3 Marks Question

Question 13 Marks

Let $A=\{1,2,4,5\}, B=\{2,3,5,6\}, C=\{4,5,6,7\}$ verify the following identity:
$A \cup(B \cap C)=[(A \cup B) \cap(A \cap C)]$

Answer

L.H.S. $=\mathrm{A} \cup(\mathrm{B} \cap \mathrm{C})$
$=\{1,2,4,5\} \cup[\{2,3,5,6\} \cap\{4,5,6,7\}]$
$=\{1,2,4,5\} \cup\{5,6\}$
$=\{1,2,4,5,6\}$
R.H.S. $=(A \cup B) \cap(A \cup C)$
$=[\{1,2,4,5\} \cup\{2,3,5,6\}] \cap[\{1,2,4,5\} \cup\{4,5,6,7\}]$
$=\{1,2,3,4,5,6\} \cap\{1,2,4,5,6,7\}$
$=\{1,2,4,5,6\}$
$\therefore$ L.H.S. $=$ R.H.S. Hence verified.

View full question & answer→

Question 23 Marks

Simplify:
$\frac{1}{1+a^{m-n}+a^{m-p}}+\frac{1}{1+a^{n-m}+a^{n-p}}+\frac{1}{1+a^{p-m}+a^{p-n}}$

Answer

$\mathrm{T}_{1}=\frac{1}{1+a^{m-n}+a^{m-p}}$
$\Rightarrow \mathrm{T}_{1}=\frac{1}{1+\frac{a^{m}}{a^{n}}+\frac{a^{m}}{a^{p}}}$
[Using $\frac{a^{m}}{a^{n}}=a^{m-n}$ ]
$\Rightarrow \mathrm{T}_{1}=\frac{a^{n+p}}{a^{n+p}+a^{m+p}+a^{m+n}}$
[Using $\mathrm{a}^{\mathrm{m}} \cdot \mathrm{a}^{\mathrm{n}}=\mathrm{a}^{\mathrm{m}+\mathrm{n}}$ ]
Similarly, let $\mathrm{T}_{2}=\frac{1}{1+a^{n-m}+a^{n-p}}$
$\Rightarrow \mathrm{T}_{2}=\frac{1}{1+\frac{a^{n}}{a^{m}}+\frac{a^{n}}{a^{p}}}$
$\Rightarrow \mathrm{T}_{2}=\frac{a^{m+p}}{a^{m+p}+a^{m+n}+a^{n+p}}$
and let $\mathrm{T}_{3}=\frac{1}{1+a^{p-m}+a^{p-n}}$
$\Rightarrow \mathrm{T}_{3}=\frac{1}{1+\frac{a^{p}}{a^{m}}+\frac{a^{p}}{a^{n}}}$
$\Rightarrow \mathrm{T}_{3}=\frac{a^{m+n}}{a^{m+n}+a^{m+p}+a^{n+p}}$
$ \begin{aligned} & \quad, \mathrm{T}_{1}+\mathrm{T}_{2}+\mathrm{T}_{3}=\frac{1}{1+a^{m-n}+a^{m-p}} \\
& +\frac{1}{1+a^{n-m}+a^{n-p}}+\frac{1}{1+a^{p-m}+a^{p-n}} \\
& =\frac{a^{n+p}}{a^{n+p}+a^{m+p}+a^{m+n}}+\frac{a^{m+p}}{a^{m+p}+a^{m+n}+a^{n+p}}+\frac{a^{m+n}}{a^{m+n}+a^{m+p}}+a^{n+p} \\
& =\frac{a^{n+p}+a^{m+p}+a^{m+n}}{a^{n+p}+a^{m+p}+a^{m+n}} \\
& =1 \end{aligned} $
Thus,
$\frac{1}{1+a^{m-n}+a^{m-p}}+\frac{1}{1+a^{n-m}+a^{n-p}}+\frac{1}{1+a^{p-m}+a^{p-n}}=1$

View full question & answer→

Question 33 Marks

Find the present value of ₹ 25,000 due 10 years hence when the interest of $8 \%$ is compounded:
i. annually
ii. semi-annually
iii. quarterly
iv. continuously

Answer

Let P be the present value of S = 25000.
i. We have, $S =25000, i =\frac{8}{100}=0.08$ and $n =10$.
$\begin{array}{l}\therefore P = S (1+ i )^{- n } \\ \Rightarrow P =25000(1+0.08)^{-10}=25000(1.08)^{-10}=25000 \times 0.46319349=11579.83\end{array}$
Hence, the present value is ₹11,579.83.
ii. We have, $S=25000, i =\frac{8}{200}=0.04$ and $n =10 \times 2=20$
$\begin{array}{l}\therefore P = S (1+ i )^{- n } \\ \Rightarrow P =25000(1+0.04)^{-20}=25000(1.04)^{-20}=25000 \times 0.45638695=11409.67\end{array}$
Hence, the present value is ₹11,409.67
iii. We have, $S=25000, i =\frac{8}{400}=0.02$ and $n =10 \times 4=40$
$\begin{array}{l}\therefore P = S (1+ i )^{- n } \\ \Rightarrow P =25000(1+0.02)^{-40}=25000(1.02)^{-40}=25000 \times 0.45289042=11322.26\end{array}$
Hence, the present value is ₹11,322.26.
iv. We have, $S =25000, r =\frac{8}{100}=0.08$ and $n =10$
$\begin{array}{l}\therefore P=S e^{-r ~ n} \\ \Rightarrow P=25000 e^{-0.8}=25000 \times 0.44933=11233.25\end{array}$
Hence, the present value is ₹11,233.25

View full question & answer→

Question 43 Marks

Find the domain and the range of the given function: $f(x)=\frac{1}{1-x^{2}}$

Answer

Given $f(x)=\frac{1}{1-x^{2}}$
For $\mathrm{D}_{\mathrm{f}}, \mathrm{f}(\mathrm{x})$ must be a real number $\Rightarrow \frac{1}{1-x^{2}}$ must be a real number
$\Rightarrow 1-\mathrm{x}^{2} \neq 0 \Rightarrow \mathrm{x} \neq-1,1$
$\Rightarrow \mathrm{D}_{\mathrm{f}}=$ set of all real numbers except -1 , i.e. $\mathrm{D}_{\mathrm{f}}=\mathbf{R}-\{-1,1\}$
For $\mathrm{R}_{\mathrm{f}}$, let $\mathrm{y}=\frac{1}{1-x^{2}}$
$\Rightarrow 1-\mathrm{x}^{2}=\frac{1}{y}, \mathrm{y} \neq 0 \Rightarrow \mathrm{x}^{2}=1-\frac{1}{y}, \mathrm{y} \neq 0$
But $\mathrm{x}^{2} \geq 0$ for all $\mathrm{x} \in \mathrm{D}_{\mathrm{f}} \Rightarrow 1-\frac{1}{y} \geq 0$ but $\mathrm{y}^{2}>0, \mathrm{y} \neq 0$ (Multiply both sides by $\mathrm{y}^{2}$, a positive real number)
$\Rightarrow \mathrm{y}^{2}\left(1-\frac{1}{y}\right) \geq 0 \Rightarrow \mathrm{y}(\mathrm{y}-1) \geq 0 \Rightarrow(\mathrm{y}-0)(\mathrm{y}-1) \geq 0$
$\Rightarrow$ either $\mathrm{y} \leq 0$ or $\mathrm{y} \geq 1$ but $\mathrm{y} \neq 0$
$\Rightarrow \mathrm{R}_{\mathrm{f}}=(-\infty, 0) \cup[1, \infty)$

View full question & answer→

Question 53 Marks

Find the equation of the line passing through the point (-1, 3) and perpendicular to the line 3x - 4y - 16 = 0

Answer

The given equation is
$3 \mathrm{x}-4 \mathrm{y}-16=0$
$\Rightarrow y=\frac{3}{4} x-4$
$\therefore$ Slope of the line, $\mathrm{m}_{1}=\frac{3}{4}$
$\therefore$ Slope of the perpendicular line, $\mathrm{m}_{2}=-\frac{4}{3}$.
Since the line passes through $(-1,3)$, so the equation of the line is
$y-y_{0}=m\left(x-x_{0}\right)$
$\Rightarrow \mathrm{y}-3=\mathrm{m}_{2}(\mathrm{x}+1)$
$\Rightarrow \mathrm{y}-3=-\frac{3}{4}(\mathrm{x}+1)$
$\Rightarrow 3 y-9=-4 x-4$
$\Rightarrow 4 x+3 y-5=0$

View full question & answer→

Question 63 Marks

Find the sum of first 'n' terms of the series 0.7 + 0.77 + 0.777 +...

Answer

We have, $0.7+0.77+0.777+\ldots$ to $n$ terms
$7 \times 0.1+7 \times 0.11+7 \times 0.111+\ldots$. to n terms
$=7\{0.1+0.11+0.111+\ldots$ to $n$ terms $\}$
$=\frac{7}{9}\{0.9+0.99+0.999+\ldots$ to $n$ terms $\}$
$=\frac{7}{9}\left\{\frac{9}{10}+\frac{99}{100}+\frac{999}{1000}+\ldots\right.$ to $n$ terms $\}$
$=\frac{7}{9}\left\{\left(1-\frac{1}{10}\right)+\left(1-\frac{1}{100}\right)+\left(1-
\frac{1}{1000}\right)+\ldots\right.$ to $n$ terms $\}$
$=\frac{7}{9}\left\{\left(1-\frac{1}{10}\right)+\left(1-\frac{1}{10^{2}}\right)+\left(1-\frac{1}{10^{3}}\right)+\ldots\left(1-\frac{1}{10^{n}}\right)\right\}$$=\frac{7}{9}\left\{n-\left(\frac{1}{10}+\frac{1}{10^{2}}+\frac{1}{10^{2}}+\ldots+\frac{1}{10^{n}}\right)\right\}$
$=\frac{7}{9}\left\{n-\frac{1}{10} \frac{\left\{1-\left(\frac{1}{10}\right)^{n}\right\}}{\left(1-\frac{1}{10}\right)}\right\}$
$=\frac{7}{9}\left\{n-\frac{1}{9}\left(1-\frac{1}{10^{n}}\right)\right\}$
$=\frac{7}{81}\left\{9 n-1+\frac{1}{10^{n}}\right\}$

View full question & answer→

Question 73 Marks

Given a G.P. with $\mathrm{a}=729$ and 7 th term $=64$, determine $\mathrm{S}_{7}$.

Answer

Let $r$ be the common ratio of G.P. Then
$a_{7}=64 \Rightarrow \mathrm{ar}^{7-1}=64 \Rightarrow 729 \mathrm{r}^{6}=64$
$\Rightarrow \quad r^{6}=\frac{64}{729} \Rightarrow r^{6}=\left(\frac{2}{3}\right)^{6}$ or $\left(-\frac{2}{3}\right)^{6}$
$\Rightarrow \quad r=\frac{2}{3}$ or $-\frac{2}{3}$
When $\mathrm{r}=\frac{2}{3}, S_{7}=\frac{a\left(1-r^{7}\right)}{1-r}=\frac{729\left(1-\left(\frac{2}{3}\right)^{7}\right)}{1-\frac{2}{3}} 3 \times 729\left(1-\left(\frac{2}{3}\right)^{7}\right)$
$=2187-128=2059$
When $\mathrm{r}=\frac{2}{3}, S_{7}=\frac{729\left(1-\left(-\frac{2}{3}\right)^{7}\right)}{1-\left(-\frac{2}{3}\right)}=\frac{3}{5} \times 729\left(1+\frac{2^{7}}{3^{7}}\right)$
$=\frac{1}{5}(2187+128)=\frac{2315}{5}=463$

View full question & answer→

Applied Maths 3 Marks Question

Test yourself on this topic