Questions

1 Marks Question

🎯

Test yourself on this topic

27 questions · timed · auto-graded

Question 11 Mark
Let A = {–1, 2, 3} and B = {1, 3}. Determine:
B × A
Answer
B × A = {(1, -1), (1, 2), (1, 3), (3, -1), (3,2), (3, 3)}
View full question & answer
Question 21 Mark
If f and g are real function defined by$ f(x) = x^2 + 7$ and $g(x) = 3x + 5,$ find following $: f(3) + g (– 5)$
Answer
Given that$: f(x) = x^2 + 7$ and $g(x) = 3x + 5$
f($3) + g(– 5) = [(3)^2 + 7] + [3(-5) + 5]$
$= (9 + 7) + (-15 + 5) = 16 -10 = 6$
Hence$, f(3) + g(-5) = 6.$
View full question & answer
Question 31 Mark
Find the domain of each of the following functions given by:
$\text{f(x)}=\frac{3\text{x}}{28-\text{x}}$
Answer
$\text{f(x)}=\frac{3\text{x}}{28-\text{x}}$
Clearly, f(x) is not defined, if 28 - x = 0
$\Rightarrow\text{x}\neq28$
$\therefore$ Domain of f = R -{28}
View full question & answer
Question 41 Mark
Let $\text{f(x)}=\sqrt{\text{x}}$ and $\text{g(x)}=\text{x}$ be two functions defined in the domain $\text{R}^+\cup\{0\}$. Find:
$\text{(fg)(x)}$
Answer
We have, $\text{f(x)}=\sqrt{\text{x}}$ and $\text{g(x)}=\text{x}$ be two functions defined in the domain $\text{R}^+\cup\{0\}.$
$(\text{f}\text{g})(\text{x})=\text{f(x)}.\text{g(x)}=\sqrt{\text{x}}.\text{x}=\text{x}^{\frac{3}{2}}$
View full question & answer
Question 51 Mark
Is the given relation a function? Give reasons for your answe $r:s = (n, n^{2 }) | n$ is a positive integer
Answer
We have$, s = (n, n^2) |$ n is a positive integer
Since the square of any positive integer is unique$,$ every element in the domain has unique image. Hence$, S$ is a function.
View full question & answer
Question 61 Mark
Is the given relation a function? Give reasons for your answer:
$\text{g}=\big\{\text{n}, \frac{1}{\text{n}}\big\},$ | n is a positive integer
Answer
We have, $\text{g}=\big\{\text{n}, \frac{1}{\text{n}}\big\},$ | n is a positive integer}
For n, it is a positive integer and $\frac{1}{\text{n}}$ is unique and distinct. Therefore,every element in the domain has unique image. So, it is a function.
View full question & answer
Question 71 Mark
Let A = {–1, 2, 3} and B = {1, 3}. Determine:
A × B
Answer
A × B = {(-1, 1), (-1, 3), (2, 1), (2, 3), (3,1), (3, 3)}
View full question & answer
Question 81 Mark
Given $\text{A} = \big\{1, 2, 3, 4, 5\big\}$$\text{S} = \text{(x, y)} : \text{x} \in\text{A}, \text{y} \in\text{A}.$ Find the ordered pairs which satisfy the condition given below:
X + Y = 5
Answer
Given that:
$\text{A} = \big\{1, 2, 3, 4, 5\big\}$
$\text{S} = \text{(x, y)} : \text{x} \in\text{A}, \text{y} \in\text{A}.$
X + Y = 5, So, the ordered pairs satisfying the given conditions are (1, 4), (4, 1), (2, 3), (3, 2).
View full question & answer
Question 91 Mark
Let A = {–1, 2, 3} and B = {1, 3}. Determine:
A × A
Answer
A × A = {(-1, -1), (-1, 2), (-1, 3), (2, -1), (2, 2), (2, 3), (3, -1), (3, 2), (3, 3)}
View full question & answer
Question 101 Mark
Is the given relation a function? Give reasons for your answer:
t = {(x, 3) | x is a real number.
Answer
We have, t = {(x, 3)| x is a real number}.
Since every element in the domain has the image 3, it is a constant function.
View full question & answer
Question 111 Mark
Find the range of the following functions given by:
$\text{f(x)}=1+3\cos2\text{x}$
Answer
We knoe that, $-1\leq\cos2\text{x}\leq1$
$\Rightarrow-3\leq3\cos2\text{x}\leq3$
$\Rightarrow-2\leq1+3\cos2\text{x}\leq4$
$\Rightarrow-2\leq\text{f(x)}\leq4$
$\therefore $ Range of $\text{f}=[-2, 4]$
View full question & answer
Question 121 Mark
Given $\text{A} = \big\{1, 2, 3, 4, 5\big\}$$\text{S} = \text{(x, y)} : \text{x} \in\text{A}, \text{y} \in\text{A}.$ Find the ordered pairs which satisfy the condition given below:
X + Y < 5
Answer
Given that:
$\text{A} = \big\{1, 2, 3, 4, 5\big\}$
$\text{S} = \text{(x, y)} : \text{x} \in\text{A}, \text{y} \in\text{A}.$
X + Y < 5, So, the ordered pairs satisfying the given conditions are (1, 1), (1, 2), (2, 1), (1, 3), (2, 2), (3, 1).
View full question & answer
Question 131 Mark
Is the given relation a function? Give reasons for your answer:
h = {(4, 6), (3, 9), (– 11, 6), (3, 11)}
Answer
We have, h = {(4, 6),(3, 9), (-11, 6), (3, 11)}.
Since pre-image 3 has two images 9 and 11, it is not a function.
View full question & answer
Question 141 Mark
If $f$ and $g$ are two real valued function defined as $f(x) = 2x + 1, g(x) = x^2 + 1,$ then find.
$f + g$
Answer
We have$, f(x) = 2x + 1, g(x) = x^2 + 1$
$(f + g)(x) = f(x) + g(x)$
$= 2x + 1 + x^2 + 1 = x^2 + 2x +2$
View full question & answer
Question 151 Mark
If $f$ and $g$ are real function defined by $f(x) = x^2 + 7$ and $g(x) = 3x + 5,$ find following $: f(t) - f (- 2)$
Answer
$\text{f}(\text{t})-\text{f}(-2)=[\text{t}^2+7]-[(-2)^2+7]=\text{t}^2+7-11$
$=\text{t}^2-4$
Hence$, \text{f}\text{(t)}+\text{f}(-2)=\text{t}^2-4.$
View full question & answer
Question 161 Mark
Find the range of the following functions given by:
f(x) = x - 3
Answer
 We know that, $|\text{x}-3|\geq0$
$\Rightarrow\text{f(x)}\geq0$
$\therefore$ Range of $\text{f} = (0, \infty]$
View full question & answer
Question 171 Mark
Let $\text{f(x)}=\sqrt{\text{x}}$ and $\text {g(x)}=\text{x}$ be two functions defined in the domain $\text{R}^+\cup\{0\}$. Find:
$\bigg(\frac{\text{f}}{\text{g}}\bigg)(\text{x})$
Answer
We have, $\text{f(x)}=\sqrt{\text{x}}$ and $\text {g(x)}=\text{x}$ be two functions defined in the domain $\text{R}^+\cup\{0\}.$
$\bigg(\frac{\text{f}}{\text{g}}\bigg)(\text{x})=\frac{\text{f(x)}}{\text{g(x)}}=\frac{\sqrt{\text{x}}}{\text{x}}=\frac{\text{1}}{\sqrt{\text{x}}}$
View full question & answer
Question 181 Mark
In following case, find a and b.
(2a + b, a – b) = (8, 3)
Answer
We have, (2a + b, a - b) = (8, 3)
$\Rightarrow$ 2a + b = 8 and a – b = 3
On solving, we get
 $\text{a}= \frac{11}{3}$ and $\text{b}= \frac{2}{3}$
View full question & answer
Question 191 Mark
Let $\text{f(x)}=\sqrt{\text{x}}$ and $\text{g(x)}=\text{x}$ be two functions defined in the domain $\text{R}^+\cup\{0\}$. Find:
$\text{(f - g)}\text{(x)}$
Answer
We have, $\text{f(x)}=\sqrt{\text{x}}$ and $\text{g(x)}=\text{x}$ be two functions defined in the domain $\text{R}^+\cup\{0\}.$
$(\text{f}-\text{g})(\text{x})=\text{f(x)}-\text{g(x)}=\sqrt{\text{x}}-\text{x}$
View full question & answer
Question 201 Mark
Find the range of the following functions given by:
f(x) = 1 - x - 2
Answer
 We know that,$|\text{x}-2|\geq0$
$\Rightarrow-|\text{x}-2|\leq0$
$\Rightarrow1-|\text{x}-2|\leq1$
$\Rightarrow\text{f(x)}\leq1$
$\therefore$ Range of f is $(-\infty, 11]$
View full question & answer
Question 211 Mark
If $f$ and $g$ are two real valued function defined as $f(x) = 2x + 1, g(x) = x^2 + 1,$ then find. $f - g$
Answer
We have$, f(x) = 2x + 1, g(x) = x^2 + 1$
$(f - g)(x) = f(x) - g(x)$
$= (2x + 1) - (x^2 + 1) = 2x + 1 - x^{2 }- 1 + 2x - x^2$​​​​​​​
View full question & answer
Question 221 Mark
Let $\text{f(x)}=\sqrt{\text{x}}$ and $\text{g(x)}=\text{x}$ be two functions defined in the domain $\text{R}^+\cup\big\{0\big\}$. Find:
$\text{(f+g)}\text{(x)}$
Answer
We have, $\text{f(x)}=\sqrt{\text{x}}$ and $\text{g(x)}=\text{x}$ be two functions defined in the domain $\text{R}^+\cup\big\{0\big\}.$
$(\text{f}+\text{g})(\text{x})=\text{f(x)}+\text{g(x)}=\sqrt{\text{x}}+\text{x}$
View full question & answer
Question 231 Mark
Is the given relation a function? Give reasons for your answer:
f = {(x, x) | x is a real number}
Answer
We have, f = {(x, x) | x is a real number}
Since every element in the domain has unique image, it is a function.
View full question & answer
Question 241 Mark
If $f$ and $g$ are two real valued function defined as $f(x) = 2x + 1,\  g(x) = x^2 + 1,$ then find. $fg$
Answer
We have$, f(x) = 2x + 1,\  g(x) = x^2 + 1$
$(fg)(x) = f(x) . g(x)$
$= (2x + 1)(x^2 + 1) = 2x^3 + 2x + x^2 + 1 = 2x^3 + x^2 + 2x + 1$
View full question & answer
Question 251 Mark
Let A = {–1, 2, 3} and B = {1, 3}. Determine:
B × B
Answer
B × B = {(1, 1), (1, 3), (3, 1), (3, 3)}
View full question & answer
Question 261 Mark
Given $\text{A} = \big\{1, 2, 3, 4, 5\big\}$$\text{S} = \text{(x, y)} : \text{x} \in\text{A}, \text{y} \in\text{A}.$ Find the ordered pairs which satisfy the condition given below:
X + Y > 5
Answer
Given that:
$\text{A} = \big\{1, 2, 3, 4, 5\big\}$
$\text{S} = \text{(x, y)} : \text{x} \in\text{A}, \text{y} \in\text{A}.$
X + Y > 5, So, the ordered pairs satisfying the given conditions are (4, 5), (5, 4), (5, 5).
View full question & answer
Question 271 Mark
If $f$ and $g$ are two real valued function defined as $f(x) = 2x + 1, \ g(x) = x^2 + 1,$ then find. $\frac{\text{f}}{\text{g}}$
Answer
We have$, f(x) = 2x + 1, g(x) = x^2 + 1$
$\frac{\text{f}}{\text{g}}\text{(x)}=\frac{\text{f(x)}}{\text{g(x)}}=\frac{2\text{x} + 1}{\text{x}^2 +1}$
View full question & answer
1 Marks Question - Maths STD 11 Science Questions - Vidyadip