Free Objective Test 01 Practice Test - 11th and 12th

Question 1

For positive integers n₁,n₂ the value of the expression $(1 + i)^{n_{1}}$ + $(1 + i^{3})^{n_{1}}$ + $(1 + i^{5})^{n_{2}}$ + $(1 + i^{7})^{n_{2}}$ where i= $\sqrt[2]{- 1}$ is a real number if and only if

$n_{1}$ = $n_{2}$ +1

$n_{1}$ = $n_{2}$ -1

$n_{1}$ = $n_{2}$

$n_{1}$ >0, $n_{2}$ >0

SOLUTION

Solution : D

Using $i^{3} = - i, i^{5} = i a n d i^{7} = - i$ , we can write the given expression as
$(1 + i)^{n_{1}} + (1 + i^{3})^{n_{1}} + (1 + i^{5})^{n_{2}} + (1 + i^{7})^{n_{2}} w h e r e i = \sqrt[2]{1} = 2 [^{n_{1}} C_{0} +^{n_{1}} C_{2} (i)^{2} +^{n_{1}} C_{4} (i)^{4} +^{n_{1}} C_{6} (i)^{6} + \dots \dots] + 2 [^{n_{2}} C_{0} +^{n_{2}} C_{2} (i)^{2} +^{n_{2}} C_{4} (i)^{4} +^{n_{2}} C_{6} (i)^{6} + \dots \dots] = 2 [^{n_{1}} C_{0} -^{n_{1}} C_{2} +^{n_{1}} C_{4} -^{n_{1}} C_{6} + \dots \dots] + 2 [^{n_{2}} C_{0} -^{n_{2}} C_{2} +^{n_{2}} C_{4} -^{n_{2}} C_{6} + \dots \dots]$
This is a real number irrespective of values of $n_{1}$ and $n_{2}$ .

Question 2

The conjugate of $\frac{(2 + i)^{2}}{3 + i}$ , in the form of a+ib, is

$\frac{13}{2} + i (\frac{15}{2})$

$\frac{13}{10} + i (\frac{- 15}{2})$

$\frac{13}{10} + i (\frac{- 9}{10})$

$\frac{13}{10} + i (\frac{9}{10})$

SOLUTION

Solution : C

z = $\frac{(2 + i)^{2}}{3 + i} = \frac{3 + 4 i}{3 + i} \times \frac{3 - i}{3 - i} = \frac{13}{10} + i \frac{9}{10}$

Conjugate = $\frac{13}{10} - i \frac{9}{10}$ .

Question 3

If (1+i)(1+2i)(1+3i)......(1+ni) = a+ib, then 2.5.10.....(1+ $n^{2}$ ) is equal to

a^{2}

b^{2}

a^{2}

b^{2}

\sqrt{a^{2} + b^{2}}

\sqrt{a^{2} - b^{2}}

SOLUTION

Solution : B

we have
(1+i)(1+2i)(1+3i)......(1+ni) = a+ib .......(i)

$\Rightarrow$ (1+i)(1+2i)(1+3i)......(1+ni) = a-ib ........(ii)

Multiplying (i) and (ii), we get 2.5.10.....(1+ $n^{2}$ ) = $a^{2}$ + $b^{2}$

Question 4

If x=-5+ $\sqrt{- 4}$ , then the value of the expression $x^{4}$ +9 $x^{3}$ +35 $x^{2}$ -x+4 is

160

-160

-60

SOLUTION

Solution : B

x+5 =4i $\Rightarrow$ $x^{2}$ +10x+25=-16

Now, $x^{4}$ +9 $x^{3}$ +35 $x^{2}$ -x+4

=( $x^{2}$ +10x+41)( $x^{2}$ -x+4)-160=-160

Question 5

$\frac{1 - i}{1 + i}$ is equal to

cos $\frac{π}{2}$ +isin $\frac{π}{2}$

cos $\frac{π}{2}$ -isin $\frac{π}{2}$

sin $\frac{π}{2}$ +icos $\frac{π}{2}$

None of these

SOLUTION

Solution : B

$\frac{1 - i}{1 + i}$ = $\frac{(1 - i) (1 - i)}{(1 + i) (1 - i)}$ = $\frac{1 + (i)^{2} - 2 i}{1 + 1}$ =-i

Which can be written as cos $\frac{π}{2}$ -isin $\frac{π}{2}$

Question 6

The values of x and y for which the numbers 3+i $x^{2}$ y and $x^{2}$ +y+4i are conjugate complex are

(-2,-1) or (2,-1)

(1,-2) or (-2,1)

(1,2) or (-1,-2)

None of these

SOLUTION

Solution : A

According to condition, 3-i $x^{2}$ y = $x^{2}$ + y + 4i

⇒ $x^{2}$ + y = 3 And $x^{2}$ y = -4 ⇒ x = ±2, y = -1

⇒ (x,y) = (2,-1) or (-2,-1)

Question 7

For the complex number z, which of the following is true?

$z + ¯ z$ is real and $z ¯ z$ is imaginary

$z + ¯ z$ is imaginary and $z ¯ z$ is real

Both $z + ¯ z$ and $z ¯ z$ are real.

Both $z + ¯ z$ and $z ¯ z$ are imaginary numbers

SOLUTION

Solution : C

Let $z = x + i y$

Here, $z + ¯ z = (x + i y) + (x - i y) = 2 x$ (Real)

And $z ¯ z = (x + i y) (x - i y) = x^{2} + y^{2}$ (Real).

Question 8

The number of solutions of the equation $z^{2}$ + $¯ z$ = 0 is

SOLUTION

Solution : D

Let z = x+iy, so that $¯ z$ = x - iy, therefore

$z^{2} + ¯ z = 0 \Leftrightarrow (x^{2} - y^{2} + x) + i (2 x y - y)$ = 0

Equating real and imaginary parts , we get

$x^{2} - y^{2} + x$ = 0 .......(i)

And 2xy - y = 0 ⇒ y = 0 or x = $\frac{1}{2}$

if y = 0 , then (i) gives $x^{2}$ + x = 0 ⇒ x = 0 or

x = -1

If x = $\frac{1}{2}$ ,

Then $x^{2} - y^{2} + x = 0 \Rightarrow y^{2} = \frac{1}{4} + \frac{1}{2} = \frac{3}{4} \Rightarrow y = \pm \frac{\sqrt{3}}{2}$

Hence, there are four solutions in all.

Question 9

If z is a complex number, then z. $¯ z$ = 0 if and only if

z=0

Re(z)=0

Im(z)=0

None of these

SOLUTION

Solution : A

Let z = x+iy, $¯ z$ = x-iy

∴ z $¯ z$ = 0 ⇒ (x+iy)(x-iy) = 0 ⇒ $x^{2} + y^{2}$ = 0

It is possible onle when x and y oth simultaneously zero

i.e, z = 0 +0i = 0

Question 10

If z is a complex number such that $z^{2}$ = ( $¯ z)^{2}$ ,then

z is purely real

z is purely imaginary

Either z is purely real or purely imaginary

None of these

SOLUTION

Solution : C

Let z = x+iy, then its coonjucate $¯ z$ = x-iy

Given that $z^{2} = (¯ z)^{2}$

⇒ $x^{2}$ - $y^{2}$ + 2ixy = $x^{2}$ - $y^{2}$ - 2ixy ⇒ 4ixy = 0

if x ≠ 0 then y = 0 and if y ≠ 0 then x = 0

Question 11

If -1+ $\sqrt{- 3}$ =r $e^{i θ}$ , then $θ$ is equal to

$\frac{π}{3}$

- $\frac{π}{3}$

$\frac{2 π}{3}$

- $\frac{2 π}{3}$

SOLUTION

Solution : C

Here -1+ $\sqrt{- 3}$ =r $e^{i θ}$ $\Rightarrow$ -1+ $i \sqrt{3}$ =r $e^{i θ}$

=rcos $θ$ =-1 And rsin $θ$ = $\sqrt{3}$

Hence tan $θ$ =- $\sqrt{3}$ $\Rightarrow$ tan $θ$ =tan $\frac{2 π}{3}$ .Hence $θ$ = $\frac{2 π}{3}$

Question 12

If z=r $e^{i θ}$ ,then | $e^{i z}$ |=

$e^{r s i n θ}$

$e^{- r s i n θ}$

$e^{- r c o s θ}$

$e^{r c o s θ}$

SOLUTION

Solution : B

if z=r $e^{i θ}$ =r(cos $θ$ + isin $θ$ )

$\Rightarrow$ iz=ir(cos $θ$ + isin $θ$ )=-rcos $θ$ + irsin $θ$

or $e^{i z}$ = $e^{- r c o s θ + i r s i n θ}$ = $e^{- s i n θ} e^{r i c o s θ}$

or | $e^{i z}$ |= $| e^{- r s i n θ} | | e^{i r c o s θ} |$

= $e^{- r s i n θ}$ $[c o s^{2} (r c o s θ) + s i n^{2} (r c o s θ)]^{2}$ = $e^{- r s i n θ}$

Question 13

Real part of $e^{e^{i θ}}$ is

$e^{c o s θ}$ [cos(sin $θ$ )]

$e^{c o s θ}$ [cos(cos $θ$ )]

$e^{s i n θ}$ [cos(sin $θ$ )]

$e^{s i n θ}$ [sin(sin $θ$ )]

SOLUTION

Solution : A

$e^{e^{i θ}} = e^{c o s θ + i s i n θ} = e^{c o s θ} e^{i s i n θ} = e^{c o s θ} [c o s (s i n θ) + i s i n (s i n θ)] ∴ R e a l p a r t o f e^{e^{i θ}} i s e^{c o s θ} [c o s (s i n θ)]$

Question 14

If $z_{1}, z_{2} a n d z_{3}$ are complex numbers such that

| $z_{1}$ |=| $z_{2}$ |=| $z_{3}$ |= $∣ ∣ \frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}} ∣ ∣ = 1$ ,

then | $z_{1}$ + $z_{2}$ + $z_{3}$ |

Equal to 1

Less than 1

Greater than 3

Equal to 3

SOLUTION

Solution : A

1= $∣ ∣ \frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}} ∣ ∣$ = $∣ ∣ ∣ \frac{z_{1} * {¯ z}_{1}}{z_{1}} + \frac{z_{2} * {¯ z}_{2}}{z_{2}} + \frac{z_{3} * {¯ z}_{3}}{z_{3}} ∣ ∣ ∣$

(hence, ${| z_{1} |}^{2}$ =1= $z_{1}$ $_{1}$ , etc)

$\Rightarrow$ | $z_{1}$ + $z_{2}$ + $z_{3}$ |=| $¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ z_{1} + z_{2} + z_{3}$ |= | $z_{1}$ + $z_{2}$ + $z_{3}$ |

(hence,| $_{1}$ |=| $z_{1}$ |)

Question 15

Find the complex number z satisfying the equations $∣ ∣ \frac{z - 12}{z - 8 i} ∣ ∣$ = $\frac{5}{3}$ , $∣ ∣ \frac{z - 4}{z - 8} ∣ ∣$ =1

$6 \pm 8 i$

6+8i, 6+17i

None of these

SOLUTION

Solution : C

We have $∣ ∣ \frac{z - 12}{z - 8 i} ∣ ∣$ = $\frac{5}{3}$ , $∣ ∣ \frac{z - 4}{z - 8} ∣ ∣$ =1

Let z=x+iy, then

$∣ ∣ \frac{z - 12}{z - 8 i} ∣ ∣$ = $\frac{5}{3}$ $\Rightarrow$ 3|z-12| = 5|z-8i|

$\Rightarrow$ $9 (x - 12)^{2}$ + $9 y^{2}$ = 25 $x^{2}$ +25 $(y - 8)^{2}$ $\dots$ $\dots$ (i)

$∣ ∣ \frac{z - 4}{z - 8} ∣ ∣$ =1 $\Rightarrow$ |z-4| = |z-8|

$\Rightarrow$ $(x - 4)^{2}$ + $y^{2}$ = $y^{2}$ + $(x - 8)^{2}$ $\Rightarrow$ x=6

Putting x=6 in (i), we get $y^{2}$ -25y+136=0

y=17, 8

Hence z=6+17i or z=6+8i

Trick: Check it with options.

Free Objective Test 01 Practice Test - 11th and 12th

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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