Question 11 Mark
Let $\text{f(x)}=\frac{\alpha\text{x}}{\text{x}+1},\text{x}\neq-1.$ Then write the value of $\alpha$ satisfying $\text{f}(\text{f(x)})=\text{x}$ for all $\text{x}\neq-1$
AnswerWe have,
$\text{f(x)}=\frac{\text{ax}}{\text{x}+1}$
Now,
$\text{f}(\text{f(x)})=\text{x}$
$\Rightarrow\text{f}\Big(\frac{\text{ax}}{\text{x}+1}\Big)=\text{x}$
$\Rightarrow\frac{\alpha\big(\frac{\text{ax}}{\text{x}+1}\big)}{\frac{\text{ax}}{\text{x}+1}+1}=\text{x}$
$\Rightarrow\frac{\frac{\alpha^2\text{x}}{\text{x}+1}}{\frac{\alpha\text{x}+\text{x}+1}{\text{x}+1}}=\text{x}$
$\Rightarrow\frac{\text{a}^2\text{x}}{\text{ax}+\text{x}+1}=\text{x}$
$\Rightarrow\frac{\alpha^2}{\text{ax}+\text{x}+1}=1$
$\Rightarrow\alpha^2=\alpha\text{x}+\text{x}+1$
$\Rightarrow\alpha^2-\alpha\text{x}-(\text{x}+1)=0$
$\Rightarrow\alpha^2-\alpha(\text{x}+1)+\alpha-(\text{x}+1)=0$
$\Rightarrow\alpha\big[(\alpha)-(\text{x}+1)\big]+1\big[\alpha-(\text{x}+1)\big]=0$
$\Rightarrow\big[\alpha-(\text{x}+1)\big]\big[\alpha+1\big]=0$
$\Rightarrow\alpha+1=0$ $\big[\because\ \alpha=\text{x}+1$ does not satisfies $\text{f}(\text{f(x)})=\text{x}\big]$
$\Rightarrow\alpha=-1$
View full question & answer→Question 21 Mark
If f is a real function satisfying $\text{f}\Big(\text{x}+\frac{1}{\text{x}}\Big)=\text{x}^2+\frac{1}{\text{x}^2}$ for all $\text{x}\in\text{R}-\{0\},$ then write the expression for f(x).
AnswerWe have,$\text{f}\Big(\text{x}+\frac{\text{1}}{{\text{x}}}\Big) = \text{x}^2+\frac{1}{\text{x}}$
Now,
$\text{x}^2+\frac{1}{\text{x}^2} = \Big(\text{x}+\frac{1}{\text{x}}\Big)^2-2$ $\big[\because(\text{a}+\text{b})^2=\text{a}^2+\text{b}^2+2\text{ab}\big]$
$\Rightarrow\text{f}\Big(\text{x}+\frac{1}{\text{x}}\Big)=\Big(\text{x}+\frac{1}{\text{x}}\Big)^2-2$
$\Rightarrow\text{f}(\text{x})=\text{x}^2-2,\text{where } |\text{x}|\geq2$
View full question & answer→Question 31 Mark
Write the domain and range of $\text{f(x)}=\sqrt{\text{x}-[\text{x}]}$
AnswerWe have,
$\text{f(x)}=\sqrt{\text{x}-[\text{x}]}$
we know that $0\leq\text{x}-[\text{x}]<1$ for all $\text{x}\in\text{R}$
$\Rightarrow\ \text{f(x)}=\sqrt{\text{x}-[\text{x}]}$ is defined for all $\text{x}\in\text{R}$
$\therefore\ \text{Domain(f)}=\text{R}$
and Range (f) = [0, 1)
View full question & answer→Question 41 Mark
Let $A$ and $B$ be two sets such that $n(A) = p$ and $n(B) = q,$ write the number of functions from $A$ to $B.$
AnswerIt is given that $A$ and $B$ are two sets such that $n(A) = p$ and $n(B) = q.$
Now, any element of set $A,$ say $\text{a}_\text{i}(1\leq\text{i}\leq\text{p}),$ is related with an element of set $B$ in $q$ ways.Similarly, other elements of set $A$ are related with an element of set $B$ in $q$ ways.
Thus, every element of set $A$ is related with every element of set B in q ways.
$\therefore$ Total number of functions from $A$ to $B = q \times q \times q \times ...... \times q (p$ times$) = q^p$
View full question & answer→Question 51 Mark
What is the fundamental difference between a relation and a function? Is every relation a function?
AnswerFunction is a type of relation. But in a function no two ordered pairs have the same first element. For eg. $R_1$ and $R_2$ are two relations.
Clearly, $R_1$ is a function, but $R_2$ is not a function because two ordered pairs $(1, 2)$ and $(1, 4)$ have the same first element.
This means every function is a relation but every relation is not a function.
View full question & answer→Question 61 Mark
Write the domain and range of function f(x) given by $\text{f(x)}=\frac{1}{\sqrt{\text{x}-|\text{x}|}}$
AnswerWe have,
$\text{f(x)}=\frac{1}{\sqrt{\text{x}-|\text{x}|}}$
We know that,
$|\text{x}|=\begin{cases}\text{x},&\text{if x}\geq0\\-\text{x},&\text{if x}<0\end{cases}$
$\Rightarrow\text{x}-|\text{x}|=|\text{x}|=\begin{cases}\text{x}-\text{x}=0,&\text{if x }\geq0\\\text{x}+\text{x}=2\text{x},&\text{if x }<0\end{cases}$
$\Rightarrow\text{x}-|\text{x}|\leq0$ for all x
$\Rightarrow\frac{1}{\sqrt{\text{x}-|\text{x}|}}$ does not take real values for any $\text{x}\in\text{R}$
$\Rightarrow\text{f(x)}$ is not defined for any $\text{x}\in\text{R}$
Hence, domain $(\text{f})=\phi=\text{Range(f)}$
View full question & answer→Question 71 Mark
Let $f$ and $g$ be two functions given by:
$f=\{(2,4),(5,6),(8,-1),(10,-3)\} \text { and } g=\{(2,5),(7,1),(8,4),(10,13),(11,-5)\}$
Find the domain of $f+g$.
AnswerIt is given that f and g are two functions such that:
$f=\{(2,4),(5,6),(8,-1),(10,-3)\}$
and $g=\{(2,5),(7,1),(8,4),(10,13),(11,-5)\}$
Now,
Domain of $f=D_f=\{2,5,8,10\}$
Domain of $g=D_g=\{2,7,8,10,11\}$
$\therefore\text{ Domain of f}+\text{g}=\text{D}_\text{f }\cap\text{D}_\text{g}=\{2,8,10\}$
View full question & answer→Question 81 Mark
If $\text{f(x)}=1-\frac{1}{\text{x}},$ then write the value of $\text{f}\Big(\text{f}\Big(\frac{1}{\text{x}}\Big)\Big)$
AnswerWe have,
$\text{f(x)}=1-\frac{1}{\text{x}}$
Now,
$\text{f}\Big(\text{f}\Big(\frac{1}{\text{x}}\Big)\Big)=\text{f}\bigg[1-\frac{1}{\frac{1}{\text{x}}}\bigg]$
$=\text{f}\big[1-\text{x}\big]$
$=1-\frac{1}{1-\text{x}}$
$=\frac{1-\text{x}-1}{1-\text{x}}$
$=\frac{-\text{x}}{1-\text{x}}$
$\text{f}\Big(\text{f}\Big(\frac{1}{\text{x}}\Big)\Big)=\frac{-\text{x}}{1-\text{x}}$
View full question & answer→Question 91 Mark
It is given that the functions $f(x) = 3x^2 - 1$ and $g(x) = 3 + x$ are equal.
Answer$\therefore f(x) = g(x)$
$\Rightarrow 3x^2 - 1 = 3 + x$
$\Rightarrow 3x^2 - x - 4 = 0$
$\Rightarrow (x + 1)(3x - 4) = 0$
$\Rightarrow x + 1 = 0$ or $3x - 4 = 0$
$\Rightarrow x = -1$ or $\text{x}=\frac{4}{3}$
View full question & answer→Question 101 Mark
Write the range of the function $\text{f(x)}=\text{e}^{\text{x}-[\text{x}]},\text{x}\in\text{R}$
AnswerWe have,$\text{f}(\text{x})=\text{e}^{\text{x}-[\text{x}]},\text{x}\in\text{R}$
we know that $0\leq\text{x}-[\text{x}]<1$ for all $\text{x}\in\text{R}$
$\therefore$ Range $\big(\text{e}^{\text{x}-[\text{x}]}\big)=[1,\text{e)}$
View full question & answer→Question 111 Mark
If f, g, h are three function defined from R to R as follows:
$\text{g(x)}=\sin\text{x}$
AnswerWe have,
$\text{g(x)}=\sin\text{x}$
Range of $\text{g(x)}=\{\text{x }\in\text{R}:-1\leq\text{x}\leq1\}$
View full question & answer→Question 121 Mark
If $\text{f(x)}=4\text{x}-\text{x}^2,\text{ x}\in\text{R},$ then write the value of $f(a + 1) - f(a - 1).$
AnswerWe have,
$f(x) = 4x - x^2$^
Now, $f(a + 1) = 4(a + 1) - (a + 1)^2$
$= 4a + 4 - a^2 - 1 - 2a$
$= -a^2+ 3 + 2a$
$\Rightarrow f(a + 1) = -a^2 + 2a + 3 ...(i)$
and, $f(a - 1) = 4(a - 1) - (a - 1)^2$
$= 4a - 4 - (a^2 + 1 - 2a)$
$= 4a - 4 - a^2 - 1 + 2a$
$= 6a - a - 5$
$f(a - 1) = -a^2 + 6a - 5 ....(ii)$
Subtracting equation (ii) from equation (i), we get
$f(a + 1) - f(a - 1)$
$= -a^2 + 2a + 3 - (-a^2 + 6a - 5)$
$= -a^2 + 2a + 3 + a^2 - 6a + 5$
$= -4a + 8 = 4(2 - a)$
View full question & answer→Question 131 Mark
Write the range of the real function f(x) = |x|.
AnswerRange $(|\text{x}|)=[0,\infty]$
View full question & answer→Question 141 Mark
Write the range of the function $\text{f(x)}=\cos[\text{x}],$ where $\frac{-\pi}{2}<\text{x}<\frac{\pi}{2}$
AnswerWe have,$\text{f}(\text{x})=\cos[\text{x}], \text{where}\frac{-\pi}{2}<\text{x}<\frac{\pi}{2}$
clearly,Range $(\cos[\text{x}])= \{1,\cos1,\cos2\}$
View full question & answer→Question 151 Mark
Define a function as a set of ordered pairs.
AnswerFunction: Let $A$ and $B$ be two non$-$empty sets. A relation $A$ to $B$
i.e., a sub$-$set of $A \times B,$ is called a function $($or a mapping a map$)$ from $A$ to $B,$ if
- For each $\text{a}\in\text{A}$ there exists $\text{b}\in\text{B}$ such that $(\text{a},\text{b})\in\text{f}$
- $(\text{a},\text{b})\in\text{f}$ and $(\text{a},\text{c})\in\text{f}$
$\Rightarrow\text{ b} =\text{c}$
If $(\text{a},\text{b})\in\text{f},$ then $'b\ '$ is called the image of $'a\ '$ under $f.$
If a function is expresed as the set of ordered pairs, the domain $f$ is the set of all first components of members of $f$ and the range of $f$ is the set of second components of members of $f.$ View full question & answer→Question 161 Mark
If $f, g, h$ are real functions given by $f(x) = x^2, \text{g(x)}=\tan\text{x}$ and $\text{h(x)}=\log_\text{e}\text{x},$ then write the value of (hogof) $\Big(\sqrt{\frac{\pi}{4}}\Big)$
AnswerWe have,
$\text{f(x)}=\text{x}^2,\text{ g(x)}=\tan\text{x}$ and $\text{h(x)}=\log_\text{e}\text{x}$
Now,
$(\text{hogof})\Big(\sqrt{\frac{\pi}{4}}\Big)=\text{h}\big[\text{g(f)}\big]\Big(\sqrt{\frac{\pi}{4}}\Big)$
$=\text{h}\bigg[\text{g}\Big(\sqrt{\frac{\pi}{4}}\Big)^2\bigg]$
$=\text{h}\Big[\text{g}\Big(\frac{\pi}{4}\Big)\Big]$
$=\text{h}\Big[\tan\frac{\pi}{4}\Big]$
$=\text{h}(1)$
$=\log_\text{e}1$
$=0$
View full question & answer→Question 171 Mark
Write the domain and range of function f(x) given by $\text{f(x)}=\sqrt{[\text{x}]-\text{x}}$
AnswerWe know,
$\text{f(x)}=\sqrt{[\text{x}]-\text{x}}$
We know that $-1<[\text{x}]<-\text{x}\leq0$ for all $\text{x}\in\text{R}$
$\Rightarrow\sqrt{[\text{x}]-\text{x}}$ is not defined for all $\text{x}\in\text{R}$
$\therefore\ \text{Domain(f)}=\phi$
and $\text{Range(f)}=\phi$
View full question & answer→Question 181 Mark
Write the range of the function $\text{f(x)}=\sin[\text{x}],$ where $\frac{-\pi}{4}\leq\text{x}\leq\frac{\pi}{4}$
AnswerWe have,
$\text{f}(\text{x})=\sin[\text{x}], $ where $\frac{-\pi}{4}\leq\text{x}\leq\frac{\pi}{4}$
Now,
$[\text{x}]= \begin{cases}0, & \text{x} = \frac{\pi}{4}\\-1, &\text{ x} = \frac{-\pi}{4}\end{cases}$
$\therefore \sin\text{{x}} = \begin{cases}0, & \text{x} = \frac{\pi}{4}\\-\sin1, &\text{ x }=\frac{-\pi}{4}\end{cases}$
$\therefore$ Range $(\sin[\text{x}])= \{{-\sin 1,0,\sin1}\}$
View full question & answer→Question 191 Mark
Let $f$ and $g$ be two real functions given by:
$f=\{(0,1),(2,0),(3,-4),(4,2),(5,1)\} \text { and } g=\{(1,0),(2,2),(3,-1),(4,4),(5,3)\}$
Find the domain of fg .
AnswerIt is given that f and g are two real functions such that,
$f=\{(0,1),(2,0),(3,-4),(4,2),(5,1)\}$
and $g=\{(1,0),(2,2),(3,-1),(4,4),(5,3)\}$
Now,
Domain of $f=D_f=\{0,2,3,4,5\}$
Domain of $\mathrm{g}=\mathrm{D}_{\mathrm{g}}=\{1,2,3,4,5\}$
$\therefore$ Domain of $\mathrm{f}+\mathrm{g}=\mathrm{D}_{\mathrm{f}} \cap \mathrm{D}_{\mathrm{g}}=\{2,3,4,5\}$
View full question & answer→Question 201 Mark
Write the domain and range of the function $\text{f(x)}=\frac{\text{x}-2}{2-\text{x}}$
AnswerWe have,
$\text{f(x)}=\frac{\text{x}-2}{2-\text{x}}$
Domain of f, Clearly, f(x) is defined for all $\text{x}\in\text{R}$ except for which,
$2-\text{x}\neq0\text{ i.e., x}\neq2$
Hence, domain (f) = R - {2}
Range of f, Let f(x) = y
$\Rightarrow\ \frac{\text{x}-2}{2-\text{x}}=\text{y}$
$\Rightarrow\ \frac{-1(2-\text{x})}{2-\text{x}}=\text{y}$
$\Rightarrow-1=\text{y}$
$\Rightarrow\text{y}=-1$
$\therefore\ \text{Range(f)}=\{-1\}$
View full question & answer→Question 211 Mark
If $\text{f(x)}=\cos\big[\pi^2\big]\text{x}+\cos\big[-\pi^2\big]\text{x},$ where[x] denotes the greatest integer less than or equal to x, then write the value of $\text{f}(\pi)$
AnswerWe have,
$\text{f}(\text{x})=\cos[\pi^2]\text{x}+\cos[-\pi^2]\text{x}$
$\therefore \text{f}(\pi)= \cos[\pi^2]\pi+\cos[-\pi^2]\pi$
$=0+0$ $[ \because\cos\text{n}\pi=0]$
$=0$
$\therefore\text{ f}(\pi)= 0$
View full question & answer→