Download App

MATHS (Each question 1 marks)

Functions · Gujarat Board (GSEB) STD 11 Science (GSEB - English Medium). 21 questions.

Questions

(Each question 1 marks)

Take a timed test

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

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 $\text{x}\neq-1$
Answer
We 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).
Answer
We 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 $\text{f(x)}=\sqrt{\text{x}-[\text{x}]}$
Answer
We 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.
Answer
It 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 × q × q × ...... × 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?
Answer
Function 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 $\text{f(x)}=\frac{1}{\sqrt{\text{x}-|\text{x}|}}$
Answer
We 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)} and g = {(2, 5), (7, 1), (8, 4), (10, 13), (11, -5)} Find the domain of f + g.
Answer
It 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 $\text{f}\Big(\text{f}\Big(\frac{1}{\text{x}}\Big)\Big)$
Answer
We 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
$\therefore f(x)=g(x) \Rightarrow 3 x^2-1=3+x \Rightarrow 3 x^2-x-4=0 \Rightarrow(x+1)(3 x-4)=0 \Rightarrow x+1=0$ or $3 x-4=0 \Rightarrow x=-1$ or $x=\frac{4}{3}$
View full question & answer→
Question 101 Mark
Write the range of the function $\text{f(x)}=\text{e}^{\text{x}-[\text{x}]},\text{x}\in\text{R}$
Answer
We 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: $\text{g(x)}=\sin\text{x}$
Answer
We 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 $f(a + 1) - f(a - 1)$.
Answer
We have, $f(x)=4 x-x^2$ Now, $f(a+1)=4(a+1)-(a+1)^2=4 a+4-a^2-1-2 a=-a^2+3+2 a \Rightarrow f(a+1)=-a^2+2 a+$ $3 \ldots(\mathrm{i})$ and, $f(a-1)=4(a-1)-(a-1)^2=4 a-4-\left(a^2+1-2 a\right)=4 a-4-a^2-1+2 a=6 a-a-5 f(a-1)=-a^2+6 a-5 \ldots$. (ii) Subtracting equation (ii) from equation (i), we get $f(a+1)-f(a-1)=-a^2+2 a+3-\left(-a^2+6 a-5\right)=-a^2+2 a+3+$ $a^2-6 a+5=-4 a+8=4(2-a)$
View full question & answer→
Question 141 Mark
Write the range of the function $\text{f(x)}=\cos[\text{x}],$ where $\frac{-\pi}{2}<\text{x}<\frac{\pi}{2}$
Answer
We 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.
Answer
Function: Let A and B be two non-empty sets. A relation A to B i.e., a sub-set of A × B, is called a function (or a mapping a map) from A to B, if
  1. For each $\text{a}\in\text{A}$ there exists $\text{b}\in\text{B}$ such that $(\text{a},\text{b})\in\text{f}$
  2. $(\text{a},\text{b})\in\text{f}$ 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) $\Big(\sqrt{\frac{\pi}{4}}\Big)$
Answer
We 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 $\text{f(x)}=\sqrt{[\text{x}]-\text{x}}$
Answer
We 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 $\frac{-\pi}{4}\leq\text{x}\leq\frac{\pi}{4}$
Answer
We 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)} and g = {(1, 0), (2, 2), (3, -1), (4, 4), (5, 3)} Find the domain of fg.
Answer
It 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 $g=D_g=\{1,2,3,4,5\}$
$\therefore\text{ Domain of f}+\text{g}=\text{D}_\text{f }\cap\text{D}_\text{g}=\{2,3,4,5\}$
View full question & answer→
Question 201 Mark
Write the domain and range of the function $\text{f(x)}=\frac{\text{x}-2}{2-\text{x}}$
Answer
We 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 $\text{f}(\pi)$
Answer
We 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→
Generate Paper FreeAll Functions topics