, where 
is ![[-\infty, 0]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-94a61bc9af7b9f196d46b44726815c11_l3.png "Rendered by QuickLaTeX.com")
Solution: Option(D) is correct. $f(x)=a x$, where $a>0$ Case 1 : When $x<0$, then ax lies between $(0,1)$ Case 2 : When $x \geq 0$, then $a x \geq 1$ Union of above two cases, gives us the...
. Then, range 
![[-1,1]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-bd9ec9b7fdf72f6f954becf26cf7129a_l3.png "Rendered by QuickLaTeX.com")
![(0,1]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3243e3012721250a044fef79f51a0c2f_l3.png "Rendered by QuickLaTeX.com")
Solution: Option (B) is correct. $\mathrm{f}(\mathrm{x})=\frac{x^{2}}{\left(1+x^{2}\right)}$ The range of $f(x)$ can be found out by putting $f(x)=y$ $\begin{array}{l}...
. Then, range 
![(-\infty, 1]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2eb654b4016c2cc85fa136277a9d7f78_l3.png "Rendered by QuickLaTeX.com")
Solution: Option(B) is correct. $f(x)=\frac{1}{\left(1-x^{2}\right)}$ The range of $f(x)$ can be found out by putting $f(x)=y$ $\begin{array}{l} \mathrm{y}=\frac{1}{\left(1-x^{2}\right)} \\...
. Then, 
![(-\infty,-1]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-17e48fe12e81cf52057d1a1ecdfc8269_l3.png "Rendered by QuickLaTeX.com")
Solution: Option (B) is correct. $f(x)=\log (1-x)+\sqrt{x^{2}-1}$ Solving inequality, $\log (1-x) \geq 0$ $\Rightarrow 1-x \geq \mathrm{e}^{0} \quad \begin{array}{c}\text { (Log taken to the...
. Then, dom (f) and range (f) are respectively 
and 
and and Solution: Option(A) is correct. $f(x)=x^{3}$ $f(x)$ can assume any value, therefore domain of $f(x)$ is $R$ Range of the function can be positive or negative Real numbers, as the cube of any number...
. Then, dom (f) and range (f) are respectively. and and and and 
Solution: Option(C) is correct. $f(x)=x_{2}$ $f(x)$ can assume any value, therefore domain of $f(x)$ is $R$ Range of the function can only be positive Real numbers, as the square of any number is...
. Then, 
![[1,2]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ab425e31cf7af5fb3d46646ff538fdcd_l3.png "Rendered by QuickLaTeX.com")
Solution: Option(C) is correct. $f(x)=\sqrt{\log \left(2 x-x^{2}\right)}$ For $f(x)$ to be defined $2 x-x^2$ should be positive. Solving inequality, (Log taken to the opposite side of the equation...
. Then, ![\left[0, \frac{\pi}{2}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-cdf7f72b896a3c8ef3745557ca17a3c9_l3.png "Rendered by QuickLaTeX.com")
![\left[\frac{3 \pi}{2}, 2 \pi\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-aa7b802c7af63a3c3f09db8b0292cb35_l3.png "Rendered by QuickLaTeX.com")
![\left[0, \frac{\pi}{2}\right] \cup\left[\frac{3 \pi}{2}, 2 \pi\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-be8b8bc855076ca9e77d0af23204c861_l3.png "Rendered by QuickLaTeX.com")
Solution: Option(C) is correct. $\mathrm{f}(\mathrm{x})=\sqrt{\cos x}$ As per graph of $\sqrt{\cos x}$ the domain is $\left[0, \frac{\pi}{2}\right] \cup\left[\frac{3 \pi}{2}, 2 \pi\right]$
. Then, 
? 
![\left[0, \frac{2}{3}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b52292076dc04c74fe44cded15544d8b_l3.png "Rendered by QuickLaTeX.com")
![\left[\frac{-2}{3}, \frac{2}{3}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c6d1d6173855bcc8e2e22c311f052282_l3.png "Rendered by QuickLaTeX.com")
Solution: Option(B) is correct. $\begin{array}{l} \mathrm{f}(\mathrm{x})=\cos ^{-1}(3 \mathrm{x}-1) \end{array}$ Domain for function $\cos ^{-1} \mathrm{x}$ is $[-1,1]$ and range is $[0, \pi]$ When...
. Then, ![\left[\frac{-1}{2}, \frac{1}{2}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2b2127e7c856ce7280a6fc7257d27da7_l3.png "Rendered by QuickLaTeX.com")
![\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d03f7933dddf8dbceee8c92c85707a25_l3.png "Rendered by QuickLaTeX.com")
![\left[\frac{-\pi}{4}, \frac{\pi}{4}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-36caf65bb0d2f985fd702c950c7c7653_l3.png "Rendered by QuickLaTeX.com")
Solution: Option(B) is correct. $f(x)=\cos ^{-1}2 x$ Domain for function $\cos ^1 \mathrm{x}$ is $[-1,1]$ and range is $[0, \pi]$ When a function is multiplied by an integer, the domain of the...
Then, 
![[-1,1]-\{0\}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0d140ee7fc9eafbc15a3b47a6ad32f55_l3.png "Rendered by QuickLaTeX.com")
Solution: Option(B) is correct. $f(x)=\frac{\sin ^{-1} x}{x}$ The domain of the function is defined for $\mathrm{x} \neq 0$ domain of $\sin ^{-1} x$ is $[-1,1]$ So, domain of...
. Then, 
? 
Solution: Option(D) is correct. $\mathrm{f}(\mathrm{x})=\frac{x}{\left(x^{2}-1\right)}$ The domain of the function is defined for $\begin{array}{l} \mathrm{x}^{2}-1 \neq 0 \\ \Rightarrow \mathrm{x}...
. Then, ? 
![(-\infty,-1] \cup(1, \infty)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f70651ff938d8644c2a7b1e4ab8dcdbc_l3.png "Rendered by QuickLaTeX.com")
Solution: Option(C) is correct. $\mathrm{f}(\mathrm{x})=e^{\sqrt{x^{2}-1}} \cdot \log (x-1)$ The domain of the function is defined for $\begin{array}{l} \mathrm{x}-1>0 \quad \text { and }...
. Then, dom (f) – ? 
![[1,4]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b466cd4c540c856fc9d44d24b88d173d_l3.png "Rendered by QuickLaTeX.com")
![(-\infty, 4]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7342c3266939896b38339bca55408448_l3.png "Rendered by QuickLaTeX.com")
![(-\infty, 1] \cup(4, \infty)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3cacb249d82a72cc2e90b22ea1598fb5_l3.png "Rendered by QuickLaTeX.com")
Solution: Option(D) is correct. $\mathrm{f}(\mathrm{x})=\sqrt{\frac{x-1}{x+4}}$ The domain of the function can be defined for $\sqrt{\frac{x-1}{x+4}} \geq 0$ $\begin{array}{l} \Rightarrow...
. Then, dom ![[-3,3]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9cbb1b08c8e67cf43d5548e7d359de56_l3.png "Rendered by QuickLaTeX.com")
![[-\infty,-3]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-697587bce5e8d6c46d3797bdabaae044_l3.png "Rendered by QuickLaTeX.com")
![(-\infty,-3] \cup(4, \infty)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b4747a453581bf4e2adb77e958e00ac9_l3.png "Rendered by QuickLaTeX.com")
Solution: Option(A) is correct. $\mathrm{f}(\mathrm{x})=\sqrt{9-x^{2}}$ The domain of the function can be defined for $\sqrt{9-x^{2}} \geq 0$ $\begin{array}{l} \Rightarrow \sqrt{9-x^{2}} \geq 0 \\...
Solution: Option(B) is correct. $\begin{array}{l} \mathrm{f}=\{(1,2),(3,5),(4,1)\} \\ \mathrm{g}=\{(2,3),(5,1),(1,3)\} \end{array}$ According to the combination of $\mathrm{f}$ and $\mathrm{g}$,...
and 
then 
Solution: Option(A) is correct. $\begin{array}{l} f(x)=x ^2 \\ g(x)=\tan x \\ h(x)=\log x \end{array}$ According to the combination of $f, g$ and $h$,...
and 
then 
Solution: Option(B) is correct. $f(x)=8 x^{3}$ $g(x)=x^{1 / 3}$ According to the combination of $\mathrm{f}$ and $\mathrm{g}$, $\operatorname{gof}(\mathrm{x})=\mathrm{f}(\mathrm{f}(\mathrm{x}))$...
then 
Solution: Option(D) is correct. $f(x)=x^2-3 x+2$ According to the combination of $\mathrm{f}$ and $\mathrm{f}$, $\operatorname{fof}(x)=f(f(x))$ Therefore, fof $(x)=f(f(x))$ $\begin{array}{l}...
![f(x)=\sqrt[3]{3-x^{3}}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6c33645c75bd40d87b10258556b03f43_l3.png "Rendered by QuickLaTeX.com")
then (f of) 
Solution: Option(B) is correct. $f(x) \sqrt[3]{3-x^{3}}$ According to the combination of $f$ and $f$, $\text { fof }(x)=f(f(x))$ Therefore, fof $(x)=f(f(x))$...
then (f of of) 
Solution: Option(C) is correct. $\mathrm{f}(\mathrm{x})=\frac{1}{(1-x)}$ According to the combination of $\mathrm{f}$ and $\mathrm{f}$, fofof $(x)=f(f(f(x)))$ Therefore, fof $(x)=f(f(f(x))$...
then 
Solution: Option(C) is correct. $\begin{array}{l} \mathrm{f}\left(\mathrm{x}+\frac{1}{x}\right)=\left(\mathrm{x} ^2+\frac{1}{x^{2}}\right) \\ \Rightarrow...
and 
then 
Solution: Option(C) is correct. $\begin{array}{l} \mathrm{f}(\mathrm{x})=(\mathrm{x}^2-1) \\ \mathrm{g}(\mathrm{x})=(2 \mathrm{x}+3) \end{array}$ According to the combination of $\mathrm{f}$ and...
then 
Solution: Option(A) is correct. $\mathrm{f}(\mathrm{x})=\frac{(4 x+3)}{(6 x-4)}, \mathrm{x} \neq \frac{2}{3}$ According to the combination of $\mathrm{f}$ and $\mathrm{f}$,...
(y) = ? 
Solution: Option(B) is correct. $\mathrm{f}: \mathrm{N} \rightarrow \mathrm{X}: \mathrm{f}(\mathrm{x})=4 \mathrm{x} 2+12 \mathrm{x}+15$ We need to find $\mathrm{f}-1$, Suppose...
. Then 
Solution: Option(A) is correct. $\text { f: } R-\left\{-\frac{4}{3}\right\} \rightarrow-\left\{\frac{4}{3}\right\}: f(x)=\frac{4 x}{(3 x+4)}$ We need to find $\mathrm{f}-1$ Suppose $f(x)=y$...
(y) = ? 
Solution: Option(C) is correct. $\mathrm{f}: \mathrm{Q} \rightarrow \mathrm{Q}: \mathrm{f}(\mathrm{x})=(2 \mathrm{x}+3)$ We need to find $\mathrm{f}-1$ Suppose $\mathrm{f}(\mathrm{x})=\mathrm{y}$...
Solution: Option(B) is correct. One-One Function Suppose $\mathrm{p}_{1}, \mathrm{p}_{2}, \mathrm{q}_{1}, \mathrm{q}_{2}$ be two arbitrary elements in $\mathrm{R}$ + Therefore, $f\left(p_{1},...
Then, 
is Solution: Option(D) is correct. $f:\mathrm{N} \rightarrow \mathrm{N}: \mathrm{f}(\mathrm{x})=$ $f: N \rightarrow N: f(x)=\left\{\begin{array}{l}\frac{1}{2}(n+1) \text {, when } n \text { is odd } \\...
and 
. Then 
is Solution: Option(B) is correct. f: $\mathrm{A} \rightarrow \mathrm{B}: \mathrm{f}(\mathrm{x})=\frac{(x-2)}{(x-3)}$ Where, $\mathrm{A}=\mathrm{R}-\{3\}$ and $\mathrm{B}=\mathrm{R}-\{1\}$ One-One...
Solution: Option() is correct. One-One function Suppose $\mathrm{p}, \mathrm{q}$ be two arbitrary elements in $\mathrm{R}+$ Therefore, $f(p)=f(q)$ $\begin{array}{l}...
Solution: Option (C) is correct. One-one function $\cos x$ graph cuts y axis repeatedly, hence it is many-one. Onto function Range of $f(x)$ is $[-1,1]$ Co-domain is $\mathrm{R}$ So here, Range of...
is Solution: Option(D) is correct. One-One function Suppose $\mathrm{p}, \mathrm{q}$ be two arbitrary elements in $\mathrm{R}+$ Therefore, $f(p)=f(q)$ $\begin{array}{l} \Rightarrow...
is Solution: Option(B) is correct. One-One function Suppose $\mathrm{p}, \mathrm{q}$ be two arbitrary elements in $\mathrm{R}$ Therefore, $f(p)=f(q)$ $\begin{array}{l} \Rightarrow...
is Solution: Option(D) is correct. One-One function Suppose $\mathrm{p}, \mathrm{q}$ be two arbitrary elements in $\mathrm{R}$ Therefore, $f(p)=f(q)$ $\begin{array}{l} \Rightarrow...
+ x + 1 is Solution: Option (B) is correct. One-One function Suppose $\mathrm{p}, \mathrm{q}$ be two arbitrary elements in $\mathrm{N}$ Therefore, $f(p)=f(q)$ $\begin{array}{l} \Rightarrow...
Solution: Option (B) is correct. One-One function Suppose $\mathrm{p}, \mathrm{q}$ be two arbitrary elements in $\mathrm{N}$ Therefore, $f(p)=f(q)$ $\begin{array}{l} \Rightarrow 2 \mathrm{p}=2...
Solution: A function is stated as the relation between the two sets, where there is exactly one element in set B, for every element of set A. A function is represented as f: A → B, which means ‘f’...
Answers: (i) Now we have, f(x) is defined for all real numbers x. Domain (f) = R If x < 0, –|x| < 0 f (x) < 0 If x ≥ 0, –x ≤ 0. –|x| ≤ 0 ⇒ f (x) ≤ 0 ∴ f (x) ≤ 0 or f (x) ∈ (–∞, 0] Range (f)...
Answers: (i) f(x) is defined for all real values of x, except when 2 – x = 0 or x = 2. Domain (f) = R – {2} f (x) = (x-2)/(2-x) f (x) = -(2-x)/(2-x) = –1 If x ≠ 2, f(x) = –1 ∴ Range (f) = {–1} (ii)...
Answers: (i) Square of a real number is not negative. f(x) takes real values only when x – 1 ≥ 0 x ≥ 1 ∴ x ∈ [1, ∞) Domain (f) = [1, ∞) If x ≥ 1 x – 1 ≥ 0 √(x-1) ≥ 0 ⇒ f (x) ≥ 0 f(x) ∈ [0, ∞) ∴...
Answers: (i) f(x) is defined for all real values of x, except when bx – a = 0 or x = a/b. Domain (f) = R – (a/b) Consider, f (x) = y (ax+b)/(bx-a) = y ax + b = y(bx – a) ax + b = bxy – ay ax – bxy...
Answers: (i) Square of a real number is not negative. f (x) takes real values only when 9 – x2 ≥ 0 9 ≥ x2 x2 ≤ 9 x2 – 9 ≤ 0 x2 – 32 ≤ 0 (x + 3)(x – 3) ≤ 0 x ≥ –3 and x ≤ 3 x ∈ [–3, 3] ∴ Domain (f) =...
Answers: (i) Square of a real number is not negative. f (x) takes real values only when x – 2 ≥ 0 x ≥ 2 ∴ x ∈ [2, ∞) ∴ Domain (f) = [2, ∞) (ii) Square of a real number is not negative. f (x) takes...
Answer: f(x) is defined for all real values of x, except when x2 – 8x + 12 = 0. x2 – 8x + 12 = 0 x2 – 2x – 6x + 12 = 0 x(x – 2) – 6(x – 2) = 0 (x – 2)(x – 6) = 0 x – 2 = 0 or x – 6 = 0 x = 2 or 6 ∴...
Answers: (i) f(x) is defined for all real values of x, except when x + 1 = 0 or x = –1. ∴ Domain of f = R – {–1} (ii) f (x) is defined for all real values of x, except when x2 – 9 = 0. x2 – 9 = 0...
Answers: (i) f (x) is defined for all real values of x, except when x = 0. ∴ Domain of f = R – {0} (ii) f (x) is defined for all real values of x, except when x – 7 = 0 or x = 7. ∴ Domain of f = R –...
Answer: f: R → R ∈ f(x) = x2 g : R → C ∈ g(x) = x2 Both the functions are equal only when the domain and codomain of both the functions are equal. So, the domain of f ≠ domain of g. ∴ f and g are...
Answer: If x = 0, 0 + y = 3 y = 3 If x = 1, 1 + y = 3 y = 2 If x = 2, 2 + y = 3 y = 1 If x = 3, 3 + y = 3 y = 0 ∴ R = {(0, 3), (1, 2), (2, 1), (3, 0)} Hence, R is a...
Answers: (i) If x = 1, y = 3(1) = 3 If x = 2, y = 3(2) = 6 If x = 3, y = 3(3) = 9 ∴ R = {(1, 3), (2, 6), (3, 9)} Hence, R is a function. (ii) If x = 1, y > 1 + 1 or y > 2 y = {4, 6} If x = 2,...
Answer: Given, f (x) = loge x ⇒ f (y) = loge y Consider, f (xy) F (xy) = loge (xy) f (xy) = loge (x × y) [since, logb (a×c) = logb a + logb c] f (xy) = loge x + loge y ∴ f (xy) = f (x) + f...
Answers: (i) Domain of f = R+ Value of logarithm to the base e (natural logarithm) can take all possible real values. ∴ The image set of f = R (ii) f(x) = –2 loge x = –2 ∴ x = e-2 [logb a = c ⇒ a =...
Answer: f(y) = –1 y2 = –1 Domain of f - R, Value of y2 i- non-negative There is no real y for which y2 = –1. ∴{y: f(y) = –1} = ∅
Answers: (i) Domain of f = R (set of real numbers) Square of a real number is always positive or zero. ∴ range of f = R+∪ {0} (ii) f(x) = 4 x2 = 4 x2 – 4 = 0 (x – 2)(x + 2) = 0 ∴ x = ± 2 ∴ {x: f(x)...
Find: f (1), f (–1), f (0), f (2). Answers: If, x > 0, f (x) = 4x + 1 Substitute, x = 1 f (1) = 4(1) + 1 f (1) = 4 + 1 f (1) = 5 If x < 0, f(x) = 3x – 2 Substitute, x = –1 f (–1) =...
Answers: A = {–2, –1, 0, 1, 2} f : A → Z f(x) = x2 – 2x – 3 (i) A - Domain of the function f. The range is the set of elements f(x) for all x ∈ A. Substitute, x = –2 in f(x) f(–2) = (–2)2 – 2(–2) –...
Consider ‘f’ as a function and R as a relation defined from set X to set Y. Domain of the relation R might be a subset of the set X, but the domain of the function f must be equal to X. This is...
Function as a correspondence between two sets: Let A and B be two non-empty sets. Then a function ‘f’ from set A to B is a rule or method or correspondence which associates elements of set A to...
Find: f (√-3) Answer: f (√-3) √-3 is not a real number [Function f(x) is defined only when x ∈ R] ∴ f (√-3) do not exist.
Find: (i) f (1) (ii) f (√3) Answers: (i) If x ≥ 1, f (x) = 1/x f (1) = 1/1 ∴ f(1) = 1 (ii) √3 = 1.732 > 1 If x ≥ 1, f (x) = 1/x ∴ f (√3) = 1/√3...
Find: (i) f (1/2) (ii) f (-2) Answers: (i) If 0 ≤ x ≤ 1, f(x) = x ∴ f (1/2) = ½ (ii) If x < 0, f(x) = x2 f (–2) = (–2)2 ∴ f (–2) = 4
Answer: f [f (x)] = f [(x+1)/(x-1)] f [f (x)] = [(x+1)/(x-1) + 1] / [(x+1)/(x-1) – 1] f [f (x)] = [[(x+1) + (x-1)]/(x-1)] / [[(x+1) – (x-1)]/(x-1)] f [f (x)] = [(x+1) + (x-1)] / [(x+1) – (x-1)] f [f...
Answer: f {f (x)} = f {1/(1 – x)} f {f (x)} = 1 / 1 – (1/(1 – x)) f {f (x)} = 1 / [(1 – x – 1)/(1 – x)] f {f (x)} = 1 / (-x/(1 – x)) f {f (x)} = (1 – x) / -x f {f (x)} = (x – 1) / x ∴ f {f (x)} = (x...
Answer: Consider, y = (ax – b) / (bx – a) [Cross multiply] y(bx – a) = ax – b bxy – ay = ax – b bxy – ax = ay – b x(by – a) = ay – b x = (ay – b) / (by – a) = f (y) ∴ x = f (y) Thus,...
Answer: f (a + b) = (a + b – a)2 (a + b – b)2 f (a + b) = (b)2 (a)2 ∴ f (a + b) = a2b2
Answer: f(x) = x2 – 3x + 4. Consider, f (x) = f (2x + 1). Let us take, f (2x + 1) = (2x + 1)2 – 3(2x + 1) + 4 f (2x + 1) = (2x) 2 + 2(2x) (1) + 12 – 6x – 3 + 4 f (2x + 1) = 4x2 + 4x + 1 – 6x + 1 f...
Function as a set of ordered pairs: Let A and B be two non-empty sets. A relation from A to B, i.e., a subset of A×B, is called a function (or a mapping) from A to B. 1. if for each a ∈ A there...
Solution: (iii) f/g We know that $ \left( f/g \right)\text{ }\left( x \right)\text{ }=\text{ }f\left( x \right)/g\left( x \right)~ $ $ \left( f/g \right)\text{ }\left( x \right)\text{ }=\text{...
Solution: It is given that f(x) = loge (1 – x) and g(x) = [x] We know, f(x) will have real values only when 1 – x > 0 Or, 1 > x x < 1, Therefore, x ∈ (–∞, 1) Domain of f = (–∞, 1)...
Solution: (vii) f2 + 7f We know that: $ \left( {{f}^{2}}~+\text{ }7f \right)\text{ }\left( x \right)\text{ }=\text{ }{{f}^{2}}\left( x \right)\text{ }+\text{ }\left( 7f \right)\left( x \right) $ $...
Solutions: (v) g/f We know, (g/f) (x) = g(x)/f(x) (g/f) (x) = √(9-x2) / √(x+1) = √[(9-x2) / (x+1)] Domain of g/f = Domain of f ∩ Domain of g = [–1, ∞) ∩ [–3, 3] = [–1, 3] However, (g/f) (x) is...
(iii) fg We know that: $ \left( fg \right)\left( x \right)=f\left( x \right)g\left( x \right) $ $ \left( fg \right)\left( x \right)=\surd \left( x+1 \right)\times \surd \left( 9-{{x}^{2}} \right) $...
Solution: We are given that f(x) = √(x+1) and g(x) = √(9-x2) It is known that the square of a real number can never be negative. So, f(x) will have real values only when x + 1 ≥ 0 x ≥ –1, x ∈ [–1,...
Solution: (iii) fg We know that: (fg)(x) = f(x)g(x) $ \left( fg \right)\left( x \right)=\text{ }\left( 2x\text{ }+\text{ }5 \right)\left( {{x}^{2}}~+\text{ }x \right) $ $ \left( fg \right)\left( x...
Solution: It is given that f(x) = 2x + 5 and g(x) = x2 + x In this case, both f(x) and g(x) are defined for all x ∈ R. So, the domain of f is equal to the domain of g which is equal to R (i) f + g...
Solution: (ii) f (x) = √(x-1) and g (x) = √(x+1) We have f(x): [1, ∞) → R+ and g(x): [–1, ∞) → R+ as real square root is defined only for non-negative numbers. (a) f + g We know, (f + g) (x) = f(x)...
Solution: (i) We are given that : f (x) = x3 + 1 and g(x) = x + 1 Relation is defined such that f(x): R → R and g(x): R → R (a) f + g We have that: (f + g) (x) = f(x) + g(x) $ \left( f+g...
Given, $a–2b+11=0$……. (i) $3a–6b+33=0$……. (ii) From equation (i), ⇒ $b=(a+11)/2$ When $a=-1$, we get $b=(-1+11)/2=5$. When $a=-3$, we get $b=(-3+11)/2=4$. Thus, we have the following table giving...