MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 11 Commerce Applied Mathematics Number Theory Quiz 10 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Number Theory
Quiz 10
A number system with a base of two is referred as ______________.
Report Question
0%
Unary number system
0%
Binary number system
0%
Octal number system
0%
None of these
If a register containing data (11001100)2 is subjected to arithmetic shift left operation, then the content of the register after 'ashl' shall be _____________.
Report Question
0%
(11001100)2
0%
(1101100)2
0%
(10011001)2
0%
(10011000)2
C it refers to a _____________.
Report Question
0%
computer language.
0%
CPU instruction.
0%
0 or 1 value.
0%
digital representation of an alphabetic character.
Which of the following is true?
Report Question
0%
Byte is a single digit in a binary number
0%
Bit represents a grouping of digital numbers
0%
Eight-digit binary number is called a bit
0%
Eight-digit binary number is called a byte
State true(T) or false(F).
The sum of primes cannot be a prime.
Report Question
0%
True
0%
False
Explanation
This statement is also false.
As consider, the numbers $$2$$ and $$3$$ then sum of $$2$$ and $$3$$ i.e. is $$5$$ is also a prime number.
State true or false:
The product of primes cannot be a prime.
Report Question
0%
True
0%
False
Explanation
The product of two primes can't be a prime because it violates the definition of the prime number as it will be divided by the prime numbers which are multiplied rather than $$1$$ and the number itself.
So the number is not prime it will become a composite number instead.
State true(T) or false(F).
Odd numbers cannot be composite.
Report Question
0%
True
0%
False
Explanation
The given statement is false.
As, consider the number $$9$$, which is an odd number and it is a composite number too.
Since $$3$$ divides the given number.
So $$3$$ is a divisor of $$9$$ other than $$1$$ and $$9$$.
State true(T) or false(F).
An even number is composite.
Report Question
0%
True
0%
False
Explanation
This statement is also false.
Since $$2$$ is the number which is both even and prime.
Hence an even number may not be composite.
Mark the correct alternative of the following.
Which of the following numbers is prime?
Report Question
0%
$$23$$
0%
$$51$$
0%
$$38$$
0%
$$26$$
The least prime is?
Report Question
0%
$$1$$
0%
$$2$$
0%
$$3$$
0%
$$5$$
Explanation
The number $$2$$ is the least prime. It is the only even prime also.
Mark the correct alternative of the following.
Which of the following are not twin-primes?
Report Question
0%
$$3, 5$$
0%
$$5, 7$$
0%
$$11, 13$$
0%
$$17, 23$$
Mark the correct alternative of the following.
Which of the following numbers are twin primes?
Report Question
0%
$$3, 5$$
0%
$$5, 11$$
0%
$$3, 11$$
0%
$$13, 17$$
Mark the correct alternative of the following.
The smallest number which is neither prime nor composite is?
Report Question
0%
$$0$$
0%
$$1$$
0%
$$2$$
0%
$$3$$
Explanation
All primes have two positive divisors. There are only two integers that are neither composite or prime and they are 1 and 0.
The number 1 has only 1 positive divisor, itself. The number 0 has an infinite number of divisors.
1 is neither prime nor composite.
Express the following complex numbers in the standard form $$ a+ib$$ :
$$ \left ( \dfrac{1}{1-4i}-\dfrac{2}{1+i} \right )\left ( \dfrac{3-4i}{5+i} \right )$$
Report Question
0%
$$ \dfrac{307}{442}+i \dfrac{599}{442}i$$
0%
$$ \dfrac{307}{442}-i \dfrac{599}{442}i$$
0%
$$ \dfrac{-307}{442}+i \dfrac{599}{442}i$$
0%
None of the above
Express the following complex numbers in the standard form $$ a+ib$$ :
$$ \dfrac{\left ( 2+i \right )^{3}}{2+3i}$$
Report Question
0%
$$ \dfrac{37}{13}-\dfrac{16}{13}i$$
0%
$$ \dfrac{-37}{13}+\dfrac{16}{13}i$$
0%
$$ \dfrac{37}{13}+\dfrac{16}{13}i$$
0%
None of the above
Find the modulus and argument of the following complex numbers and hence express each of them in the polar form:
$$1-i$$
Report Question
0%
$$\sqrt{2}(cos\,\pi /4+i\, sin\, \pi /4)$$
0%
$$\sqrt{2}(cos\,\pi /3-i\, sin\, \pi /3)$$
0%
$$\sqrt{2}(cos\,\pi /4-i\, sin\, \pi /4)$$
0%
$$\sqrt{2}(cos\,\pi /3+i\, sin\, \pi /3)$$
Explanation
Let $$z=1-i$$. Then, $$\left | z \right |=\sqrt{1^{2}+(-1)^{2}}=\sqrt{2}$$.
Let $$\alpha $$ be the acute angle given by $$tan\, \alpha =\dfrac{\left | Im(z) \right |}{\left | Re(z) \right |}$$.
Then,
$$tan\, \alpha =\dfrac{|-1|}{|1|}=1\Rightarrow \alpha =\dfrac{\pi }{4}$$
Clearly, z lies in the fourth quadrant. Therefore, $$arg(z)= -\alpha =-\dfrac{\pi }{4}$$.
Express the following complex numbers in the standard form $$ a+ib$$ :
$$ \dfrac{3-4i}{\left ( 4-2i \right )\left ( 1+i \right )}$$
Report Question
0%
$$ \dfrac{1}{4}+\dfrac{3}{4}i$$
0%
$$ \dfrac{1}{4}-\dfrac{3}{4}i$$
0%
$$ \dfrac{-1}{4}-\dfrac{3}{4}i$$
0%
None of the above
Express the following complex numbers in the standard from $$ a+ib$$ :
$$ \dfrac{5+\sqrt{2}i}{1-\sqrt{2}i}$$
Report Question
0%
$$ 1-2\sqrt{2}i$$
0%
$$ 1+\sqrt{2}i$$
0%
$$ 1+2\sqrt{2}i$$
0%
$$ 1-\sqrt{2}i$$
Explanation
$$ \dfrac{5+\sqrt{2}i}{1-\sqrt{2}i}$$
Rationalizing the denominator, we get
$$ = \dfrac{5+\sqrt{2}i}{1-\sqrt{2}i} \times \dfrac{1+\sqrt{2}i}{1+\sqrt{2}i}$$
$$ = \dfrac{5+2i^2+(5+1)\sqrt{2}i}{1+2}$$
$$ = \dfrac{3+6\sqrt{2}i}{1+2}$$......................$$(\because i^2=-1)$$
$$ = 1+2\sqrt{2}i$$
The real part of $$(i - \sqrt{3})^{13}$$ is
Report Question
0%
$$2^{-10}\sqrt3$$
0%
$$-2^{12}\sqrt3$$
0%
$$2^{-12}\sqrt3$$
0%
$$-2^{-12}\sqrt3$$
0%
$$-2^{10}\sqrt3$$
Explanation
$$w^2=\dfrac{-1-i\sqrt3}{2}$$
$$\Rightarrow iw^2=\dfrac{-i+\sqrt3}{2}$$
$$\Rightarrow -2iw^2=i-\sqrt3$$
$$\Rightarrow (-2iw^2)^{13}=(i-\sqrt3)^{13}$$
$$\Rightarrow -2^{13}i^{13}w^{26}=(i-\sqrt3)^{13}$$
$$\Rightarrow -2^{13}iw^2=(i-\sqrt3)^{13}$$
$$\Rightarrow -2^{13}\left({\dfrac{-i+\sqrt3}{2}}\right)=(i-\sqrt3)^{13}$$
$$\Rightarrow -2^{12}(-i+\sqrt3)=(i-\sqrt3)^{13}$$
$$\therefore (i-\sqrt3)^{13}=-2^{12}\sqrt3+i2^{12}$$
Let z be a complex number such that $$\left|\dfrac{z-i}{z+2i}\right|=1$$ and $$|z|=\dfrac{5}{2}$$. Then the value of $$|z+3i|$$ is?
Report Question
0%
$$\dfrac{7}{2}$$
0%
$$\dfrac{15}{4}$$
0%
$$2\sqrt{3}$$
0%
$$\sqrt{10}$$
Explanation
Given $$\left|\dfrac{z - i}{z + 2 i } \right| = 1$$
$$\Rightarrow |z - i| = |z + 2i|$$
Let $$z = x + iy$$
So, $$x^2 + (y- 1)^2 = x^2 + (y + 2)^2$$
$$-2y + 1 = 4y + 4$$
$$6y = -3 \Rightarrow y = \dfrac{-1}{2}$$
$$x^2 + y^2 = \dfrac{25}{4} \Rightarrow x^2 = \dfrac{24}{4} = 6$$
$$z = \pm \sqrt{6} - \dfrac{i}{2}$$
$$|z + 3i| = \sqrt{6 + \dfrac{25}{4}} = \sqrt{\dfrac{49}{4}} = \dfrac{7}{2}$$
$$\therefore$$ $$\boxed{|2 + 3i| = \dfrac{7}{2}}.....Answer$$
Hence option $$'A'$$ is the answer.
Mark against the correct answer in each of the following .
$$i^{91}=$$?
Report Question
0%
$$1$$
0%
$$-1$$
0%
$$i$$
0%
$$-i$$
Explanation
$$i^{91}\\=(1^4)^{22}\times i^3 \\=1\times i^2 \times i\\=1\times (-1)\times i\\=-i$$
$$(1-\sqrt{-1})(1+\sqrt{-1})(5-\sqrt{-7})(5+\sqrt{-7})=?$$
Report Question
0%
$$(25+7i)$$
0%
$$(32+5i)$$
0%
$$(29-3i)$$
0%
$$none\ of\ these$$
Explanation
Given expression
$$=(1-i)(1+i)(5-\sqrt 7 i)(5+\sqrt 7i)$$
$$=(1-i^2)(25-7i^2)\\=(1+1)(25+7)\\=(2\times 32)=64$$.
Mark against the correct answer in each of the following .
$$i^{326}=$$?
Report Question
0%
$$1$$
0%
$$-1$$
0%
$$i$$
0%
$$-i$$
Explanation
$$i^{326}\\=(i^4)^{81}\times i^2 =(1)^{81}\times (-1)\\=1\times (-1)=-1$$
If $$\mid z^2 - 3\mid = 3\mid z\mid$$ then the maximum value of $$\mid z\mid$$ is
Report Question
0%
1
0%
$$\dfrac{3+\sqrt {21}}{2}$$
0%
$$\dfrac{\sqrt {21} - 3}{2}$$
0%
none of these
Explanation
$$ \mid z^{2} - 3 \mid \geq \mid z \mid^{2} - 3$$
$$\Rightarrow 3 \mid z \mid \geq \mid z \mid ^{2} - 3$$
$$\Rightarrow \mid z \mid ^{2} - 3 \mid z \mid -3 \leq 0$$
$$\Rightarrow 0 < \mid z \mid \leq \dfrac{3 + \sqrt{21}}{2}$$
If $$z_1$$ and $$z_2$$ are any two complex numbers then
$$|z_1 +\sqrt {z_1^2 -z_2^2}| + |z_1 -\sqrt {z_1^2 -z_2^2}|$$ is equal to
Report Question
0%
$$|z_1|$$
0%
$$|z_2|$$
0%
$$|z_1 + z_2|$$
0%
$$None\ of\ these$$
Explanation
We know that,
$$|z_1+z_2|^2+|z_1-z_2|^2=2[|z_1|^2+|z_2|^2]$$ ---- ( 1 )
Now,
$$\left[|z_1+\sqrt{z_1^2-z_2^2}|+|z_1-\sqrt{z_1^2-z_2^2}|\right]^2$$ $$=\left|z_1+\sqrt{z_1^2-z_2^2}\right|^2+\left|z_1-\sqrt{z_1^2-z_2^2}\right|^2+2\left|z_1^2-(z_1^2-z_2^)\right|$$
$$=2|z_1|^2+2|z_1^2-z_2^2|+2|z_2^2|$$ [ From ( 1 ) ]
$$=2|z_1|^2+2|z_2|^2+2|z_1^2-z_2^2|$$
$$=|z_1+z_2|^2+|z_1-z_2|^2+2|z_1+z_2||z_1-z_2|$$
$$=\left(\left|z_1+z_2\right|+\left|z_1-z_2\right|\right)^2$$
Taking square root on both sides we get,
$$|z_1+\sqrt{z_1^2-z_2^2}|+|z_1-\sqrt{z_1^2-z_2^2}|$$
$$=\left|z_1+z_2\right|+\left|z_1-z_2\right|$$
$$(2-3i)(-3+4i)=?$$
Report Question
0%
$$(6+17i)$$
0%
$$(6-17i)$$
0%
$$(-6+17i)$$
0%
$$none\ of\ these$$
Explanation
$$(2-3i)(-3+4i)\\=(-6+8i+9i-12i^2)\\=(-6+17i+12)\\=(6+17i)$$.
Mark against the correct answer in each of the following .
$$i^{273}=$$?
Report Question
0%
$$i$$
0%
$$-i$$
0%
$$1$$
0%
$$-1$$
Explanation
$$i^{273}=(i^4)^{68}\times i=(1)^{68}\times i=(1\times i)=i$$
Compare List I with List II and choose the correct answer using codes given below:
List I (Complex number)
List II (Its modulus)
$$(4-3i)$$
$$10$$
$$(8+6i)$$
$$\dfrac{1}{5}$$
$$\dfrac{1}{(3+4i)}$$
$$1$$
$$\dfrac{(3-4i)}{(3+4i)}$$
$$5$$
Report Question
0%
$$(i)-(p), (ii)-(s), (iii)-(r), (iv)-(q)$$
0%
$$(i)-(s), (ii)-(p), (iii)-(q), (iv)-(r)$$
0%
$$(i)-(s), (ii)-(p), (iii)-(r), (iv)-(q)$$
0%
$$(i)-(r), (ii)-(p), (iii)-(s), (iv)-(q)$$
Which of the following is a composite number?
Report Question
0%
$$23$$
0%
$$29$$
0%
$$32$$
0%
none of these
Explanation
(a) $$23$$
Since it cannot be broken into factors
Hence $$23$$ is not a composite number
(b) $$29$$
Since it cannot be broken into factors
Hence $$29$$ is not a composite number
(c) $$32$$
Since it can be broken into factors i.e.$$2 \times 2 \times 2 \times 2 \times 2$$
Hence $$32$$ is a composite number
Option (c) is the correct answer
State whether the following statements are true (T) or false (F) :
There are infinitely many prime numbers.
Report Question
0%
True
0%
False
State whether the following statements are true (T) or false (F):
A natural number is called a composite number if it has at least one more factor other than $$ 1$$ and the number itself.
Report Question
0%
True
0%
False
Explanation
Given statement is true as a natural number is composite if it has at least one factor other than 1 and the number itself.
Which of the following is an odd composite number ?
Report Question
0%
$$ 7 $$
0%
$$ 9 $$
0%
$$ 11 $$
0%
$$ 12 $$
Explanation
$$ 9 = 3 \times 3 $$ , is an odd composite number.
The modulus of $$\overline { 6+{ i }^{ 3 } } +\overline { 6+{ i } }+\overline { 6+{ i }^{ 2 } } $$ is
Report Question
0%
$$17$$
0%
$$\sqrt{533}$$
0%
$$\sqrt{456}$$
0%
$$49$$
Given $$z$$ is a complex number with modulus $$1$$. Then the equation in $$a$$, $$\left(\dfrac{1+ia}{1-ia}\right )^4=z$$ has
Report Question
0%
all roots real and distinct.
0%
two real and two imaginary.
0%
three roots real and one imaginary.
0%
one root real and three imaginary.
Explanation
$$\left (\dfrac{1+ia}{1-ia}\right )^4 = z$$
$$\left (\dfrac{-i(i-a)}{-i(i+a)}\right )^4 = z$$
$$\left (\dfrac{a-i}{a+i}\right )^4= z$$
$$\left |\dfrac{a-i}{a+i}\right |^4= |z| =1$$
$$|a-i| ^4 = |a+i|^4$$
$$|a-i| = |a+i|$$
$$\therefore a$$ lies on the perpendicular bisector of $$i$$ and $$-i$$.
$$a $$ lies on real axis. Hence the roots all are real and distinct.
If $$ \displaystyle z_{0}=\frac{1-i}{2}$$, then $$ \displaystyle \left (1+z_{0} \right )\left (1+z_{0}^{{2}^{1}} \right )\left (1+z_{0}^{{2}^{2}} \right ).......... \left (1+z_{0}^{2^n} \right )$$ must be
Report Question
0%
$$(1-i)(1+\dfrac{1}{2^{2^n}})$$ for $$n>1$$
0%
$$(1-i)(1-\dfrac{1}{2^{2^n}})$$ for $$n>1$$
0%
$$\dfrac{1+i}{2}$$ for $$n>1$$
0%
$$(1-i)(1-\dfrac{1}{2^{2^{n+1}}})$$ for $$n>1$$
Explanation
Let $$P=(1+z_0)(1+{z_0}^{2^1})+(1+{z_0}^{2^2})....(1+{z_0}^{2^n})$$
$$\Rightarrow (1-z_0)P=(1-z_0)(1+z_0)(1+{z_0}^{2^1})+(1+{z_0}^{2^2})....(1+{z_0}^{2^n})$$
$$\Rightarrow (1-z_0)P=(1-{z_0}^{2^2})(1+{z_0}^{2^2}).......(1+{z_0}^{2^n})$$
$$\Rightarrow (1-z_0)P=1-{z_0}^{2^{n+1}}$$
$$\Rightarrow P=\dfrac{1-{z_0}^{2^{n+1}}}{1-{z_0}}$$
$$\Rightarrow P=\dfrac{1-({z_0}^2)^{2n}}{1-z_0}$$
$$n>1$$
Since, $$z_0=\dfrac{1-i}{2}$$
$${z_0}^2=\dfrac{1-1-2i}{4}$$
$$\Rightarrow {z_0}^2=\dfrac{-i}{2}$$
$$\therefore \ P=\dfrac{1-\left(\dfrac{-i}{2}\right)^{2^n}}{1-z_0}$$
$$\Rightarrow P=\dfrac{1-\dfrac{1}{2^{2^n}}}{\dfrac{1+i}{2}}=\dfrac{\left(1-\dfrac{1}{2^{2^n}}\right)2(1-i)}{1+1}$$
$$\Rightarrow P=(1-i)\left(1-\dfrac{1}{2^{2^n}}\right)$$
Dividing f(z) by $$z-i$$, we obtain the remainder $$i$$ and dividing it by $$z+i$$, we get the remainder $$1+i$$, then remainder upon the division of f(z) by $$z^2+1$$ is
Report Question
0%
$$\displaystyle \frac {1}{2}(z+1)+i$$
0%
$$\displaystyle \frac {1}{2}(iz+1)+i$$
0%
$$\displaystyle \frac {1}{2}(iz-1)+i$$
0%
$$\displaystyle \frac {1}{2}(z+i)+1$$
Explanation
$$\textbf{Step -1: Using remainder theorem to get the following equations.}$$
$$\text{Given, Dividing }f(z) \text{ by } z - i \text{, we get remainder }i$$
$$\text{Therefore by remainder theorem, }f(i) = i$$ $$\dots \text{(i)}$$
$$\text{Given, Dividing }f(z) \text{ by } z + i \text{, we get remainder }1 + i$$
$$\text{Therefore by remainder theorem, }f(-i) = 1+i$$ $$\dots \text{(ii)}$$
$$\text{Let }R(x) \text{ be the remainder when }f(z)\text{is divided by }z^2 + 1$$
$$\text{degree of R(z) is 1, hence }R(z) = az+b$$
$$\text{we can simplify }z^2 + 1 = (z -i) (z+i)$$
$$\therefore f(z) = Q(z)(z-i)(z+i) + R(z)$$
$$\textbf{Step -2: Solving for a and b.}$$
$$\text{From (i) and (ii)}$$
$$f(i) = i = ai + b$$
$$\Rightarrow b = i - ai$$
$$f(-i) = 1+i = -ai +b$$
$$\Rightarrow -ai + i - ai = 1+i$$
$$\Rightarrow -2ai = 1$$
$$\Rightarrow a = -\dfrac{1}{2i}$$
$$= -\dfrac{1}{2i} \times \dfrac{i}{i}$$
$$= -\dfrac{i}{2(-1)}$$
$$\therefore a = \dfrac{i}{2}$$
$$ b = i - ai$$
$$\Rightarrow b = i - \dfrac{i}{2} (i)$$
$$\Rightarrow b = i - \dfrac{(-1)}{2}$$
$$\Rightarrow b = i + \dfrac{1}{2}$$
$$\therefore R(z) = az+b$$
$$= \dfrac{i}{2}z + i + \dfrac{1}{2}$$
$$R(z) = \dfrac{1}{2}(iz + 1) + i$$
$$\textbf{Hence option B is correct}$$
The number of solutions of $$log _{\frac{1}{5}}log_{\frac{1}{2}}(\left | z \right |^{2}+4\left | z \right |+3)< 0$$ is/are?
Report Question
0%
$$0$$
0%
$$2$$
0%
$$4$$
0%
infinite
Explanation
The glven inequality becomes $$log_{\frac{1}{2}}|\mathrm{z}|^{2}+4|\mathrm{z}|+3>1$$, since $$\displaystyle \frac{1}{5}<1$$
$$\displaystyle \Rightarrow|\mathrm{z}|^{2}+4|\mathrm{z}|+3<\frac{1}{2}$$
$$\Rightarrow(|\mathrm{z}|+2)^{2}$$ + 1 $$<\displaystyle \frac{1}{2}$$
$$(|\displaystyle \mathrm{z}|+2)^{2}<\frac{-1}{2}$$
This is not possible. Hence, there is no solution.
Let $$z$$ be a complex number and $$c$$ be a real number $$\geq $$ 1 such that z + $$c\left | z+1 \right |+i=0 ,$$ then $$c$$ belongs to
Report Question
0%
$$[2, 3]$$
0%
$$(3, 4)$$
0%
$$[1,\sqrt{2}]$$
0%
None of these
Explanation
Since $$c$$ $$\left | z+1 \right |$$ is real
$$z + i$$ is real
let $$z-i = x $$, where $$x$$ is real
$$z+i=-c$$ $$\left | z+1 \right |$$ (given)
$$x^{2}=c^{2}\left \{ x+1 \right \}^{2}+1^{2})$$
$$(c^{2}-1)x^{2}+2c^{2}x+2c^{2}=0$$
As $$x$$ is real $$4c^{4}-8c^{2}(c^{2}-1)\geq 0$$
$$4c^{2}-(c^{2}-2)\leq 0$$
$$c^{2}\leq 2$$
$$c\leq \sqrt{2}$$
But $$c\geq1$$
$$\Rightarrow c\, \, \epsilon [1,\sqrt{2}]$$
lf $$\displaystyle \log_{\tan 30^{\circ}}\left(\frac{2|Z|^{2}+2|Z|-3}{|z|+1}\right) <-2$$ then
Report Question
0%
$$|\displaystyle \mathrm{z}|<\frac{3}{2}$$
0%
$$|z|>\displaystyle \frac{3}{2}$$
0%
$$|z|>2$$
0%
$$|z|<2$$
Explanation
we have,
$$\Rightarrow log_{(\dfrac{1}{\sqrt{3}})}(\dfrac{2|z|^{2}+2|z|-3}{|z|+1})<-2$$
$$\Rightarrow \dfrac{2|z|^{2}+2|z|-3}{|z|+1}>(\dfrac{1}{\sqrt{3}})^{-2}=(\sqrt{3})^{2}=3$$ [as$$\dfrac{1}{\sqrt{3}}<1$$ so equality sign changes]
$$\Rightarrow 2|z|^{2}+2|z|-3>3|z|+3$$
$$\Rightarrow 2|z|^{2}-|z|-6>0$$
$$\Rightarrow 2|z|^{2}-4|z|+3|z|-6>0$$
$$\Rightarrow 2|z| (|z|-2)+3 (|z|-2)>0$$
$$\Rightarrow (2|z|+3) (|z|-2)>0$$
$$\Rightarrow |z|>2$$
If $$x = 2 + 5i($$where $$1 i = \sqrt{-1})$$ and $$2(\displaystyle \frac{1}{1! 9!} + \frac{1}{3! 7!}) + \frac{1}{5! 5!} = \frac{2^{a}}{b!}$$ then $$ x^{3}-5x^{2}+33x-10 = $$
Report Question
0%
$$a + b$$
0%
$$b - a$$
0%
$$a-b$$
0%
$$-a-b$$
0%
$$(a - b)(a + b)$$
Explanation
$$ \displaystyle \frac{10! 2^{a}}{b!} = 2[^{10}C_{1} + ^{10}C_{3}] + ^{10} C_{5}$$
$$ \displaystyle \frac{10! 2^{a}}{b!} = ^{10}C_{1} + ^{10}C_{3} + ^{10}C_{5} + ^{10}C_7 + ^{10}C_{9} = 2^{9}$$
$$ \therefore a = 9, b = 10$$
$$ x = 2 + 5i$$
$$ (x - 2)^{2} = - 25$$
$$x^{2} - 4x + 29 = 0$$
$$ \therefore x^{3} - 5x^{2} + 33x - 10 = (x - 1)(x^{2} - 4x + 29) + 19$$
$$\Rightarrow 19 = a + b$$
If $$\left | \log_{\sqrt{3}} \frac{\left | z \right |^{2}-\left | z \right |+1}{2+{}\left | z \right |}\right |< 2$$, then
Report Question
0%
$$|\displaystyle \mathrm{z}|<\frac{1}{3}$$
0%
$$|\mathrm{z}|=1$$
0%
$$|\mathrm{z}|=5$$
0%
$$1<|\mathrm{z}|<5$$
Explanation
$$\left|log_{{\sqrt{3}}}(\dfrac{|z|^{2}-|z|+1}{|z|+2})\right|<2$$
$$\Rightarrow -2<\log_{{\sqrt{3}}}(\dfrac{|z|^{2}-|z|+1}{|z|+2})<2$$
$$\Rightarrow \dfrac{1}{3}<\dfrac{|z|^{2}-|z|+1}{|z|+2}<3$$
$$\Rightarrow |z|+2<3|z|^{2}-3|z|+3$$
$$\Rightarrow 3|z|^{2}-4|z|+1>0$$
$$\Rightarrow (3|z|-1)(|z|-1)>0$$
Hence
$$|z|<\dfrac{1}{3}$$ and $$|z|>1$$ ...(i)
Similarly
$$\dfrac{|z|^{2}-|z|+1}{|z|+2}<3$$
$$\Rightarrow |z|^{2}-|z|+1<3|z|+6$$
$$\Rightarrow |z|^{2}-4|z|-5<0$$
$$\Rightarrow (|z|+1)(|z|-5)<0$$
$$\Rightarrow 1<|z|<5$$ ...(ii)
From (i) and (ii), we get
$$ 1<|z|<5$$
If z be a complex number satisfying$$\displaystyle\ z^{4}+z^{3}+2z^{2}+z+1=0$$ then $$\displaystyle\ |z|$$ is
Report Question
0%
$$\displaystyle\ \frac{1}{2}$$
0%
$$\displaystyle\ \frac{3}{4}$$
0%
$$\displaystyle\ 1$$
0%
None of these
Explanation
$${ z }^{ 4 }+{ z }^{ 3 }+2{ z }^{ 2 }+z+1=0$$
$$\left[ { z }^{ 4 }+2{ z }^{ 2 }+1 \right] +\left[ { z }^{ 3 }+z \right] =0$$
$$\left[ { z }^{ 2 }+1 \right] \left[ { z }^{ 2 }+1+z \right] =0$$
$${ z }^{ 2 }+1=0\Longrightarrow z=\pm i$$
$$\left| z \right| =1$$
or
$${ z }^{ 2 }+z+1=0$$
$$z=\cfrac { -1\pm \sqrt { 1-4 } }{ 2 } =\cfrac { -1\pm \sqrt { 3i } }{ 2 } $$
$$\left| z \right| =\sqrt { \cfrac { 1 }{ 4 } +\cfrac { 3 }{ 4 } } =1$$
$$\therefore \boxed { \left| z \right| =1 } $$
If the expression $${(1+ir)}^{3}$$ is of the form of $$s(1+i)$$ for some real $$s$$ where $$r$$ is also real and $$i=\sqrt{-1}$$, then the value of $$r$$ can be
Report Question
0%
$$\cot{\cfrac{\pi}{8}}$$
0%
$$\sec{\pi}$$
0%
$$\tan{\cfrac{\pi}{12}}$$
0%
$$\tan{\cfrac{5\pi}{12}}$$
Find the value of $$x$$ such that $$\displaystyle \frac{(x + \alpha)^2 - (x + \beta)^2}{ \alpha + \beta} = \frac{sin 2 \theta}{sin^2 \theta}$$. when $$\alpha$$ and $$\beta $$ are the roots of $$t^2 - 2t + 2 = 0$$
Report Question
0%
$$x = icot \, \, \, \theta - 1$$
0%
$$x = -(icot \, \, \, \theta + 1$$)
0%
$$x = icot \, \, \, \theta $$
0%
$$x = itan \, \, \, \theta - 1$$
Explanation
$$t^{2}-2t+2=0$$
$$(t-1)^{2}-1+2=0$$
$$(t-1)^{2}=-1$$
$$t=1\pm i$$
Hence
$$\alpha+\beta=2$$
$$\alpha-\beta=2i$$
Now let $$n=2$$
We get
$$\dfrac{2x(\alpha-\beta)+\alpha^{2}-\beta^{2}}{2}=\dfrac{sin2\theta}{sin^{2}\theta}$$
Hence
$$\dfrac{4ix+2i-(-2i)}{2}=\dfrac{2sin\theta.cos\theta}{sin^{2}\theta}$$
$$2ix+2i=2cot\theta$$
$$ix=cot\theta-i$$
$$-x=icot\theta+1$$
$$x=-(1+icot\theta)$$
If $$z_1, z_2$$ be two non zero complex numbers satisfying the equation $$\displaystyle \left | \frac{z_1 + z_2}{z_1 - z_2} \right | = 1$$ then $$\displaystyle \frac{z_1}{z_2} + \left ( \frac{z_1}{z_2} \right )$$ is
Report Question
0%
zero
0%
1
0%
purely imaginary
0%
2
Explanation
let $$\frac{z_{1}}{z_{2}} =x$$
given $$|\frac{x+1}{x-1}| = 1$$
$$\Rightarrow |x+1|=|x-1|$$
Let $$x=a+ib$$
We get $$|a+1+ib|=|a-1+ib|$$
$$\Rightarrow a=0$$
Therefore $$x=ib$$ and $$\overline { x } =-ib$$
Therefore $$x+\overline { x } =ib-ib=0$$
So the correct option is $$A$$
Find the range of real number $$\alpha$$ for which the equation $$z + \alpha |z - 1| + 2i = 0; z= x + iy$$ has a solution. Find the solution.
Report Question
0%
$$\displaystyle x = 5/2, y = - 2$$
0%
$$\displaystyle x = -2, y = 5/2$$
0%
$$\displaystyle x = -5/2, y = 2$$
0%
$$\displaystyle x = 2, y = -5/2$$
Explanation
$$x+iy+\alpha(\sqrt{(x-1)^{2}+y^{2}})+2i=0$$
$$x+i(y+2)=\alpha(\sqrt{(x-1)^{2}+y^{2}}$$
$$x^{2}-(y+2)^{2}+i2x(y+2)=\alpha^{2}((x-1)^{2}+y^{2})$$
Real part
$$x^{2}-\alpha(x-1)^{2}-\alpha(y^{2})=0$$ and
Imaginary part
$$2x(y+2)=0$$
$$x(y+2)=0$$
Either $$x=0 $$or $$y=-2$$ ...if the above equations have a solution.
Considering y=-2,
Hence
$$\alpha\epsilon[-\dfrac{\sqrt{5}}{2},\dfrac{\sqrt{5}}{2}]$$
Hence
$$x=\dfrac{5}{2}$$ if $$|\alpha|\neq1$$.
If n is a natural number $$\geq 2$$, such that $$z^n=(z+1)^n$$, then
Report Question
0%
Roots of equation lie on a straight line parallel to the $$y-axis$$
0%
Roots of equation lie on a straight line parallel to the $$x-axis$$
0%
Sum of the real parts of the roots is $$-[(n-1)/2]$$
0%
None of these
Explanation
Given,
$$z^n=(z+1)^n\rightarrow |z^n|=|(z+1)^n|$$
$$\therefore |z|^n = |z+1|^n\Rightarrow |z|=|z+1|$$
$$\Rightarrow |z|^2=|z+1|^2$$
$$\Rightarrow x^2+y^2=(x+1)^2+y^2$$, where $$z=x-iy$$
$$\Rightarrow x=-\dfrac {1}{2}$$
Hence, $$z$$ lies on the line $$x=-1/2$$.
Hence, sum of real parts of the roots is $$-(n-1)/2$$ (since equation has $$n-1$$ roots).
Let $$\displaystyle\ z_{1}= a+ib, z_{2}= p+iq$$ be two unimodular complex numbers such that $$\displaystyle\ Im(z_{1}z_{2})=1$$. If$$\displaystyle\ \omega_{1}= a+ip, \omega_{2}=b+iq$$ then
Report Question
0%
$$\displaystyle\ Re(\omega_{1}\omega_{2})=1$$
0%
$$\displaystyle\ Im(\omega_{1}\omega_{2})=1$$
0%
$$\displaystyle\ Rm(\omega_{1}\omega_{2})=0$$
0%
$$\displaystyle\ Im(\omega_{1}\bar{\omega_{2}})=0$$
Explanation
$$z_1 = a + ib$$ and $$z_2 = p + iq$$
Given that $$Im(z_1z_2) = 1$$, so $$aq + bp = 1$$
Now, $$\omega_1 = a + ip, \omega_2 = b + iq$$
$$\Rightarrow \omega_1\omega_2 = ab - pq + i(aq + bp)$$
$$\Rightarrow Im(\omega_1\omega_2) = 1$$
Find the regions of the z-plane for which $$\displaystyle \left | \frac{z - a}{z + \overline a} \right | < 1, = 1$$ or $$> 1$$. when the real part of a is positive.
Report Question
0%
The required regions are the right half of the z-pane, the imaginary axis and the left half of the z-plane respectively.
0%
The required regions are the left half of the z-pane, the imaginary axis and the right half of the z-plane respectively.
0%
The required regions are the right half of the z-pane the real axis and the left half of the z-plane respectively.
0%
The required regions are the left half of the z-pane the real axis and the right half of the z-plane respectively.
Explanation
Let
$$z=x+iy$$ and $$a$$ be a fixed complex number $$(a+ib)$$.
Hence
$$|\dfrac{z-a-ib}{z+a-ib}|=1$$
$$|(x-a)+i(y-b)|=|(x+a)+i(y-b)|$$
$$(x-a)^{2}=(x+a)^{2}$$
$$2x(a)=0$$
$$x=0$$ ... imaginary axis.
If
$$|\dfrac{z-a-ib}{z+a-ib}|<1$$
Then
$$(x-a)^{2}<(x+a)^{2}$$
Or
$$x>0$$ ... positive real axis.
If
$$|\dfrac{z-a-ib}{z+a-ib}|>1$$
$$(x-a)^{2}>(x+a)^{2}$$
$$x<0$$ ... negative real axis.
Find all complex numbers satisfying the equation $$2|z|^2 + z^2 - 5 + i \sqrt{3} = 0$$
Report Question
0%
$$\displaystyle \pm \left ( \frac{\sqrt{6}}{2} + \frac{1}{\sqrt{2}} i \right ); \pm \left ( \frac{1}{\sqrt{6}} + \frac{3}{2} i \right )$$
0%
$$\displaystyle \pm \left ( \frac{\sqrt{6}}{2} - \frac{1}{\sqrt{2}} i \right ); \pm \left ( \frac{2}{\sqrt{6}} - \frac{3}{2} i \right )$$
0%
$$\displaystyle \pm \left ( \frac{\sqrt{6}}{2} - \frac{1}{\sqrt{3}} i \right ); \pm \left ( \frac{1}{\sqrt{6}} - \frac{3}{2} i \right )$$
0%
$$\displaystyle \pm \left ( \frac{\sqrt{6}}{2} - \frac{1}{\sqrt{2}} i \right ); \pm \left ( \frac{1}{\sqrt{6}} - \frac{3}{\sqrt{2}} i \right )$$
Explanation
$$Let\quad z=a+ib\\ Thus:\quad 2({ a }^{ 2 }+{ b }^{ 2 })+({ a }^{ 2 }-{ b }^{ 2 }+2abi)-5+i\sqrt { 3 } =0\\ =>3{ a }^{ 2 }+{ b }^{ 2 }-5=0\quad and\quad 2ab+\sqrt { 3 } =0=>a=-\frac { \sqrt { 3 } }{ b } \\ Substituting\quad in\quad the\quad first:\quad \frac { 9 }{ 4{ b }^{ 2 } } +{ b }^{ 2 }=5=>{ b }^{ 2 }=\frac { 9 }{ 2 } \quad or\quad \frac { 1 }{ 2 } \\ =>b=\pm \frac { 3 }{ \sqrt { 2 } } \quad or\quad \pm \frac { 1 }{ \sqrt { 2 } } =>a=\mp \frac { 1 }{ \sqrt { 6 } } \quad or\quad \mp \frac { \sqrt { 3 } }{ \sqrt { 2 } } =\mp \frac { \sqrt { 6 } }{ 2 } \\ =>z=\pm \left( \frac { \sqrt { 6 } }{ 2 } -\frac { i }{ \sqrt { 2 } } \right) \quad or\quad \pm \left( \frac { 1 }{ \sqrt { 6 } } -\frac { 3i }{ \sqrt { 2 } } \right) \\ $$
Hence, (d) is correct.
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
0
Answered
0
Not Answered
0
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page
