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Maths M.C.Q (1 Marks)

Model Paper 7 · CBSE STD 11 Science (CBSE - English Medium). 18 questions.

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M.C.Q (1 Marks)

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18 questions · timed · auto-graded

MCQ 11 Mark
The total number of 9 digit numbers which have all different digits is
  • A
    91
  • B
    $10 \times 10!$
  • C
    10!
  • D
    $9 \times 9$ !
Answer
(d) $9 \times 9$ !
Explanation: We have to form 9 digit numbers from 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 and we know that 0 can not be put on extremely left place. Therefore, first place from the left can be filled in 9 ways.
Now repetition is not allowed. Therefore, the remaining 8 places can be filled in 9! 
$\therefore$ The required number of ways $=9 \times 9$ !
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MCQ 21 Mark
If $y=\frac{\sin x+\cos x}{\sin x-\cos x}$, then $\frac{d y}{d x}$ at $x =0$ is equal to
  • A
    $0$
  • ✓
    $-2$
  • C
    $\frac{1}{2}$
  • D
    Does not exist
Answer
Correct option: B.
$-2$
Given, $y=\frac{\sin x+\cos x}{\sin x-\cos x}$
$\frac{d y}{d x}=\frac{(\sin x-\cos x)(\cos x-\sin x)-(\sin x+\cos x)(\cos x+\sin x)}{(\sin x-\cos x)^2}$
$=\frac{-(\sin x-\cos x)^2-(\sin x+\cos x)^2}{(\sin x-\cos x)^2}$
$=\frac{-\left[\sin ^2 x+\cos ^2 x-2 \sin x \cos x+\sin ^2 x+\cos ^2 x+2 \sin x \cos x\right]}{(\sin x-\cos x)^2}$
$=\frac{-2}{(\sin x-\cos x)^2}$
$\therefore\left(\frac{d y}{d x}\right)_{\text {at } x=0}=\frac{-2}{(\sin 0-\cos 0)^2}$
$=\frac{-2}{(-1)^2}$
$=-2$
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MCQ 31 Mark
$\cos 15^{\circ}=?$
  • A
    $\frac{(\sqrt{3}+1)}{\sqrt{2}}$
  • B
    $\frac{(\sqrt{3}+1)}{2 \sqrt{2}}$
  • C
    $\frac{(\sqrt{3}-1)}{2 \sqrt{2}}$
  • D
    $\frac{(\sqrt{3}-1)}{\sqrt{2}}$
Answer
(b) $\frac{(\sqrt{3}+1)}{2 \sqrt{2}}$
Explanation: $\cos 15^{\circ}=\cos \left(45^{\circ}-30^{\circ}\right)=\cos 45^{\circ} \cos 30^{\circ}+\sin 45^{\circ} \sin 30^{\circ}$ $=\left(\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}\right)+\left(\frac{1}{\sqrt{2}} \times \frac{1}{2}\right)=\frac{(\sqrt{3}+1)}{2 \sqrt{2}}$
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MCQ 41 Mark
If A = {1, 2, 3, 4, 5, 6} then the number of proper subsets is
  • A
    63
  • B
    36
  • C
    64
  • D
    25
Answer
(a) 63
Explanation: 63
The no. of proper subsets $=2^{ n }-1$
Here $n(A)=6$
In case of the proper subset, the set itself is excluded that's why the no. of the subset is 63 . But if it is asked no. of improper or just no. of subset then you may write 64
So no. of proper subsets $=63$
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MCQ 51 Mark
Solve the system of inequalities: $\frac{x+7}{x-8}>2, \frac{2 x+1}{7 x-1}>5$
  • A
    (4, 8)
  • B
    (3, 6)
  • C
    no solution
  • D
    (2, 5)
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MCQ 61 Mark
$\sum_{r=0}^n 4^r .^n C_r$ is equal to
  • A
    $6^n$
  • B
    $5^{-n}$
  • C
    $4^n$
  • ✓
    $5^n$
Answer
Correct option: D.
$5^n$
$\sum_{r=0}^n 4^r .^n C_r=4^0 \cdot n C_0+4^1 \cdot n C_1+4^2{ }^n C_2+\ldots+4^n \cdot{ }^n C_n$
$=1+4 \cdot{ }^n C_1+4^2 \cdot{ }^n C_2+\ldots .+4^n \cdot{ }^n C_n$
$=(1+4)^n$
$=5^n$
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MCQ 71 Mark
The integral part of $(\sqrt{2}+1)^6$ is
  • A
    $98$
  • B
    $96$
  • ✓
    $99$
  • D
    $100$
Answer
Correct option: C.
$99$
We have $(1+ x )^{ n }=1+{ }^{ n } C_1(x)+{ }^{ n } C_2(x)^2+\ldots .+( x )^{ n }$
Hence $(\sqrt{2}+1)^6=1+{ }^6 C_1(\sqrt{2})+{ }^6 C_2(\sqrt{2})^2+{ }^6 C_3(\sqrt{2})^3+{ }^6 C_4(\sqrt{2})^4+{ }^6 C_5(\sqrt{2})^5+(\sqrt{2})^6$
$\Rightarrow(\sqrt{2}+1)^6=1+6(\sqrt{2})+15 \times 2+20 \times 2(\sqrt{2})+15 \times 4+6 \times 4(\sqrt{2})+8$
$=99+70 \sqrt{2}$
Hence integral part of $(\sqrt{2}+1)^6=99$`
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MCQ 81 Mark
The set of all prime numbers is
  • A
    an infinite set
  • B
    a singleton set
  • C
    a multi set
  • D
    a finite set
Answer
(a) an infinite set
Explanation: Set A = {2, 3, 5, 7,...} so it is infinite.
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MCQ 91 Mark
$\tan 15^{\circ}=?$
  • A
    $\frac{(\sqrt{2}+1)}{(\sqrt{2}-1)}$
  • B
    $\frac{(\sqrt{3}+1)}{(\sqrt{3}-1)}$
  • C
    $\frac{(\sqrt{3}-1)}{(\sqrt{3}+1)}$
  • D
    $\frac{(\sqrt{2}-1)}{(\sqrt{2}+1)}$
Answer
(c) $\frac{(\sqrt{3}-1)}{(\sqrt{3}+1)}$
Explanation: $\tan 15^{\circ}=\tan \left(45^{\circ}-30^{\circ}\right)=\frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ} \tan 30^{\circ}}=\frac{\left(1-\frac{1}{\sqrt{3}}\right)}{\left(1+\frac{1}{\sqrt{3}}\right)}=\frac{(\sqrt{3}-1)}{(\sqrt{3}+1)}$
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MCQ 101 Mark
$\lim _{x \rightarrow \infty} \frac{\sin x}{x}=$
  • A
    $2$
  • B
    $1$
  • C
    $\infty$
  • ✓
    $0$
Answer
Correct option: D.
$0$
Explanation: $\lim _{x \rightarrow \infty} \frac{\sin x}{x}$
$\text { Let } x=\frac{1}{y}$
$x \rightarrow \infty$
$\therefore y \rightarrow 0$
$=\underset{{y \rightarrow 0}}{\lim} \frac{\sin \frac{1}{y}}{\frac{1}{y}}$
$=\underset{{y \rightarrow 0}}{\lim} y \sin \frac{1}{y}$
$=\underset{{y \rightarrow 0}}{\lim} y \times \underset{{y \rightarrow 0}}{\lim} \sin \frac{1}{y}$
$=0 \times \underset{{y \rightarrow 0}}{\lim} \sin \frac{1}{y}$
$=0$
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MCQ 121 Mark
If $z=\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^5+\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^5$, then
  • A
    Re(z)=0
  • B
    $\left.\operatorname{Re}( z )>0, I _{ m } z \right)>$
  • C
    $\operatorname{Re}( z )>0, I _{ rs }( z )<0$
  • D
    $I _{ m }( z )=0$
Answer
(d) $I _{ m }( z )=0$
Explanation: From given,
$z=2^5 C _0{\frac{\sqrt{3}^2}{2}}^5+{ }^5 C _2 \frac{\sqrt{3}}{2}^3 \frac{i^2}{2}+{ }^5 C _4 \frac{\sqrt{3}}{2} \frac{i}{2}$
Since $t^2=-1$ and $t^4=1$, will not contain any $i$ and hence $I_m(z)=0$
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MCQ 131 Mark
A plane is parallel to yz-plane so it is perpendicular to:
  • A
    y-axis
  • B
    none of these
  • C
    z-axis
  • D
    x-axis
Answer
(d) x-axis
Explanation: Any plane parallel to yz-plane is perpendicular to x-axis.  
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MCQ 141 Mark
The two lines ax + by = c and a′x + b′y = c′ are perpendicular if
  • A
    ab′ = ba′
  • B
    aa′ + bb′ = 0
  • C
    ab + a′b′ = 0
  • D
    ab′ + ba′ = 0
Answer
(b) $aa ^{\prime}+ bb  ^{\prime}=0$
Explanation: We know that Slope of the line $a x+b y=c$ is $\frac{-a}{b}$, and the slope of the line $a^{\prime} x+b^{\prime} y=c^{\prime}$ is $\frac{-a^{\prime}}{b^{\prime}}$ The lines are perpendicular if $\tan \theta=\frac{3}{5-x}$ (1)
$\frac{-a}{b} \frac{-a^{\prime}}{b}=-1 \text { or } a a^{\prime}+b b^{\prime}=0$
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MCQ 151 Mark
If $y =\sqrt{x+\sqrt{x+\sqrt{x+\ldots+\text { to }} \infty \infty}}$ then $\frac{d y}{d x}=$
  • A
    $\frac{1}{2 y+1}$
  • ✓
    $\frac{1}{2 y-1}$
  • C
    $\frac{x}{y+1}$
  • D
    $\sqrt{\frac{x}{y+1}}$
Answer
Correct option: B.
$\frac{1}{2 y-1}$
Explanation: $y =\sqrt{(x+y)}$
$y^2=x+y$
$2 yy^{\prime}=1+y^{\prime}$
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MCQ 161 Mark
A batsman scores runs in 10 innings as 38, 70, 48, 34, 42, 55, 63, 46, 54 and 44. The mean deviation about mean is
  • A
    7.6
  • B
    6.4
  • C
    8.6
  • D
    10.6
Answer
(c) 8.6
Explanation: First we arrange score in the ascending order
Then scores are : $34,38,42,44,46,48,54,55,63,70$
As there are 10 items in this data,
So, median will be the mean of fifth and sixth term
$\therefore \text { Median }=\frac{46+48}{2}=47$
Now, deviation from median for each value
$d_i=13,9,5,3,1,1,7,8,16,23$
$\therefore$ Required Mean deviation $=\frac{\sum d_i}{10}=\frac{86}{10}=8.6$
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MCQ 171 Mark
Let $f(x)=x^3$. Then, dom (f) and range ( $f$ ) are respectively
  • A
    R and $R ^{+}$
  • B
    $R ^{+}$and R
  • C
    $R ^{+}$and $R ^{+}$
  • D
    R and R
Answer
(d) R and R
Explanation: $f ( x )= x ^3$
$f(x)$ can assume any value, so domain of $f(x)$ is $R$
The Range of the function can be positive or negative Real numbers, as the cube of any number depends on the sign of the number, So Range of $f(x)$ is $R$
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MCQ 181 Mark
$\frac{\sin 3 x-\sin x}{\cos x-\cos 3 x}$ is equal to
  • A
    $\cot 2 x$
  • B
    $-\cot 2 x$
  • C
    $-\tan 2 x$
  • D
    $\tan 2 x$
Answer
(a) $\cot 2 x$
Explanation: Using $\sin C -\sin D =2 \cos \frac{(C+D)}{2} \sin \frac{(C-D)}{2}$ 
and $\cos C -\cos D =-2 \sin \frac{(C+D)}{2} \sin \frac{(C-D)}{2}$, we get 
$\frac{\sin 3 x-\sin x}{\cos x-\cos 3 x}=\frac{2 \cos \left(\frac{t}{2}\right) \sin \left(\frac{2 z}{2}\right)}{2 \sin \left(\frac{4 z}{2}\right) \sin \left(\frac{2 z}{2}\right)}=\frac{\cos 2 x \sin x}{\sin 2 x \sin x}=\cot 2 x$.
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M.C.Q (1 Marks) - Maths STD 11 Science Questions - Vidyadip