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Maths M.C.Q (1 Marks)

Model Paper 4 · CBSE STD 11 Science (CBSE - English Medium). 18 questions.

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M.C.Q (1 Marks)

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18 questions · 6 auto-graded MCQ + 12 self-marked written.

MCQ 11 Mark
How many diagonals are there in an octagon?
  • A
    24
  • B
    28
  • C
    20
  • D
    36
Answer
(c) 20
Explanation: No. of diagonals in a polygon of n sides $=\frac{1}{2} n(n-3)$,
Put n = 8, we get 20.
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MCQ 21 Mark
Mark the correct answer for: $i ^{-38}=$ ?
  • A
    i
  • B
    -i
  • C
    -1
  • D
    1
Answer
(c)-1
Explanation: $i ^{-38}=\frac{1}{i^{38}} \times \frac{i^2}{i^2}=\frac{-1}{i^{-40}}=\frac{-1}{\left(i^4\right)^{10}}=\frac{-1}{i^{10}}=\frac{-1}{1}=-1$
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MCQ 31 Mark
If $\sin x=\frac{-2 \sqrt{6}}{5}$ and $x$ lies in quadrant III, then $\cot x$ ?
  • A
    $\frac{3}{2 \sqrt{6}}$
  • B
    $\frac{1}{2 \sqrt{6}}$
  • C
    $\frac{-1}{2 \sqrt{6}}$
  • D
    $\frac{-3}{2 \sqrt{6}}$
Answer
(b) $\frac{1}{2 \sqrt{6}}$
Explanation: we know that $\cos ^2 x=\left(1-\sin ^2 x\right)=\left(1-\frac{24}{25}\right)=\frac{1}{25} \Rightarrow \cos x=\frac{-1}{5} \quad$ [In quadrant III, $\cos x$ is negative]
$\therefore \cot x=\frac{\cos x}{\sin x}=\frac{-1}{5} \times \frac{5}{-2 \sqrt{6}}=\frac{1}{2 \sqrt{6}}$
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MCQ 41 Mark
If $A$ and $B$ are two sets, then $A \cap(A \cup B)$ equals
  • A
    $B$
  • B
    $\phi$
  • ✓
    $A$
  • D
    $A \cap B$
Answer
Correct option: C.
$A$
Let us assume that $x \in A \cap(A \cup B)$
$\Rightarrow x \in A$ and $x \in(A \cup B)$
$\Rightarrow x \in A$ and $(x \in A$ or $x \in B)$
$\Rightarrow(x \in A$ and $x \in A)$ or $(x \in A$ and $x \in B)$
$\Rightarrow x \in A $ or $x \in A \cap B$
$\Rightarrow x \in A$
Therefore, $A \cap(A \cup B)=A$
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MCQ 61 Mark
$\sum_{r=0}^n 4^r \cdot{ }^n C_r$ is equal to
  • A
    $6^{ n }$
  • B
    $5^{-n}$
  • C
    $4^{ n }$
  • ✓
    $5^n$
Answer
Correct option: D.
$5^n$
$\sum^n 4^r \cdot{ }^n C_r=4^0 \cdot{ }^n C_0+4^1 \cdot{ }^n C_1+4^2 \cdot{ }^n C_2+\ldots+4^n \cdot{ }^n C_n$
$=1+4 \cdot{ }^n C_1+4^2 \cdot{ }^n C_2+\ldots .+4^n \cdot{ }^n C_n$
$=(1+4)^{ n }$
$=5^{ n }$
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MCQ 71 Mark
The arithmetic mean of two numbers is $34$ and their geometric mean is $16$. The numbers are
  • A
    $56$ and $12$
  • ✓
    $64 $ and $4$
  • C
    $60$ and $8$
  • D
    $52$ and $16$
Answer
Correct option: B.
$64 $ and $4$
Explanation : Let the required numbers be $a$ and $b$.
Then,
$\frac{a+b}{2}=34 $
$\Rightarrow a+b=68) $ and $\sqrt{a b}=16$
$ \Rightarrow a b=(16)^2=256$
$( a - b )^2=( a + b )^2-4 ab $
$=(68)^2-4 \times 256$
$=(4624-1024)=3600$
$\Rightarrow a-b=\sqrt{3600}=60$
On solving $a + b =68$ and $a - b =60$, we obtain $a =64, \ b=4$.
$\therefore$ the required numbers are $64$ and $4$ .
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MCQ 81 Mark
If $A=\{x: x$ is a multiple of $3, x$ natural no., $x<30\}$ and $B=\{x: x$ is a multiple of $5, x$ is natural no., $x<30\}$ then $A - B$ is
  • A
    $\{3,6,9,12,15,18,21,24,27,30\}$
  • B
    $\{3,6,9,12,18,21,24,27\}$
  • C
    $\{3,5,6,9,10,12,15,18,20,21,25,27$,
  • D
    $\{3,6,9,12,18,21,24,27,30\}$ $30\}$
Answer
(b) $\{3,6,9,12,18,21,24,27\}$
Explanation: Since set $B$ represent multiple of 5 so from Set $A$ common multiple of 3 and 5 are excluded.
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MCQ 91 Mark
$\sin 75^{\circ}=?$
  • A
    $\frac{(\sqrt{2}-1)}{2 \sqrt{2}}$
  • B
    $\frac{(\sqrt{3}-1)}{2 \sqrt{2}}$
  • C
    $\frac{(\sqrt{3}+1)}{2 \sqrt{2}}$
  • D
    $\frac{(\sqrt{2}+1)}{2 \sqrt{2}}$
Answer
(c) $\frac{(\sqrt{ } 1+1)}{2 \sqrt{2}}$
Explanatien: $\sin 75^{\circ}=\sin \left(90^{\circ}-15^{\circ}\right)=\cos 15^{\circ}=\frac{(\sqrt{3}+1)}{2 \sqrt{2}}$
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MCQ 111 Mark
Let A = {a, b, c}, then the range of the relation R = {(a, b), (a, c), (b, c)} defined on A is
  • A
    {b,c}
  • B
    {C}
  • C
    {a,b}
  • D
    {a,b,c}
Answer
(a) $(b, c)$
Explanation: Since the range is represented by the $y$-coordinate of the ordered pair ( $x, y$ ). Therefore, the range of the given relation is $\{b, c\}$.
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MCQ 121 Mark
The multiplicative inverse of $(3+2 i)^2$ is
  • A
    $\left(\frac{-4}{169}+\frac{12}{169} i\right)$
  • ✓
    $\left(\frac{5}{169}-\frac{12}{169} i\right)$
  • C
    $\left(\frac{-5}{169}+\frac{12}{169} i\right)$
  • D
    $\left(\frac{5}{169}+\frac{12}{169} i\right)$
Answer
Correct option: B.
$\left(\frac{5}{169}-\frac{12}{169} i\right)$
$z=(3+2 i)^2=\left(9+4 i^2+12 i\right)$
$=(9-4+12 i)=(5+120)$
$\Rightarrow z^{-1}=\frac{1}{(5+12 i)} \times \frac{(5-12 i)}{(5-12 i)}=\frac{(5-12 i)}{\left(25-144 i^2\right)}=\frac{(5-12 i)}{(25+144)}=\frac{(5-12 i)}{(16 i)}$
$\Rightarrow z^{-1}=\left(\frac{5}{169}-\frac{12}{169} i\right)$
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MCQ 131 Mark
Given the sets $A=\{1,2,3\}, B=\{3,4\}, C=\{4,5,6\}$, then $A \cup(B \cap C)$ is
  • A
    $(1, 2, 3)$
  • B
    $(3)$
  • C
    $(1,2,3,4,5,6)$
  • ✓
    $(1,2,3,4,5)$
Answer
Correct option: D.
$(1,2,3,4,5)$
Given $A=\{1,2,3\}, B=\{3,4\}$ and $C=\{4,5,6\}$
$B \cap C=(4)$
$A \cup(B \cap C)=\{1,2,3,4\}$
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MCQ 141 Mark
The equations of the lines through $(-1,-1)$ and making angles of $45^{\circ}$ with the line $x+y=0$ are
  • A
    x - 1 = 0, y - 1 = 0
  • B
    x + 1 = 0, y + 1 = 0
  • C
    x – 1 = 0, y – x = 0
  • D
    x + y = 0, y + 1 = 0
Answer
(b) x + 1 = 0, y + 1 = 0 
Explanation: The lines x + 1 = 0 and y + 1 = 0 are perpendicular to each other. 
The slope of the line $x+y=0$ is -1
Hence the angle made by this line with respect to $X$-axis is $45^{\circ}$
In other words, the angle made by this line with $x+1=0$ is $45^{\circ}$
Clearly the other line with which it can make $45^{\circ}$ is $y+1=0$
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MCQ 161 Mark
If two squares are chosen at random on a chess board, the probability that they have a side common is
  • A
    $\frac{2}{7}$
  • B
    $\frac{1}{18}$
  • C
    $\frac{1}{9}$
  • D
    $\frac{4}{9}$
Answer
(b) $\frac{1}{18}$
Explanation: Total number of squares $=64$.
Two squares can be selected in ${ }^{64} C_2$ ways.
In each column, there are 7 pairs of adjacent squares where each pair share 1 side in common.
Total such pairs $=8 \times 7=56$.
In each row, there are 7 pairs of adjacent squares where each pair share 1 side in common.
Total such pairs $=8 \times 7=56$.
Therefore, favourable cases $=56+56=112$.
Required probability $=\frac{112}{{ }^{64} C_2}=\frac{112}{2018}=\frac{1}{18}$
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MCQ 171 Mark
If R is a relation from a non – empty set A to a non – empty set B, then
  • A
    $R \subseteq A \times B$
  • B
    $R=A \cap B$
  • C
    $R=A \cup B$
  • D
    $R=A \times B$
Answer
(a) $R \subseteq A \times B$
Explanation: Let $A$ and $B$ be two sets. Then a relation $R$ from set $A$ to set $B$ is a subset of $A \times B$. Thus, $R$ is a relation from $A$ to $B \Leftrightarrow R \subseteq A \times B$.
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MCQ 181 Mark
$\cos 20^{\circ} \cos 40^{\circ} \cos 80^{\circ}=?$
  • A
    $\frac{1}{16}$
  • ✓
    $\frac{1}{8}$
  • C
    $\frac{\sqrt{3}}{8}$
  • D
    $\frac{\sqrt{3}}{16}$
Answer
Correct option: B.
$\frac{1}{8}$
Given exp. $=\frac{1}{2}\left(2 \cos 20^{\circ} \cos 80^{\circ}\right) \cos 40^{\circ}$
$=\frac{1}{2}\left[\cos \left(80^{\circ}+20^{\circ}\right)+\cos \left(80^{\circ}-20^{\circ}\right)\right] \cos 40^{\circ}$
$=\frac{1}{2}\left[\left(\cos 100^{\circ}+\cos 60^{\circ}\right) \cos 40^{\circ}\right]$
$=\frac{1}{2}\left[\left(\cos 100^{\circ}+\frac{1}{2}\right) \cos 40^{\circ}\right]$
$=\frac{1}{4}\left(2 \cos 100^{\circ} \cos 40^{\circ}\right)+\frac{1}{4} \cos 40^{\circ}$
$\left.=\frac{1}{4} \cos \left(100^{\circ}+40^{\circ}\right)+\cos \left(100^{\circ}-40^{\circ}\right)\right]+\frac{1}{4} \cos 40^{\circ}$
$\left.=\frac{1}{4} \cos 140^{\circ}+\cos 60^{\circ}\right)+\frac{1}{4} \cos 40^{\circ}$
$=\frac{1}{4}\left(\cos 140^{\circ}+\cos 40^{\circ}\right)+\left(\frac{1}{4} \times \frac{1}{2}\right)$
$=\frac{1}{4}\left[\cos \left(180^{\circ}-40^{\circ}\right)+\cos 40^{\circ}\right]+\frac{1}{8}$
$=\frac{1}{4}\left(-\cos 40^{\circ}+\cos 40^{\circ}\right)+\frac{1}{8}$
$=\frac{1}{8}$
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