $(a, b)$ where $b$ is exactly divisible by $a$.

$R = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)\}$
Domain of $R = \{1, 2, 3, 4, 6\}$
Range of $R = \{1, 2, 3, 4, 6\}$

View full question & answer→

Question 43 Marks

The function f is defined by $\begin{equation} f(x)=\left\{\begin{array}{ll} {1-x,} & {x<0} \\ {1} & {, x=0} \\ {x+1,} & {x>0} \end{array}\right. \end{equation}$
Draw the graph of f(x).

Answer

Here it is given that, f(x) = 1 – x, x < 0, this gives
f(– 4) = 1 – (– 4) = 5;
f(– 3) =1 – (– 3) = 4,
f(– 2) = 1 – (– 2) = 3
f(–1) = 1 – (–1) = 2; etc, and f(1) = 2, f (2) = 3, f (3) = 4
f(4) = 5 and so on for f(x) = x + 1, x > 0.
Thus, the graph of the given function f(x) is as shown in figure given below.

View full question & answer→

Question 53 Marks

Let R be the set of real numbers. Define the real function f: R $\rightarrow$ R by f(x) = x + 10 and sketch the graph of this function.

Answer

Here we have f(x) = x + 10
Here f(0) = 10, f(1) = 11, f(2) = 12, ..., f(10) = 20, etc., and f(–1) = 9, f(–2) = 8, ..., f(–10) = 0 and so on.
Therefore, the shape of the graph of the given function assumes the form as shown in figure below:

View full question & answer→

Question 63 Marks

Define the real valued function f : R – {0} $\rightarrow$ R defined by $\begin{equation} f(x)=\frac{1}{x} \end{equation}$, x $\in$R – {0}. Complete the Table given below using this definition. What is the domain and range of this function?

x	–2	–1.5	–1	–0.5	0.25	0.5	1	1.5	2
$\begin{equation} y=\frac{1}{x} \end{equation}$	...	...	...	...	...	...	...	...	...

Answer

The completed table is given by:

x	–2	–1.5	–1	–0.5	0.25	0.5	1	1.5	2
$\begin{equation} y=\frac{1}{x} \end{equation}$	– 0.5	– 0.67	–1	– 2	4	2	1	0.67	0.5

The domain is all real numbers except zero and its range is also all real numbers except zero. The graph of f is given in figure shown below:

View full question & answer→

Question 73 Marks

Draw the graph of the function $f : R \rightarrow R$ defined by $f (x) = x^3, x \in R.$

Answer

Given function is, $f (x) = x^3, x \in R.$
We have $f(0) = 0, f(1) = 1, f(–1) = –1, f(2) = 8, f(–2) = –8, f(3) = 27; f(–3) = –27,$ etc.
Therefore, $f =\{(x, x^3): x \in R\}$. The graph of $f$ is given in figure showjbelow:

View full question & answer→

Question 83 Marks

Define the function $f: R \rightarrow R$ by $y = f(x) = x^2, x \in R$. Complete the Table given below by using this definition. What is the domain and range of this function? Draw the graph of f.

$x$	$– 4$	$–3$	$–2$	$–1$	$0$	$1$	$2$	$3$	$4$
$y = f(x) = x^2$

Answer

The completed table is given below:

$x$	$– 4$	$–3$	$–2$	$–1$	$0$	$1$	$2$	$3$	$4$
$y = f (x) = x^2$	$16$	$9$	$4$	$1$	$0$	$1$	$4$	$9$	$16$

Domain of $f = \{x : x \in R\}$. Range of $f = \{x^2: x \in R\}$. The graph of f is given by the figure shown below:

View full question & answer→

Maths 3 Marks Question

Take a timed test