Question 11 Mark
Find $f(x)$,
if$g(x)=1+\sqrt{x}$ and $f[g(x)]=3+2 \sqrt{x}+x$.
Answer$ g(x)=1+\sqrt{x}$
$f(g(x))=3+2 \sqrt{x}+x$
$=x+2 \sqrt{ } x+1+2$
$=(\sqrt{ } x+1)^2+2$
$f(\sqrt{ } x+1)=(\sqrt{ } x+1)^2+2$
$\therefore f(x)=x^2+2 $
View full question & answer→Question 21 Mark
Find the range of the following function : $f(x)=1+2 x+4 x$
Answer$f(x)=1+2 x+4 x$
Since, $2 x>0,4 x>0$
$
\therefore \mathrm{f}(\mathrm{x})>1
$
$\therefore$ Range of $f=(1, \infty)$
View full question & answer→Question 31 Mark
Find the range of the following function : $f(x)=[x]-x$
Answer$f(x)=[x]-x=-\{x\}$
$\therefore$ Range of $f=(-1,0] \ldots . .[0 \leq\{x\}<1]$
View full question & answer→Question 41 Mark
Find the range of the following function : $f(x)=\frac{1}{1+\sqrt{x}}$
Answer$f(x)=\frac{1}{1+\sqrt{x}}=y$, (say)
$ \therefore \sqrt{ } \mathrm{x} y+\mathrm{y}=1$
$\therefore \sqrt{\mathrm{x}}=\frac{1-y}{y} \geq 0$
$\therefore \frac{y-1}{y} \leq 0$
$\therefore 0<\mathrm{y} \leq 1 $
$\therefore$ Range of $f=(0,1]$
View full question & answer→Question 51 Mark
Find the range of the following function : $f(x)=\frac{x}{9+x^2}$
Answer$f(x)=\frac{x}{9+x^2}=y$ (say)
$\therefore \mathrm{x}^2 \mathrm{y}-\mathrm{x}+9 \mathrm{y}=0$
For real $x$, Discriminant $>0$
$ \therefore 1-4(y)(9 y) \geq 0$
$\therefore y^2 \leq \frac{1}{36}$
$\therefore \frac{-1}{6} \leq y \leq \frac{1}{6}$
$\therefore \text { Range of } f=\left[\frac{-1}{6}, \frac{1}{6}\right] $
View full question & answer→Question 61 Mark
Find the range of the following function : $f(x)=|x-5|$
Answerf(x) = |x – 5|
∴ Range of f = [0, ∞)
View full question & answer→Question 71 Mark
Find the domain of the following functions.$f(x)={ }^{5-x} P_{x-1}$
Answer$ f(x)={ }^{5-x} P_{x-1}$
$5-x>0, x-1 \geq 0, x-1 \leq 5-x$
$\therefore x<5, x \geq 1 \text { and } 2 x \leq 6$
$\therefore x \leq 3$
$\therefore \text { Domain of } f=\{1,2,3\} $
View full question & answer→Question 81 Mark
Find the domain of the following functions.$f(x)=x$ !
Answer$f(x)=x$ !
$\therefore$ Domain of $f=$ set of whole numbers $(W)$
View full question & answer→Question 91 Mark
Find the domain of the following functions.$f(x)=\sqrt{x-3}+\frac{1}{\log (5-x)}$
Answer$f(x)=\sqrt{x-3}+\frac{1}{\log (5-x)}$
For $f$ to be defined,
$ x-3 \geq 0,5-x>0 \text { and } 5-x \neq 1$
$x \geq 3, x<5 \text { and } x \neq 4$
$\therefore \text { Domain of } f=[3,4) \cup(4,5) $
View full question & answer→Question 101 Mark
Find the domain of the following functions.$f(x)=\frac{x^2+4 x+4}{x^2+x-6}$
Answer$f(x)=\frac{x^2+4 x+4}{x^2+x-6}=\frac{x^2+4 x+4}{(x+3)(x-2)}$
For $f$ to be defined, $x \neq-3,2$
$\therefore$ Domain of $\mathrm{f}=(-\infty,-3) \cup(-3,2) \cup(2, \infty)$
View full question & answer→Question 111 Mark
Solve the following for $x$, where $|x|$ is modulus function, $[x]$ is the greatest integer function, $\{x\}$ is a fractional part function.$\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]=\frac{5 x}{6}$
Answer$\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]=\frac{5 x}{6}$
L.H.S. = an integer
R.H.S. $=$ an integer
$\therefore \mathrm{x}=6 \mathrm{k}$, where $\mathrm{k}$ is an integer
View full question & answer→Question 121 Mark
Solve the following for $x$, where $|x|$ is modulus function, $[x]$ is the greatest integer function, $\{x\}$ is a fractional part function.$[x-2]+[x+2]+\{x\}=0$
Answer$ [x-2]+[x+2]+\{x\}=0$
$\therefore[x]-2+[x]+2+\{x\}=0$
$\therefore[x]+x=0 \ldots \ldots[\{x\}+[x]=x]$
$\therefore x=0$
View full question & answer→Question 131 Mark
Solve the following for $x$, where $|x|$ is modulus function, $[x]$ is the greatest integer function, $\{x\}$ is a fractional part function.$[\mathrm{x}] 2-5[\mathrm{x}]+6=0$
Answer$ [x]^2-5[x]+6=0$
$\therefore([x]-3)([x]-2)=0$
$\therefore[x]=3 \text { or } 2$
$\text { If }[x]=2 \text {, then } 2 \leq x<3$
$\text { If }[x]=3 \text {, then } 3 \leq x<4$
$\therefore \text { Solution set }=[2,4) $
View full question & answer→Question 141 Mark
Solve the following for $x$, where $|x|$ is modulus function, $[x]$ is the greatest integer function, $\{x\}$ is a fractional part function.$2[2 x-5]-1=7$
Answer$2[2 x-5]-1=7$
$ \therefore[2 x-5]=\frac{7+1}{2}=4$
$\therefore[2 x]-5=4$
$\therefore[2 x]=9$
$\therefore 9 \leq 2 x<10$
$\therefore \frac{9}{2} \leq x<5$
$\therefore$ Solution set $=\left[\frac{9}{2}, 5\right)$
View full question & answer→Question 151 Mark
Solve the following for $x$, where $|x|$ is modulus function, $[x]$ is the greatest integer function, $\{x\}$ is a fractional part function.$-2<[\mathrm{x}] \leq 7$
Answer$ -2<[\mathrm{x}] \leq 7$
$\therefore-2<\mathrm{x}<8$
$\therefore \text { Solution set }=(-2,8) $
View full question & answer→Question 161 Mark
Solve the following for $x$, where $|x|$ is modulus function, $[x]$ is the greatest integer function, $\{x\}$ is a fractional part function.$\left|x^2-x-6\right|=x+2$
Answer$\left|x^2-x-6\right|=x+2 \ldots$.(i)
R.H.S. must be non-negative
$ \therefore x \geq-2 \ldots \text {.(ii) }$
$|(x-3)(x+2)|=x+2$
$\therefore(x+2)|x-3|=x+2 \text { as } x+2 \geq 0$
$\therefore|x-3|=1 \text { if } x \neq-2$
$\therefore x-3= \pm 1$
$\therefore x=4 \text { or } 2 $
$\therefore \mathrm{x}=-2$ also satisfies the equation
$\therefore$ Solution set $=\{-2,2,4\}$
View full question & answer→Question 171 Mark
Solve the following for $x$, where $|x|$ is modulus function, $[x]$ is the greatest integer function, $\{x\}$ is a fractional part function.$1<|x-1|<4$
Answer$1<|x-1|<4$
$ \therefore-4<\mathrm{x}-1<-1 \text { or } 1<\mathrm{x}-1<4$
$\therefore-3<\mathrm{x}<0 \text { or } 2<\mathrm{x}<5$
$\therefore \text { Solution set }=(-3,0) \cup(2,5) $
View full question & answer→Question 181 Mark
Show that, $\log _y x^3 \cdot \log _z y^4 \cdot \log _x z^5=60$.
Answer$
\begin{aligned}
\text { L.H.S. } & =\log _y\left(x^3\right) \log _z\left(y^4\right) \log _x\left(z^5\right) \\
& =\left(3 \log _y x\right)\left(4 \log _z y\right)\left(5 \log _x z\right) \\
& =60\left(\frac{\log x}{\log y}\right)\left(\frac{\log y}{\log z}\right)\left(\frac{\log z}{\log x}\right) \\
& =60=\text { R.H.S. }
\end{aligned}
$
View full question & answer→Question 191 Mark
If $\log _3\left[\log _2\left(\log _3 x\right)\right]=1$, show that $x=6561$.
Answer$ \log _3\left[\log _2\left(\log _3 x\right)\right]=1$
$\therefore \log _2\left(\log _3 x\right)=3^1$
$\therefore \log _3 x=2^3$
$\therefore \log _3 x=8$
$\therefore x=3^8$
$\therefore x=6561 $
View full question & answer→Question 201 Mark
If $\log \left(\frac{a+b}{2}\right)=\frac{1}{2}(\log a+\log b)$, then show that $a=b$.
Answer$ \log \left(\frac{a+b}{2}\right)=\frac{1}{2}(\log a+\log b)$
$\therefore 2 \log \left(\frac{a+b}{2}\right)=\log a+\log b$
$\therefore \log \left(\frac{a+b}{2}\right)^2=\log a b$
$\therefore \frac{(a+b)^2}{4}=a b$
$\therefore a^2+2 a b+b^2=4 a b$
$\therefore a^2+2 a b-4 a b+b^2=0$
$\therefore a^2-2 a b+b^2=0$
$\therefore(a-b)^2=0$
$\therefore a-b=0$
$\therefore a=b$
View full question & answer→Question 211 Mark
Simplify $\log \left(\log x^4\right)-\log (\log x)$
Answer$ \log \left(\log x^4\right)-\log (\log x)$
$=\log (4 \log x)-\log (\log x) \ldots . .\left[\log m^n=n \log m\right]$
$=\log 4+\log (\log x)-\log (\log x) \ldots . .[\log (m n)=\log m+\log n]$
$=\log 4 $
View full question & answer→Question 221 Mark
Show that $\log \frac{\mathrm{a}^2}{\mathrm{bc}}+\log \frac{\mathrm{b}^2}{\mathrm{ca}}+\log \frac{\mathrm{c}^2}{\mathrm{ab}}=0$
Answer$
\begin{aligned}
\text { L.H.S. } & =\log \frac{a^2}{b c}+\log \frac{b^2}{c a}+\log \frac{c^2}{a b} \\
& =\log \left(\frac{a^2}{b c} \times \frac{b^2}{c a} \times \frac{c^2}{a b}\right) \\
& =\log \left(\frac{a^2 b^2 c^2}{a^2 b^2 c^2}\right)=\log 1=0=\text { R.H.S. }
\end{aligned}
$
View full question & answer→Question 231 Mark
Find $x$, if $x=3^{3 \log _3 2}$.
Answer$\mathrm{X}=3^{3 \log _3 2}$
$=3^{\log 3\left(2^3\right)}$
$=2^3 \ldots\left[a^{\log _a b}=\mathrm{b}\right]$
$=8 $
View full question & answer→Question 241 Mark
For any base show that $\log (1+2+3)=\log 1+\log 2+\log 3$
Answer$ \text { L.H.S. }=\log (1+2+3)=\log 6$
$\text { R.H.S. }=\log 1+\log 2+\log 3$
$=0+\log (2 \times 3)$
$=\log 6$
$\therefore \text { L.H.S. }=\text { R.H.S. }$
View full question & answer→Question 251 Mark
Let $f: R-\{2\} \rightarrow R$ be defined by $f(x)=\frac{x^2-4}{x-2}$ and $g: R \rightarrow R$ be defined by $g(x)=$ $x+2$. Examine whether $f=g$ or not.
Answer$ f(x)=\frac{x^2-4}{x-2}, x \neq 2$
$\therefore f(x)=x+2, x \neq 2 \text { and } g(x)=x+2, $
The domain of $f=R-\{2\}$
The domain of $g=R$
Here, $f$ and $g$ have different domains.
$\therefore \mathrm{f} \neq \mathrm{g}$
View full question & answer→Question 261 Mark
Find fog and gof:$f(x)=256 x^4, g(x)=\sqrt{x}$
Answer$f(x)=256 x^4, g(x)=\sqrt{ } x$
(fog) $(x)=f(g(x))=f(\sqrt{x})=256(\sqrt{x})^4=256 x^2$
$(g \circ f)(x)=g(f(x))=g\left(256 x^4\right)=\sqrt{256 x^4}=16 x^2$
View full question & answer→Question 271 Mark
Find fog and gof:$f(x)=3 x-2, g(x)=x^2$
Answer$ f(x)=3 x-2, g(x)=x^2$
$\text { (fog) }(x)=f(g(x))=f\left(x^2\right)=3 x^2-2$
$\text { (gof) }(x)=g(f(x))$
$=g(3 x-2)$
$=(3 x-2)^2$
$=9 x^2-12 x+4$
View full question & answer→Question 281 Mark
Find fog and gof:$f(x)=x^2+5, g(x)=x-8$
Answer$ f(x)=x^2+5, g(x)=x-8$
$\text { (fog) }(x)=f(g(x))$
$=f(x-8)$
$=(x-8)^2+5$
$=x^2-16 x+64+5$
$=x^2-16 x+69$
$(g \circ f)(x)=g(f(x))$
$=g\left(x^2+5\right)$
$=x^2+5-8$
$=x-3 $
View full question & answer→Question 291 Mark
If $f(x)=3 x+a$ and $f(1)=7$, find $a$ and $f(4)$.
Answer$ f(x)=3 x+a, f(1)=7$
$\therefore 3(1)+a=7$
$\therefore a=7-3=4$
$\therefore f(x)=3 x+4$
$\therefore f(4)=3(4)+4=12+4=16 $
View full question & answer→Question 301 Mark
A function $f$ is defined as $f(x)=5-x$ for $0 \leq x \leq 4$. Find the values of $x$ such that$f(x)=5$
Answer$ f(x)=5$
$\therefore 5-x=5$
$\therefore x=0 $
View full question & answer→Question 311 Mark
A function $f$ is defined as $f(x)=5-x$ for $0 \leq x \leq 4$. Find the values of $x$ such that$f(x)=3$
Answer$ f(x)=3$
$\therefore 5-x=3$
$\therefore x=5-3=2$
View full question & answer→Question 321 Mark
A function $f$ is defined as $f(x)=4 x+5$, for $-4 \leq x<0$. Find the values of $f(-1)$, $f(-2), f(0)$, if they exist.
Answer$f(x)=4 x+5,-4 \leq x<0$
$f(-1)=4(-1)+5=-4+5=1$
$f(-2)=4(-2)+5=-8+5=-3$
$x=0 \notin \text { domain of } f$
$\therefore f(0) \text { does not exist. } $
View full question & answer→Question 331 Mark
Find whether the following functions are onto or not.$f: R \rightarrow R$ defined by $f(x)=x^2+3$ for all $x \in R$
Answer$f(x)=x^2+3=y$ (say)
$
(x, y \in R)
$
Clearly $y \geq 3 \ldots . .\left[x^2 \geq 0\right]$
$\therefore$ All the real numbers less than 3 from codomain $\mathrm{R}$, have not been preassigned any element from the domain $\mathrm{R}$.
$\therefore \mathrm{f}$ is not onto.
View full question & answer→Question 341 Mark
Find whether the following functions are onto or not.$f: Z \rightarrow Z$ defined by $f(x)=6 x-7$ for all $x \in Z$
Answer$ \text { (i) } f(x)=6 x-7=y \text { (say) }$
$(x, y \in Z)$
$\therefore x=\frac{7+y}{6}$
Since every integer $\mathrm{y}$ does not give integer $\mathrm{x}, \mathrm{f}$ is not onto.
View full question & answer→Question 351 Mark
Find whether the following functions are one-one.$f: R \rightarrow R$ defined by $f(x)=x^2+5$
Answer$f: R \rightarrow R$, defined by $f(x)=x^2+5$
Note that $f(-x)=f(x)=x^2+5$
$\therefore \mathrm{f}$ is not one-one (i.e., many-one) function.
View full question & answer→Question 361 Mark
Which of the following relations are functions? If it is a function determine its domain and range.
{(2, 1), (3, 1), (5, 2)}
Answer{(2, 1), (3, 1), (5, 2)}
Every element of set A has been assigned a unique element in set B.
∴ Given relation is a function.
Domain = {2, 3, 5}, Range = {1, 2}
View full question & answer→Question 371 Mark
Which of the following relations are functions? If it is a function determine its domain and range.
{(0, 0), (1, 1), (1, -1), (4, 2), (4, -2) (9, 3), (9, -3), (16, 4), (16, -4)}
Answer{(0, 0), (1, 1), (1, -1), (4, 2), (4, -2) (9, 3), (9, -3) (16, 4), (16, -4)}
∵ (1, 1), (1, -1) ∈ the relation
∴ Given relation is not a function.
As element 1 of the domain has not been assigned a unique element of co-domain.
View full question & answer→Question 381 Mark
Which of the following relations are functions? If it is a function determine its domain and range.
{(2, 1), (4, 2), (6, 3), (8, 4), (10, 5) (12, 6), (14, 7)}
Answer{(2, 1), (4, 2), (6, 3), (8, 4), (10, 5) (12, 6), (14, 7)}
Every element of set A has been assigned a unique element in set B
∴ Given relation is a function
Domain = {2, 4, 6, 8, 10, 12, 14}, Range = {1, 2, 3, 4, 5, 6, 7}
View full question & answer→Question 391 Mark
Solve the following for $x$, where $|x|$ is modulus function, $[x]$ is the greatest integer function, $\{x\}$ is a fractional part function.$2\{x\}=x+[x]$
Answer$ 2\{x\}=x+[x]$
$=[x]+\{x\}+[x] \ldots . .[x=[x]+\{x\}]$
$\therefore\{x\}=2[x] $
R.H.S. is an integer
$\therefore$ L.H.S. is an integer
$ \therefore\{x\}=0$
$\therefore[x]=0$
$\therefore x=0 $
View full question & answer→Question 401 Mark
Solve the following for $x$, where $|x|$ is modulus function, $[x]$ is the greatest integer function, $\{x\}$ is a fractional part function.$\{x\}=0.5$
Answer$\{x\}=0.5$
$
\therefore x=\ldots . .,-2.5,-1.5,-0.5,0.5,1.5, \ldots .
$
$\therefore$ The solution set $=\{x: x=n+0.5, n \in Z\}$
View full question & answer→Question 411 Mark
Solve the following for $x$, where $|x|$ is modulus function, $[x]$ is the greatest integer function, $\{x\}$ is a fractional part function.$\{x\}=0$
Answer$\{\mathrm{x}\}=0$
$\therefore \mathrm{x}$ is an integer
$\therefore$ The solution set is $\mathrm{Z}$.
View full question & answer→Question 421 Mark
Solve the following for $x$, where $|x|$ is modulus function, $[x]$ is the greatest integer function, $\{x\}$ is a fractional part function.$\{x\}>4$
Answer$\{x\}>4$
This is a meaningless statement as $0 \leq\{\mathrm{X}\}<1$
$\therefore$ The solution set $=\{\}$ or $\Phi$
View full question & answer→Question 431 Mark
Solve the following for $x$, where $|x|$ is modulus function, $[x]$ is the greatest integer function, $\{x\}$ is a fractional part function.$[x+[x+[x]]]=9$
Answer$[x+[x+[x]]]=9$
$\therefore[\mathrm{x}+[\mathrm{x}]+[\mathrm{x}]]=9 \ldots \ldots .[\mathrm{x}+\mathrm{n}]=[\mathrm{x}]+\mathrm{n}$, if $\mathrm{n}$ is an integer $]$
$\therefore[\mathrm{x}+2[\mathrm{x}]]=9$
$\therefore[x]+2[x]=9 \ldots .[[2[x]$ is an integer $]]$
$\therefore[\mathrm{x}]=3$
$\therefore \mathrm{x} \in[3,4)$
View full question & answer→Question 441 Mark
Solve the following for $x$, where $|x|$ is modulus function, $[x]$ is the greatest integer function, $\{x\}$ is a fractional part function.$2|x|=5$
Answer$2|x|=5$
$\therefore|x|=\frac{5}{2}$
$\therefore x= \pm \frac{5}{2}$
View full question & answer→Question 451 Mark
Solve the following for $x$, where $|x|$ is modulus function, $[x]$ is the greatest integer function, $\{x\}$ is a fractional part function.$|x| \leq 3$
Answer$|x| \leq 3$ The solution set of $|x| \leq a$ is $-a \leq x \leq a$
$\therefore$ The required solution is $-3 \leq \mathrm{x} \leq 3$
$\therefore$ The solution set is $[-3,3]$
View full question & answer→Question 461 Mark
Solve the following for $x$, where $|x|$ is modulus function, $[x]$ is the greatest integer function, $\{x\}$ is a fractional part function.$x^2+7|x|+12=0$
Answer$x^2+7|x|+12=0$
$\therefore(|\mathrm{x}|)^2+7|\mathrm{x}|+12=0$
$\therefore(|x|+3)(|x|+4)=0$
$\therefore$ There is no $\mathrm{x}$ that satisfies the equation.
The solution set $=\{\}$ or $\Phi$
View full question & answer→Question 471 Mark
If $f(x)=4[x]-3$, where $[x]$ is greatest integer function of $x$, then find$\mathrm{f}(2 \pi)$, where $\pi=3.14$
Answer$ \text { (iv) } f(2 \pi)=4[2 \pi]-3$
$=4[6.28]-3 \ldots . .[\because \pi=3.14]$
$=4(6)-3 \ldots \ldots .[\because 6 \leq 6.28<7,[6.28]=6]$
$=21 $
View full question & answer→Question 481 Mark
If $f(x)=4[x]-3$, where $[x]$ is greatest integer function of $x$, then find.
$f\left(-\frac{5}{2}\right)$
Answer$ f\left(-\frac{5}{2}\right)=\mathrm{f}(-2.5)$
$=4[-2.5]-3$
$=4(-3)-3 \ldots \ldots .[\because-3 \leq-2.5 \leq-2,[-2.5]=-3]$
$=-15$
View full question & answer→Question 491 Mark
If $f(x)=4[x]-3$, where $[x]$ is greatest integer function of $x$, then find$f(0.5)$
Answer$ f(0.5)=4[0.5]-3$
$=4(0)-3 \ldots \ldots \ldots .[\because 0 \leq 0.5<1,[0.5]=0]$
$=-3$
View full question & answer→Question 501 Mark
If $f(x)=4[x]-3$, where $[x]$ is greatest integer function of $x$, then find$f(7.2)$
Answer$ f(x)=4[x]-3$
$f(7.2)=4[7.2]-3$
$=4(7)-3 \ldots \ldots . .[\because 47.2<8,[7.2]=7]$
$=25 $
View full question & answer→Question 511 Mark
If $f(x)=2|x|+3 x$, then find$f(-5)$
Answer$ \text { (ii) } f(-5)=2|-5|+3(-5)$
$=2(5)-15 \ldots .[\because|x|=-x, x<0]$
$=10-15$
$=-5$
View full question & answer→Question 521 Mark
If $f(x)=2|x|+3 x$, then find$f(2)$
Answer$ f(x)=2|x|+3 x$
$\text { (i) } f(2)=2|2|+3(2)$
$=2(2)+6 \ldots .[\because|x|=x, x>0]$
$=10$
View full question & answer→Question 531 Mark
If $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc}4 x-2, & x \leq-3 \\ 5, & -3<x<3 \\ x^2, & x \geq 3\end{array}\right.$, then find$f(5)$
View full question & answer→Question 541 Mark
If $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc}4 x-2, & x \leq-3 \\ 5, & -3<x<3 \\ x^2, & x \geq 3\end{array}\right.$, then find$f(1)$
View full question & answer→Question 551 Mark
If $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc}4 x-2, & x \leq-3 \\ 5, & -3$\mathrm{f}(-3)$
Answer$ f(-3)=4(-3)-2$
$=-12-2$
$=-14$
View full question & answer→Question 561 Mark
If $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc}4 x-2, & x \leq-3 \\ 5, & -3$f(-4)$
Answer$ f(x)=4 x-2, x \leq-3$
$=5,-3=x^2, x \geq 3$
$\text { (i) } f(-4)=4(-4)-2$
$=-16-2$
$=-18 $
View full question & answer→Question 571 Mark
If $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{ll}x^2+3, & x \leq 2 \\ 5 x+7, & x>2 \end{array}\right.$},then find$f(0)$
View full question & answer→Question 581 Mark
If $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{ll}x^2+3, & x \leq 2 \\ 5 x+7, & x>2 \end{array}\right.$},then find$\mathrm{f}(\mathrm{C})$
Answer$f(2)=2^2+3$
$ =4+3$
$=7 $
View full question & answer→Question 591 Mark
If $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{ll}x^2+3, & x \leq 2 \\ 5 x+7, & x>2 \end{array}\right.$},then find$f(3)$
Answer$ f(x)=x^2+3, x \leq 2$
$=5 x+7, x>2$
$\text { (i) } f(3)=5(3)+7$
$=15+7$
$=22$
View full question & answer→Question 601 Mark
Check if the following functions have an inverse function. If yes, find the inverse function.$f(x)=8$
Answerf(x) = 8 = y (say)
For every value of x, the value of the function f is the same.
∴ f is not one-one i.e. (many-one) function.
∴ f does not have the inverse.
View full question & answer→Question 611 Mark
If $f(x)=2 x^2+3, g(x)=5 x-2$, then findgog
Answer$(g \circ g)(x)=g(g(x))$
$=g(5 x-2)$
$=5(5 x-2)-2$
$=25 x-12 $
View full question & answer→Question 621 Mark
If $f(x)=2 x^2+3, g(x)=5 x-2$, then find fof
Answer$ \text { (fof) }(x)=f(f(x))$
$=f\left(2 x^2+3\right)$
$=2\left(2 x^2+3\right)^2+3$
$=2\left(4 x^4+12 x^2+9\right)+3$
$=8 x^4+24 x^2+21 $
View full question & answer→Question 631 Mark
If $f(x)=2 x^2+3, g(x)=5 x-2$, then findgof
Answer$ (g \circ f)(x)=g(f(x))$
$=g\left(2 x^2+3\right)$
$=5(2 x+3)-2$
$=10 x^2+13 $
View full question & answer→Question 641 Mark
If $f(x)=2 x^2+3, g(x)=5 x-2$, then findfog
Answer$ f(x)=2 x^2+3, g(x)=5 x-2$
$\text { (i) (fog) }(x)=f(g(x))$
$=f(5 x-2)$
$=2(5 x-2)^2+3$
$=2\left(25 x^2-20 x+4\right)+3$
$=50 x^2-40 x+11 $v
View full question & answer→Question 651 Mark
If $f(x)=3 x+5, g(x)=6 x-1$, then find( $\mathrm{f} / \mathrm{g})(\mathrm{x})$ and its domain
Answer$\left(\frac{\mathrm{f}}{\mathrm{g}}\right)(x)=\frac{\mathrm{f}(x)}{\mathrm{g}(x)}=\frac{3 x+5}{6 x-1}, x \neq \frac{1}{6}$ Domain $=R-\left\{\frac{1}{6}\right\}$
View full question & answer→Question 661 Mark
If $f(x)=3 x+5, g(x)=6 x-1$, then find $(fg) (3)$
Answer$ (\mathrm{fg})(3)=f(3) g(3)$
$=[3(3)+5][6(3)-1]$
$=(14)(17)$
$=238 $
View full question & answer→Question 671 Mark
If $f(x)=3 x+5, g(x)=6 x-1$, then find$(f-g)(2)$
Answer$(f-g)(2)=f(2)-g(2)$
$=[3(2)+5]-[6(2)-1]$
$=6+5-12+1$
$=0$
View full question & answer→Question 681 Mark
If $f(x)=3 x+5, g(x)=6 x-1$, then find$(f+g)(x)$
Answer$ f(x)=3 x+5, g(x)=6 x-1$
$(i)(f+g)(x)=f(x)+g(x)$
$=3 x+5+6 x-1$
$=9 x+4$
View full question & answer→Question 691 Mark
Given that $\log 2=a$ and $\log 3=b$, write $\log \sqrt{96}$ terms of $a$ and $b$.
Answer$\log 2=a \text { and } \log 3=b$
$\log \sqrt{ } 96=\frac{1}{2} \log (96)$
$=\frac{1}{2} \log \left(2^5 \times 3\right)$
$=\frac{1}{2}\left(\log 2^5+\log 3\right) \ldots . .[\because \log m n=\log m+\log n]$
$=\frac{1}{2}(5 \log 2+\log 3) \ldots \ldots\left[\cdot\left[\log m^n=n \log m\right]\right.$
$=\frac{5 a+b}{2} $
View full question & answer→Question 701 Mark
Write the following expressions as a single logarithm.$\ln (x+2)+\ln (x-2)-3 \ln (x+5)$
Answer$\ln (x+2)+\ln (x-2)-3 \ln (x+5)$
$=\ln [(x+2)(x-2)]-\ln (x+5)^3$
$\ldots\left[\begin{array}{l} \log m+\log n=\log m n \\ n \log m=\log m^n \end{array}\right]$
$=\ln \left(x^2-4\right)-\ln (x+5)^3$
$=\ln \left(\frac{x^2-4}{(x+5)^3}\right) \ldots\left[\log m-\operatorname{logn}=\log \frac{m}{n}\right]$
View full question & answer→Question 711 Mark
Write the following expressions as a single logarithm.$\frac{1}{3} \log (x-1)+\frac{1}{2} \log (x)$
Answer$ \frac{1}{3} \log (x-1)+\frac{1}{2} \log x$
$=\log \left((x-1)^{\frac{1}{3}}\right)+\log \left(x^{\frac{1}{2}}\right)$
$=\log (\sqrt[3]{x-1} \sqrt{x}) $
$\ldots\left[n \log m=\log m^n\right]$
$=\log (\sqrt[3]{x-1} \sqrt{x})$$[\log m+\log n=\log m n]$
View full question & answer→Question 721 Mark
Write the following expressions as a single logarithm.
$5 \log x+7 \log y-\log z$
Answer$5 \log x+7 \log y-\log z$
$=\log \left(x^5\right)+\log \left(y^7\right)-\log z$
$\ldots\left[n \log m=\log m^n\right]$
$=\log \left(x^5 y^7\right)-\log \mathrm{z}$
$\ldots[\log m+\log n=\log m n]$
$=\log \left(\frac{x^5 y^7}{z}\right)$
$\ldots\left[\log m-\log n=\log \frac{m}{n}\right]$
View full question & answer→Question 731 Mark
Write the following expressions as sum or difference of logarithms:$\ln \left(\frac{a^3(a-2)^2}{\sqrt{b^2+5}}\right)$
Answer$ \ln \left(\frac{a^3(a-2)^2}{\sqrt{b^2+5}}\right)$
$=\ln \left(a^3(a-2)^2\right)-\ln \sqrt{b^2+5}$
$\quad \ldots\left[\log \frac{m}{n}=\log m-\log n\right]$
$ =\ln a^3+\ln (a-2)^2-\ln \left(b^2+5\right)^{\frac{1}{2}}$
$\qquad \ldots[\log m n=\log m+\log n]$
$=3 \ln a+2 \ln (a-2)-\frac{1}{2} \ln \left(b^2+5\right) $
$\ldots\left[\log m^n=n \log m\right]$
View full question & answer→Question 741 Mark
Write the following expressions as sum or difference of logarithms:$\log (\sqrt{x} \sqrt[3]{y})$
Answer$
\begin{aligned}
\log (\sqrt{x} \sqrt[3]{y})= & \log (\sqrt{x})+\log (\sqrt[3]{y}) \\
& \ldots[\log mn =\log m +\log n ] \\
= & \log x^{\frac{1}{2}}+\log y^{\frac{1}{3}} \\
= & \frac{1}{2} \log x+\frac{1}{3} \log y
\end{aligned}
$
... [ $\left.\log m ^{ n }= nlog m \right]$
View full question & answer→Question 751 Mark
Write the following expressions as sum or difference of logarithms:
$\log \left(\frac{p q}{r s}\right)$
Answer$\log \left(\frac{ pq }{ rs }\right)=\log ( pq )-\log ( rs )$
$\left[\log \frac{m}{n}=\log m-\log n\right]$
$ =\log p+\log q-(\log r+\log s)$
$\quad \ldots[\log m n=\log m+\log n]$
$=\log p+\log q-\log r-\log s$
View full question & answer→Question 761 Mark
Find the domain of$f(x)=\log 10\left(x^2-5 x+6\right)$
Answer$f(x)=\log _{10}\left(x^2-5 x+6\right)$
$
x^2-5 x+6=(x-2)(x-3)
$
$f$ is defined, when $(x-2)(x-3)>0$
$
\therefore x <2 \text { or } x >3
$
Solution of $(x-a)(x-b)>0$ is $x<a$ or $x>b$ where $a<b$
$\therefore$ Domain of $f=(-\infty, 2) \cup(3, \infty)$
View full question & answer→Question 771 Mark
Find the domain of$f(x)=\ln (x-5)$
Answer$f(x)=\ln (x-5)$
$f$ is defined, when $x-5>0$
$
\therefore x >5
$
$\therefore$ Domain of $f=(5, \infty)$
View full question & answer→Question 781 Mark
Express the following logarithmic equations in exponential form:$\ln \frac{1}{2}=-0.693$
Answer$\ln \left(\frac{1}{2}\right)=-0.693$
$\therefore \quad \frac{1}{2}= e ^{-0.693}$,i.e., $e ^{-0.693}=\frac{1}{2}$
View full question & answer→Question 791 Mark
Express the following logarithmic equations in exponential form:$\ln e =1$
Answer$\quad \ln e =1$
$\therefore \quad e = e ^1$,i.e., $e ^{ l }= e$
View full question & answer→Question 801 Mark
Express the following logarithmic equations in exponential form:$\ln 1=0$
Answer$\quad \ln 1=0 \quad$
$\therefore \quad 1= e ^0$, i.e., $e ^0=1$
View full question & answer→Question 811 Mark
Express the following logarithmic equations in exponential form:$\log _{\frac{1}{2}}(8)=-3$
Answer$\log _{\frac{1}{2}}(8)=-3$
$\therefore \quad 8=\left(\frac{1}{2}\right)^{-3}$,i.e., $\left(\frac{1}{2}\right)^{-3}=8$
View full question & answer→Question 821 Mark
Express the following logarithmic equations in exponential form:$\log _{10} 0.001=-3$
Answer$\log _{10}(0.001)=-3$
$\therefore \quad 0.001=10^{-3}$, i.e., $10^{-3}=0.001$
View full question & answer→Question 831 Mark
Express the following logarithmic equations in exponential form:$\log _5 \frac{1}{25}=-2$
Answer$\quad \log _5\left(\frac{1}{25}\right)=-2$
$\therefore \quad \frac{1}{25}=5^{-2}$,i.e., $5^{-2}=\frac{1}{25}$
View full question & answer→Question 841 Mark
Express the following logarithmic equations in exponential form:$\log _2 64=6$
Answer$\log _2 64=6$
$\therefore 64=2^6$, i.e., $2^6=64$
View full question & answer→Question 851 Mark
Express the following exponential equations in logarithmic form:$e^{-x}=6$
Answer$\quad e ^{-x}=6$
$\therefore \quad-x=\log _e 6$
[By definition of logarithm]
i.e., $\log _e 6=-x$
View full question & answer→Question 861 Mark
Express the following exponential equations in logarithmic form:$e^{\frac{1}{2}}=1.6487$
Answer$e ^{\frac{1}{2}}=1.6487$
$\therefore \quad \frac{1}{2}=\log _e(1.6487) \ldots$ [By definition of logarithm]
i.e., $\log _e(1.6487)=\frac{1}{2}$
View full question & answer→Question 871 Mark
Express the following exponential equations in logarithmic form:$e ^2=7.3890$
Answer$e ^2=7.3890$
$\therefore \quad 2=\log _e(7.3890) \ldots$ [By definition of logarithm] i.e., $\log _e(7.3890)=2$
(e is a mathematical constant, whose value is approximately 2.71828 )
View full question & answer→Question 881 Mark
Express the following exponential equations in logarithmic form:$10^{-2}=0.01$
Answer$10^{-2}=0.01$
$\therefore \quad-2=\log _{10}(0.01) \ldots$ [By definition of logarithm] i.e., $\log _{10}(0.01)=-2$
View full question & answer→Question 891 Mark
Express the following exponential equations in logarithmic form:$3^{-4}=\frac{1}{81}$
Answer$\quad 3^{-4}=\frac{1}{81}$
$\therefore \quad-4=\log _3\left(\frac{1}{81}\right) \ldots$ [By definition of logarithm] i.e., $\log _3\left(\frac{1}{81}\right)=-4$
View full question & answer→Question 901 Mark
Express the following exponential equations in logarithmic form:$9^{\frac{3}{2}}=27$
Answer$\quad 9^{\frac{3}{2}}=27$
$\therefore \quad \frac{3}{2}=\log _9 27$
..[By definition of logarithm]
i.e., $\log _9 27=\frac{3}{2}$
View full question & answer→Question 911 Mark
Express the following exponential equations in logarithmic form:$23^1=23$
Answer$23^1=23$
$\therefore \quad 1=\log _{23} 23$
[By definition of logarithm]
i.e., $\log _{23} 23=1$
View full question & answer→Question 921 Mark
Express the following exponential equations in logarithmic form:$54^0=1$
Answer$54^0=1$
$\therefore \quad 0=\log _{54} 1$
..[By definition of logarithm]
i.e., $\log _{54} 1=0$
View full question & answer→Question 931 Mark
Express the following exponential equations in logarithmic form:$2^5=32$
Answer$\begin{array}{ll}
& 2^5=32 \\
\therefore & 5=\log _2 32 \quad \ldots \text { [By definition of logarithm] } \\
& \text { i.e., } \log _2 32=5
\end{array}
$
View full question & answer→Question 941 Mark
If $f(x)=3\left(4^{x+1}\right)$, find $f(-3)$.
Answer$ f(x)=3\left(4^{x+1}\right)$
$\therefore f(-3)=3\left(4^{-3+1}\right)$
$=3\left(4^{-2}\right)$
$=\frac{3}{16} $
View full question & answer→Question 951 Mark
Express the area $\mathrm{A}$ of a square as a function of itsperimeter $\mathrm{P}$
Answerperimeter $(P)=4 \mathrm{~s}$
$\therefore \mathrm{s}=\frac{\mathrm{P}}{4}$
Area $(\mathrm{A})=s^2=\left(\frac{\mathrm{P}}{4}\right)^2$
$\therefore \mathrm{A}=\frac{\mathrm{P}^2}{16}$
View full question & answer→Question 961 Mark
Express the area $\mathrm{A}$ of a square as a function of itsside $s$
View full question & answer→Question 971 Mark
Find the domain and range of the following functions.$f(x)=\sqrt[3]{x+1}$
Answer$
f(x)=\sqrt[3]{x+1}
$
$f$ is defined for all real $x$ and the values of $f(x) \in R$
$\therefore$ Domain of $f=R$, Range of $f=R$
View full question & answer→Question 981 Mark
If $\mathrm{f}(\mathrm{x})=\frac{a-x}{b-x}, \mathrm{f}(2)$ is undefined, and $\mathrm{f}(3)=5$, find $a$ and $\mathrm{b}$.
Answer$f(x)=\frac{a-x}{b-x}$
Given that,
$f(2)$ is undefined
$ b-2=0$
$\therefore b=2 \ldots .(i)$
$f(3)=5$
$\therefore \frac{a-3}{b-3}=5$
$\therefore \frac{a-3}{2-3}=5 \ldots . .[\text { [From (i)] }$
$\therefore a-3=-5$
$\therefore a=-2$
$\therefore a=-2, b=2 $
View full question & answer→Question 991 Mark
Find $x$, if $g(x)=0$ where
$g(x)=x^3-2 x^2-5 x+6$
Answer$ g(x)=x^3-2 x^2-5 x+6$
$=(x-1)\left(x^2-x-6\right)$
$=(x-1)(x+2)(x-3)$
$g(x)=0$
$\therefore(x-1)(x+2)(x-3)=0$
$\therefore x-1=0 \text { or } x+2=0 \text { or } x-3=0$
$\therefore x=1,-2,3$
View full question & answer→Question 1001 Mark
Find $x$, if $g(x)=0$ where
$g(x)=6 x^2+x-2$
Answer$g(x)=6 x^2+x-2$
$g(x)=0$
$\therefore 6 x^2+x-2=0$
$\therefore(2 x-1)(3 x+2)=0$
$\therefore 2 x-1=0 \text { or } 3 x+2=0$
$\therefore x=\frac{1}{2} \text { or } x=\frac{-2}{3}$
View full question & answer→Question 1011 Mark
Find $x$, if $g(x)=0$ where$g(x)=\frac{18-2 x^2}{7}$
Answer$ g(x)=\frac{18-2 x^2}{7}$
$g(x)=0$
$\frac{18-2 x^2}{7}=0$
$\therefore 18-2 x^2=0$
$\therefore x^2=9$
$\therefore x= \pm 3 $
View full question & answer→Question 1021 Mark
Find $x$, if $g(x)=0$ where
$g(x)=\frac{5 x-6}{7}$
Answer$ g(x)=\frac{5 x-6}{7}$
$g(x)=0$
$\therefore \frac{5 x-6}{7}=0$
$\therefore x=\frac{6}{5} $
View full question & answer→Question 1031 Mark
If $f(m)=m^2-3 m+1$, find
$\left(\frac{\mathbf{f}(2+h)-f(2)}{h}\right), \mathrm{h} \neq 0$.
Answer$ \left(\frac{\mathbf{f}(2+h)-f(2)}{h}\right)$
$=\frac{(2+h)^2-3(2+h)+1-\left(2^2-3(2)+1\right)}{h}$
$=\frac{\mathrm{h}^2+\mathrm{h}}{\mathrm{h}}$
$=\mathrm{h}+1$
View full question & answer→Question 1041 Mark
If $f(m)=m^2-3 m+1$, find$f(-x)$
Answer$f(-x)=(-x) 2-3(-x)+1=x 2+3 x+1$
View full question & answer→Question 1051 Mark
If $f(m)=m^2-3 m+1$, find
$f(x+1)$
Answer$f(x+1)=(x+1)^2-3(x+1)+1$
$=x^2+2 x+1-3 x-3+1$
$=x^2-x-1$
View full question & answer→Question 1061 Mark
If $f(m)=m^2-3 m+1$, find
$f\left(\frac{1}{2}\right)$
Answer$ f\left(\frac{1}{2}\right)=\left(\frac{1}{2}\right)^2-3\left(\frac{1}{2}\right)+1$
$=\frac{1}{4}-\frac{3}{2}+1$
$=\frac{1-6+4}{4}$
$=-\frac{1}{4}$
View full question & answer→Question 1071 Mark
If $f(m)=m^2-3 m+1$, find
$f(-3)$
Answer$ f(-3)=(-3)^2-3(-3)+1$
$=9+9+1$
$=19$
View full question & answer→Question 1081 Mark
If $f(m)=m^2-3 m+1$, find$f(0)$
Answer$f(m)=m^2-3 m+1$
$f(0)=0^2-3(0)+1=1$
View full question & answer→Question 1091 Mark
Check if the relation given by the equation represents $y$ as function of $x$.$3 x-6=21$
Answer$3 x-6=21$
$\therefore \mathrm{x}=9$
$\therefore \mathrm{x}=9$ represents a point on the $\mathrm{X}$-axis.
There is no $y$ involved in the equation.
So the given equation does not represent a function.
View full question & answer→Question 1101 Mark
Check if the relation given by the equation represents $y$ as function of $x$.$2 y+10=0$
Answer$2 y+10=0$
$\therefore \mathrm{y}=-5$
$\therefore$ For every value of $\mathrm{x}$, there is a unique value of $\mathrm{y}$.
$\therefore \mathrm{y}$ is a function of $\mathrm{x}$.
View full question & answer→Question 1111 Mark
Check if the relation given by the equation represents $y$ as function of $x$.$x^2-y=25$
Answer$x^2-y=25$
$\therefore \mathrm{y}=\mathrm{x}^2-25$
$\therefore$ For every value of $\mathrm{x}$, there is a unique value of $\mathrm{y}$.
$\therefore \mathrm{y}$ is a function of $\mathrm{x}$.
View full question & answer→Question 1121 Mark
Check if the relation given by the equation represents $y$ as function of $x$.$x+y^2=9$
Answer$\mathrm{x}+\mathrm{y}^2=9$
$\therefore \mathrm{y}^2=9-\mathrm{x}$
$\therefore \mathrm{y}= \pm \sqrt{9-x}$
$\therefore$ For one value of $\mathrm{x}$, there are two values of $\mathrm{y}$.
$\therefore \mathrm{y}$ is not a function of $\mathrm{x}$.
View full question & answer→Question 1131 Mark
Check if the relation given by the equation represents $y$ as function of $x$.$2 x+3 y=12$
Answer$2 x+3 y=12$
$\therefore \mathrm{y}=\frac{12-2 x}{3}$
$\therefore$ For every value of $\mathrm{x}$, there is a unique value of $y$.
$\therefore \mathrm{y}$ is a function of $\mathrm{x}$.
View full question & answer→Question 1141 Mark
Which sets of ordered pairs represent functions from A = {1, 2, 3, 4} to B = {-1, 0, 1, 2, 3}? Justify.
{(1, 1), (2, 1), (3, 1), (4, 1)}
Answer{(1, 1), (2, 1), (3, 1), (4, 1)} represents a function
Reason: Every element of set A has been assigned a unique image in set B.
View full question & answer→Question 1151 Mark
Which sets of ordered pairs represent functions from A = {1, 2, 3, 4} to B = {-1, 0, 1, 2, 3}? Justify.
{(1, 3), (4, 1), (2, 2)}
Answer{(1, 3), (4, 1), (2, 2)} does not represent a function.
Reason:
3 ∈ A does not have an image in set B.
View full question & answer→Question 1161 Mark
Which sets of ordered pairs represent functions from A = {1, 2, 3, 4} to B = {-1, 0, 1, 2, 3}? Justify.
{(1, 2), (2, -1), (3, 1), (4, 3)}
Answer{(1, 2), (2, -1), (3, 1), (4, 3)} represents a function.
Reason: Every element of set A has been assigned a unique image in set B.
View full question & answer→Question 1171 Mark
Which sets of ordered pairs represent functions from A = {1, 2, 3, 4} to B = {-1, 0, 1, 2, 3}? Justify.
{(1, 0), (3, 3), (2, -1), (4, 1), (2, 2)}
Answer{(1, 0), (3, 3), (2, -1), (4, 1), (2, 2)} does not represent a function.
Reason: (2, -1), (2, 2), show that element 2 ∈ A has been assigned two images -1 and 2 from set B.
View full question & answer→Question 1181 Mark
Check if the following relations are functions :
AnswerNo, Reason: Not every element of set A has been assigned an image from set B.
View full question & answer→Question 1191 Mark
Check if the following relations are functions:
AnswerNo, Reason: An element of set A has been assigned more than one element from set B.
View full question & answer→Question 1201 Mark
Check if the following relations are functions :
AnswerYes, Reason: Every element of set A has been assigned a unique element in set B.
View full question & answer→