MCQ 11 Mark
How many even numbers can be formed by using all the digits 2, 3, 4, 5, 6?
- A72
- B36
- C120
- D24
Questions
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18 questions · 8 auto-graded MCQ + 10 self-marked written.
MCQ 21 Mark
If $\frac{3+2 \sin \theta}{1-2 \sin \theta}$ is a real number and $0<\theta<2 \pi$, then $\theta=$
- A$\frac{\pi}{3}$
- B$\frac{\pi}{2}$
- ✓$\pi$
- D$\frac{\pi}{6}$
Answer
View full question & answer→Correct option: C.
$\pi$
Given:
$\frac{3+2 i \sin \theta}{1-2 i \sin \theta}$ is a real number
On rationalising,
we get, $3+2 i \sin \theta$
$1-2 i \sin \theta \times 1+2 i \sin \theta$
$1+2 i \sin \theta$
$=\frac{(3+2 i \sin \theta)(1+2 i \sin \theta)}{(1)^2-(2 i \sin \theta)^2}$
$=\frac{3+2 i \sin \theta+6 i \sin \theta+4 i^2 \sin ^2 \theta}{1+4 \sin ^2 \theta}$
$=\frac{3-4 \sin ^2 \theta+8 i \sin \theta}{1+4 \sin ^2 \theta}\left[\because i^2=-1\right]$
$=3-4 \sin ^2 \theta$
$1+4 \sin ^2 \theta+i 8 \sin \theta$
$1+4 \sin ^2 \theta$ For the above term to be real, the imaginary part has to be zero.
$\therefore \frac{8 \sin \theta}{1+4 \sin ^2 \theta}=0$
$\Rightarrow 8 \sin \theta=0$
For this to be zero,
$\sin \theta=0$
$\Rightarrow \theta=0$
$\pi, 2 \pi, 3 \pi \ldots$
But
$0<\theta<2 \pi$
Hence,
$\theta=\pi$
$\frac{3+2 i \sin \theta}{1-2 i \sin \theta}$ is a real number
On rationalising,
we get, $3+2 i \sin \theta$
$1-2 i \sin \theta \times 1+2 i \sin \theta$
$1+2 i \sin \theta$
$=\frac{(3+2 i \sin \theta)(1+2 i \sin \theta)}{(1)^2-(2 i \sin \theta)^2}$
$=\frac{3+2 i \sin \theta+6 i \sin \theta+4 i^2 \sin ^2 \theta}{1+4 \sin ^2 \theta}$
$=\frac{3-4 \sin ^2 \theta+8 i \sin \theta}{1+4 \sin ^2 \theta}\left[\because i^2=-1\right]$
$=3-4 \sin ^2 \theta$
$1+4 \sin ^2 \theta+i 8 \sin \theta$
$1+4 \sin ^2 \theta$ For the above term to be real, the imaginary part has to be zero.
$\therefore \frac{8 \sin \theta}{1+4 \sin ^2 \theta}=0$
$\Rightarrow 8 \sin \theta=0$
For this to be zero,
$\sin \theta=0$
$\Rightarrow \theta=0$
$\pi, 2 \pi, 3 \pi \ldots$
But
$0<\theta<2 \pi$
Hence,
$\theta=\pi$
MCQ 31 Mark
If $\alpha$ and $\beta$ are acute angles satisfying $\cos 2 \alpha=\frac{3 \cos 2 \beta-1}{3-\cos 2 \beta}$, then $\alpha$ is
- A$\sqrt{2} \cot \beta$
- ✓$\sqrt{2} \tan \beta$
- C$\frac{1}{\sqrt{2}} \cot \beta$
- D$\frac{1}{\sqrt{2}} \tan \beta$
Answer
View full question & answer→Correct option: B.
$\sqrt{2} \tan \beta$
$\cos 2 \alpha=\frac{3 \cos 2 \beta-1}{3-\cos 2 \beta}$
$\Rightarrow \frac{\cos 2 \alpha-1}{\cos 2 \alpha+1}=\frac{(3 \cos 2 \beta-1)-(3-\cos 2 \beta)}{(3 \cos 2 \beta-1)+(3-\cos 2 \beta)} \text { [Using compounds and dividendo] }$
$\Rightarrow \frac{\cos 2 \alpha-1}{\cos 2 \alpha+1}=\frac{4 \cos 2 \beta-4}{2 \cos 2 \beta+2}$
$\Rightarrow-\frac{1-\cos 2 \alpha}{1+\cos 2 \alpha}=\frac{-4(1-\cos 2 \beta)}{2(1+\cos 2 \beta)}$
$\Rightarrow \frac{1-\cos 2 \alpha}{1+\cos 2 \alpha}=\frac{2(1-\cos 2 \beta)}{(1+\cos 2 \beta)}$
$\Rightarrow \frac{2 \sin ^2 \alpha}{2 \cos ^2 \alpha}=\frac{2\left(2 \sin ^2 \beta\right)}{\left(2 \cos ^2 \beta\right)}$
$\Rightarrow \tan ^2 \alpha=2 \tan ^2 \beta$
$\therefore \tan \alpha=\sqrt{2} \tan \beta$
$\Rightarrow \frac{\cos 2 \alpha-1}{\cos 2 \alpha+1}=\frac{(3 \cos 2 \beta-1)-(3-\cos 2 \beta)}{(3 \cos 2 \beta-1)+(3-\cos 2 \beta)} \text { [Using compounds and dividendo] }$
$\Rightarrow \frac{\cos 2 \alpha-1}{\cos 2 \alpha+1}=\frac{4 \cos 2 \beta-4}{2 \cos 2 \beta+2}$
$\Rightarrow-\frac{1-\cos 2 \alpha}{1+\cos 2 \alpha}=\frac{-4(1-\cos 2 \beta)}{2(1+\cos 2 \beta)}$
$\Rightarrow \frac{1-\cos 2 \alpha}{1+\cos 2 \alpha}=\frac{2(1-\cos 2 \beta)}{(1+\cos 2 \beta)}$
$\Rightarrow \frac{2 \sin ^2 \alpha}{2 \cos ^2 \alpha}=\frac{2\left(2 \sin ^2 \beta\right)}{\left(2 \cos ^2 \beta\right)}$
$\Rightarrow \tan ^2 \alpha=2 \tan ^2 \beta$
$\therefore \tan \alpha=\sqrt{2} \tan \beta$
MCQ 41 Mark
For two sets $A \cup B=A$ if
- A$A=B$
- B$A \neq B$
- C$B \subseteq A$
- D$A \subseteq B$
Answer
View full question & answer→(c) $B \subseteq A$
Explanation: The union of two sets is a set of all those elements that belong to A or to B or to both A and B .
If $A \cup B=A$, then $B \subseteq A$
Explanation: The union of two sets is a set of all those elements that belong to A or to B or to both A and B .
If $A \cup B=A$, then $B \subseteq A$
MCQ 51 Mark
The solution set for $|3 x-2| \leq \frac{1}{2}$
- A$\left[\frac{5}{6}, \frac{2}{3}\right]$
- B$\left[\frac{2}{3}, \frac{2}{3}\right]$
- ✓$\left[\frac{1}{2}, \frac{5}{6}\right]$
- D$\left[\frac{5}{6}, \frac{1}{2}\right]$
Answer
View full question & answer→Correct option: C.
$\left[\frac{1}{2}, \frac{5}{6}\right]$
$|3 x-2| \leq \frac{1}{2}$
$\Rightarrow \frac{-1}{2} \leq 3 x-2 \leq \frac{1}{2}$
$\Rightarrow \frac{-1}{2}+2 \leq 3 x-2+2 \leq \frac{1}{2}+2$
$\Rightarrow \frac{3}{2} \leq 3 x \leq \frac{5}{2} \quad[\because|x| \leq a \Leftrightarrow-a \leq x \leq a]$
$\Rightarrow \frac{3}{2} \cdot \frac{1}{3} \leq 3 x \cdot \frac{1}{3} \leq \frac{5}{2} \cdot \frac{1}{3}$
$\Rightarrow \frac{1}{2} \leq x \leq \frac{5}{6}$
$\Rightarrow x \in\left[\frac{1}{2}, \frac{5}{6}\right]$
$\Rightarrow \frac{-1}{2} \leq 3 x-2 \leq \frac{1}{2}$
$\Rightarrow \frac{-1}{2}+2 \leq 3 x-2+2 \leq \frac{1}{2}+2$
$\Rightarrow \frac{3}{2} \leq 3 x \leq \frac{5}{2} \quad[\because|x| \leq a \Leftrightarrow-a \leq x \leq a]$
$\Rightarrow \frac{3}{2} \cdot \frac{1}{3} \leq 3 x \cdot \frac{1}{3} \leq \frac{5}{2} \cdot \frac{1}{3}$
$\Rightarrow \frac{1}{2} \leq x \leq \frac{5}{6}$
$\Rightarrow x \in\left[\frac{1}{2}, \frac{5}{6}\right]$
MCQ 61 Mark
In Pascal’s triangle, each row begins with 1 and ends in
- A-1
- B$0$
- C2
- D1
Answer
View full question & answer→(d)1
Explanation:
The pascal's triangle is given by
Explanation:
The pascal's triangle is given by
MCQ 71 Mark
If $S$ be the sum, $P$ the product and $R$ be the sum of the reciprocals of $n$ terms of a $GP,$ then $P^2$ is equal to
- A$\left(\frac{R}{S}\right)^n$
- B$\frac{S}{R}$
- C$\frac{R}{S}$
- ✓$\left(\frac{S}{R}\right)^n$
Answer
View full question & answer→Correct option: D.
$\left(\frac{S}{R}\right)^n$
$\left(\frac{S}{R}\right)^n$
Explanation: According to the question,
Sum of $n$ terms of the $G.P., S =\frac{a\left(r^n-1\right)}{(r-1)}$
Product of $n$ terms of the $G.P., P = a ^{ n } r \left[\frac{n(n-1)}{2}\right]$
Sum of the reciprocals of $n$ terms of the $G.P., R =\frac{\left[\frac{1}{r^n}-1\right]}{a\left(\frac{1}{r}-1\right)}=\frac{\left(r^n-1\right)}{a r^{(n-1)}(r-1)}$
$\begin{array}{l}\therefore P^2=\left\{a^2 r^{\frac{2(n-1)}{2}}\right\}^n \\ \Rightarrow P^2=\left\{\frac{\frac{a\left(r^n-1\right)}{(r-1)}}{\frac{\left(r^{n-1}\right)}{a r^{(n-1)(r-1)}}}\right\}^n \\ \Rightarrow P^2=\left\{\frac{S}{R}\right\}^n\end{array}$
Sum of n terms, $S =\frac{a\left(r^n-1\right)}{r-1}$
Product of the $G.P., P = a ^{ n } r^{\frac{n(n+1)}{2}}$
Sum of the reciprocals of $n$ terms, $R =\frac{\left(\frac{1}{r^{\infty}-1}\right)}{a\left(\frac{1}{r-1}\right)}=\frac{\left(\frac{1-r^n}{r^n}\right)}{a\left(\frac{1-r}{r}\right)}$
$\begin{array}{l} p ^2=\left\{a^2 r^{\frac{(n+1)}{2}}\right\}^n \\ p ^2=\left\{\begin{array}{l}\frac{a\left(r^{n-1}-1\right)}{r-1} \\ \frac{\left(\frac{1+n}{r^n}\right)}{a\left(\frac{1+}{r}\right)}\end{array}\right\}=\left\{\frac{S}{R}\right\}^n\end{array}$
Explanation: According to the question,
Sum of $n$ terms of the $G.P., S =\frac{a\left(r^n-1\right)}{(r-1)}$
Product of $n$ terms of the $G.P., P = a ^{ n } r \left[\frac{n(n-1)}{2}\right]$
Sum of the reciprocals of $n$ terms of the $G.P., R =\frac{\left[\frac{1}{r^n}-1\right]}{a\left(\frac{1}{r}-1\right)}=\frac{\left(r^n-1\right)}{a r^{(n-1)}(r-1)}$
$\begin{array}{l}\therefore P^2=\left\{a^2 r^{\frac{2(n-1)}{2}}\right\}^n \\ \Rightarrow P^2=\left\{\frac{\frac{a\left(r^n-1\right)}{(r-1)}}{\frac{\left(r^{n-1}\right)}{a r^{(n-1)(r-1)}}}\right\}^n \\ \Rightarrow P^2=\left\{\frac{S}{R}\right\}^n\end{array}$
Sum of n terms, $S =\frac{a\left(r^n-1\right)}{r-1}$
Product of the $G.P., P = a ^{ n } r^{\frac{n(n+1)}{2}}$
Sum of the reciprocals of $n$ terms, $R =\frac{\left(\frac{1}{r^{\infty}-1}\right)}{a\left(\frac{1}{r-1}\right)}=\frac{\left(\frac{1-r^n}{r^n}\right)}{a\left(\frac{1-r}{r}\right)}$
$\begin{array}{l} p ^2=\left\{a^2 r^{\frac{(n+1)}{2}}\right\}^n \\ p ^2=\left\{\begin{array}{l}\frac{a\left(r^{n-1}-1\right)}{r-1} \\ \frac{\left(\frac{1+n}{r^n}\right)}{a\left(\frac{1+}{r}\right)}\end{array}\right\}=\left\{\frac{S}{R}\right\}^n\end{array}$
MCQ 81 Mark
If $A \subset B$, then
- A$A^e \subset B^e$
- B$B^e \not \subset A^c$
- C$A^c=B^c$
- ✓$B^c \subset A^c$
Answer
View full question & answer→Correct option: D.
$B^c \subset A^c$
Let $A \subset$ B
To prove $B^C \subset A^c$, it is enough to show that $x \in B^c \Rightarrow x \in A^c$
$\text { Let } x \in B^C$
$\Rightarrow x \notin B$
$\Rightarrow x \notin A \text { since } A \subset B$
$\Rightarrow x \in A^c$
Hence $B ^{ C } \subset A ^{ C }$
To prove $B^C \subset A^c$, it is enough to show that $x \in B^c \Rightarrow x \in A^c$
$\text { Let } x \in B^C$
$\Rightarrow x \notin B$
$\Rightarrow x \notin A \text { since } A \subset B$
$\Rightarrow x \in A^c$
Hence $B ^{ C } \subset A ^{ C }$
MCQ 91 Mark
$2 \sin 22 \frac{1}{2}^{\circ} \cos 22 \frac{1}{2}^{\circ}=?$
- A$\sqrt{2}$
- B$\frac{1}{\sqrt{2}}$
- C$\frac{1}{2}$
- D1
Answer
View full question & answer→(b) $\frac{1}{\sqrt{2}}$]
Explanation: Using $2 \sin A \cos A =\sin 2 A$, we get
$2 \sin 22 \frac{1}{2}^{\circ} \cos 22 \frac{1}{2}^{\circ}=\sin \left(2 \times \frac{45}{2}\right)^{\circ}=\sin 45^{\circ}=\frac{1}{\sqrt{2}}$
Explanation: Using $2 \sin A \cos A =\sin 2 A$, we get
$2 \sin 22 \frac{1}{2}^{\circ} \cos 22 \frac{1}{2}^{\circ}=\sin \left(2 \times \frac{45}{2}\right)^{\circ}=\sin 45^{\circ}=\frac{1}{\sqrt{2}}$
MCQ 101 Mark
A man wants to cut three lengths from a single piece of board of length $91 \ cm.$ The second length is to be $3 \ cm$ longer than the shortest and third length is to be twice as long as the shortest. What are the possible lengths for the shortest board if the third piece is to be at least $5 \ cm$ longer than the second?
- A$3 \leq x \leq 91$
- B$3 \leq x \leq 5$
- C$5 \leq x \leq 91$
- ✓$8 \leq x \leq 22$
Answer
View full question & answer→Correct option: D.
$8 \leq x \leq 22$
Let the length of the shortest piece be $x \ cm$. Then we have the length of the second and third pieces are $x +3$ and $2 x$ centimeters respectively.
According to the question,
$x+(x+3)+2 x \leq 91$
$\Rightarrow 4 x+3 \leq 91$
$\Rightarrow 4 x \leq 88$
$\Rightarrow x \leq 22$
Also $2 x \geq(x+3)+5$
$\Rightarrow 2 x \geq x+8$
$\Rightarrow x \geq 8$
$\Rightarrow 8 \leq x \leq 22$
Hence the shortest piece may be atleast $8 \ cm$ long but it cannot be more than $22\ cm$ in length.
According to the question,
$x+(x+3)+2 x \leq 91$
$\Rightarrow 4 x+3 \leq 91$
$\Rightarrow 4 x \leq 88$
$\Rightarrow x \leq 22$
Also $2 x \geq(x+3)+5$
$\Rightarrow 2 x \geq x+8$
$\Rightarrow x \geq 8$
$\Rightarrow 8 \leq x \leq 22$
Hence the shortest piece may be atleast $8 \ cm$ long but it cannot be more than $22\ cm$ in length.
MCQ 111 Mark
$R$ is a relation from $\{11,12,13\}$ to $\{8,10,12\}$ defined by $y=x-3$. Then, $R^{-1}$ is
- A{(10,13),(12,10)}
- B{(10,13), (8,11), (12,10)}
- C{(11,8), (13,10)}
- D{(8,11), (10,13)}
Answer
View full question & answer→(d) $\{(8,11),(10,13)\}$
Explanation: Since, $y=x-3$
Therefore, for $x=11, y=8$.
For $x =12, y =9$. [ But the value $y =9$ does not exist in the given set.]
For $x=13, y=10$.
So, we have $R =\{(11,8),(13,10)\}$
Now, $R^{-1}=\{(8,11),(10,13)\}$
Explanation: Since, $y=x-3$
Therefore, for $x=11, y=8$.
For $x =12, y =9$. [ But the value $y =9$ does not exist in the given set.]
For $x=13, y=10$.
So, we have $R =\{(11,8),(13,10)\}$
Now, $R^{-1}=\{(8,11),(10,13)\}$
MCQ 121 Mark
Let $x, y \in R$, then $x + iy$ is a non real complex number if
- A$y=0$
- B$x \neq 0$
- C$x=0$
- D$y \neq 0$
Answer
View full question & answer→(d) $y \neq 0$
Explanation: If a complex number has to be a non real complex number then its imaginary part should not be zero $\Rightarrow i y \neq 0 \Rightarrow y \neq 0$
Explanation: If a complex number has to be a non real complex number then its imaginary part should not be zero $\Rightarrow i y \neq 0 \Rightarrow y \neq 0$
MCQ 131 Mark
Which set is the subset of all given sets?
- A
- B{0}
- C
- D{ }
Answer
View full question & answer→(d) { }
Explanation: { } denoted as null set and Null set is subset of all sets.
Explanation: { } denoted as null set and Null set is subset of all sets.
MCQ 141 Mark
The coordinates of the foot of perpendiculars from the point $(2,3)$ on the line $y=3 x+4$ is given by
- A$\frac{2}{3},-\frac{1}{3}$
- B$\frac{10}{37},-10$
- C$\frac{37}{10}, \frac{-1}{10}$
- ✓$\frac{-1}{10}, \frac{37}{10}$
Answer
View full question & answer→Correct option: D.
$\frac{-1}{10}, \frac{37}{10}$
Given equation is $y=3 x+4 \ldots (i)$
Since, this equation is in $y = mx + b$ form.
Thus, the slope $\left(m_1\right)$ of the given equation is $3$
Suppose equation of any line passing through the point $(2,3)$ is
$y-y_1=m\left(x-x_1\right)$
$\Rightarrow y-3=m(x-2) \ldots \text { (ii) }$
Given that eq. $(i)$ is perpendicular to eq. $(ii)$
And we know that, if two lines are perpendicular then,
$m_1 m_2=-1$
$\Rightarrow 3 \times m_2=-1$
$\Rightarrow m_2=-\frac{1}{3}$
$\therefore$ the slope of the required line $=-\frac{1}{3}$
Substituting the value of slope in eq. $(ii),$ we obtain
$y-3=-\frac{1}{3}(x-2)$
$\Rightarrow 3 y-9=-x+2$
$\Rightarrow x+3 y-9-2=0$
$\Rightarrow x+3 y-11=0 \ldots \text { (iii) }$
Now, we have to find the coordinates of foot of the perpendicular.
Solving eq. $(i)$ and $(iii),$ we obtain
$x+3(3 x+4)-11=0[$ from $(i)]$
$\Rightarrow x+9 x+12-11=0$
$\Rightarrow 10 x+1=0$
$\Rightarrow x=-\frac{1}{10}$
Substituting the value of $x$ in Eq $(i),$ we obtain
$y=3\left(-\frac{1}{10}\right)+4$
$\Rightarrow y =-\frac{3}{10}+4$
$\Rightarrow y =\frac{-3+40}{10}$
$\Rightarrow y =\frac{37}{10}$
So, the required coordinates are $\left(-\frac{1}{10}, \frac{37}{10}\right)$
Since, this equation is in $y = mx + b$ form.
Thus, the slope $\left(m_1\right)$ of the given equation is $3$
Suppose equation of any line passing through the point $(2,3)$ is
$y-y_1=m\left(x-x_1\right)$
$\Rightarrow y-3=m(x-2) \ldots \text { (ii) }$
Given that eq. $(i)$ is perpendicular to eq. $(ii)$
And we know that, if two lines are perpendicular then,
$m_1 m_2=-1$
$\Rightarrow 3 \times m_2=-1$
$\Rightarrow m_2=-\frac{1}{3}$
$\therefore$ the slope of the required line $=-\frac{1}{3}$
Substituting the value of slope in eq. $(ii),$ we obtain
$y-3=-\frac{1}{3}(x-2)$
$\Rightarrow 3 y-9=-x+2$
$\Rightarrow x+3 y-9-2=0$
$\Rightarrow x+3 y-11=0 \ldots \text { (iii) }$
Now, we have to find the coordinates of foot of the perpendicular.
Solving eq. $(i)$ and $(iii),$ we obtain
$x+3(3 x+4)-11=0[$ from $(i)]$
$\Rightarrow x+9 x+12-11=0$
$\Rightarrow 10 x+1=0$
$\Rightarrow x=-\frac{1}{10}$
Substituting the value of $x$ in Eq $(i),$ we obtain
$y=3\left(-\frac{1}{10}\right)+4$
$\Rightarrow y =-\frac{3}{10}+4$
$\Rightarrow y =\frac{-3+40}{10}$
$\Rightarrow y =\frac{37}{10}$
So, the required coordinates are $\left(-\frac{1}{10}, \frac{37}{10}\right)$
MCQ 151 Mark
$\lim _{\theta \rightarrow 0} \frac{1-\cos 4 \theta}{1-\cos 6 \theta}$ is equal to
- ✓$4 / 9$
- B$1 / 2$
- C$-1$
- D$-1 / 2$
Answer
View full question & answer→Correct option: A.
$4 / 9$
$4 / 9$
Given, $\lim _{\theta \rightarrow 0} \frac{1-\cos 4 \theta}{1-\cos 6 \theta}$
$=\lim _{\theta \rightarrow 0} \frac{2 \sin ^2 2 \theta}{2 \sin ^2 3 \theta} \ \left[\because 1-\cos \theta=2 \sin ^2 \frac{\theta}{2}\right]$
$=\lim _{\theta \rightarrow 0} \frac{\sin ^2 2 \theta}{\sin ^2 3 \theta}$
$=\lim _{\theta \rightarrow 0}\left[\frac{\sin 2 \theta}{\sin 3 \theta}\right]^2$
$=\lim _{\substack{\theta \rightarrow 0\ 3 \theta \rightarrow 0}}\left[\frac{\frac{2 \theta}{3 \theta} \times 2 \theta}{\frac{\sin 3 \theta}{3 \theta} \times 3 \theta}\right]^2$
$=\left[\frac{2 \theta}{3 \theta}\right]^2=\left(\frac{2}{3}\right)^2=\frac{4}{9}$
Given, $\lim _{\theta \rightarrow 0} \frac{1-\cos 4 \theta}{1-\cos 6 \theta}$
$=\lim _{\theta \rightarrow 0} \frac{2 \sin ^2 2 \theta}{2 \sin ^2 3 \theta} \ \left[\because 1-\cos \theta=2 \sin ^2 \frac{\theta}{2}\right]$
$=\lim _{\theta \rightarrow 0} \frac{\sin ^2 2 \theta}{\sin ^2 3 \theta}$
$=\lim _{\theta \rightarrow 0}\left[\frac{\sin 2 \theta}{\sin 3 \theta}\right]^2$
$=\lim _{\substack{\theta \rightarrow 0\ 3 \theta \rightarrow 0}}\left[\frac{\frac{2 \theta}{3 \theta} \times 2 \theta}{\frac{\sin 3 \theta}{3 \theta} \times 3 \theta}\right]^2$
$=\left[\frac{2 \theta}{3 \theta}\right]^2=\left(\frac{2}{3}\right)^2=\frac{4}{9}$
MCQ 161 Mark
Three digits are chosen at random from 1, 2, 3, 4, 5, 6, 7, 8 and 9 without repeating any digit. What is the probability that the product is odd?
- A$\frac{5}{108}$
- B$\frac{5}{42}$
- C$\frac{2}{3}$
- D$\frac{7}{48}$
Answer
View full question & answer→(b) $\frac{5}{42}$
Explanation: Here, $n ( S )={ }^9 C _3$, Let favourable event $= E$
Explanation: Here, $n ( S )={ }^9 C _3$, Let favourable event $= E$
MCQ 171 Mark
Let A and B be finite sets containing m and n elements respectively. The number of relations that can be defined from A to B is
- A$2^{m+n}$
- B$0$
- C$2^{ mn }$
- Dmn
Answer
View full question & answer→(c) $2^{ mn }$
Explanation: We have $n ( A )= m , n ( B )= n$.
$\therefore$ Number of relations defined from $A$ to $B$
$=$ number of possible subsets of $A \times B =2^{ n ( A \times B)}=2^{ mn }$
Explanation: We have $n ( A )= m , n ( B )= n$.
$\therefore$ Number of relations defined from $A$ to $B$
$=$ number of possible subsets of $A \times B =2^{ n ( A \times B)}=2^{ mn }$
MCQ 181 Mark
$\operatorname{cosec} 150^{\circ}=?$
- A-2
- B$-\sqrt{2}$
- C2
- D$\sqrt{2}$
Answer
View full question & answer→(c) 2E
Explanation: $\operatorname{cosec} 150^{\circ}=\operatorname{cosec}\left(180^{\circ}-30^{\circ}\right)=\operatorname{cosec} 30^{\circ}=2$
Explanation: $\operatorname{cosec} 150^{\circ}=\operatorname{cosec}\left(180^{\circ}-30^{\circ}\right)=\operatorname{cosec} 30^{\circ}=2$