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Maths 2 Marks Questions

Relations and Functions · CBSE STD 11 Science (CBSE - English Medium). 18 questions.

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18 questions · timed · auto-graded

Question 12 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 $\text{a}-2\text{b}=6+\text{b}$
$\Rightarrow0-2\text{b}=6+\text{b}$
$\Rightarrow\text{b}=-2$
$\therefore \ \text{a}=0, \text{b}=-2$
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Question 22 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 $f = R^+$
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Question 32 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 $x^2 - 1 = 0$
$\Rightarrow (x - 1)(x + 1) = 0$
$\Rightarrow x = -1, 1$
$\therefore$ Domain of $f = R - {-1, 1}$
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Question 42 Marks
If $f$ and $g$ are real function defined by $f(x) = x^2 + 7$ 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$, \text{f}(-2)+\text{g}(-1)=13$
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Question 52 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.
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Question 62 Marks
If $f$ and $g$ are real function defined by $f(x) = x^2 + 7$ 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$, \frac{\text{f}\text{(t)}-\text{f}(5)}{\text{t}-5},\text{t}\neq5=\text{t}+5.$
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Question 72 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, $\text{f}\Big(\frac{1}{\text{x}}\Big)=-\text{f(x)}$
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Question 82 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.
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Question 92 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 
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Question 102 Marks
If $f$ and $g$ are real function defined by $f(x) = x^2 + 7$ 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$, \text{f}\Big(\frac{1}{2}\Big)\times\text{g}(14)=\frac{1363}{4}$
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Question 112 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 $\text{f(x)}=|\text{x}-2|+|2+\text{x}|, -3\leq\text{x}\leq3.$
$\begin{cases} -(\text{x}-2)-(2+\text{x}), -3\leq\text{x}<-2\\-(\text{x}-2)-(2+\text{x}), -2\leq\text{x}<2\$\text{x}-2)+(2+\text{x}),2\leq\text{x}\leq3 \end{cases}$
$\therefore\text{f(x)}=\begin{cases}-2\text{x}-3\leq\text{x}-2\\4,-2\leq\text{x}<2\\2\text{x},2\leq\text{x}\leq3\end{cases}$
$\text{f(x)}=\begin{cases}-2\text{x}-3\leq\text{x}-2\\4,-2\leq\text{x}<2\\2\text{x},2\leq\text{x}\leq3\end{cases}$
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Question 122 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 $(\text{P} \cap \text{Q})=\big\{1, 2\big\}$
$(\text{P} \cup \text{Q}) \times (\text{P} \cap \text{Q})=\big\{0, 1, 2\big\}\times\big\{1, 2\big\}$
$=\big\{(0, 1), (0, 2), (1, 1), (1, 2), (2, 1), (2, 2)\big\}$
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Question 132 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}
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Question 142 Marks
If $R_1 = (x, y) | y = 2x + 7,$ where $\text{x} \in \text{R}$ and $– 5 \leq \times \leq 5$ is a relation. Then find the domain and Range of $R_1.$
Answer
We have, $R_1 = (x, y) | y = 2x + 7,$ where $\text{x} \in \text{R}$ and $– 5 \leq \times \leq 5$
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 $[-3,17]$
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Question 152 Marks
If $R_3= (x, |x| ) |x$ is a real number is a relation. Then find domain and range of $R_3.$
Answer
We have $, R_3 = (x, |x|) | x$ is real number
Clearly$,$ domain of $R_3 = R$
Now$, \text{x}\in\text{R}$ and $|\text{x}| \geq 0.$
Range of $R_3$ is $(0,\infty).$
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Question 162 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, $\text{f}\Big(-\frac{1}{\text{x}}\Big)=\frac{-1}{\text{f(x)}}.$
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Question 172 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 $\big\{2\text{n}\pi: \text{n}\in\text{Z}\big\}$
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Question 182 Marks
If $R_2 = (x, y) | x$ and $y$ are integers and $x^2 + y^{2 }= 64$ is a relation. Then find $R_2.$
Answer
Given that$: x^{2 }+ y^{2 }= 64, \text{y}\in\text{z}$
The sum of the squares of two integers is $64$
$\therefore$ For $x = 0, \text{y}=\pm8$
For $\text{x}=\pm8, y = 0$
Hence$, R_2{(0, 8), (0, -8), (8, 0), (-8, 0)}$
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2 Marks Questions - Maths STD 11 Science Questions - Vidyadip