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Maths M.C.Q (1 Marks)

Model Paper 8 · CBSE STD 11 Science (CBSE - English Medium). 18 questions.

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M.C.Q (1 Marks)

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18 questions · 7 auto-graded MCQ + 11 self-marked written.

MCQ 11 Mark
The value of $\left({ }^7 C_0+{ }^7 C_1\right)+\left({ }^7 C_1+{ }^7 C_2\right)+\ldots .+\left({ }^7 C_6+{ }^7 C_7\right)$ is
  • ✓
    $2^8-2$
  • B
    $2^8=1$
  • C
    $2^7-1$
  • D
    $2^8$
Answer
Correct option: A.
$2^8-2$
Explanation: $\left({ }^7 C_0+{ }^7 C_1\right)+\left({ }^7 C_1+{ }^7 C_2\right)+\left({ }^7 C_2+{ }^7 C_3\right)+\left({ }^7 C_4+{ }^7 C_5\right)+\left({ }^7 C_5+{ }^7 C_6\right)+\left({ }^7 C_6+{ }^7 C_7\right)$
$=1+2 \times{ }^7 C_1+2 \times{ }^7 C_2+2 \times{ }^7 C_3+2 \times{ }^7 C_5+2 \times{ }^7 C_6+1$
$=1+2 \times{ }^7 C_1+2 \times{ }^7 C_2+2 \times{ }^7 C_3+2 \times{ }^7 C_2+2 \times{ }^7 C_6+1$
$=2+2^2\left({ }^7 C_1+{ }^7 C_2+{ }^7 C_3\right)$
$=2+2^2\left(7+\frac{7}{2} \times 6+\frac{7}{3} \times \frac{6}{2} \times 5\right)$
$=2+252$
$=254$
$=2^8-2$
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MCQ 21 Mark
$\lim _{x \rightarrow \pi} \frac{\sin x}{x-\pi}$ is equal to
  • A
    1
  • B
    -1
  • C
    2
  • D
    -2
Answer
(b) -1
Explanation: Given, $\lim _{x \rightarrow \pi} \frac{\sin x}{x-\pi}=\lim _{x \rightarrow \pi} \frac{\sin (\pi-x)}{-(\pi-x)}$
$=-1\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right.$ and $\left.\pi-x \rightarrow 0 \Rightarrow x \rightarrow \pi\right]$
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MCQ 31 Mark
If $3 \sin x + 4 \cos x = 5,$ then $4 \sin x - 3 \cos x =$
  • A
    $1$
  • B
    $5$
  • C
    $3$
  • ✓
    $0$
Answer
Correct option: D.
$0$
$\frac{3}{5} \sin x+\frac{4}{5} \cos x=1$
Let $\cos \alpha=\frac{3}{5}$ and $\sin \alpha=\frac{4}{5}$
$\therefore \cos \alpha \sin x+\sin \alpha \cos x=1$
$\Rightarrow \sin (\alpha+x)=\sin \frac{\pi}{2}$
$\Rightarrow \alpha+x=\pi$
$\Rightarrow x=\frac{\pi}{2}-\alpha \ldots . \text { (i) }$
We have to find the value of $4 \sin x - 3 \cos x$
$4 \sin \left(\frac{\pi}{2}-\alpha\right)-3 \cos \left(\frac{\pi}{2}-\alpha\right) \ldots$ From eq. $(i)$
$=4 \cos \alpha-3 \sin \alpha$
$=4 \times \frac{3}{5}-3 \times \frac{4}{5}\left(\because \cos \alpha=\frac{3}{5} \text { and } \sin \alpha=\frac{4}{5}\right)$
$0$
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MCQ 41 Mark
Which of the following is a set?
A. A collection of vowels in English alphabets is a set.
B. The collection of most talented writers of India is a set.
C. The collection of most difficult topics in Mathematics is a set.
D. The collection of good cricket players of India is a set.
  • A
    B
  • B
    D
  • C
    A
  • D
    C
Answer
(c) A
Explanation: The set is {a, e, i, o, u}
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MCQ 61 Mark
If $\left(1-x+x^2\right)^n=a_0+a_1 x+a_2 x^2+\ldots+a_{2 n} x^{2 n}$, then $a_0+a_2+a_4+\ldots+a_{2 n}$ equals.
  • A
    $3^n+\frac{1}{2}$
  • B
    $\frac{3^n+1}{2}$
  • C
    $\frac{3^n-1}{2}$
  • D
    $\frac{1-3^n}{2}$
Answer
(b) $\frac{x^2+1}{2}$
Explanation: $\left(1-x+x^2\right)^n=a_0+a_1 x+a_2 x^2+\ldots+a_{2 n} x^{2 n} . .(1)$
Put x=1 in (1),we get
$1=a_0+a_1+a_2+a_3+\ldots+a_{2 n} . .(2)$
Put x=-1 in(1),we get
3^n=a_0-a_1+a_2-a_3+\ldots+a_{2 n}
Adding(1) and(2),we get
$3^n+1=2\left(a_0+a_2+a_4+\ldots+a_{2 n}\right)$
Thus, $a_0+a_2+a_4+\ldots+a_{2 n}=\frac{3^n+1}{2}$
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MCQ 71 Mark
In the expansion of $(x+a)^n$, if the sum of odd terms be $P$ and the sum of even terms be $Q$, then $4 P Q=$ ?
  • A
    $(x+a)^n=(x-a)^n$
  • B
    $(x+a)^{2 n}-(x-a)^{2 n}$
  • C
    $(x+a)^n+(x-a)^n$
  • D
    $(x+a)^{2 n}+(x-a)^{2 n}$
Answer
(b) $(x+a)^{2 n}-(x-a)^{2 n}$
Explanation: $P + Q =( x + a )^n$ and $P - Q =( x - a )^n$
$\Rightarrow 4 P Q=(P+Q)^2-(P-Q)^2=(x+a)^{2 n}-(x-a)^{2 n}$
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MCQ 81 Mark
Each set $X _{ r }$ contains 5 elements and each set $Y _{ r }$ contains 2 elements and $\bigcup_{r=1}^{20} x_r=S=\bigcup_{r=1}^n Y_r$. If each element of $S$ belong to exactly 10 of the $X_r$ 's and to exactly 4 of the $Y_r$ 's, then $n$ is
  • A
    10
  • B
    20
  • C
    50
  • D
    100
Answer
(b) 20
Explanation: The correct answer is (B)
Since, $n \left( X _{ r }\right)=5, \bigcup_{r=1}^{20} X_r= S$, we obtain $n ( S )=100$
But each element of S belong to exactly 10 of the X ’s
Thus, $\frac{100}{10}=10$ are the number of distinct elements in S .
Also each element of $S$ belong to exactly 4 of the $Y_{r^{\prime}}$ 's and each $Y_{r^{\prime} s}$ contain 2 elements. If $S$ has $n$ number of $Y_r$ in it.
Then $\frac{2 n}{4}=10$
which gives n = 20
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MCQ 91 Mark
If $\frac{3 \pi}{4}<\alpha<\pi$, then $\sqrt{2 \cot \alpha+\frac{1}{\sin ^2 \alpha}}$ is equal to
  • A
    $-1+\cot \alpha$
  • ✓
    $-1-\cot \alpha$
  • C
    $1-\cot \alpha$
  • D
    $1+\cot \alpha$
Answer
Correct option: B.
$-1-\cot \alpha$
We have:
$\sqrt{2 \cot \alpha+\frac{1}{\sin ^2 \alpha}}$
$=\sqrt{\frac{2 \cos \alpha}{\sin \alpha}+\frac{1}{\sin ^2 \alpha}}$
$=\sqrt{\frac{2 \sin \alpha \cos \alpha+1}{\sin ^2 \alpha}}$
$=\sqrt{\frac{2 \sin \alpha \cos \alpha+\sin ^2 \alpha+\cos ^2 \alpha}{\sin ^2 \alpha}}$
$=\sqrt{\frac{(\sin \alpha+\cos \alpha)^2}{\sin ^2 \alpha}}$
$=\sqrt{(1+\cot \alpha)^2}$
$=|1+\cot \alpha|$
$=-(1+\cot \alpha)\left[\text { When } \frac{3 \pi}{4}<\alpha<\pi, \cot \alpha<-1 \Rightarrow \cot \alpha+1<0\right]$
$=-1-\cot \alpha$
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MCQ 101 Mark
If $y =1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots$ then $\frac{d y}{d x}=$
  • A
    $y^2$
  • B
    $y + 1$
  • ✓
    $y$
  • D
    $y - 1$
Answer
Correct option: C.
$y$
Differentiating both sides with respect to $x$,
we get $\frac{d y}{d x}=\frac{d}{d x}\left(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^2}{3!}+\ldots\right)$
$=\frac{d}{d x}(1)+\frac{d}{d w}\left(\frac{x}{11}\right)+\frac{d}{d w}\left(\frac{x^2}{2!}\right)+\frac{d}{d w}\left(\frac{x^3}{3!}\right)+\frac{d}{d x}\left(\frac{x^4}{4!}\right)+\ldots$
$=\frac{d}{d x}(1)+\frac{1}{1!} \frac{d}{d x}(x)+\frac{1}{2!} \frac{d}{d w}\left(x^2\right)+\frac{1}{3!} \frac{d}{d w}\left(x^3\right)+\frac{1}{4!} \frac{d}{d w}\left(x^4\right)+\ldots$
$=0+\frac{1}{1!} \times 1+\frac{1}{2!} \times 2 \alpha+\frac{1}{3!} \times 3 \alpha^2+\frac{1}{4!} \times 4 \alpha^3+\ldots\left(y=\alpha^2 \Rightarrow \frac{d y}{d \alpha}=n \alpha^{n-1}\right)$
$=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots\left[\frac{x}{n!}=\frac{1}{(n-1)!}\right]$
$=y$
$\therefore \frac{d y}{d x}=y$
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MCQ 111 Mark
A fair dice is rolled n times. The number of all the possible outcomes is
  • A
    6n
  • B
    $n^6$
  • C
    $6^{ n }$
  • D
    6+n
Answer
(c) $6^n$
Explanation: Each time there are 6 possibilities, therefore for n times there are $6^{ n }$ possibilities.
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MCQ 121 Mark
Mark the correct answer for $3 i^{34}+5 i^{27}-2 i^{38}+5 i^{41}=$ ?
  • A
    $1$
  • ✓
    $-1$
  • C
    $-4i$
  • D
    $10i$
Answer
Correct option: B.
$-1$
$3^{34}+5 i^{27}-2 i^{38}+5 i^{41}$
$=3 \times\left(i^4\right)^8 \times i^2+5 \times\left(i^4\right)^6 \times i^3-2 \times\left(i^4\right)^9 \times i^2+5 \times\left(i^4\right)^{10} \times i$
$=3 \times 1 \times(-1)+5 \times 1 \times(-i)-2 \times 1 \times(-1)+5 \times 1 \times i$
$=-3-5 i+2+5 i$
$=-1$
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MCQ 131 Mark
The length of the foot of perpendicular drawn from the point P (3, 4, 5) on y-axis is
  • A
    $\sqrt{34}$
  • B
    10
  • C
    $\sqrt{113}$
  • D
    $5 \sqrt{2}$
Answer
(a) $\sqrt{34}$
Explanation: Let l be the foot of the perpendicular from point P on the y -axis. Therefore, its x and z -coordinates are zero, i.e., $(0,4,0)$. Therefore, the distance between the points $(0,4,0)$ and $(3,4,5)$ is $\sqrt{9+25}=\sqrt{34}$.
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MCQ 141 Mark
The coordinates of the foot of perpendicular from (0, 0) upon the line x + y = 2 are
  • A
    (1, 1)
  • B
    (1, -2)
  • C
    (-1, 2)
  • D
    (1, 2)
Answer
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MCQ 151 Mark
If $f(x)=x \sin x$, then $f^{\prime}\left(\frac{\pi}{2}\right)$ is equal to
  • A
    1
  • B
    $\frac{1}{2}$
  • C
    -1
  • D
    $0$
Answer
(a) 1
Explanation: f′ (x) = x cosx + sinx 
So, $f^{\prime}\left(\frac{\pi}{2}\right)=\frac{\pi}{2} \cos \frac{\pi}{2}+\sin \frac{\pi}{2}=1$
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MCQ 161 Mark
The mean of the series $x_1, x_2, \ldots, x_n$ is $\bar{X}$. If $x_2$ is replaced by $\lambda$, then what is the new mean?
  • A
    $\frac{\bar{X}-x_2-\lambda}{n}$
  • ✓
    $\frac{n \bar{X}-x_2-\lambda}{n}$
  • C
    $\frac{\bar{X}-x_2+\lambda}{n}$
  • D
    $\bar{X}-x_2+\lambda$
Answer
Correct option: B.
$\frac{n \bar{X}-x_2-\lambda}{n}$
We know, $\bar{X}=\frac{x_1+x_2+\ldots+x_n}{n}$
$\Rightarrow x _1+ x _2+\ldots+ x _{ n }= n \bar{X}$
$\Rightarrow x _1+ x _2+\ldots+ x _{ n }$
$= n \overline{ X }- x _2$
$\Rightarrow x _1+ x _3+\ldots+ x _{ n }+\lambda$
$= n \overline{ X }- x _2+\lambda$
$\Rightarrow \text { Mean }=\frac{\text { Sum of all values }}{\text { Total mumber of values }}$
$=\frac{x_1+x_3+\ldots+x_n+\lambda}{n}$
$=\frac{n \bar{X}-x_2-\lambda}{n}$
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MCQ 171 Mark
If $n ( A )=10, n ( B )=6$ and $n ( C )=5$ for three disjoint sets $A , B$ and $C ,$ then $n(A \cup B \cup C)=$
  • A
    $11$
  • ✓
    $21$
  • C
    $1$
  • D
    $9$
Answer
Correct option: B.
$21$
Since $A, B, C$ are disjoint
$\therefore n(A \cup B \cup C)$
$= n ( a )+ n ( B )+ n ( C )$
$=10+6+5$
$=21$
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MCQ 181 Mark
$\sin \frac{\pi}{12}=?$
  • A
    $\frac{(\sqrt{3}+1)}{2 \sqrt{2}}$
  • B
    $\frac{(\sqrt{3}-1)}{2 \sqrt{2}}$
  • C
    $\frac{(2 \sqrt{3}+1)}{2 \sqrt{5}}$
  • D
    $\frac{-(\sqrt{3}-1)}{2}$
Answer
(b) $\frac{(\sqrt{3}-1)}{2 \sqrt{2}}$
Explanation: $\sin \frac{\pi}{12}=\sin \left(\frac{\pi}{4}-\frac{\pi}{6}\right)=\sin \frac{\pi}{4} \cos \frac{\pi}{6}-\cos \frac{\pi}{4} \sin \frac{\pi}{6}$ 
$=\left(\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}\right)-\left(\frac{1}{\sqrt{2}} \times \frac{1}{2}\right)=\frac{(\sqrt{3}-1)}{2 \sqrt{2}}$
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