M.C.Q (1 Marks) - MATHS STD 11 Science Questions - Vidyadip

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M.C.Q (1 Marks)

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18 questions · self-marked practice — reveal the answer and mark yourself.

MCQ 11 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
$\sqrt{2} \cot \beta$
B
$\sqrt{2} \tan \beta$
C
$\frac{1}{\sqrt{2}} \cot \beta$
D
$\frac{1}{\sqrt{2}} \tan \beta$

Answer

(b) $\sqrt{2} \tan \beta$
Explanation: $\cos 2 \alpha=\frac{3 \cos 2 \beta-1}{3-\cos 2 \beta}$
$\begin{array}{l}\Rightarrow \frac{\cos 2 \alpha-1}{\cos 2 \alpha+1}=\frac{(3 \cos 2 \beta-1)-(3-\cos 2 \beta)}{(3 \cos 2 \beta-1)+(3-\cos 2 \beta)} \text { [Using compounds and dividendo] } \\ \Rightarrow \frac{\cos 2 \alpha-1}{\cos 2 \alpha+1}=\frac{4 \cos 2 \beta-4}{2 \cos 2 \beta+2} \\ \Rightarrow-\frac{1-\cos 2 \alpha}{1+\cos 2 \alpha}=\frac{-4(1-\cos 2 \beta)}{2(1+\cos 2 \beta)} \\ \Rightarrow \frac{1-\cos 2 \alpha}{1+\cos 2 \alpha}=\frac{2(1-\cos 2 \beta)}{(1+\cos 2 \beta)} \\ \Rightarrow \frac{2 \sin ^2 \alpha}{2 \cos ^2 \alpha}=\frac{2\left(2 \sin ^2 \beta\right)}{\left(2 \cos ^2 \beta\right)} \\ \Rightarrow \tan ^2 \alpha=2 \tan ^2 \beta \\ \therefore \tan \alpha=\sqrt{2} \tan \beta\end{array}$

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MCQ 21 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
$\frac{R}{S}$
D
$\left(\frac{S}{R}\right)^n$

Answer

(d) $\left(\frac{S}{R}\right)^n$
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, $R =\frac{\left(\frac{1}{r^{\infty}-1}\right)}{a\left(\frac{1}{r-1}\right)}=\frac{\left(\frac{1-r^n}{r^n}\right)}{a\left(\frac{1-r}{r}\right)}$
$\begin{array}{l} p ^2=\left\{a^2 r^{\frac{(n+1)}{2}}\right\}^n \\ p ^2=\left\{\begin{array}{l}\frac{a\left(r^{n-1}-1\right)}{r-1} \\ \frac{\left(\frac{1+n}{r^n}\right)}{a\left(\frac{1+}{r}\right)}\end{array}\right\}=\left\{\frac{S}{R}\right\}^n\end{array}$

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MCQ 31 Mark

If $A \subset B$, then

A
$A^e \subset B^e$
B
$B^e \not \subset A^c$
C
$A^c=B^c$
D
$B^e \subset A^c$

Answer

(d) $B^c \subset A^c$
Explanation: Let A $\subset$ B
To prove $B^C \subset A^c$, it is enough to show that $x \in B^c \Rightarrow x \in A^c$
$\begin{array}{l}\text { Let } x \in B^C \\
\Rightarrow x \notin B \\
\Rightarrow x \notin A \text { since } A \subset B\end{array}$
$\Rightarrow x \in A^c$
Hence $B ^{ C } \subset A ^{ C }$

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MCQ 41 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
$5 \leq x \leq 91$
D
$8 \leq x \leq 22$

Answer

(d) $8 \leq x \leq 22$
Explanation: 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,
$\begin{array}{l}x+(x+3)+2 x \leq 91 \\ \Rightarrow 4 x+3 \leq 91 \\ \Rightarrow 4 x \leq 88 \\ \Rightarrow x \leq 22\end{array}$
Also $2 x \geq(x+3)+5$
$\begin{array}{l}\Rightarrow 2 x \geq x+8 \\ \Rightarrow x \geq 8 \\ \Rightarrow 8 \leq x \leq 22\end{array}$
Hence the shortest piece may be atleast 8 cm long but it cannot be more than 22cm in length.

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MCQ 51 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
$\frac{37}{10}, \frac{-1}{10}$
D
$\frac{-1}{10}, \frac{37}{10}$

Answer

(d) $\frac{-1}{10}, \frac{37}{10}$
Explanation: 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
$\begin{array}{l}y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-3=m(x-2) \ldots \text { (ii) }\end{array}$
Given that eq. (i) is perpendicular to eq. (ii)
And we know that, if two lines are perpendicular then,
$\begin{array}{l}m_1 m_2=-1 \\ \Rightarrow 3 \times m_2=-1 \\ \Rightarrow m_2=-\frac{1}{3}\end{array}$
$\therefore$ the slope of the required line $=-\frac{1}{3}$
Substituting the value of slope in eq. (ii), we obtain
$y-3=-\frac{1}{3}(x-2)$
$\begin{array}{l}\Rightarrow 3 y-9=-x+2 \\
\Rightarrow x+3 y-9-2=0 \\
\Rightarrow x+3 y-11=0 \ldots \text { (iii) }\end{array}$
Now, we have to find the coordinates of foot of the perpendicular.
Solving eq. (i) and (iii), we obtain
$\begin{array}{l}x+3(3 x+4)-11=0[\text { from }(i)] \\
\Rightarrow x+9 x+12-11=0\end{array}$
$\begin{array}{l}\Rightarrow 10 x+1=0 \\
\Rightarrow x=-\frac{1}{10}\end{array}$
Substituting the value of $x$ in Eq (i), we obtain
$\begin{array}{l}y=3\left(-\frac{1}{10}\right)+4 \\
\Rightarrow y =-\frac{3}{10}+4 \\
\Rightarrow y =\frac{-3+40}{10} \\
\Rightarrow y =\frac{37}{10}\end{array}$
So, the required coordinates are $\left(-\frac{1}{10}, \frac{37}{10}\right)$

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MCQ 61 Mark

$\lim _{\theta \rightarrow 0} \frac{1-\cos 4 \theta}{1-\cos 6 \theta}$ is equal to

A
$4 / 9$
B
$1 / 2$
C
-1
D
$-1 / 2$

Answer

(a) $4 / 9$
Explanation: Given, $\lim _{\theta \rightarrow 0} \frac{1-\cos 4 \theta}{1-\cos 6 \theta}=\lim _{\theta \rightarrow 0} \frac{2 \sin ^2 2 \theta}{2 \sin ^2 3 \theta}\left[\because 1-\cos \theta=2 \sin ^2 \frac{\theta}{2}\right]$
$\begin{array}{l}=\lim _{\theta \rightarrow 0} \frac{\sin ^2 2 \theta}{\sin ^2 3 \theta}=\lim _{\theta \rightarrow 0}\left[\frac{\sin 2 \theta}{\sin 3 \theta}\right]^2 \\ =\lim _{\substack{\theta \rightarrow 0 \\ 3 \theta \rightarrow 0}}\left[\frac{\frac{2 \theta}{3 \theta} \times 2 \theta}{\frac{\sin 3 \theta}{3 \theta} \times 3 \theta}\right]^2=\left[\frac{2 \theta}{3 \theta}\right]^2=\left(\frac{2}{3}\right)^2=\frac{4}{9}\end{array}$

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MCQ 71 Mark

$\operatorname{cosec} 150^{\circ}=?$

A
-2
B
$-\sqrt{2}$
C
2
D
$\sqrt{2}$

Answer

(c) 2E
Explanation: $\operatorname{cosec} 150^{\circ}=\operatorname{cosec}\left(180^{\circ}-30^{\circ}\right)=\operatorname{cosec} 30^{\circ}=2$

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MCQ 81 Mark

How many even numbers can be formed by using all the digits 2, 3, 4, 5, 6?

A
72
B
36
C
120
D
24

Answer

(a) 72
Explanation: To form an even number the last number can only be an even digit, therefore the number of impossibility for
the last digit of number = 3
Now the ten's place can be filled by any of the remaining 4 digits, and hence the no. of ways for ten's place = 4
Then there remain three digits, so no. of ways of filling hundred's place = 3
Similarly no. of ways of filling thousand's place = 2 and of ten thousand = 1
Therefore, the total possibilities are = 3 x 4 x 3 x 2 x 1 = 72

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MCQ 91 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
$\frac{\pi}{2}$
C
$\pi$
D
$\frac{\pi}{6}$

Answer

(c) $\pi$
Explanation: $\pi$
Given:
$\frac{3+2 i \sin \theta}{1-2 i \sin \theta}$ is a real number
On rationalising, we get,
$\begin{array}{l}3+2 i \sin \theta \\ 1-2 i \sin \theta\end{array} \times \begin{array}{l}1+2 i \sin \theta \\ 1+2 i \sin \theta\end{array}$
$\begin{array}{l}=\frac{(3+2 i \sin \theta)(1+2 i \sin \theta)}{(1)^2-(2 i \sin \theta)^2} \\ =\frac{3+2 i \sin \theta+6 i \sin \theta+4 i^2 \sin ^2 \theta}{1+4 \sin ^2 \theta} \\ =\frac{3-4 \sin ^2 \theta+8 i \sin \theta}{1+4 \sin ^2 \theta}\left[\because i^2=-1\right]\end{array}$
$=\begin{array}{l}3-4 \sin ^2 \theta \\ 1+4 \sin ^2 \theta\end{array}+i \begin{array}{c}8 \sin \theta \\ 1+4 \sin ^2 \theta\end{array}$ 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,
$\begin{array}{l}\sin \theta=0 \\
\Rightarrow \theta=0\end{array}$
$\pi, 2 \pi, 3 \pi \ldots$
But
$0<\theta<2 \pi$
Hence,
$\theta=\pi$

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MCQ 101 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

(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 $B \subseteq A$

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MCQ 111 Mark

The solution set for $|3 x-2| \leq \frac{1}{2}$

A
$\left[\frac{5}{6}, \frac{2}{3}\right]$
B
$\left[\frac{2}{3}, \frac{2}{3}\right]$
C
$\left[\frac{1}{2}, \frac{5}{6}\right]$
D
$\left[\frac{5}{6}, \frac{1}{2}\right]$

Answer

(c) $\left[\frac{1}{2}, \frac{5}{6}\right]$
Explanation: $|3 x-2| \leq \frac{1}{2}$
$\begin{array}{l}\Rightarrow \frac{-1}{2} \leq 3 x-2 \leq \frac{1}{2} \\ \Rightarrow \frac{-1}{2}+2 \leq 3 x-2+2 \leq \frac{1}{2}+2 \\ \Rightarrow \frac{3}{2} \leq 3 x \leq \frac{5}{2} \quad[\because|x| \leq a \Leftrightarrow-a \leq x \leq a] \\ \Rightarrow \frac{3}{2} \cdot \frac{1}{3} \leq 3 x \cdot \frac{1}{3} \leq \frac{5}{2} \cdot \frac{1}{3} \\ \Rightarrow \frac{1}{2} \leq x \leq \frac{5}{6} \\ \Rightarrow x \in\left[\frac{1}{2}, \frac{5}{6}\right]\end{array}$

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MCQ 121 Mark

$2 \sin 22 \frac{1}{2}^{\circ} \cos 22 \frac{1}{2}^{\circ}=?$

A
$\sqrt{2}$
B
$\frac{1}{\sqrt{2}}$
C
$\frac{1}{2}$
D
1

Answer

(b) $\frac{1}{\sqrt{2}}$]
Explanation: Using $2 \sin A \cos A =\sin 2 A$, we get
$2 \sin 22 \frac{1}{2}^{\circ} \cos 22 \frac{1}{2}^{\circ}=\sin \left(2 \times \frac{45}{2}\right)^{\circ}=\sin 45^{\circ}=\frac{1}{\sqrt{2}}$

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MCQ 131 Mark

In Pascal’s triangle, each row begins with 1 and ends in

A
-1
B
$0$
C
2
D
1

Answer

(d)1
Explanation:
The pascal's triangle is given by

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MCQ 141 Mark

Which set is the subset of all given sets?

A
{1}
B
{0}
C
{1,2,3,4}
D
{ }

Answer

(d) { }
Explanation: { } denoted as null set and Null set is subset of all sets.

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MCQ 151 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

(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, $R^{-1}=\{(8,11),(10,13)\}$

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MCQ 161 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

(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 $\Rightarrow i y \neq 0 \Rightarrow y \neq 0$

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MCQ 171 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

(b) $\frac{5}{42}$
Explanation: Here, $n ( S )={ }^9 C _3$, Let favourable event $= E$

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MCQ 181 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
$2^{ mn }$
D
mn

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 $A \times B =2^{ n ( A \times B)}=2^{ mn }$

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