Mathematics Test

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Mathematics Test - 6

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Weekly Quiz Competition

Question 1
3 / -1

If A=[aij]_3×3 is a square matrix so that
aij=i²⁻j², then A is a

A
unit matrix

B
symmetric matrix

C
skew symmetric matrix

D
orthogonal matrix

Solution

Given \(A=\left[a_{i j}\right]_{(3 \times 3)}\)
where, \(a_{i j}=i^{2}-j^{2}\)
\(\therefore a_{i j}=0\) if \(i=j\)
Now, \(a_{12}=1^{2}-2^{2}=-3\)
\(a_{13}=1^{2}-3^{2}=-8\)
\(a_{21}=2^{2}-1^{2}=3\)
\(a_{23}=2^{2}-3^{2}=-5\)
\(a_{31}=3^{2}-1^{2}=8\)
\(a_{32}=3^{2}-2^{2}=5\)
\(\therefore A=\left[\begin{array}{ccc}0 & -3 & -8 \\ 3 & 0 & -5 \\ 8 & 5 & 0\end{array}\right]\)
Here, \(A^{T}=-A\)
\(\therefore A\) is a skew-symmetric matrix.
Hence option C is correct.
Question 2
3 / -1

Directions: The following question has four choices, out of which ONE or MORE can be correct.

For \(0<\theta<\frac{\pi}{2},\) the solution \((s)\) of \(\sum_{m=1}^{6} \operatorname{cosec}\left(\theta+\frac{(m-1) \pi}{4}\right) \operatorname{cosec}\left(\theta+\frac{m \pi}{4}\right)=4 \sqrt{2}\) is / are

A
π/4

B
π/6

C
π/12

D
5π/6

Solution
Question 3
3 / -1

If \(A=\left[\begin{array}{cc}2 & 2 \\ -3 & 2\end{array}\right], B=\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]\) then \(\left(B^{-1} A^{-1}\right)^{-1}=\)

A

\(\left[\begin{array}{cc}2 & -2 \\ 2 & 3\end{array}\right]\)

B

\(\left[\begin{array}{cc}2 & 2 \\ -2 & 3\end{array}\right]\)

C

\(\left[\begin{array}{cc}2 & -3 \\ 2 & 2\end{array}\right]\)

D

\(\left[\begin{array}{cc}1 & -1 \\ -2 & 3\end{array}\right]\)

Solution

\(A=\left[\begin{array}{cc}2 & 2 \\ -3 & 2\end{array}\right] \quad B=\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]\)
\(A^{-1}=\frac{1}{10}\left[\begin{array}{cc}2 & -2 \\ 3 & 2\end{array}\right] \quad B^{-1}=\left[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right]\)
Now. \(B^{-1} A^{-1}=\left[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right] \cdot \frac{1}{10}\left[\begin{array}{cc}2 & -2 \\ 3 & 2\end{array}\right]\)
\(=\frac{1}{10}\left[\begin{array}{cc}0+3 & 0+2 \\ -2+0 & 2+0\end{array}\right]\)
\(=\frac{1}{10}\left[\begin{array}{rr}3 & 2 \\ -2 & 2\end{array}\right]=\left[\begin{array}{rr}3 / 10 & 2 / 10 \\ -2 / 10 & 2 / 10\end{array}\right]\)
Now. \(\left(B^{-1} A^{-1}\right)^{-1}=\left[\begin{array}{rr}2 & -2 \\ 2 & 3\end{array}\right]\)
Hence option A is correct.
Question 4
3 / -1

If \(0 \leq x \leq 2 \pi\) and \(|\cos x| \leq \sin x,\) then

A

The set values of \(x\) is \(\left[\frac{\pi}{4}, \frac{2 \pi}{4}\right]\)

B

then number of solutions thta are integral multiple of \(\frac{\pi}{4}\) is four.

C
the sum of the largest and the smallest solution is π

D

the set of all values of \(x\) is \(x \in\left[\frac{\pi}{4}, \frac{\pi}{2}\right) \cup\left(\frac{\pi}{2}, 3 \pi 4\right]\)

Solution
Question 5
3 / -1

Directions: The following question has four choices, out of which one or more is/are correct.

For any two events A and B in a sample space

A

\(P\left(\frac{A}{B}\right) \geq \frac{P(A)+P(B)-1}{P(B)}, P(B) \neq 0\) is alway

B

\(P(A \cap B)=P(A)-P(A \cap B)\) does not hold

C

\(P(A \cup B)=1-P(\bar{A}) P(B),\) if \(A\) and \(B\) are disjointed

D
None of these

Solution
Question 6
3 / -1

\(A=\left|\begin{array}{cc}1 & 1 \\ 2 & 2 \\ 3 & -1\end{array}\right|\) and \(B=\left|\begin{array}{cc}2 & 1 \\ 1 & 2 \\ -1 & 0\end{array}\right|,\) then find matrix \(C\) if \(A+2 B+C=0\)

A

\(\left|\begin{array}{cc}-5 & -3 \\ -4 & -6 \\ -1 & 1\end{array}\right|\)

B

\(\left|\begin{array}{cc}-5 & -9 \\ -4 & -6 \\ -1 & 1\end{array}\right|\)

C

\(\left|\begin{array}{cc}-5 & -3 \\ -4 & -6 \\ -1 & 9\end{array}\right|\)

D

\(\left|\begin{array}{cc}-5 & -7 \\ -4 & -6 \\ -1 & 1\end{array}\right|\)

Solution

\(2 B=2 \begin{array}{cc}2 & 1 \\ 1 & 2 \\ -1 & 0\end{array}|=| \begin{array}{cc}4 & 2 \\ 2 & 4 \\ -2 & 0\end{array} \mid\)
\(A+2 B=\left|\begin{array}{cc}1 & 1 \\ 2 & 2 \\ 3 & -1\end{array}\right|+\left|\begin{array}{cc}4 & 2 \\ 2 & 4 \\ -2 & 0\end{array}\right|=\left|\begin{array}{cc}1+4 & 2+1 \\ 2+2 & 4+2 \\ 3+(-2) & (-1)+0\end{array}\right|=\left|\begin{array}{cc}5 & 3 \\ 4 & 6 \\ 1 & -1\end{array}\right|\)
\(C=-(A+2 B)=\left|\begin{array}{cc}5 & 3 \\ 4 & 6 \\ 1 & -1\end{array}\right| \Rightarrow\left|\begin{array}{cc}-5 & -3 \\ -4 & -6 \\ -1 & 1\end{array}\right|\)
\(\therefore A+2 B+C=0\)
Hence option A is correct.
Question 7
3 / -1

If \(A=[x, y], B=\left[\begin{array}{ll}a & h \\ h & b\end{array}\right], C=\left[\begin{array}{l}x \\ y\end{array}\right]\)
then \(A B C=\)

A
(ax+hy+bxy)

B
(ax²+2hxy+by²)

C
(ax²−2hxy+by²)

D
(bx²−2hxy+ay²)

Solution
Question 8
3 / -1

Directions: The following question has four choices, out of which ONE or MORE can be correct.

Let z₁ and z₂ be complex numbers such that z₁ ≠ z₂ and | z₁| = | z₂ |, If z₁ has positive real part and z₂ has negative imaginary part, then (z₁+z₂)/(z₁-z₂) may be

A
zero

B
real and positive

C
real and negative

D
None of these

Solution

Given: \(\left|z_{1}\right|=\left|z_{2}\right|\)
\(N o w, \frac{z_{1}+z_{2}}{z_{1}-z_{2}} \times \frac{z_{1}}{z_{1}}-z_{2}\)
\(=\frac{z_{1} z_{1}-z_{1} z_{2}+z_{2} z_{1}-z_{2} z_{2}}{\left|z_{1}-z_{2}\right|^{2}}\)
\(=\frac{\left|z_{1}\right|^{2}+\left(z_{2} \bar{z}_{1}-z_{1} \bar{z}_{2}\right)-\left|z_{2}\right|^{2}}{\left|z_{1}-z_{2}\right|^{2}}\)
\(=\frac{z_{2} z_{1}-z_{1} z_{2}}{\left|z_{1}-z_{2}\right|^{2}}\left(\because\left|z_{1}\right|^{2}=\left|z_{2}\right|^{2}\right)\)
As, we know \(z-\bar{z}=2 i \operatorname{lm}(z)\)
\(\therefore z_{2} z_{1}-z_{1} z_{2}=2 i \operatorname{lm}\left(z_{2} z_{1}\right)\)
\(\frac{z_{1}+z_{2}}{z_{1}-z_{2}}=\frac{2 i \operatorname{lm}\left(z_{2} z_{1}\right)}{\left|z_{1}-z_{2}\right|^{2}} ;\) zero.

Hence option A is correct.
Question 9
3 / -1

Using elementary transformations, find the inverse of matrix, \(A=\left[\begin{array}{ll}1 & 3 \\ 2 & 7\end{array}\right]\)

A

\(\left[\begin{array}{cc}7 & -3 \\ -2 & 2\end{array}\right]\)

B

\(\left[\begin{array}{cc}8 & -3 \\ -2 & 1\end{array}\right]\)

C

\(\left[\begin{array}{cc}7 & -3 \\ -7 & 1\end{array}\right]\)

D

\(\left[\begin{array}{cc}7 & -3 \\ -2 & 1\end{array}\right]\)

Solution
Question 10
3 / -1

Directions: The following question has four choices out of which ONE or MORE can be correct.

The probabilities that a student passes in Mathematics, Physics and Chemistry are m, p and c respectively. Of these subjects, the student has a 75% chance of passing in at least one, 50% chance of passing in at least two, and 40% chance of passing in exactly two. Which of the following relations is/are true?

A
p + m + c = (19/20)

B
p + m + c = (27/20)

C
pmc = (1/4)

D
None of these

Solution

Let A, B and C, respectively, denote the events that the student passes in Maths, Physics and Chemistry.

It is given:

P (A) = m, P(B) = p and P(C) = c

Probability of at least one success:

p + m + c - pm - pc - cm + pmc = 0.75 ------(i)

Probability of at least two successes:

mc + mp + pc - 2mcp = 0.5 ------------(ii)

Probability of exactly two successes:

mc + mp + pc - 3mcp = 0.4 ------------(iii)

From eq. (ii) and (iii),

mcp = 0.5 - 0.4 = 0.1 = 1/10

mc + mp + pc = 0.5 + 0.2 = 0.7 = 7/10

From (i),

p + m + c = 27/20

Hence option B is correct.
Question 11
3 / -1

Directions: The following question has four choices, out of which ONE or MORE can be correct.

If \(I_{n}=\int_{-\pi}^{\pi} \frac{\sin n x}{\left(1+\pi^{x}\right) \sin x} d x, n=0,1,2, \ldots,\) then

A
I_n = I_n _{+ 2}

B

\(\sum_{m=1}^{10} I_{2 m+1}=10 \pi\)

C

\(\sum_{m=1}^{10} I_{2 m}=0\)

D
All of the above

Solution

Hence option D is correct.
Question 12
3 / -1

If A, B are two idempotent matrices and

AB = BA = 0, then A + B is

A
Scalar matrix

B
Idempotent matrix

C
Diagonal matrix

D
Nilptent matrix

Solution

Given A,B are Idempotent matrices

A²=A, B²=B

(A+B)²=A²+B²+AB+BA

(A+B)²=A+B+AB+BA

(A+B)²=A+B (since AB=BA=0)

Hence A+B is Idempotent

Hence option B is correct.
Question 13
3 / -1

If \(\left[\begin{array}{ll}x+3 & 2 y+x \\ z-1 & 4 a-z\end{array}\right]=\left[\begin{array}{ll}0 & -7 \\ 3 & 2 a\end{array}\right]\), then
\((x+y+z+a) i s\)

A
-1

B
0

C
1

D
8

Solution

If \(A=B\), then all the elements of \(A\) and \(B\) should be same. \(\therefore x+3=0\)
\(2 y+x=-7\)
\(z-1=3\)
\(z-4 a=-2 a\)
\(x=-3\)
\(y=-2\)
\(z=4\)
\(a=2\)
\(x+y+z+a=-3-2+4+2\)
\(=1\)
Hence option C is correct.
Question 14
3 / -1

Find the inverse of \(A=\left|\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right|\) if
\(A^{-1}=\left[\begin{array}{ccc}1 / 2 & -1 / 2 & 1 / 2 \\ a & 3 & b \\ c & -3 / 2 & 1 / 2\end{array}\right]\)
\(|a b c| ?\)

A
1/2

B
3

C
10

D
12

Solution

Given \(A=\left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right]\)
\(\therefore A \mid=0 .(2-3)-1(1-9)+2(1+6)\)
\(=0+8-10\)
\(=-2 \neq 0\)
If \(C\) be the matrix of cofactors of the elements in \(|A| \therefore \quad C=\left[\begin{array}{lll}C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33}\end{array}\right]\)
\(C_{11}=\left|\begin{array}{ll}2 & 3 \\ 1 & 1\end{array}\right|=-1 ; \quad C_{12}=-\left|\begin{array}{ll}1 & 3 \\ 3 & 1\end{array}\right|=8 ; \quad C_{13}=\left|\begin{array}{ll}1 & 2 \\ 3 & 1\end{array}\right|=-5\)
\(C_{21}=-\left|\begin{array}{ll}1 & 2 \\ 1 & 1\end{array}\right|=1 ; \quad C_{22}=\left|\begin{array}{ll}0 & 2 \\ 3 & 1\end{array}\right|=-6 ; \quad C_{23}=-\left|\begin{array}{ll}0 & 1 \\ 3 & 1\end{array}\right|=3\)
\(C_{31}=\left|\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right|=-1 ; \quad C_{32}=-\left|\begin{array}{ll}0 & 2 \\ 1 & 3\end{array}\right|=2 ; \quad C_{33}=\left|\begin{array}{ll}0 & 1 \\ 1 & 2\end{array}\right|=-1\)
\(\therefore C=\left[\begin{array}{ccc}-1 & 8 & -5 \\ 1 & -6 & 3 \\ -1 & 2 & -1\end{array}\right]\)
\(\therefore \quad \operatorname{Adj}(A)=C^{\prime}=\left[\begin{array}{ccc}-1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1\end{array}\right]\)
Hence, \(A^{-1}=\frac{A d j A}{|A|}=-\frac{1}{2}\left[\begin{array}{ccc}-1 & 1 & -1 \\ 8 & -6 & -2 \\ -5 & 3 & -1\end{array}\right]\)
\(=\left[\begin{array}{ccc}1 / 2 & -1 / 2 & 1 / 2 \\ -4 & 3 & -1 \\ 5 / 2 & -3 / 2 & 1 / 2\end{array}\right]\)
\(\Rightarrow|a b c|=10\)
Hence option C is correct.
Question 15
3 / -1

If \(A=\left[\begin{array}{ll}2 & 5 \\ 4 & 4\end{array}\right], \quad B=\left[\begin{array}{ll}0 & 5 \\ 1 & 6\end{array}\right],\) find \(5 A^{\prime}+3 B^{\prime}\)

A

\(\left[\begin{array}{ll}10 & 23 \\ 40 & 38\end{array}\right]\)

B

\(\left[\begin{array}{ll}10 & 63 \\ 40 & 38\end{array}\right]\)

C

\(\left[\begin{array}{ll}10 & 23 \\ 40 & 48\end{array}\right]\)

D

\(\left[\begin{array}{cc}90 & 23 \\ 40 & 38\end{array}\right]\)

Solution

\(A=\left[\begin{array}{ll}2 & 5 \\ 4 & 4\end{array}\right] \quad B=\left[\begin{array}{ll}0 & 5 \\ 1 & 6\end{array}\right]\)
\(A^{T}=\left[\begin{array}{ll}2 & 4 \\ 5 & 4\end{array}\right]\) (Transposing is inter changing the rows \(1 \&\) columns)
\(B^{T}=\left[\begin{array}{ll}0 & 1 \\ 5 & 6\end{array}\right]\)
\(\therefore \quad 5 A^{T}+3 B^{T}\)
\(=5\left[\begin{array}{ll}2 & 4 \\ 5 & 4\end{array}\right]+3\left[\begin{array}{ll}0 & 1 \\ 5 & 6\end{array}\right]\)
\(=\left[\begin{array}{ll}10 & 20 \\ 25 & 20\end{array}\right]+\left[\begin{array}{cc}0 & 3 \\ 15 & 18\end{array}\right]\)
\(=\left[\begin{array}{ll}10 & 23 \\ 40 & 38\end{array}\right]\)
Hence option A is correct.

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Mathematics Test - 6

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Question 1

unit matrix

symmetric matrix

skew symmetric matrix

orthogonal matrix

Solution

Question 2

π/4

π/6

π/12

5π/6

Solution

Question 3

\(\left[\begin{array}{cc}2 & -2 \\ 2 & 3\end{array}\right]\)

\(\left[\begin{array}{cc}2 & 2 \\ -2 & 3\end{array}\right]\)

\(\left[\begin{array}{cc}2 & -3 \\ 2 & 2\end{array}\right]\)

\(\left[\begin{array}{cc}1 & -1 \\ -2 & 3\end{array}\right]\)

Solution

Question 4

The set values of \(x\) is \(\left[\frac{\pi}{4}, \frac{2 \pi}{4}\right]\)

then number of solutions thta are integral multiple of \(\frac{\pi}{4}\) is four.

the sum of the largest and the smallest solution is π

the set of all values of \(x\) is \(x \in\left[\frac{\pi}{4}, \frac{\pi}{2}\right) \cup\left(\frac{\pi}{2}, 3 \pi 4\right]\)

Solution

Question 5

\(P\left(\frac{A}{B}\right) \geq \frac{P(A)+P(B)-1}{P(B)}, P(B) \neq 0\) is alway

\(P(A \cap B)=P(A)-P(A \cap B)\) does not hold

\(P(A \cup B)=1-P(\bar{A}) P(B),\) if \(A\) and \(B\) are disjointed

None of these

Solution

Question 6

\(\left|\begin{array}{cc}-5 & -3 \\ -4 & -6 \\ -1 & 1\end{array}\right|\)

\(\left|\begin{array}{cc}-5 & -9 \\ -4 & -6 \\ -1 & 1\end{array}\right|\)

\(\left|\begin{array}{cc}-5 & -3 \\ -4 & -6 \\ -1 & 9\end{array}\right|\)

\(\left|\begin{array}{cc}-5 & -7 \\ -4 & -6 \\ -1 & 1\end{array}\right|\)

Solution

Question 7

(ax+hy+bxy)

(ax2+2hxy+by2)

(ax2−2hxy+by2)

(bx2−2hxy+ay2)

Solution

Question 8

zero

real and positive

real and negative

None of these

Solution

Question 9

\(\left[\begin{array}{cc}7 & -3 \\ -2 & 2\end{array}\right]\)

\(\left[\begin{array}{cc}8 & -3 \\ -2 & 1\end{array}\right]\)

\(\left[\begin{array}{cc}7 & -3 \\ -7 & 1\end{array}\right]\)

\(\left[\begin{array}{cc}7 & -3 \\ -2 & 1\end{array}\right]\)

Solution

Question 10

p + m + c = (19/20)

p + m + c = (27/20)

pmc = (1/4)

None of these

Solution

Question 11

In = In + 2

\(\sum_{m=1}^{10} I_{2 m+1}=10 \pi\)

\(\sum_{m=1}^{10} I_{2 m}=0\)

All of the above

Solution

Question 12

Scalar matrix

Idempotent matrix

Diagonal matrix

Nilptent matrix

Solution

(ax²+2hxy+by²)

(ax²−2hxy+by²)

(bx²−2hxy+ay²)

I_n = I_n _{+ 2}