If z is any complex number, then $\frac{z-\bar{z}}{2 i}$ is
A
either 0 or purely imaginary
B
purely imaginary
C
purely real
D
either 0 or purely real
Answer
(c) purely real Explanation: Let $z = x + i y$ Then $\bar{z}= x - iy$ $\therefore z-\overline{z}=(x+iy)-(x-iy)=2 iy$ Now $\frac{z-\bar{z}}{2 i}=y$ Hence $\frac{z-\bar{z}}{2 i}$ is purely real.
If $A$ and $B$ are two given sets, then $A \cap(A \cap B)^C$ is equal to
A
B
B
A
C
$A \cap B ^{ C }$
D
$\phi$
Answer
(c) $A \cap B ^{ C }$ Explanation: $A \cap B ^{ C }$ $A$ and $B$ are two sets. $A \cap B$ is the common region in both the sets. ( $A \cap B^C$ ) is all the region in the universal set except $A \cap B$ Now, $A \cap(A \cap B)^c=A \cap B^C$
$(x+3)+4>-2 x+5$
$\Rightarrow x +7>-2 x +5$
$\Rightarrow x +7+2 x >-2 x +5+2 x$
$\Rightarrow 3 x +7>5$
$\Rightarrow 3 x +7-7>5-7$
$\Rightarrow 3 x >-2$
$\Rightarrow x >\frac{-2}{3}$
$\Rightarrow x \in\left(\frac{-2}{3}, \infty\right)$
We have, $C _0+3 C _1+5 C _2+\ldots+(2 n +1) C _{ n }$
$=\left(C_0+C_1+C_2+\ldots+C_n\right)+2\left(C_1+2 C_2+\ldots+n C_n\right)$
$=2^n+2\left(n \cdot 2^{n-1}\right)$
$=(n+1) \cdot 2^n$
If $a , b , c$ are in $GP$ and $a^{\frac{1}{z}}=b^{\frac{1}{y}}=c^{\frac{1}{3}}$ then $x , y , z$ are in
A
$GP$
✓
$AP$
C
$H.M.$
D
$HP$
Answer
Correct option: B.
$AP$
Let $a^{\frac{1}{x}}=b^{\frac{1}{ y }}=c^{\frac{1}{z}}=k$ Then, $a = k ^{ x }, b = k ^{ y }$ and $c = k ^{ z }$.
Since, $a, b, c$ are in $GP$
$\Rightarrow b^2=a c$
$\Rightarrow\left(k^y\right)^2=\left(k^x \times k^z\right)$
$\Rightarrow k^{2 y}=k^{( x + z )}$
$\Rightarrow 2 y = x + z$
$\Rightarrow x , y , z$ are in $AP.$
The reflection of the point (4, -13) about the line 5x + y + 6 = 0 is
A
(1, 2)
B
(0, 0)
C
(3, 4)
D
(-1, -14)
Answer
(d) (-1, -14) Explanation: Suppose (h, k) be the point of reflection of the given point (4, – 13) about the line 5x + y + 6 = 0. The mid-point of the line segment joining points $( h , k )$ and $(4,-13)$ is given by $\frac{h+4}{2}, \frac{k-13}{2}$ This point lies on the given line, thus we have $5 \frac{h+4}{2}+\frac{k-13}{2}+6=0$ or $5 h+k+19=0 \ldots$ Again the slope of the line joining points $( h , k )$ and $(4,-13)$ is given by $\frac{k+13}{h-4}$. This line is perpendicular to the given line and therefore, $(-5) \frac{k+3}{h-4}=-1$ This gives $5 k+65=h-4$ or $h-5 k-69=0 \ldots$ On solving (1) and (2), we obtain $h =-1$ and $k =-14$. Therefore thethe point $(-1,-14)$ is the reflection of the given point.
Two dice each numbered from $1$ to $6$ are thrown together.
Let $A$ and $B$ be two events given by
$A:$ even number on the first die
$B:$ number on the second die is greater than $4 $
What is the value of $P ( A \cup B )$ ?
A
$\frac{1}{4}$
B
$\frac{1}{2}$
C
$\frac{1}{6}$
✓
$\frac{2}{3}$
Answer
Correct option: D.
$\frac{2}{3}$
Let $S$ be the sample space.
$\therefore n ( S )=36$
$A.$ even number on the first die
$B.$ number on the second die is greater that 4
$\therefore n ( A )=18, n ( B )=12$
$ P ( A )=\frac{18}{36}=\frac{1}{2}$ and $P ( B )=\frac{12}{36}=\frac{1}{3}$
$Also , A \cap B =\{(2,5),(2,6),(4,5),(4,6),(6,5),(6,6)\}$
$P ( A \cap B )=\frac{6}{36}=\frac{1}{6}$
$P ( A \cup B )= P ( A )+ P ( B )- P ( A \cap B )$
$=\frac{1}{2}+\frac{1}{3}-\frac{1}{6}$
$=\frac{3+2-1}{6}$
$=\frac{4}{6}$
$=\frac{2}{3}$
Let A and B be two sets containing four and two elements respectively.Then,the number of subsets of A×B,each having at least three elements is:
A
256
B
219
C
510
D
275
Answer
(b) 219 Explanation: $n(A)=4, n(B)=2$ $n(A \times B)=8$ $\therefore$ Number of subsets having at least 3 elements $=2^8-\left(1+{ }^8 C_1+{ }^8 C_2\right)=219$
If $\cot \theta=\frac{1}{2}$ and $\sec \phi=\frac{-5}{3}$, where $\theta$ lies in quadrant $III$ and $\phi$ lies in quadrant $II,$ then $\tan (\theta+\phi)=$ ?