3 Marks Question

Question 13 Marks

$\text { If } u =\{1,2,3,4,5,6,7,8,9,10,12,24\}$
$A =\{ x : x \text { is prime and } x \leq 10\}$
$B =\{ x : x \text { is a factor of } 24\}$
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, $A’ = \{1,4,6,8,9,10,12,24\} B’ = \{5,7,9,10\}$
$A \cup B=\{1,2,3,4,5,6,7,8,12,24\}$
$(A \cup B)^{\prime}=\{9,10\}$
$A \cap B=\{2,3\}(A \cup B)^{\prime}=\{1,4,5,6,7,8,9,10,12,24\}$
$\text { (i) } A-B=A \cap B^{\prime}$
$\text { L.H.S }=A-B=\{2,3,5,7\}-\{1,2,3,4,6,8,12,24\}=\{5,7\}$
$\text { R.H.S }=A \cap B^{\prime}=\{2,3,5,7\} \cap\{5,7,9,10\}=\{5,7\}$
$\therefore \text { L.H.S }=\text { R.H.S, }$
$\text { (ii) }(A \cup B)^{\prime}=A \cap B '$
$\text { L.H.S }=(A \cup B)^{\prime}=\{9,10\}$
$\text { R.H.S = A' } \cap B^{\prime}=\{1,4,6,8,9,10,12,24\} \cap\{5,7,9,10\}$
$=\{9,10\}$
$\therefore \text { L.H.S = R.H.S, }$
$\text { (iii) }(A \cap B)^{\prime}=A^{\prime} \cap B^{\prime}$
$\text { L.H.S }=(A \cap B)^{\prime}=\{1,4,5,6,7,8,9,10,12,24\}$
$\text { R.H.S }=A^{\prime} \cap B^{\prime}=\{1,4,6,8,9,10,12,24\} \cap\{5,7,9,10\}$
$=\{1,4,5,6,7,8,9,10,12,34\}$
$\therefore \text { L.H.S = R.H.S }$

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Question 23 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
$\frac{1-i x+1+i x}{1-i x-1-i x}=\frac{a-i b+1}{a-i b-1}$
$\Rightarrow \frac{2}{-2 i x}=\frac{1-a-i b}{-(1-a+i b)}$
$\Rightarrow \frac{1}{i x}=\frac{1+a-i b}{1-a+i b}$
$\Rightarrow i x=\frac{1-a+i b}{1+a-i b} \times \frac{1+a+i b}{1+a+i b}$
$\Rightarrow i x=\frac{1-a^2-b^2+2 i b}{(1+a)^1-i^2 b^1}$
$\Rightarrow i x=\frac{1-a^2-b^2+2 i b}{(1+a)^2+b^2}$
$=\frac{1-a^2-b^2}{(1+a)^2+b^2}+\frac{2 b}{(1+a)^2+b^2} i$
If $a^2+b^2=1$ then
$x=\frac{2 b}{(1+a)^2+b^2}$ which is real.

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Question 33 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$]$
$\Rightarrow x+i y=a^3+i^3 b^3+3 i a b(a+i b)$
$\Rightarrow x+i y=a^3-i b^3+i 3 a^2 b-3 a b^2$
$\Rightarrow x+i y=a^3-3 a b^2+i\left(3 a^2 b-b^3\right)$
On equating real and imaginary parts from both sides,
we get $x = a ^3-3 ab ^2$ and $y =3 a ^2 b- b ^3$
$\Rightarrow \frac{x}{a}= a ^2-3 b^2$ and $\frac{y}{b}=3 a ^2- b ^2$
Now, $\frac{x}{a}-\frac{y}{b}=a^2-3 b^2-3 a^2+b^2$
$=-2 a^2-2 b^2=-2\left(a^2+b^2\right)$

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Question 43 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

We have $T_1={ }^n C_0 a^n b^0=729 \ldots \text { (i) }$
$T_2={ }^n C_1 a^{n-1} b=7290 \ldots \text { (ii) }$
$T_3={ }^n C_2 a^{n-2} b^2=30375 \ldots \text { (iii) }$
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
$\frac{n(n-1) a^{2 n-2} b^2}{2 n^2 a^{2 n-2} b^2}=\frac{729 \times 30375}{7290 \times 7290}$
$\Rightarrow \frac{(n-1)}{2 n}=\frac{30375}{72900}$
$\Rightarrow \frac{n-1}{2 n}=\frac{5}{12}$
$\Rightarrow 12 n-12=10 n$
$\Rightarrow 2 n=12$
$\Rightarrow n=6$
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 53 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
$\left(\frac{2}{x}-\frac{x}{2}\right)^5={ }^5 C_0\left(\frac{2}{x}\right)^5+{ }^5 C_1\left(\frac{2}{x}\right)^4\left(\frac{-x}{2}\right)+{ }^5 C_2\left(\frac{2}{x}\right)^3\left(\frac{-x}{2}\right)^2+{ }^5 C_3\left(\frac{2}{x}\right)^2\left(\frac{-x}{2}\right)^3 +{ }^5 C_4\left(\frac{2}{x}\right)\left(\frac{-x}{2}\right)^4+{ }^5 C_5\left(\frac{-x}{2}\right)^5$
$=\frac{32}{x^5}+5 \cdot \frac{16}{x^4} \cdot \frac{-x}{2}+10 \cdot \frac{8}{x^3} \cdot \frac{x^2}{4}+10 \cdot \frac{4}{x^2} \cdot \frac{-x^3}{8}+5 \cdot \frac{2}{x} \cdot \frac{x^4}{16}+\frac{-x^5}{32}$
$=\frac{32}{x^5}-\frac{40}{x^3}+\frac{20}{x}-5 x+\frac{5}{8} x^3-\frac{x^5}{32}$

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Question 63 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
$A B=\sqrt{[2-(-3)]^2+(4-7)^2+(-1-2)^2}$
$=\sqrt{(2+3)^2+(4-7)^2+(-1-2)^2}=\sqrt{25+9+9}=\sqrt{43} \text { units }$
$(iii)$ Let $A(-1, 3, -4)$ and $B(1, -3, 4)$ be two points. Then
$A B=\sqrt{[1-(-1)]^2+(-3-3)^2+[4-(-4)]^2}$
$=\sqrt{4+36+64}=\sqrt{104}=2 \sqrt{26}$
$(iv)$ Let $A(2, -1, 3)$ and $B(-2, 1, 3)$ be two points. Then
$A B=\sqrt{(-2-2)^2+[1-(-1)]^2+(3-3)^2}$
$=\sqrt{(-2-2)^2 (1+1)^2+(3-3)^2}$
$=\sqrt{16+4+0}=\sqrt{20}=2 \sqrt{5} \text { units }$

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Question 73 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.
$\text { Then } A B=\sqrt{(1+2)^2+(2-3)^2+(3-5)^2}=\sqrt{9+1+4}=\sqrt{14}$
$B C=\sqrt{(7-1)^2+(0-2)^2+(-1-3)^2}=\sqrt{36+4+16}=\sqrt{56}=2 \sqrt{14}$
$A C=\sqrt{(7+2)^2+(0-3)^2+(-1-5)^2}=\sqrt{81+9+36}=\sqrt{126}=3 \sqrt{14}$
Now $AC = AB + BC$
Therefore$,A,B,C$ are collinear.

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Question 83 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 $= x + 3x + 3x - 2$
$= (7x - 2)\ cm$
Now $7 x-2 \geqslant 61$
$\Rightarrow 7 x \geqslant 61+2$
$\Rightarrow 7 x \geqslant 63$
$\Rightarrow x \geqslant 9$
Thus the minimum length of shortest side $= 9 \ cm$

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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 $b = 5, 6, 7, 8, 9, 10$
$\therefore \text { Domain of } R=\{0,1,2,3,4,5\}$
$\text { Range of } R=\{5,6,7,8,9,10\}$

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