Q: Explain why 7 x 11 x 13 + 13 and 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 are composite numbers.

Given \$ 7 \times 11\times 13 +13\$\$=13 \times (7 \times 11 +1)=3 \times 78\$ This number is multiple of two integers.Hence it has more than two factors.Hence it is a composite number.similarly in\$7 \times 6 \$\$ \times 5 \$\$ \times 4 \$\$ \times 3 \$ \$ \times 2 \$\$ \times 1 +5\$ \$= 5\$(\$7 \times 6 \$\$ \times 4 \$\$ \times 3 \$ \$ \times 2 \$\$ \times 1 +1 )=5 \times 1009\$ This number is multiple of two integers.Hence it has more than two factors.Hence it is a composite number.

Q: Explain why \$3\times 5 \times 7+7\$ is a composite number.

\$3 \times 5 \times 7 + 7 = 105 + 7 = 112\$. \$112\$ is an even number and is therefore a composite number. Also, \$112\$ is divisible by other numbers such as \$2, 4, 7, 8, 14, 16, 28\$ and \$56\$ apart from itself \$112\$ and \$1\$Hence \$3\times5\times 7+7\$ is a composite number

Q: If \$x -iy =\sqrt{\frac{a -ib}{c -id}}\$ prove that \$(x^{2} +y^{2})^{2} =\frac{a^{2} + b^{2}}{c^{2} + d^{2}}\$.

Given that,\$x - iy = \sqrt{\dfrac{a - ib}{c - id}} \$\$\implies (x - iy)^2 = \dfrac{a-ib}{c-id} \times \dfrac{c+id}{c+id} = \dfrac{(ac + bd) - i(bc - ad)}{c^2 + d^2}\$\$\implies (x^2 - y^2) - i(2xy) = \left(\dfrac{ac+bd}{c^2 + d^2} \right)- i \left(\dfrac{bc - ad}{c^2 + d^2}\right)\$Equating real and imaginary parts on both sides, we get\$x^2 - y^2 = \dfrac{ac+bd}{c^2 + d^2}\$ and \$2xy = \dfrac{bc - ad}{c^2 + d^2}\$Now, \$(x + iy)^2 = (x^2 - y^2) + i(2xy) = \left(\dfrac{ac + bd}{c^2 + d^2}\right) + i \left(\dfrac{bc - ad}{c^2 + d^2}\right)\$\$\implies (x + iy)^2= \dfrac{(ac + bd) + i(bc - ad)}{c^2 + d^2} = \dfrac{(a + ib)(c - id)}{(c + id)(c - id)} = \dfrac{a + ib}{c + id}\$\$\implies x + iy = \sqrt{\dfrac{a + ib}{c + id}}\$\$LHS = (x^2 + y^2)^2 = [(x - iy)(x + iy)]^2 =(x - iy)^2 (x + iy)^2\$ \$= \left(\dfrac{a-ib}{c-id}\right)\left(\dfrac{a+ib}{c+id}\right)\$ \$= \dfrac{a^2 + b^2}{c^2 + d^2} \$ \$= RHS\$Hence proved.

Q: If \$(x+iy)(2-3i)=4+i\$ then (x, y) =

\$(x+iy)(2-3i)=4+i\$\$2x-(3x)i+(2y)i-3yi^{2}=4+i\$\$\underbrace{2x+3y}_{Real }+\underbrace{(2y-3x)}_{Imaginary }i=4+i\$Comparing the real & imaginary parts,\$2x+3y=4\$--------------------------(1)\$2y-3x=1\$----------------------------(2)Solving eq(1) & eq(2),\$4x+6y=8\$\$-9x+6y=3\$\$13x=5\Rightarrow x=\dfrac{5}{13}\$\$y=\dfrac{14}{13}\$\$\therefore (x,y)=\left (\dfrac{5}{13},\dfrac{14}{13} \right )\$

Q: If \$ z_1 , z_2 \$ and \$ z_3 \$ are complex numbers such that \$ | z_1 | = | z_2 | = | z_3 | = \left| \dfrac {1}{z_1} + \dfrac {1}{z_2} + \dfrac {1}{z_3} \right| = 1 \$ then \$ | z_1 + z_2 + z_3 | \$ is :

\$ 1 = \left| \dfrac {1}{z_1} + \dfrac {1}{z_2} + \dfrac {1}{z_3} \right| = \left| \dfrac {z_1\bar{z_1}}{z_1} + \dfrac {z_2\bar{z_2}}{z_2} + \dfrac {z_3\bar{z_3}}{z_3} \right| \$\$ [ \because | \bar{z_1}^2 = 1 = z_1 \bar{z_1} ] \$\$ = | \bar{z_1} + \bar{z_2} + \bar{z_3} | = | \overline{z_1 + z_2 + z_3 } | = | z_1 + z_2 + z_3 | \$ \$ \quad \quad \quad [ \because | \bar{z_1} | = | z_1 | ] \$

Q: If \$(x+iy)^3=u+iv\$, then prove that \$\displaystyle \frac{u}{x}+\frac{v}{y} =4(x^2 - y^2)\$.

Given, \$(x+iy)^3=u+iv\$\$x^3+(iy)^3 + 3. x. iy(x+iy) = u+iv\$\$x^3+i^3y^3+3x^2yi+3xy^2 i^2=u+iv\$\$x^3-iy^3+3x^2yi-3xy^2=u+iv\$\$(x^3-3xy^2)=i(3x^2y-y^3)=u+iv\$On equating real and imaginary parts, we get\$u=x^3-3xy^2, v=3x^2y-y^3\$\$\displaystyle \therefore \frac{u}{x}+\frac{v}{y}=\frac{x^3-3xy^2}{x}+\frac{3x^2y-y^3}{y}\$\$\displaystyle=\frac{x(x^2-3y^2)}{x}+\frac{y(3x^2-y^2)}{y}\$\$=x^2-3y^2+3x^2-y^2=4x^2-4y^2=4(x^2-y^2)\$

Q: Number of prime numbers from 1 to 50 are

There are 15 prime numbers from 1 to 50. They are \$2, 3, 5, 7, 11, 13, 17,19, 23, 29, 31,37,41,43,47.\$

Q: Find the conjugate and modulus of \$7-24i\$

Let \$z = 7 - 24i\$\$\left| z \right| = \sqrt {{{\left( 7 \right)}^2} + {{\left( { - 24} \right)}^2}} \$\$ = \sqrt {49 + 576} \$\$ = \sqrt {625} \$\$ = 24\$Also \$\bar{z} = 7 + 24i\$

Question 1

Explain why 7 x 11 x 13 + 13 and 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 are composite numbers.

Accepted Answer

Given $ 7 \times 11\times 13 +13$$=13 \times (7 \times 11 +1)=3 \times 78$ This number is multiple of two integers.Hence it has more than two factors.Hence it is a composite number.similarly in$7 \times 6 $$ \times 5 $$ \times 4  $$ \times 3 $ $ \times 2 $$ \times 1 +5$   $= 5$($7 \times 6 $$ \times 4  $$ \times 3 $ $ \times 2 $$ \times 1 +1 )=5 \times 1009$   This number is multiple of two integers.Hence it has more than two factors.Hence it is a composite number.

Question 2

If $\alpha$ and $\beta$ are different complex numbers with $|\beta|=1$, then find $\left|\dfrac{\beta -\alpha}{1-\bar{\alpha}\beta}\right|$.

Accepted Answer

Toppr Ask Answer

Question 3

Explain why $3\times 5 \times 7+7$ is a composite number.

Accepted Answer

$3 \times 5 \times 7 + 7 = 105 + 7 = 112$. $112$ is an even number and is therefore a composite number. Also, $112$ is divisible by other numbers such as $2, 4, 7, 8, 14, 16, 28$ and $56$ apart from itself $112$ and $1$Hence $3\times5\times 7+7$ is a composite number

Question 4

If $x -iy =\sqrt{\frac{a -ib}{c -id}}$ prove that $(x^{2} +y^{2})^{2} =\frac{a^{2} + b^{2}}{c^{2} + d^{2}}$.

Accepted Answer

Given that,$x - iy = \sqrt{\dfrac{a - ib}{c - id}} $$\implies (x - iy)^2 = \dfrac{a-ib}{c-id} \times \dfrac{c+id}{c+id} = \dfrac{(ac + bd) - i(bc - ad)}{c^2 + d^2}$$\implies (x^2 - y^2) - i(2xy) = \left(\dfrac{ac+bd}{c^2 + d^2} \right)- i \left(\dfrac{bc - ad}{c^2 + d^2}\right)$Equating real and imaginary parts on both sides, we get$x^2 - y^2 = \dfrac{ac+bd}{c^2 + d^2}$  and $2xy = \dfrac{bc - ad}{c^2 + d^2}$Now, $(x + iy)^2 = (x^2 - y^2) + i(2xy) = \left(\dfrac{ac + bd}{c^2 + d^2}\right) + i \left(\dfrac{bc - ad}{c^2 + d^2}\right)$$\implies (x + iy)^2= \dfrac{(ac + bd) + i(bc - ad)}{c^2 + d^2} = \dfrac{(a + ib)(c - id)}{(c + id)(c - id)} = \dfrac{a + ib}{c + id}$$\implies x + iy = \sqrt{\dfrac{a + ib}{c + id}}$$LHS = (x^2 + y^2)^2 = [(x - iy)(x + iy)]^2 =(x - iy)^2 (x + iy)^2$           $= \left(\dfrac{a-ib}{c-id}\right)\left(\dfrac{a+ib}{c+id}\right)$           $= \dfrac{a^2 + b^2}{c^2 + d^2} $           $= RHS$Hence proved.

Question 5

If $(x+iy)(2-3i)=4+i$ then (x, y) =

Accepted Answer

$(x+iy)(2-3i)=4+i$$2x-(3x)i+(2y)i-3yi^{2}=4+i$$\underbrace{2x+3y}_{Real }+\underbrace{(2y-3x)}_{Imaginary }i=4+i$Comparing the real & imaginary parts,$2x+3y=4$--------------------------(1)$2y-3x=1$----------------------------(2)Solving eq(1)  &  eq(2),$4x+6y=8$$-9x+6y=3$$13x=5\Rightarrow x=\dfrac{5}{13}$$y=\dfrac{14}{13}$$\therefore (x,y)=\left (\dfrac{5}{13},\dfrac{14}{13} \right )$

Question 6

If $ z_1 , z_2 $ and $ z_3 $ are complex numbers such that $ | z_1 | = | z_2 | = | z_3 | = \left| \dfrac {1}{z_1} + \dfrac {1}{z_2} + \dfrac {1}{z_3} \right| = 1 $ then $ | z_1 + z_2 + z_3 | $ is :

Accepted Answer

$ 1 = \left| \dfrac {1}{z_1} + \dfrac {1}{z_2} + \dfrac {1}{z_3} \right| = \left| \dfrac {z_1\bar{z_1}}{z_1} + \dfrac {z_2\bar{z_2}}{z_2} + \dfrac {z_3\bar{z_3}}{z_3} \right| $$ [ \because | \bar{z_1}^2 = 1 = z_1 \bar{z_1} ] $$ = | \bar{z_1}  + \bar{z_2} + \bar{z_3} | = | \overline{z_1 + z_2 + z_3 } | = | z_1 + z_2 + z_3 | $ $ \quad \quad \quad [ \because | \bar{z_1} | = | z_1 | ] $

Question 7

If $(x+iy)^3=u+iv$, then prove that $\displaystyle \frac{u}{x}+\frac{v}{y} =4(x^2 - y^2)$.

Accepted Answer

Given, $(x+iy)^3=u+iv$$x^3+(iy)^3 + 3. x. iy(x+iy) = u+iv$$x^3+i^3y^3+3x^2yi+3xy^2 i^2=u+iv$$x^3-iy^3+3x^2yi-3xy^2=u+iv$$(x^3-3xy^2)=i(3x^2y-y^3)=u+iv$On equating real and imaginary parts, we get$u=x^3-3xy^2, v=3x^2y-y^3$$\displaystyle \therefore \frac{u}{x}+\frac{v}{y}=\frac{x^3-3xy^2}{x}+\frac{3x^2y-y^3}{y}$$\displaystyle=\frac{x(x^2-3y^2)}{x}+\frac{y(3x^2-y^2)}{y}$$=x^2-3y^2+3x^2-y^2=4x^2-4y^2=4(x^2-y^2)$

Question 8

Number of prime numbers from 1 to 50 are

Accepted Answer

There are 15 prime numbers from 1 to 50. They are $2, 3, 5, 7, 11, 13, 17,19, 23, 29, 31,37,41,43,47.$

Question 9

Which of the following pair of numbers is co-prime?

Accepted Answer

In number theory, Co-prime numbers are those numbers if the only positive factor that divides them is $1$.Here,Option A: $(17,15)$ Factor that divides both of them is $1$.Option B: $(10,16)$ Factors that divides both oh them are $1,2.$Option C: $(20,24)$ Factors that divide both of them are $ 1,2,4.$Option C: $(31,62)$ Factors that divide both oh them are $1,31.$In reference to above definition $(17,15)$ are co-prime numbers.Hence Option A is correct.

Question 10

Find the conjugate and modulus of $7-24i$

Accepted Answer

Let $z = 7 - 24i$$\left| z \right| = \sqrt {{{\left( 7 \right)}^2} + {{\left( { - 24} \right)}^2}} $$ = \sqrt {49 + 576} $$ = \sqrt {625} $$ = 24$Also $\bar{z} = 7 + 24i$