Download App

MATHS 2 Marks Questions

Relations and Functions · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - English Medium). 18 questions.

Questions

2 Marks Questions

Take a timed test

18 questions · self-marked practice — reveal the answer and mark yourself.

Question 12 Marks
If f and g are real function defined by f(x) = x2 + 7 and g(x) = 3x + 5, find following:
$\text{f}\Big(\frac{1}{2}\Big)\times\text{g}(14)$
Answer
$\text{f}\Big(\frac{1}{2}\Big)\times\text{g}(14)=\Big[\Big(\frac{1}{2}\Big)^2+7\Big]\times[3\times14+5]$
$\Big(\frac{1}{4}+7\Big)\times[42+5]=\frac{29}{4}\times47=\frac{1363}{4}$
 Hence, $\text{f}\Big(\frac{1}{2}\Big)\times\text{g}(14)=\frac{1363}{4}$
View full question & answer→
Question 22 Marks
Redefine the function $\text{f(x)}=|\text{x}-2|+|2+\text{x}|, -3\leq\text{x}\leq3.$
Answer
Given that: $\text{f(x)}=|\text{x}-2|+|2+\text{x}|, -3\leq\text{x}\leq3$
since $|\text{x}-2|=-(\text{x}-2),\text{x}<2$
and $|\text{x}-2|=(\text{x}-2),\text{x}\geq2$
$|2+\text{x}|=-(2+\text{x}),\text{x}<-2$
$|2+\text{x}|=(2+\text{x}),\text{x}\geq-2$
Now $\text{f(x)}=|\text{x}-2|+|2+\text{x}|, -3\leq\text{x}\leq3.$
$\begin{cases} -(\text{x}-2)-(2+\text{x}), -3\leq\text{x}<-2\\-(\text{x}-2)-(2+\text{x}), -2\leq\text{x}<2\$\text{x}-2)+(2+\text{x}),2\leq\text{x}\leq3 \end{cases}$
$\therefore\text{f(x)}=\begin{cases}-2\text{x}-3\leq\text{x}-2\\4,-2\leq\text{x}<2\\2\text{x},2\leq\text{x}\leq3\end{cases}$
$\text{f(x)}=\begin{cases}-2\text{x}-3\leq\text{x}-2\\4,-2\leq\text{x}<2\\2\text{x},2\leq\text{x}\leq3\end{cases}$
View full question & answer→
Question 32 Marks
If R1 = {(x, y) | y = 2x + 7, where $\text{x} \in \text{R}$ and – 5 ≤ x ≤ 5} is a relation. Then find the domain and Range of R1.
Answer
We have, R1 = {(x, y) | y = 2x + 7, where $\text{x} \in \text{R}$ and – 5 ≤ x ≤ 5}
Domain of $\text{R}_1=\big\{-5\leq\text{x}\leq5, \text{x}\in\text{R}\big\}=[5, 5]$
$\text{x}\in[-5, 5]$
$\Rightarrow2\text{x}\in[-10, 10]$
$\Rightarrow2\text{x}+7\in[-3, 17]$
Range is $[-3,17]$
View full question & answer→
Question 42 Marks
If R3 = {(x, |x| ) |x is a real number} is a relation. Then find domain and range of R3.
Answer
We have, R3 = {(x, |x|) | x is real number}
Clearly, domain of R3 = R
Now, $\text{x}\in\text{R}$ and $|\text{x}| \geq 0.$
Range of R3 is $[0,\infty).$
View full question & answer→
Question 52 Marks
Find the domain following function given by:
$\text{f(x)}=\frac{1}{\sqrt{1-\cos\text{x}}}$
Answer
We have, $\text{f(x)}=\frac{1}{\sqrt{1-\cos\text{x}}}$
Now $-1\leq\cos\text{x}\leq1$
$\Rightarrow-1\leq-\cos\text{x}\leq1$
$\Rightarrow0\leq-\cos\text{x}\leq1$
$1-\cos\text{x}\neq0$
So, f(x) is defined, if $1-\cos\text{x}\neq0$
$\therefore \cos\text{x}\neq1$
$\therefore\text{x}\neq2\text{n}\pi, \text{n}\in\text{Z}$
$\therefore$ Domain of f is R $\big\{2\text{n}\pi: \text{n}\in\text{Z}\big\}$
View full question & answer→
Question 62 Marks
In following case, find a and b.
$\big(\frac{\text{a}}{4}, \text{a}-2\text{b}\big)=(0, 6+\text{b})$
Answer
$\big(\frac{\text{a}}{4}, \text{a}-2\text{b}\big)=(0, 6+\text{b})$
$\Rightarrow \frac{\text{a}}{4}=0\Rightarrow\text{a}=0$
and $\text{a}-2\text{b}=6+\text{b}$
$\Rightarrow0-2\text{b}=6+\text{b}$
$\Rightarrow\text{b}=-2$
$\therefore \ \text{a}=0, \text{b}=-2$
View full question & answer→
Question 72 Marks
Find the domain of function given by:
$\text{f}\text{(x)}=\frac{1}{\sqrt{\text{x}+|\text{x}|}}$
Answer
We have, $\text{f}\text{(x)}=\frac{1}{\sqrt{\text{x}+|\text{x}|}}$
If x > 0, x + |x| = x + x = 2x > 0
If x < 0, x + |x| = x -x = 0
Clearly, x = 0 is not possible.
$\therefore$ Domain of f = R+
View full question & answer→
Question 82 Marks
If f and g are real function defined by f(x) = x2 + 7 and g(x) = 3x + 5, find following:
f(-2) + g(-1)
Answer
$\text{f}(-2)+\text{g}(-1)=[(-2)^2+7]+[3(-1)+5]$
$=(4+7)+(-3+5)=11+2=13$
Hence, $\text{f}(-2)+\text{g}(-1)=13$
View full question & answer→
Question 92 Marks
If $\text{f(x)}=\frac{\text{x}-1}{\text{x}+1},$ then show that.
$\text{f}\Big(\frac{1}{\text{x}}\Big)=-\text{f(x)}$
Answer
Given that: $\text{f(x)}=\frac{\text{x}-1}{\text{x}+1}$
$\text{f}\Big(\frac{1}{\text{x}}\Big)=\frac{\frac{1}{\text{x}}-1}{\frac{1}{\text{x}}+1}=\frac{1-\text{x}}{1+\text{x}}=\frac{-(\text{x}-1)}{\text{x}+1}=-\text{f(x)}$
Hence, $\text{f}\Big(\frac{1}{\text{x}}\Big)=-\text{f(x)}$
View full question & answer→
Question 102 Marks
If $\text{f(x)}=\frac{\text{x}-1}{\text{x}+1},$ then show that.
$\text{f}\Big(\frac{-1}{\text{x}}\Big)=\frac{-1}{\text{f(x)}}$
Answer
Given that: $\text{f(x)}=\frac{\text{x}-1}{\text{x}+1}$
$\text{f}\Big(\frac{-1}{\text{x}}\Big)=\frac{-\frac{1}{\text{x}}-1}{-\frac{1}{\text{x}}+1}=\frac{-\big(\frac{1}{\text{x}}+1\big)}{-\big(\frac{1}{\text{x}}-1\big)}=\frac{1+\text{x}}{1-\text{x}}=\frac{1}{\frac{1-\text{x}}{1+\text{x}}}$
$=\frac{1}{-\big(\frac{\text{x}-1}{\text{x}+1}\big)}=\frac{-1}{\text{f(x)}}$
Hence, $\text{f}\Big(-\frac{1}{\text{x}}\Big)=\frac{-1}{\text{f(x)}}.$
View full question & answer→
Question 112 Marks
Find the domain of following function given by:
$\text{f(x)}=\frac{\text{x}^3-\text{x}+3}{\text{x}^2-1}$
Answer
We have, $\text{f(x)}=\frac{\text{x}^3-\text{x}+3}{\text{x}^2-1}$
f(x) is not defined, if x2 - 1 = 0
⇒ (x - 1)(x + 1) = 0
⇒ x = -1, 1
$\therefore$ Domain of f = R - {-1, 1}
View full question & answer→
Question 122 Marks
Let f and g be real functions defined by f(x) = 2x + 1 and g(x) = 4x - 7.
For what real numbers x, f(x) < g(x)?
Answer
Given that: f(x) = 2x + 1 and g(x) = 4x - 7
for f(x) < g(x), We get
2x + 1 < 4x - 7
⇒ 2x - 4x < -1 - 7 ⇒ -2x < -8 ⇒ 2x > 8
$\therefore$ x > 4
Hence, the required real number is x > 4.
View full question & answer→
Question 132 Marks
Let f and g be real functions defined by f(x) = 2x + 1 and g(x) = 4x - 7.
For what real numbers x, f(x) = g(x)?
Answer
Given that: f(x) = 2x + 1 and g(x) = 4x - 7
for f(x) = g(x), We get
2x + 1 = 4x - 7 ⇒ 2x - 4x = -7 -1
⇒ -2x = -8
⇒ x = 4. Hence, the required real number is 4.
View full question & answer→
Question 142 Marks
If f and g are real function defined by f(x) = x2 + 7 and g(x) = 3x + 5, find following:
$\frac{\text{f}\text{(t)}-\text{f}(5)}{\text{t}-5},$ if $\text{t}\neq5$
Answer
$\frac{\text{f}\text{(t)}-\text{f}(5)}{\text{t}-5},\text{t}\neq5=\frac{(\text{t}^2+7)-((5)^27)}{\text{t}-5}=\frac{(\text{t}^2+7)-(5^2+7)}{\text{t}-5}$
$=\frac{\text{t}^2+7-32}{\text{t}-5}=\frac{\text{t}^2-25}{\text{t}-25}=\text{t}+5$
$\frac{\text{t}^2+7-32}{\text{t}-5}=\frac{\text{t}^2-25}{\text{t}-5}=\frac{(\text{t}-5)(\text{t}+5)}{\text{t}-5}=\text{t}+5$
Hence, $\frac{\text{f}\text{(t)}-\text{f}(5)}{\text{t}-5},\text{t}\neq5=\text{t}+5.$
View full question & answer→
Question 152 Marks
If $\text{P}= \big\{\text{x} : \text{x} < 3, \text{x}\in\text{N}\big\},$ $\text{Q}= \big\{\text{x} : \text{x} \leq 2, \text{x}\in\text{W}\big\}.$ Find $(\text{P} \cup \text{Q}) \times (\text{P} \cap \text{Q}),$ where W is the set of whole numbers.
Answer
We have, $\text{P}= \big\{\text{x} : \text{x} < 3, \text{x}\in\text{N}\big\}=\big\{1, 2\big\}$
And $\text{Q}= \big\{\text{x} : \text{x} \leq 2, \text{x}\in\text{W}\big\}=\big\{0, 1, 2\big\}$
$(\text{P} \cup \text{Q}) =\big\{0, 1, 2\big\}$ And $(\text{P} \cap \text{Q})=\big\{1, 2\big\}$
$(\text{P} \cup \text{Q}) \times (\text{P} \cap \text{Q})=\big\{0, 1, 2\big\}\times\big\{1, 2\big\}$
$=\big\{(0, 1), (0, 2), (1, 1), (1, 2), (2, 1), (2, 2)\big\}$
View full question & answer→
Question 162 Marks
Given $\text{R}=\big\{\text{(x, y)}:\text{x, y}\in\text{W}, \text{x}^2+\text{y}^2=25\big\}.$ Find the domain and Range of R.
Answer
We have, $\text{R}=\big\{\text{(x, y)}:\text{x, y}\in\text{W}, \text{x}^2+\text{y}^2=25\big\}.$
= {(0, 5), (3, 4), (4, 3), (5,0)}
Domain of R = Set of first element of ordered pairs in R = {0, 3, 4, 5}
Range of R = Set of second element of ordered pairs in R = {5, 4, 3, 0}
View full question & answer→
Question 172 Marks
If R2 = {(x, y) | x and y are integers and x2 + y2 = 64} is a relation. Then find R2.
Answer
Given that: x2 + y2 = 64, $\text{y}\in\text{z}$
The sum of the squares of two integers is 64
$\therefore$ For x = 0, $\text{y}=\pm8$
For $\text{x}=\pm8,$ y = 0
Hence, R2 {(0, 8), (0, -8), (8, 0), (-8, 0)}
View full question & answer→
Question 182 Marks
Find the domain of following function given by:
f(x) = x |x|
Answer
 We have, f (x) = x |x|
We konw that 'x' and '|x|' are defined for all real values.
Clearly, f(x) is defined for and $\text{x}\in\text{R}.$
$\therefore$ Domain of f = R 
View full question & answer→
Generate Paper FreeAll Relations and Functions topics