3 Marks Question

Question 13 Marks

Find the distance between the following pairs of points:
(i) (2, 3, 5) and (4, 3, 1)
(ii) (-3, 7, 2) and (2, 4, -1)
(iii) (-1, 3, -4) and (1, -3, 4)
(iv) (2, -1, 3) and (-2, 1, 3)

Answer

Let A(2, 3, 5) and B(4, 3, 1) be two points. Then
$A B=\sqrt{(4-2)^2+(3-3)^2+(1-5)^2}=\sqrt{4+0+16}=\sqrt{20}=2 \sqrt{5}$ units
(ii) Let $A (-3,7,2)$ and $B (2,4,-1)$ be two points. Then
$\begin{array}{l}A B=\sqrt{[2-(-3)]^2+(4-7)^2+(-1-2)^2} \\ =\sqrt{(2+3)^2+(4-7)^2+(-1-2)^2}=\sqrt{25+9+9}=\sqrt{43} \text { units }\end{array}$
(iii) Let A(-1, 3, -4) and B(1, -3, 4) be two points. Then
$\begin{array}{l}A B=\sqrt{[1-(-1)]^2+(-3-3)^2+[4-(-4)]^2} \\
=\sqrt{4+36+64}=\sqrt{104}=2 \sqrt{26}\end{array}$
(iv) Let A(2, -1, 3) and B(-2, 1, 3) be two points. Then
$\begin{array}{l}A B=\sqrt{(-2-2)^2+[1-(-1)]^2+(3-3)^2} \\ =\sqrt{(-2-2)^2
(1+1)^2+(3-3)^2} \\ =\sqrt{16+4+0}=\sqrt{20}=2 \sqrt{5} \text { units }\end{array}$

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Question 23 Marks

The longest side of a triangle is 3 times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm. Find the minimum length of the shortest side.

Answer

Let the length of the shortest side be x cm.
Then length of longest side = 3x cm
length of third side = (3x - 2)cm
Perimeter of triangle = x + 3x + 3x - 2
= (7x - 2)cm
$\begin{array}{l}\text { Now } 7 x-2 \geqslant 61 \\ \Rightarrow 7 x \geqslant 61+2 \Rightarrow 7 x \geqslant 63 \Rightarrow x \geqslant 9\end{array}$
Thus the minimum length of shortest side = 9 cm

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Question 33 Marks

$\begin{array}{l}\text { If } u =\{1,2,3,4,5,6,7,8,9,10,12,24\} \\ A =\{ x : x \text { is prime and } x \leq 10\} \\ B =\{ x : x \text { is a factor of } 24\}\end{array}$
Verify the following result
i. $A - B = A \cap B^{\prime}$
ii. $(A \cup B)^{\prime}=A^{\prime} \cap B^{\prime}$
iii. $(A \cap B)^{\prime}=A^{\prime} \cup B^{\prime}$

Answer

Given, U = {1,2,3,4,5,6,7,8,9,10,12,24}
A = {2,3,5,7} B = {1,2,3,4,5,6,8,12,24}
Now, A’ = {1,4,6,8,9,10,12,24} B’ = {5,7,9,10}
$\begin{array}{l}A \cup B=\{1,2,3,4,5,6,7,8,12,24\} \\ (A \cup B)^{\prime}=\{9,10\} \\ A \cap B=\{2,3\}(A \cup B)^{\prime}=\{1,4,5,6,7,8,9,10,12,24\}\end{array}$
$\begin{array}{l}\text { (i) } A-B=A \cap B^{\prime} \\ \text { L.H.S }=A-B=\{2,3,5,7\}-\{1,2,3,4,6,8,12,24\}=\{5,7\} \text { R.H.S }=A \cap B^{\prime}=\{2,3,5,7\} \cap\{5,7,9,10\}=\{5,7\} \\ \therefore \text { L.H.S }=\text { R.H.S, }\end{array}$
$\begin{array}{l}\text { (ii) }(A \cup B)^{\prime}=A \cap B ' \\ \text { L.H.S }=(A \cup B)^{\prime}=\{9,10\} \\ \text { R.H.S = A' } \cap B^{\prime}=\{1,4,6,8,9,10,12,24\} \cap\{5,7,9,10\} \\ =\{9,10\} \\ \therefore \text { L.H.S = R.H.S, }\end{array}$
$\begin{array}{l}\text { (iii) }(A \cap B)^{\prime}=A^{\prime} \cap B^{\prime} \\ \text { L.H.S }=(A \cap B)^{\prime}=\{1,4,5,6,7,8,9,10,12,24\} \\ \text { R.H.S }=A^{\prime} \cap B^{\prime}=\{1,4,6,8,9,10,12,24\} \cap\{5,7,9,10\} \\ =\{1,4,5,6,7,8,9,10,12,34\} \\ \therefore \text { L.H.S = R.H.S }\end{array}$

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Question 43 Marks

If $(x+i y)^{1 / 3}=a+i b$, where $x, y$, $a, b \in R$, then show that $\frac{x}{a}-\frac{y}{b}=-2\left(a^2+b^2\right)$.

Answer

We have, $(x+i y)^{1 / 3}=a+i b$
$\Rightarrow x + iy =( a + ib )^3$ [cubing on both sides]
$\begin{array}{l}\Rightarrow x+i y=a^3+i^3 b^3+3 i a b(a+i b) \\ \Rightarrow x+i y=a^3-i b^3+i 3 a^2 b-3 a b^2 \\ \Rightarrow x+i y=a^3-3 a b^2+i\left(3 a^2 b-b^3\right)\end{array}$
On equating real and imaginary parts from both sides, we get
$\begin{array}{l} x = a ^3-3 ab ^2 \text { and } y =3 a ^2 b- b ^3 \\ \Rightarrow \frac{x}{a}= a ^2-3 b^2 \text { and } \frac{y}{b}=3 a ^2- b ^2\end{array}$
$\begin{array}{l}\text { Now, } \frac{x}{a}-\frac{y}{b}=a^2-3 b^2-3 a^2+b^2 \\
=-2 a^2-2 b^2=-2\left(a^2+b^2\right)\end{array}$
Hence proved.F

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Question 53 Marks

Show that a real value of x will satisfy the equation $\frac{1-i x}{1+i x}=a-i$ if $a ^2+ b ^2=1$ where a and b are real

Answer

Here $\frac{1-i x}{1+i x}=a-i b$
By componendo and dividendo, we have
$\frac{1-i x+1+i x}{1-i x-1-i x}=\frac{a-i b+1}{a-i b-1}$
$\begin{array}{l}\Rightarrow \frac{2}{-2 i x}=\frac{1-a-i b}{-(1-a+i b)} \\ \Rightarrow \frac{1}{i x}=\frac{1+a-i b}{1-a+i b} \\ \Rightarrow i x=\frac{1-a+i b}{1+a-i b} \times \frac{1+a+i b}{1+a+i b} \\ \Rightarrow i x=\frac{1-a^2-b^2+2 i b}{(1+a)^1-i^2 b^1} \\ \Rightarrow i x=\frac{1-a^2-b^2+2 i b}{(1+a)^2+b^2} \\ =\frac{1-a^2-b^2}{(1+a)^2+b^2}+\frac{2 b}{(1+a)^2+b^2} i\end{array}$
If $a^2+b^2=1$ then
$x=\frac{2 b}{(1+a)^2+b^2}$ which is real.

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Question 63 Marks

Find $a , b$ and n in the expansion of $( a + b )^{ n }$ if the first three terms of the expansion are 729,7290 and 30375 respectively.

Answer

$\begin{array}{l}\text { We have } T_1={ }^n C_0 a^n b^0=729 \ldots \text { (i) } \\ T_2={ }^n C_1 a^{n-1} b=7290 \ldots \text { (ii) } \\ T_3={ }^n C_2 a^{n-2} b^2=30375 \ldots \text { (iii) }\end{array}$
From (i) $a^{ n }=729 \ldots$ (iv)
From (ii) $na ^{ n -1} b=7290 \ldots$ (v)
From (iii) $\frac{n(n-1)}{2} a^{n-2} b^2=30375 \ldots$(vi)
Multiplying (iv) and (vi), we get
$\frac{n(n-1)}{2} a^{2 n-2} b^2=729 \times 30375 \ldots($ vii $)$
Squaring both sides of (v) we get
$n ^2 a ^{2 n -2} b^2=(7290)(7290)($ viii $)$
Dividing (vii) by (viii), we get
$\begin{array}{l}\frac{n(n-1) a^{2 n-2} b^2}{2 n^2 a^{2 n-2} b^2}=\frac{729 \times 30375}{7290 \times 7290} \\ \Rightarrow \frac{(n-1)}{2 n}=\frac{30375}{72900} \Rightarrow \frac{n-1}{2 n}=\frac{5}{12} \Rightarrow 12 n-12=10 n \\ \Rightarrow 2 n=12 \Rightarrow n=6\end{array}$
From (iv) $a^6=729 \Rightarrow a^6=(3)^6 \Rightarrow a=3$
From (v) $6 \times 3^5 \times b=7290 \Rightarrow b=5$
Thus $a =3, b=5$ and $n =6$.

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Question 73 Marks

Expand the given expression $\left(\frac{2}{x}-\frac{\pi}{2}\right)^5$

Answer

Using binomial theorem for the expansion of $\left(\frac{2}{x}-\frac{x}{2}\right)^5$ we have
$\begin{array}{l}\left(\frac{2}{x}-\frac{x}{2}\right)^5={ }^5 C_0\left(\frac{2}{x}\right)^5+{ }^5 C_1\left(\frac{2}{x}\right)^4\left(\frac{-x}{2}\right)+{ }^5 C_2\left(\frac{2}{x}\right)^3\left(\frac{-x}{2}\right)^2+{ }^5 C_3\left(\frac{2}{x}\right)^2\left(\frac{-x}{2}\right)^3 \\ +{ }^5 C_4\left(\frac{2}{x}\right)\left(\frac{-x}{2}\right)^4+{ }^5 C_5\left(\frac{-x}{2}\right)^5 \\ =\frac{32}{x^5}+5 \cdot \frac{16}{x^4} \cdot \frac{-x}{2}+10 \cdot \frac{8}{x^3} \cdot \frac{x^2}{4}+10 \cdot \frac{4}{x^2} \cdot \frac{-x^3}{8}+5 \cdot \frac{2}{x} \cdot \frac{x^4}{16}+\frac{-x^5}{32} \\ =\frac{32}{x^5}-\frac{40}{x^3}+\frac{20}{x}-5 x+\frac{5}{8} x^3-\frac{x^5}{32}\end{array}$

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Question 83 Marks

Show that the points (-2, 3, 5), (1, 2, 3) and (7, 0, -1) are collinear.

Answer

Let A(-2, 3, 5), B (1, 2, 3) and C(7, 0, -1) be three given points.
$\begin{array}{l}
\text { Then } A B=\sqrt{(1+2)^2+(2-3)^2+(3-5)^2}=\sqrt{9+1+4}=\sqrt{14} \\
B C=\sqrt{(7-1)^2+(0-2)^2+(-1-3)^2}=\sqrt{36+4+16}=\sqrt{56}=2 \sqrt{14} \\
A C=\sqrt{(7+2)^2+(0-3)^2+(-1-5)^2}=\sqrt{81+9+36}=\sqrt{126}=3 \sqrt{14}\end{array}$
Now $AC = AB + BC$
Therefore,A,B,C are collinear.

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Question 93 Marks

Determine the domain and range of the relation $R$ defined by $R=\{(x, x+5): x \in(0,1,2,3,4,5)\}$

Answer

Here $R =\{( x , x +5): x \in(0,1,2,3,4,5)\}$
$=\{(a, b): a=0,1,2,3,4,5\}$
Now a = x and b = x + 5
Putting a = 0, 1, 2, 3, 4, 5 we get b = 5, 6, 7, 8, 9, 10
$\begin{array}{l}\therefore \text { Domain of } R=\{0,1,2,3,4,5\} \\
\text { Range of } R=\{5,6,7,8,9,10\}\end{array}$

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