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MATHS M.C.Q (1 Marks)

Model Paper 8 · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - English Medium). 18 questions.

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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
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 21 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
  • A
    $2^8-2$
  • B
    $2^8=1$
  • C
    $2^7-1$
  • D
    $2^8$
Answer
(a) $2^8-2$
Explanation: $\left({ }^7 C_0+{ }^7 C_1\right)+\left({ }^7 C_1+{ }^7 C_2\right)+\left({ }^7 C_2+{ }^7 C_3\right)+\left({ }^7 C_4+{ }^7 C_5\right)+\left({ }^7 C_5+{ }^7 C_6\right)+\left({ }^7 C_6+{ }^7 C_7\right)$
$\begin{array}{l}=1+2 \times{ }^7 C_1+2 \times{ }^7 C_2+2 \times{ }^7 C_3+2 \times{ }^7 C_5+2 \times{ }^7 C_6+1 \\ =1+2 \times{ }^7 C_1+2 \times{ }^7 C_2+2 \times{ }^7 C_3+2 \times{ }^7 C_2+2 \times{ }^7 C_6+1 \\ =2+2^2\left({ }^7 C_1+{ }^7 C_2+{ }^7 C_3\right) \\ =2+2^2\left(7+\frac{7}{2} \times 6+\frac{7}{3} \times \frac{6}{2} \times 5\right) \\ =2+252 \\ =254 \\ =2^8-2\end{array}$
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MCQ 31 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 51 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 61 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 71 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 81 Mark
If $\frac{3 \pi}{4}<\alpha<\pi$, then $\sqrt{2 \cot \alpha+\frac{1}{\sin ^2 \alpha}}$ is equal to
  • A
    $-1+\cot \alpha$
  • B
    $-1-\cot \alpha$
  • C
    $1-\cot \alpha$
  • D
    $1+\cot \alpha$
Answer
(b) - 1 - cot
Explanation: We have:
$\begin{array}{l}\sqrt{2 \cot \alpha+\frac{1}{\sin ^2 \alpha}} \\ =\sqrt{\frac{2 \cos \alpha}{\sin \alpha}+\frac{1}{\sin ^2 \alpha}} \\ =\sqrt{\frac{2 \sin \alpha \cos \alpha+1}{\sin ^2 \alpha}} \\ =\sqrt{\frac{2 \sin \alpha \cos \alpha+\sin ^2 \alpha+\cos ^2 \alpha}{\sin ^2 \alpha}} \\ =\sqrt{\frac{(\sin \alpha+\cos \alpha)^2}{\sin ^2 \alpha}} \\ =\sqrt{(1+\cot \alpha)^2} \\ =|1+\cot \alpha| \\ =-(1+\cot \alpha)\left[\text { When } \frac{3 \pi}{4}<\alpha<\pi, \cot \alpha<-1 \Rightarrow \cot \alpha+1<0\right] \\ =-1-\cot \alpha\end{array}$
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MCQ 101 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 111 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 121 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 131 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 141 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
  • B
    21
  • C
    1
  • D
    9
Answer
(b) 21
Explanation: Since A, B, C are disjoint
$\begin{array}{l}\therefore n(A \cup B \cup C)= n ( a )+ n ( B )+ n ( C ) \\ =10+6+5=21\end{array}$
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MCQ 151 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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MCQ 161 Mark
If 3 sin x + 4 cos x = 5, then 4 sin x - 3 cos x =
  • A
    1
  • B
    5
  • C
    3
  • D
    $0$
Answer
(d) 0
Explanation: 3 sinx + 4 cosx = 5
$\frac{3}{5} \sin x+\frac{4}{5} \cos x=1$
Let $\cos \alpha=\frac{3}{5}$ and $\sin \alpha=\frac{4}{5}$
$\therefore \cos \alpha \sin x+\sin \alpha \cos x=1$
$\begin{array}{l}\Rightarrow \sin (\alpha+x)=\sin \frac{\pi}{2} \\ \Rightarrow \alpha+x=\pi \\
\Rightarrow x=\frac{\pi}{2}-\alpha \ldots . \text { (i) }\end{array}$
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)
$\begin{array}{l}=4 \cos \alpha-3 \sin \alpha \\ =4 \times \frac{3}{5}-3 \times \frac{4}{5}\left(\because \cos \alpha=\frac{3}{5} \text { and } \sin \alpha=\frac{4}{5}\right) \\ 0\end{array}$
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MCQ 171 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
  • C
    y
  • D
    y - 1
Answer
(c) y
Explanation: $y=1+\frac{x}{11}+\frac{x^2}{x}+\frac{x^3}{x}+\ldots$
Differentiating both sides with respect to $x$, we get $\frac{d y}{d x}=\frac{d}{d x}\left(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^2}{3!}+\ldots\right)$
$\begin{array}{l}=\frac{d}{d x}(1)+\frac{d}{d w}\left(\frac{x}{11}\right)+\frac{d}{d w}\left(\frac{x^2}{2!}\right)+\frac{d}{d w}\left(\frac{x^3}{3!}\right)+\frac{d}{d x}\left(\frac{x^4}{4!}\right)+\ldots \\ =\frac{d}{d x}(1)+\frac{1}{1!} \frac{d}{d x}(x)+\frac{1}{2!} \frac{d}{d w}\left(x^2\right)+\frac{1}{3!} \frac{d}{d w}\left(x^3\right)+\frac{1}{4!} \frac{d}{d w}\left(x^4\right)+\ldots \\ =0+\frac{1}{1!} \times 1+\frac{1}{2!} \times 2 \alpha+\frac{1}{3!} \times 3 \alpha^2+\frac{1}{4!} \times 4 \alpha^3+\ldots\left(y=\alpha^2 \Rightarrow \frac{d y}{d \alpha}=n \alpha^{n-1}\right) \\ =1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots\left[\frac{x}{n!}=\frac{1}{(n-1)!}\right] \\ =y\end{array}$
$\therefore \frac{d y}{d x}=y$

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MCQ 181 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
    $\frac{\bar{X}-x_2-\lambda}{n}$
  • B
    $\frac{n \bar{X}-x_2-\lambda}{n}$
  • C
    $\frac{\bar{X}-x_2+\lambda}{n}$
  • D
    $\bar{X}-x_2+\lambda$
Answer
(b) $\frac{n \bar{x}-x_2-\lambda}{n}$
Explanation: We know, $\bar{X}=\frac{x_1+x_2+\ldots+x_n}{n} \Rightarrow x _1+ x _2+\ldots+ x _{ n }= n \bar{X}$
$\begin{array}{l}\Rightarrow x _1+ x _2+\ldots+ x _{ n }= n \overline{ X }- x _2 \\ \Rightarrow x _1+ x _3+\ldots+ x _{ n }+\lambda= n \overline{ X }- x _2+\lambda \\ \Rightarrow \text { Mean }=\frac{\text { Sum of all values }}{\text { Total mumber of values }}=\frac{x_1+x_3+\ldots+x_n+\lambda}{n} \\ =\frac{n \bar{X}-x_2-\lambda}{n}\end{array}$
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