M.C.Q (1 Marks) - Page 4 - MATHS STD 12 Science Questions - Vidyadip

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M.C.Q (1 Marks)

MCQ 1511 Mark

If $\text{A}=\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix},$ then $A^5 =$

A
$5A$
B
$10A$
✓
$16A$
D
$32A$

Answer

Correct option: C.

$16A$

$\text{A}=\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$
$\Rightarrow\text{A}=2\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$\Rightarrow\text{A}=2\text{I}$
$\Rightarrow\text{A}^5=(2\text{I})^5$
$\Rightarrow\text{A}^5=16\times2\text{I}$
$\Rightarrow\text{A}^5=16\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$
$\Rightarrow\text{A}^5=16\text{A}$

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Question 1521 Mark

Find the minor of the element 1 in the determinant $\triangle=\begin{bmatrix}1&5\\3&8\end{bmatrix}$ is:

5
1
8
3

Answer

8

Solution:
The minor of the element 1 can be obtained by deleting the first row and the first column
$\therefore\text{ M}_{11}=8$

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Question 1531 Mark

Which of the following is not correct?

$|\text{A}|=|\text{A}^{\text{T}}|,$ where $\text{A}=[\text{a}_{\text{ij}}]_{3\times3}$
$|\text{kA}|=|\text{k}^3|,$ where $\text{A}=[\text{a}_{\text{ij}}]_{3\times3}$
If a is a skew-symmetric of odd order, then |A| = 0
$\begin{vmatrix}\text{a}&\text{c}\\\text{e}&\text{g} \end{vmatrix}+\begin{vmatrix}\text{b}&\text{c}\\\text{f}&\text{g} \end{vmatrix}+\begin{vmatrix}\text{a}&\text{d}\\\text{e}&\text{h}\end{vmatrix}+\begin{vmatrix}\text{b}&\text{d}\\\text{f}&\text{h}\end{vmatrix}$

Answer

$\begin{vmatrix}\text{a}&\text{c}\\\text{e}&\text{g} \end{vmatrix}+\begin{vmatrix}\text{b}&\text{c}\\\text{f}&\text{g} \end{vmatrix}+\begin{vmatrix}\text{a}&\text{d}\\\text{e}&\text{h}\end{vmatrix}+\begin{vmatrix}\text{b}&\text{d}\\\text{f}&\text{h}\end{vmatrix}$

Solution:
$\begin{vmatrix}\text{a}+\text{b}&\text{c}+\text{d}\\\text{e}+\text{f}&\text{g}+\text{h} \end{vmatrix}=\begin{vmatrix}\text{a}+\text{b}&\text{c}\\\text{e}+\text{f}&\text{h} \end{vmatrix}+\begin{vmatrix}\text{a}+\text{b}&\text{d}\\\text{e}+\text{f}&\text{h}\end{vmatrix}$
$=\begin{vmatrix}\text{a}&\text{c}\\\text{e}&\text{g} \end{vmatrix}+\begin{vmatrix}\text{b}&\text{c}\\\text{f}&\text{g} \end{vmatrix}+\begin{vmatrix}\text{a}&\text{d}\\\text{e}&\text{h}\end{vmatrix}+\begin{vmatrix}\text{b}&\text{d}\\\text{f}&\text{h}\end{vmatrix}$

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Question 1541 Mark

Let $\text{P}$ and $\text{Q}$ be $3\times3$ matrices with $\text{P}\neq\text{Q}.$ If $\text{P}^3=\text{Q}^3$ and $\text{P}^2\text{Q}=\text{Q}^2\text{P}$ then determinant of $(\text{P}^2+\text{Q}^2)$ is equal to:

-2
1
0
-1

Answer

0

Solution:
$\text{P}^3=\text{Q}^3$

$\Rightarrow\text{P}^3- \text{P}^2\text{Q}=\text{Q}^3- \text{Q}^2\text{P}$

$\Rightarrow\text{P}^2(\text{P- Q})=\text{Q}^2(\text{Q- P})$

$\Rightarrow \text{P}^2(\text{P- Q})-\text{Q}^2(\text{Q- P})=0$

$\Rightarrow (\text{P}^2+\text{Q}^2)(\text{P}-\text{Q})=0\Rightarrow|\text{P}^2+\text{Q}^2|=0$

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Question 1551 Mark

Find the determinant of the matrix $\text{A}=\begin{bmatrix}-\cos\theta&-tan\theta\\\cot\theta&\cos\theta\end{bmatrix}$ is:

$\sin^2\theta$
$\sin\theta$
$-\sin\theta$
$-\sin^2\theta$

Answer

$\sin^2\theta$

Solution:
Given that, $\text{A}=\begin{bmatrix}-\cos\theta&-tan\theta\\\cot\theta&\cos\theta\end{bmatrix}$
$|\text{A}|=\begin{bmatrix}-\cos\theta&-tan\theta\\\cot\theta&\cos\theta\end{bmatrix}$
$|\text{A}|=-\cos⁡\theta (\cos⁡\theta )-\cot\theta(-\tan⁡\theta) $
$|\text{A}|=-\cos^2\theta⁡+1=\sin^2⁡\theta.$

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Question 1561 Mark

Evaluate $\begin{bmatrix}\sqrt{3}&\sqrt{2}\\-1&2\sqrt{3}\end{bmatrix}$ is:

$6-3\sqrt{2}$
$6-\sqrt{2}$
$6+3\sqrt{2}$
$6+\sqrt{2}$

Answer

$6+\sqrt{2}$

Solution:
$\triangle=\begin{bmatrix}\sqrt{3}&\sqrt{2}\\-1&2\sqrt{3}\end{bmatrix}$
$\triangle=(\sqrt{3}\times2\sqrt{3})+\sqrt{2}$
$\triangle=6+\sqrt{2}$

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Question 1571 Mark

$\begin{vmatrix}\log_3512&\log_43\\\log_38&\log_49\end{vmatrix}\times\begin{vmatrix}\log_23&\log_83\\\log_34&\log_34\end{vmatrix}$

7
10
1
17

Answer

10

Solution:
$\begin{vmatrix}\log_3512&\log_43\\\log_38&\log_49\end{vmatrix}\times\begin{vmatrix}\log_23&\log_83\\\log_34&\log_34\end{vmatrix}$
$=\begin{vmatrix}\log_32^9&\log_{2^{2}}3\\\log_32^3&\log_{2^2}3^3\end{vmatrix}\times\begin{vmatrix}\log_23&\log_{2^{3}}3\\\log_32^3&\log_32^2\end{vmatrix}$
$=\begin{vmatrix}9\log_32&\frac{1}{2}\log_23\\3\log_32&\frac{1}{2}\times2\log_23\end{vmatrix}\times\begin{vmatrix}\log_23&\frac{1}{3}\log_23\\2\log_32&2\log_32\end{vmatrix}$
$=\Big(\big(9\log_32\times\log_23\big)-\big(3\log_32\times\frac{1}{2}\log_23\big)\Big)\times\Big(\big(\log_23\times2\log_32\big)\\-\Big(\frac{1}{3}\log_23\times2\log_32\Big)\Big)$
$=\Big(9-\frac{3}{2}\Big)\times\Big(2-\frac{2}{3}\Big)$
$=\frac{15}{2}\times\frac{4}{3}$
$=10$

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Question 1581 Mark

If $\text{D}_\text{k}=\begin{vmatrix}1&\text{n}&\text{n}\\2\text{k}&\text{n}^2+\text{n}+2&\text{n}^2+\text{n}\\2\text{k}-1&\text{n}^2&\text{n}^2+\text{n}+2\end{vmatrix} $ and $\sum\limits_{\text{k}=1}^\text{n}\text{D}_\text{k}=48,$ then n equals:

4
6
8
None of these.

Answer

4

Solution:
$\text{D}_\text{k}=\begin{vmatrix}1&\text{n}&\text{n}\\2\text{k}&\text{n}^2+\text{n}+2&\text{n}^2+\text{n}\\2\text{k}-1&\text{n}^2&\text{n}^2+\text{n}+2\end{vmatrix} $
$\text{D}_\text{k}=\begin{vmatrix}1&\text{n}&\text{n}\\1&\text{n}+2&-2\\2\text{k}-1&\text{n}^2&\text{n}^2+\text{n}+2\end{vmatrix}\ \text{Applying}\ \text{R}_2\rightarrow\text{R}_2-\text{R}_3$
$\text{D}_\text{k}=\begin{vmatrix}1&\text{n}&\text{n}\\0&2&-2-\text{n}\\2\text{k}-1&\text{n}^2&\text{n}^2+\text{n}+2\end{vmatrix}\ \text{Applying}\ \text{R}_2\rightarrow\text{R}_2-\text{R}_1$
Now,
$\sum\limits_{\text{k}=1}^\text{n}\text{D}_\text{k}=\begin{vmatrix}1&\text{n}&\text{n}\\0&2&-2-\text{n}\\1&\text{n}^2&\text{n}^2+\text{n}+2\end{vmatrix}+\begin{vmatrix}1&\text{n}&\text{n}\\0&2&-2-\text{n}\\3&\text{n}^2&\text{n}^2+\text{n}+2\end{vmatrix}+.......+\begin{vmatrix}1&\text{n}&\text{n}\\0&2&-2-\text{n}\\2\text{n}-1&\text{n}^2&\text{n}^2+\text{n}+2\end{vmatrix}$
$\sum\limits_{\text{k}=1}^\text{n}\text{D}_\text{k}=\Big[1\Big(2(\text{n}^2+\text{n}+2)+\text{n}^2(2+\text{n})\Big)+\text{n}\Big(0+1(-2-\text{n})\Big)+\text{n}\Big(0+2\Big)\Big]\\+\Big[1\Big(2(\text{n}^2+\text{n}+2)+\text{n}^2(2+\text{n})\Big)+\text{n}\Big(0+3(-2-\text{n})\Big)+\text{n}\Big(0+6\Big)\Big]+......\\+\Big[1\Big(2(\text{n}^2+\text{n}+2)+\text{n}^2(2+\text{n})\Big)+\text{n}\Big(0+(2\text{n}-1)(-2-\text{n})\Big)+\text{n}\Big(0+2(2\text{n}-1)\Big)\Big]$
$\sum\limits_{\text{k}=1}^\text{n}\text{D}_\text{k}=\Big[1\Big(2(\text{n}^2+\text{n}+2)+\text{n}^2(2+\text{n})\Big)+\text{n}\Big(-2-\text{n}\Big)-2\text{n}\Big]\Big(1+3+5+.....+\text{n}\Big)$
$\sum\limits_{\text{k}=1}^\text{n}\text{D}_\text{k}=\Big[1\Big(2(\text{n}^2+\text{n}+2)+\text{n}^2(2+\text{n})\Big)+\text{n}\Big(-2-\text{n}\Big)-2\text{n}\Big]\Big(\text{n}^2\Big)$
$\Rightarrow\ 2\text{n}^2+4\text{n}=48$
$\Rightarrow\ (\text{n}+6)(\text{n}-4)=0$
$\Rightarrow\ \text{n}=4$

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Question 1591 Mark

Evaluate $\begin{bmatrix}\sqrt{3}&\sqrt{2}\\-1&2\sqrt{3}\end{bmatrix}$ is:

$6-3\sqrt{2}$
$6-\sqrt{2}$
$6+3\sqrt{2}$
$6+\sqrt{2}$

Answer

$6+\sqrt{2}$

Solution:
$\triangle=\begin{bmatrix}\sqrt{3}&\sqrt{2}\\-1&2\sqrt{3}\end{bmatrix}$
$\triangle=(\sqrt{3}\times2\sqrt{3})+\sqrt{2}$
$\triangle=6+\sqrt{2}$

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MCQ 1601 Mark

The value of the determinant $\begin{vmatrix}\text{a}-\text{b}&\text{b}+\text{c}&\text{c}\\\text{b}-\text{c}&\text{c}+\text{b}&\text{b}\\\text{c}+\text{a}&\text{a}+\text{b}&\text{c}\end{vmatrix}$ is:

A
$a^3 + b^3 + c^3$
B
$3bc$
✓
$a^3 + b^3 + c^3 - 3abc$
D
None of these

Answer

Correct option: C.

$a^3 + b^3 + c^3 - 3abc$

$\begin{vmatrix}\text{a}-\text{b}&\text{b}+\text{c}&\text{a}\\\text{b}-\text{c}&\text{c}+\text{b}&\text{b}\\\text{c}-\text{a}&\text{a}+\text{b}&\text{c}\end{vmatrix}$
$=\begin{vmatrix}-\text{b}&\text{b}+\text{c}+\text{a}&\text{a}\\-\text{c}&\text{c}+\text{a}+\text{b}&\text{b}\\-\text{a}&\text{a}+\text{b}+\text{c}&\text{c}\end{vmatrix} [$Applying $C_1 \rightarrow C_1 - C_3$ and $C_2 \rightarrow C_2 + C_3]$
$=(-1)(\text{a}+\text{b}+\text{c})\begin{vmatrix}\text{b}&1&\text{a}\\\text{c}&1&\text{b}\\\text{a}&1&\text{c}\end{vmatrix} [$Taking $(-1)$ common from $C_1$ and $(a + b + c)$ common from$ C_2]$
$=(-1)(\text{a}+\text{b}+\text{c})\begin{vmatrix}\text{b}&1&\text{a}\\\text{c}-\text{b}&0&\text{b}-\text{a}\\\text{a}-\text{b}&0&\text{c}-\text{a}\end{vmatrix} [$Applying $R_2 \rightarrow R_2 - R_1$ and $R_{3 }\rightarrow R_3 - R_1]$
$= (-1)(a + b + c)[-(c - b)(c - a) + (b - a)(a - b)]$
$= (-1)(a + b + c)[-c^2 + ac + bc - ab + ba - b^2 - a^2 + ab]$
$= (-1)(a + b + c)(-a^2 - b^2 - c^2 + ab + bc + ac)$
$= (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ac)$
$= a^3 + ab^2 + ac^2 - a^2b - abc - a^2c + ba^2 + b^3 + bc^2 - ab^2-b^2c - abc + ca^2 + cb^2 + c^3 - acb - bc^2 - ac^2$
$= a^3 + b^3 + c^3- 3abc$
Hence, the correct option is $(c)$

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Question 1611 Mark

Evaluate $ |\text{A}|^2-5|\text{A}|+1,$ if $\text{A}=\begin{bmatrix}7&4\\5&5\end{bmatrix}$ is:

161
251
150
151

Answer

151

Solution:
Given that, $\text{A}=\begin{bmatrix}7&4\\5&5\end{bmatrix}$
$|\text{A}|=(7(5)-5(4))=35-20=15$
$|\text{A}|^2-5|\text{A}|+1=(15)^2-5(15)+1=225-75+1=151.$

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Question 1621 Mark

If $\left|\begin{array}{lll}<\text{br}> &1 &\text{amp; } 0 &\text{amp; } 0\\ <\text{br}>&2 &\text{amp; } 3 &\text{amp; } 4\\ <\text{br}>&5 &\text{amp; } -6 &\text{amp; x}<\text{br}> \end{array}\right|=45$ then $\text{x}=$

4
7
-5
-7

Answer

7

Solution:
Given, $\left|\begin{array}{lll}<\text{br}> &1 &\text{amp; } 0 &\text{amp; } 0\\ <\text{br}>&2 &\text{amp; } 3 &\text{amp; } 4\\ <\text{br}>&5 &\text{amp; } -6 &\text{amp; x}<\text{br}> \end{array}\right|=45$
By operation of matrix (5),
$1(3\text{x}+24)=45$
$3\text{x}=21$
$\Rightarrow \text{x}=7$

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MCQ 1631 Mark

Choose the correct answer from given four options in each of the Exercise:
If $x, y, z$ are all different from zero and $\begin{vmatrix}1+\text{x}&1&1\\1&1+\text{y}&1\\1&1&1+\text{z}\end{vmatrix}=0,$ then the value of $x^{-1} + y^{-1} + z^{-1}$ is:

A
$xyz$
B
$x^{-1} + y^{-1} + z^{-1}$
C
$-x - y - z$
✓
$-1$

Answer

Correct option: D.

$-1$

We have, $\begin{vmatrix}1+\text{x}&1&1\\1&1+\text{y}&1\\1&1&1+\text{z}\end{vmatrix}=0$
$\text{Applying C}_1\rightarrow\text{C}_1-\text{C}_3\text{ and C}_2\rightarrow\text{C}_2-\text{C}_3,$
$\Rightarrow\ \begin{vmatrix}\text{x}&0&1\\0&\text{y}&1\\-\text{z}&-\text{z}&1+\text{z}\end{vmatrix}=0$
Expanding along $R_1,$
$\text{x}\big[\text{y}(1+\text{z})+\text{z}\big]-0+1(\text{yz})=0$
$\Rightarrow\text{x}(\text{y}+\text{yz}+\text{z})+\text{zy}=0$
$\Rightarrow\text{xy}+\text{xyz}+\text{xz}+\text{xz}=0$
$\Rightarrow\frac{\text{xy}}{\text{xyz}}+\frac{\text{xyz}}{\text{xyz}}+\frac{\text{xz}}{\text{xyz}} +\frac{\text{yz}}{\text{xyz}} =0 [$On dividing by $(xyz)$ from both sides$]$
$\Rightarrow\ \frac{1}{\text{x}}+\frac{1}{\text{y}}+\frac{1}{\text{z}}+1=0$
$\Rightarrow\ \frac{1}{\text{x}}+\frac{1}{\text{y}}+\frac{1}{\text{z}}=-1$
$\therefore\text{x}^{-1}+\text{y}^{-1}+\text{z}^{-1}=-1$

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Question 1641 Mark

A set of points which do not lie on the same line are called as

collinear
non-collinear
concurrent
square

Answer

non-collinear

Solution:
A set of points which do not lie on the same line are called as non collinear points

solution

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Question 1651 Mark

$ \begin{bmatrix}1 & \text{x} & \text{x}^2 \\1 & \text{y} & \text{y}^2 \\1 & \text{z} & \text{z}^2\end{bmatrix}$

(x - y) (y + z)(z + x)
(x + y) (y - z)(z - x)
(x - y) (y - z)(z + x)
(x - y) (y - z) (z - x)

Answer

(x - y) (y - z) (z - x)

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Question 1661 Mark

If A is a 3 × 3 matrix and det(3A) = k(detA), then k =

9
6
1
27

Answer

27

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Question 1671 Mark

The equations in terms of x and y are:

x – y = 50, 2x – y = 550
x – y = 50, 2x + y = 550
x + y = 50, 2x + y = 550
x + y = 50, 2x – y = 550

Answer

x – y = 50, 2x + y = 550

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Question 1681 Mark

How much is the area of rectangular field:

60000 sq.m
30000 sq.m
30000m
3000m

Answer

30000 sq.m

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MCQ 1691 Mark

The maximum value of $\triangle=\begin{vmatrix}1&1&1\\1&1+\sin\theta&1\\1+\cos\theta&1&1\end{vmatrix}$ is $(\theta$ is real$):$

✓
$\frac{1}{2}$
B
$\frac{\sqrt{3}}{2}$
C
$\sqrt{2}$
D
$-\frac{\sqrt{3}}{2}$

Answer

Correct option: A.

$\frac{1}{2}$

$\triangle=\begin{vmatrix}1&1&1\\1&1+\sin\theta&1\\1+\cos\theta&1&1\end{vmatrix}$
$=\begin{vmatrix}1&1&1\\0&\sin\theta&0\\\cos\theta&0&0\end{vmatrix} [$Applying $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1]$
$=-\sin\theta\cos\theta$
$=-\frac{\sin2\theta}{2}$
Now, maximum and minimum value of $\sin\theta$ is $1$ and $-1.$
So, the maximum value of $-\sin\theta$ is $1.$
So, the maximum value of $-\sin2\theta$ is $1.$
Therefore, the maximum value of $-\frac{\sin2\theta}{2}$ is $\frac{1}{2}$
Hence, the correct option is $(a)$

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MCQ 1701 Mark

The value of the determinant $\begin{vmatrix}\text{x}&\text{x}+\text{y}&\text{x}+2\text{y}\\\text{x}+2\text{y}&\text{x}&\text{x}+\text{y}\\\text{x}+\text{y}&\text{x}+2\text{y}&\text{x}\end{vmatrix}$ is:

A
$9x^2(x + y)$
✓
$9y^2(x + y)$
C
$3y^2(x + y)$
D
$7x^2(x + y)$

Answer

Correct option: B.

$9y^2(x + y)$

$\begin{vmatrix}\text{x}&\text{x}+\text{y}&\text{x}+2\text{y}\\\text{x}+2\text{y}&\text{x}&\text{x}+\text{y}\\\text{x}+\text{y}&\text{x}+2\text{y}&\text{x}\end{vmatrix}$
$=\begin{vmatrix}-2\text{y}&\text{y}&\text{y}\\\text{x}+2\text{y}&\text{x}&\text{x}+\text{y}\\-\text{y}&2\text{y}&-\text{y}\end{vmatrix} [$Applying $R_1 \rightarrow R_1 - R_2$ and $R_{3 }\rightarrow R_3 - R_2]$
$=\text{y}^2\begin{vmatrix}-2&1&1\\\text{x}+2\text{y}&\text{x}&\text{x}+\text{y}\\-1&2&-1\end{vmatrix} [$Taking $(y)$ common from $R_1$ and from $R_3]$
$=\text{y}^2\begin{vmatrix}-2&-3&3\\\text{x}+2\text{y}&3\text{x}+4\text{y}&-\text{y}\\-1&0&0\end{vmatrix} [$Applying $C_2 \rightarrow C_2 + 2C_1$ and $C_3 \rightarrow C_3 - C_1]$
$=\text{y}^2\big[-1(3\text{y}-9\text{x}-12\text{y})\big]$
$=\text{y}^2[9\text{y}+9\text{x}]$
$=9\text{y}^2(\text{y}+\text{x})$
Hence, the correct option is $(b)$

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M.C.Q (1 Marks) - Page 4 - MATHS STD 12 Science Questions - Vidyadip