2 Marks Questions - MATHS STD 11 Science Questions - Vidyadip

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2 Marks Questions

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21 questions · self-marked practice — reveal the answer and mark yourself.

Question 12 Marks

If $\left(\frac{1+i}{1-i}\right)^{m} = 1$ then find the least positive integral value of $m.$

Answer

Given $\left(\frac{1+i}{1-i}\right)^{m} = 1$
Now, $\left(\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right)^{m}=1 [$multiply divide numerator and denominator by $1+i]$
$\Rightarrow$$\left[\frac{(1+i)^{2}}{1^{2}-i^{2}}\right]^{m}=1$
$\Rightarrow$$\left(\frac{1^{2}+\mathrm{i}^{2}+2 \mathrm{i}}{1+1}\right)^{\mathrm{m}}=1$
$\Rightarrow$$\left(\frac{1-1+2 i}{2}\right)^{m}=1$
$\Rightarrow$$\left(\frac{2 i}{2}\right)^{m}=1$
$\Rightarrow i^m = 1$
We can also write, $i^m = i^{4k}$
On equating the powers,
Thus, $m = 4k,$ Where k is some integer.
$\therefore 1$ is the least positive integer.
Least positive integral value of m is $4 \times 1 = 4$

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Question 22 Marks

If $(a + ib) (c + id) (e+ if) (g + ih) = A + iB$ then show that
$(a^2 + b^2)(c^2 + d^2)(e^2 + f^2)(g^2 + h^2) = A^2 + B^2$

Answer

Here $(a + ib) (c + id) (e+ if) (g + ih) = A + iB$
Taking modulus on both sides
$| (a + ib) (c + id) (e+ if) (g + ih) | = |A + iB|$
$ \Rightarrow |a + ib | |c + id | | e + if | |g + ih| = | A + iB|$
$ \Rightarrow \left( {\sqrt {{a^2} + {b^2}} } \right)\left( {\sqrt {{c^2} + {d^2}} } \right)\left( {\sqrt {{e^2} + {f^2}} } \right)$$\left( {{g^2} + {h^2}} \right) = \sqrt {{A^2} + {B^2}} $
Squaring both sides
$(a^2 + b^2)(c^2 + d^2)(e^2 + f^2)(g^2 + h^2) = A^2 + B^2$

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Question 32 Marks

Find the number of non-zero integral solutions of the equation ${\left| {1 - i} \right|^x} = {2^x}$.

Answer

Here ${\left| {1 - i} \right|^x} = {2^x}$
$ \Rightarrow {\left[ {\sqrt {{{(1)}^2} + {{( - 1)}^2}} } \right]^x} = {2^x}$$ \Rightarrow {(\sqrt 2 )^x} = {2^x}$
$ \Rightarrow {2^{\frac{x}{2}}} = {2^x} \Rightarrow \frac{x}{2} = x \Rightarrow \frac{x}{2} - x = 0$$ \Rightarrow \frac{{ - x}}{2} = 0$
$ \Rightarrow x = 0$
Thus the given equation has no non-zero integral solution.

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Question 42 Marks

Find the modulus of $\frac{{1 + i}}{{1 - i}} - \frac{{1 - i}}{{1 + i}}$.

Answer

$\left| {\frac{{1 + i}}{{1 - i}} - \frac{{1 - i}}{{1 + i}}} \right| = \left| {\frac{{{{(1 + i)}^2} - {{(1 - i)}^2}}}{{(1 - i)(1 + i)}}} \right|$
$ = \left| {\frac{{1 + {i^2} + 2i - 1 - {i^2} + 2i}}{{1 - {i^2}}}} \right|$
$\left| {\frac{{4i}}{2}} \right| = |2i| = \sqrt 4 = 2$.

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Question 52 Marks

Let $z_1 = 2 - i, z_2 = -2 + i.$ Find $\operatorname{Re} \left( {\frac{{{z_1}{z_2}}}{{{{\overline z }_1}}}} \right)$

Answer

Here $z_1 = 2 - i $ and $z_2 = -2 + i$
$\therefore \overline {{z_1}} = 2 + i$Now $z_1z_2 = (2 -i)(-2 + i)$
$= - 4 + 2i + 2i - i^2 = (-4 + 1) + 4i$
$= - 3 + 4i$
$\therefore \frac{{{z_1}{z_2}}}{{\overline {{z_1}} }} = \frac{{ - 3 + 4i}}{{2 + i}} \times \frac{{2 - i}}{{2 - i}}$$ = \frac{{ - 6 + 3i + 8i - 4{i^2}}}{{4 - {i^2}}}$
$ = \frac{{( - 6 + 4) + 11i}}{{4 + 1}} = \frac{{ - 2 + 11i}}{5} = \frac{{ - 2}}{5} + \frac{{11}}{5}i$
$\therefore \operatorname{Re} \left( {\frac{{{z_1}{z_2}}}{{\overline {{z_1}} }}} \right) = \frac{{ - 2}}{5}$

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Question 62 Marks

Express the complex number ${\left( {\frac{1}{3} + 3i} \right)^3}$ in the form of $a + ib.$

Answer

$\left(\frac13+3i\right)^3\;=\left(\frac13\right)^3+{(3i)^3}+3\times\left(\frac13\right)(3i)\left(\frac13+3i\right)$
$=\frac1{27}+27i^3+i+3i\left(\frac13+3i\right)$
$=\frac1{27}+27(-i)+i+9i^2$
$\left[ {\because {i^3} = - i\;and\;{i^2} = - 1} \right]$
$ =\frac1{27}-27i+i-9$
$=\left(\frac1{27}-9\right)-26i\;$
$=\frac{-242}{27}-26i$

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Question 72 Marks

Express the following in the form of a + ib.
$\frac { ( 3 + \sqrt { 5 } i ) ( 3 - \sqrt { 5 } i ) } { ( \sqrt { 3 } + \sqrt { 2 } i ) - ( \sqrt { 3 } - \sqrt { 2 }i ) }$

Answer

We have, $\frac { ( 3 + \sqrt { 5 }i ) ( 3 - \sqrt { 5 }i ) } { ( \sqrt { 3 } + \sqrt { 2 } i ) - ( \sqrt { 3 } - \sqrt { 2 }i ) }$
= $\frac { 9 - 3 \sqrt { 5 }i+3 \sqrt {5 }i - \sqrt { 5 }i \sqrt { 5 }i } { \sqrt { 3 } + \sqrt { 2 } i - \sqrt { 3 } + \sqrt { 2 } i }$
= $\frac { 9 + 5 } { 2 \sqrt { 2 } i } = \frac { 14 } { 2 \sqrt { 2 } i } = \frac { 7 } { \sqrt { 2 } i } \times \frac { \sqrt { 2 } i } { \sqrt { 2 } i }$
= $\frac { 7 \sqrt { 2 } i } { 2 i ^ { 2 } } = \frac { 7 \sqrt { 2 } i } { - 2 }$ = 0 - i $\frac { 7 \sqrt { 2 } } { 2 }$ = a + ib [say]
where, a = 0 and b = $\frac { - 7 \sqrt { 2 } } { 2 }$

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Question 82 Marks

Find the multiplicative inverse of the complex numbers $ = \sqrt 5 + 3i$

Answer

M.I. of $ = \sqrt 5 + 3i$
$ = \frac{1}{{\sqrt 5 + 3i}} = \frac{1}{{\sqrt 5 + 3i}} \times \frac{{\sqrt 5 - 3i}}{{\sqrt 5 - 3i}}$
$ = \frac{{\sqrt 5 - 3i}}{{{{(\sqrt 5 )}^2} - {{(3i)}^2}}}$
$ = \frac{{\sqrt 5 - 3i}}{{5 - 9{i^2}}} = \frac{{\sqrt 5 - 3i}}{{5 + 9}} = \frac{1}{{14}}(\sqrt 5 - 3i)$

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Question 92 Marks

Find the multiplicative inverse of the complex numbers 4 - 3i

Answer

M.I. of (4 - 3i)$ = \frac{1}{{4 - 3i}} = \frac{1}{{4 - 3i}} \times \frac{{4 + 3i}}{{4 + 3i}} = \frac{{4 + 3i}}{{{{(4)}^2} - {{(3i)}^2}}}$
$ = \frac{{4 + 3i}}{{16 - 9{i^2}}} = \frac{{4 + 3i}}{{16 + 9}} = \frac{1}{{25}}(4 + 3i)$

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Question 102 Marks

Express the complex number ${\left( { - 2 - \frac{1}{3}i} \right)^3}$ in the form of a + ib.

Answer

${\left( { - 2 - \frac{1}{3}i} \right)^3}$$ = - {\left( {2 + \frac{1}{3}i} \right)^3}$
$=-\left[{(2)}^3+\left(\frac13i\right)^3+3\times{(2)}^2\times\frac13i\right.$$\left. { + 3 \times 2 \times {{\left( {\frac{1}{3}i} \right)}^2}} \right]$
$=-\left[8+\frac1{27}i^3+4i+\frac23i^2\right]$$ = - \left[ {8 - \frac{1}{{27}}i + 4i - \frac{2}{3}} \right]\left[ {\begin{array}{*{20}{c}} {\because {i^3} = - i} \\ {{i^2} = - 1} \end{array}} \right]$
$ = \left[ {\left( {8 - \frac{2}{3}} \right) + \left( {4 - \frac{1}{{27}}} \right)i} \right.$
$ = - \left[ {\frac{{22}}{3} + \frac{{107}}{{27}}i} \right] = \frac{{ - 22}}{3} - \frac{{107}}{{27}}i$

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Question 112 Marks

Solve $x^2 + 2 = 0$

Answer

We have $x^2 + 2 = 0$
or $x^2 = -2$ taking squareroot both sides
i.e.,$x =\pm \sqrt{-2} = \pm \sqrt2 i$

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Question 122 Marks

Convert the complex number $\frac {-16}{1+\sqrt3i}$ into polar form.

Answer

Given complex number $\frac {-16}{1+\sqrt3i}$ convert the complex number in $x +iy$ form
$= \frac {-16}{1+\sqrt3i} \times \frac {1-\sqrt3i}{1-\sqrt3i}$
$= \frac {-16(1-\sqrt3i)}{1-(\sqrt3i)^2} = \frac {-16(1-\sqrt3i)}{1+3}$
$= -4 (1 - \sqrt3i) = -4 + 4\sqrt3i$
Let $-4 = r\; \cos \;\theta, 4\sqrt3 = r\; \sin \;\theta$
By squaring and adding, we get
$16 + 48 = r^2(\cos^2 \theta + \sin^2 \theta)$
which gives $r^2 = 64,$ i.e., $r = 8$
Hence $\cos \theta = - \frac 12, \sin \theta = \frac {\sqrt3}2$
$\theta = \pi - \frac {\pi}3 = \frac {2\pi}{3}$
Thus, the required polar form is $r(\cos \theta + i \sin \theta ) = 8\left(\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} \right)$

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Question 132 Marks

Represent the complex number $z = 1 + i \sqrt { 3 }$ in the polar form.

Answer

We have, $z = 1 + i \sqrt { 3 }$
Let $1 + i \sqrt { 3 } = r ( \cos \theta + i \sin \theta) ...(i)$
On equating real and imaginary parts both sides, we get
$r \cos \theta = 1$ and$ r \sin \theta = \sqrt { 3 } ...(ii)$
On squaring and adding Eqs. (i) and (ii), we get
$r^2 (\cos^2 \theta + \sin^2 \theta) = 1 + 3$
$\Rightarrow r^2 = 4$
$\Rightarrow r = 2$
$\therefore \cos \theta = \frac { 1 } { 2 }$ and $\sin \theta = \frac { \sqrt { 3 } } { 2 }$
Since, both $\cos \theta$ and $\sin \theta$ are positive.
So, $\theta$ lies in first quadrant.
$\therefore \theta = \frac { \pi } { 3 }$
On putting $r = 2$ and $\theta = \frac { \pi } { 3 }$ in Eq. $(i),$ we get
polar form of $z = 2 \left( \cos \frac { \pi } { 3 } + i \sin \frac { \pi } { 3 } \right)$

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Question 142 Marks

Find the multiplicative inverse of $2 - 3i.$

Answer

Let $z = 2 - 3i$
Then, $\overline { z }= 2 + 3i$
and $|z|^2 = 2^2 + (- 3)^2 = 4 + 9 = 13$
Therefore, the multiplicative inverse of $2 - 3i$ is given by
$z^{-1} = \frac { \overline { z } } { | z | ^ { 2 } } = \frac { 2 + 3 i } { 13 } = \frac { 2 } { 13 } + \frac { 3 } { 13 } i$

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Question 152 Marks

Express $(-\sqrt3 + \sqrt{-2})(2\sqrt3 - i)$ in the form of a + ib.

Answer

Let z= $(-\sqrt3 + \sqrt{-2})(2\sqrt3 - i)$ = $(-\sqrt3 + \sqrt{2i^2})(2\sqrt3 - i)$ = $(-\sqrt3 + \sqrt{2}i)(2\sqrt3 - i)$
z $= -6 + \sqrt3 i + 2\sqrt6 i - \sqrt2 i^2$
z $= (-6 + \sqrt2)+\sqrt3(1 + 2\sqrt2)i$

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Question 162 Marks

Express $(5 - 3i)^3$ in the form $a + ib.$

Answer

We have, $(5-3i)^3= 5^3 - 3 \times 5^2 \times (3i) + 3 \times 5 (3i)^2 - (3i)^3$
$ [(a-b)^3= a^3 - 3a^2b+ 3b^2a - b^3]$
$= 125 - 225i - 135 + 27i = -10 - 198i$

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Question 172 Marks

Find the modulus and argument of complex number $\frac { 1 } { 1 + i }$.

Answer

We have, $\frac { 1 } { 1 + i } = \frac { 1 } { 1 + i } \times \frac { 1 - i } { 1 - i } = \frac { 1 - i } { 1 ^ { 2 } - i ^ { 2 } } = \frac { 1 - i } { 1 + 1 }$
$= \frac { 1 - i } { 2 } = \frac { 1 } { 2 } - \frac { i } { 2 }$
Let $r \cos \theta = \frac { 1 } { 2 } ...(i)$
and $r \sin \theta = - \frac { 1 } { 2 } ...(ii)$
On squaring and adding Eqs. (i) and (ii), we get
$r^2 \cos^2 \theta + r^2 \sin^2 \theta = \left( \frac { 1 } { 2 } \right) ^ { 2 } + \left( - \frac { 1 } { 2 } \right) ^ { 2 }$
$\Rightarrow r^2 =\frac { 1 } { 4 } + \frac { 1 } { 4 } = \frac { 1 } { 2 } \Rightarrow r = \frac { 1 } { \sqrt { 2 } } [\because r$ is positive$]$
On putting the value of r in Eqs. $(i)$ and $(ii),$ we get
$\cos \theta = \frac { 1 } { \sqrt { 2 } }$ and $\sin \theta = - \frac { 1 } { \sqrt { 2 } }$
Since, $\cos \theta$ is positive and \sin\theta is negative.
So, $\theta$ lies in IV quadrant.
$\therefore \theta = - \frac { \pi } { 4 }$
Hence, modulus of $\frac { 1 } { 1 + i }$ is $\frac { 1 } { \sqrt { 2 } }$ and argument is $\frac { - \pi } { 4 }.$

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Question 182 Marks

Find the conjugate of $\frac{(3-2 i)(2+3 i)}{(1+2 i)(2-i)}$

Answer

We have $\frac{(3-2 i)(2+3 i)}{(1+2 i)(2-i)}$
$=\frac{6+9 i-4 i+6}{2-i+4 i+2}=\frac{12+5 i}{4+3 i} \times \frac{4-3 i}{4-3 i}$
$=\frac{48-36 i+20 i+15}{16+9}=\frac{63-16 i}{25}=\frac{63}{25}-\frac{16}{25} i$
Therefore, conjugate of $\frac{(3-2 i)(2+3 i)}{(1+2 i)(2-i)}$is $\frac{63}{25}+\frac{16}{25} i$ [If z = x + iy then then conjugate is x - iy]

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Question 192 Marks

Solve $\sqrt{5} x^{2}+x+\sqrt{5}=0$

Answer

Here, a= $\sqrt 5$ = c and b = 1

the discriminant of the equation is
$1^{2}-4 \times \sqrt{5} \times \sqrt{5}=1-20=-19$
Therefore, the solutions are
= $\frac{-1 \pm \sqrt{-19}}{2 \sqrt{5}}=\frac{-1 \pm \sqrt{19} i}{2 \sqrt{5}}$

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Question 202 Marks

Solve $x^2 + x + 1= 0$

Answer

Here $a = 1, b = 1$ and $c = 1x = \{-b \pm \sqrt{b^2-4ac} \over 2a\}$
$b^{2}-4 a c=1^{2}-4 \times 1 \times 1=1-4=-3$
Therefore, the solutions are given by $x=\frac{-1 \pm \sqrt{-3}}{2 \times 1}=\frac{-1 \pm \sqrt{3} i}{2}$

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Question 212 Marks

If 4x + i(3x - y) = 3 + i (-6), where x and y are real numbers, then find the values of x and y.

Answer

We have,
4x + i (3x - y) = 3 - 6i
$\Rightarrow$ 4x + i (3x - y) = 3 + i (- 6)
On equating real and imaginary parts from both sides, we get
4x = 3 $\Rightarrow$ x = $\frac { 3 } { 4 }$ and 3x - y = - 6
$\Rightarrow$ 3 $\left( \frac { 3 } { 4 } \right)$ - y = - 6 [$\because$ x = $\frac { 3 } { 4 }$]
$\Rightarrow$ $\frac { 9 } { 4 }$ - y = - 6
$\Rightarrow$ y = $\frac { 9 } { 4 } + 6 = \frac { 33 } { 4 }$
$\therefore$ x = $\frac { 3 } { 4 }$ and y = $\frac { 33 } { 4 }$

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