M.C.Q (1 Marks)

MCQ 1511 Mark

Choose the correct answer. Let $x, y \in R,$ then $x + iy$ is a non real complex number if:

A
$x = 0$
B
$y = 0$
C
$x \neq 0$
✓
$y \neq 0$

Answer

Correct option: D.

$y \neq 0$

$x + yi$ is a non$-$real complex number if $y ≠ 0.$ If $x, y \in R$

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MCQ 1521 Mark

Determine the values of $p$ for which the quadratic equation $2x^2+ px + 8 = 0$ has equal roots.

A
$p = ±64$
✓
$p = ±8$
C
$p = ±4$
D
$p = ±16$

Answer

Correct option: B.

$p = ±8$

In $a x^2+b x+c=0$
Discriminant $D=b^2-4 a c$
$D=0$, for the roots to be real and equal
In $2 x^2+p x+8=0$
Thus, we have $a=2, b=p$ and $c=8$
Then $D=b^2-4 a c=0$
$p^2-4 \times 2 \times 8=0$
$\Rightarrow p^2-64=0$
$\Rightarrow \mathrm{p}= \pm 8$

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MCQ 1531 Mark

If $z_1=2+3 i$ and $z_2=5+2 i$, then find $z_1-z_2$.

✓
$-3 + 1i$
B
$3 - i$
C
$7 + 5i$
D
$7 - 5i$

Answer

Correct option: A.

$-3 + 1i$

In subtracting one complex number from other,
difference of corresponding parts of two complex numbers is calculated.
So, $z_1-z_2=(2-5)+(3-2) i=-3+1 {~i}$.

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MCQ 1541 Mark

$1 + 0 i$ is, $...........$ for complex number $z.$

A
additive inverse
B
additive identity element
✓
multiplicative identity element
D
multiplicative inverse

Answer

Correct option: C.

multiplicative identity element

On multiplying one $( 1 + 0 i )$ to a complex number, we get same complex number so $1 + 0i$ is multiplicative identity element for complex number $z.$
i.e. $z \times 1 = z.$

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MCQ 1551 Mark

If $\text{x}+\text{iy}=\frac{3+5\text{i}}{7-6\text{i}},$ then $\text{y}=$

A
$\frac{9}{85}$
B
$\frac{-9}{85}$
✓
$\frac{53}{85}$
D
none of these

Answer

Correct option: C.

$\frac{53}{85}$

$\text{x}+\text{iy}=\frac{3+5\text{i}}{7-6\text{i}}$
$\Rightarrow\text{x}+\text{iy}=\frac{3+5\text{i}}{7-6\text{i}}\times\frac{7+6\text{i}}{7+6\text{i}}$
$\Rightarrow\text{x}+\text{iy}=\frac{21+53\text{i}+30\text{i}^2}{49-36\text{i}^2}$
$\Rightarrow\text{x}+\text{iy}=\frac{21-30+53\text{i}}{49+36}$
$\Rightarrow\text{x}+\text{iy}=\frac{-9}{85}+\text{i}\frac{53}{85}$
On comparing both the sides:
$\text{y}=\frac{53}{85}$

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MCQ 1561 Mark

$\frac{1+2\text{i}+3\text{i}^2}{1-2\text{i}+3\text{i}^2}$ equals:

A
$i$
B
$-1$
✓
$-i$
D
$4$

Answer

Correct option: C.

$-i$

Let $\text{z}=\frac{1+2\text{i}+3\text{i}^2}{1-2\text{i}+3\text{i}^2}$
$\Rightarrow\text{z}=\frac{1+2\text{i}-3}{1-2\text{i}-3}$
$\Rightarrow\text{z}=\frac{-2+2\text{i}}{-2-2\text{i}}\times\frac{-2+2\text{i}}{-2+2\text{i}}$
$\Rightarrow\text{z}=\frac{(-2+2\text{i})^2}{(-2)^2-(2\text{i})^2}$
$\Rightarrow\text{z}=\frac{4+4\text{i}^2-8\text{i}}{4+4}$
$\Rightarrow\text{z}=\frac{4-4-8\text{i}}{8}$
$\Rightarrow\text{z}=\frac{-8\text{i}}{8}$
$\Rightarrow\text{z}=-\text{i}$

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MCQ 1571 Mark

What is the number of solution$(s)$ of the equation $|\sqrt{\text{x}-2}|+\sqrt{\text{x} (\sqrt{\text{x}-4})}+2=0$

✓
$2$
B
$4$
C
No solution
D
Infinitely many solutions

Answer

Correct option: A.

$2$

We have $|\sqrt{\text{x}-2}|+\sqrt{\text{x}}(\sqrt{\text{x}-4})+2=0$
$|\sqrt{\text{x}} – 2| + \sqrt{(\text{x})^2} – 4\sqrt{\text{x}} + 2 = 0$
$|\sqrt{\text{x}} – 2| + |\sqrt{\text{x}} -2|^2 – 2 = 0$
$|\sqrt{\text{x}} – 2| = -2,\text{ 1}$
Thus, $\sqrt{\text{x}}-2=+1,-1$ or $\text{x}=1,\text{ 9}$

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MCQ 1581 Mark

If $\text{z}=\Big(\frac{1+\text{i}}{1-\text{i}}\Big),$ then $z^4$ equals:

✓
$1$
B
$-1$
C
$0$
D
none of these.

Answer

Correct option: A.

$1$

Let $\text{z}=\frac{1+\text{i}}{1-\text{i}}$
Rationalising the denominator:
$\text{z}=\frac{1+\text{i}}{1-\text{i}}\times\frac{1+\text{i}}{1+\text{i}}$
$\Rightarrow\text{z}=\frac{1+\text{i}^2+2\text{i}}{1-\text{i}^2}$
$\Rightarrow\text{z}=\frac{1-1+2\text{i}}{1+1}$
$\Rightarrow\text{z}=\frac{2\text{i}}{2}$
$\Rightarrow\text{z}=\text{i}$
$\Rightarrow\text{z}^4=\text{i}^4$
Since $\text{i}^2=-1,$ we have:
$\Rightarrow\text{z}^4=\text{i}^2\times\text{i}^2$
$\Rightarrow\text{z}^4=1$

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MCQ 1591 Mark

Choose the correct answer.
The real value of $\theta$ for which the expression $\frac{1+\text{i}\cos\theta}{1-2\text{i}\cos\theta}$ is a real number is:

A
$\text{n}\pi+\frac{\pi}{4}$
B
$\text{n}\pi+(-1)^{\text{n}}\frac{\pi}{4}$
✓
$2\text{n}\pi\pm\frac{\pi}{2}$
D
None of these.

Answer

Correct option: C.

$2\text{n}\pi\pm\frac{\pi}{2}$

Let $\text{z}=\frac{1+\text{i}\cos\theta}{1-2\text{i}\cos\theta}=\frac{1+\text{i}\cos\theta}{1-2\text{i}\cos\theta}\times\frac{1+2\text{i}\cos\theta}{1+2\text{i}\cos\theta}$
$=\frac{1+2\text{i}\cos\theta+\text{i}\cos\theta+2\text{i}^2\cos^2\theta}{1-4\text{i}^2\cos^2\theta}$
$=\frac{1-3\text{i}\cos\theta-2\cos^2\theta}{1+4\cos^2\theta}$
$=\frac{1-2\cos^2\theta}{1+4\cos^2\theta}+\frac{3\cos\theta}{1+4\cos^2\theta}\text{i}$
If $z$ is real number, then
$\frac{3\cos\theta}{1+4\cos^2\theta}=0$
$\Rightarrow3\cos\theta=0$
$\Rightarrow\cos\theta=0$
$\therefore\ \theta=(2\text{n}+1)\frac{\pi}{2},\text{ n}\in\text{N}$

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MCQ 1601 Mark

For what value of $\theta, 1$ lies between the roots of the quadratic equation $3\text{x}^2 – 3\sin\theta \text{x} – 2\cos^2\theta = 0$

✓
$2\text{n}\pi+\frac{\pi}{6}<\theta<2\text{n}\pi+\frac{5\pi}{6}$
B
$2\text{n}\pi+\frac{\pi}{3}<\theta<2\text{n}\pi+\frac{5\pi}{3}$
C
$2\text{n}\pi+\frac{\pi}{6}<\theta\leq2\text{n}\pi+\frac{5\pi}{6}$
D
$2\text{n}\pi+\frac{\pi}{3}<\theta\leq2\text{n}\pi+\frac{5\pi}{3}$

Answer

Correct option: A.

$2\text{n}\pi+\frac{\pi}{6}<\theta<2\text{n}\pi+\frac{5\pi}{6}$

Let, $\text{f}(\text{x})=3\text{x}^2-3\sin\theta\text{x}-2\cos^2\theta$
The coefficient of $\text{x}^2>0$
$\text{f}(1)<0$
So, $3-3\sin\theta-2\cos^2\theta<0$
$\Rightarrow2\sin^2\theta-3\sin\theta+1<0$
$\Rightarrow(2\sin\theta-1)(\sin\theta-1)<0$
$\Rightarrow\frac{1}{2}2\text{n}\pi+\frac{\pi}{6}<\theta<2\text{n}\pi+\frac{5\pi}{6}$

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MCQ 1611 Mark

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

✓
$\frac{-1\pm\text{i}\sqrt{7}}{4}$
B
$\frac{1\pm\text{i}\sqrt{7}}{4}$
C
$\frac{1\pm\sqrt{7}}{4}$
D
$\frac{-1\pm\sqrt{7}}{4}$

Answer

Correct option: A.

$\frac{-1\pm\text{i}\sqrt{7}}{4}$

$2\text{x}^2+\text{x}+1=0$
$\text{D}=1^2-4\times2\times1=1-8=-7\leq0$
Since $\text{D}\leq0,$ imaginary roots are there.
$\Rightarrow\text{x}=\frac{-1\pm\sqrt{1^2-42.1}}{2.2}=\frac{-1\pm\text{i}\sqrt{7}}{4}$

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MCQ 1621 Mark

Roots of a quadratic equation are imaginary when discriminant is $..............?$

A
zero
B
greater than zero
✓
less than zero
D
greater than or equal to zero

Answer

Correct option: C.

less than zero

For a quadratic equation, $ax^2 + bx + c = 0$, discriminant is $b^2$-4ac.
Roots are $\frac{-\text{b}\pm\sqrt{b^2 -4\text{ac}}}{2\text{a}}$ for imaginary roots, radical is negative
i.e. discriminant should be less than zero.

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MCQ 1631 Mark

The argument of $\frac{(1-\text{i}\sqrt{3})}{(1+\text{i}\sqrt{3})}$ is:

✓
$60^\circ$
B
$120^\circ$
C
$210^\circ$
D
$240^\circ$

Answer

Correct option: A.

$60^\circ$

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MATHS M.C.Q (1 Marks)