Mathematics Test

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Mathematics Test - 69

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Weekly Quiz Competition

Question 1
1 / -0

Find the value of\(\left|\begin{array}{ccc}1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3}\end{array}\right|\).

A
\((a-b)(b-c)(c-a)(a+b+c)\)

B
\((a-b)(b-c)(c-a)(a-b-c)\)

C
\((a+b)(b-c)(c-a)(a-b+c)\)

D
\((a+b)(b+c)(c+a)(a+b+c)\)

Solution

Given,

\(\left|\begin{array}{ccc}1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3}\end{array}\right|\)

Applying \(\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1}, \mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{1}\), we get

\(=\left|\begin{array}{ccc}1 & 0 & 0 \\ a & b-a & c-a \\ a^{3} & b^{3}-a^{3} & c^{3}-a^{3}\end{array}\right|\)

As we know,

\(a^3 – b^3 = (a – b) (a^2 + ab + b^2 ) \)

\(=\left|\begin{array}{ccc}1 & 0 & 0 \\ a & b-a & c-a \\ a^{3} & (b-a)(b^2+ab+a^2) & (c-a)(c^2+ca+a^2)\end{array}\right|\)

Taking out \( (b - a), (c - a) \) common from \(C_{2}\) and \(C_{3}\) respectively, we get

\(=(b-a)(c-a)\left|\begin{array}{ccc}1 & 0 & 0 \\ a & 1 & 1 \\ a^{3} & b^{2}+a b+a^{2} & c^{2}+a c+a^{2}\end{array}\right|\)

\(=-(a-b)(c-a)\left[1\left(c^{2}+a c+a^{2}-b^{2}-a b-a^{2}\right) -0+0\right]\)

\(=-(a-b)(c-a)\left(c^{2}+a c-b^{2}-a b\right)\)

\(=-(a-b)(c-a)\left[-\left(b^{2}-c^{2}\right)-a(b-c)\right]\)

\(=-(a-b)(c-a)[(b-c)(-b-c-a)]\)

\(=(a-b)(b-c)(c-a)(a+b+c)\)
Question 2
1 / -0

What is the probability of solving a given problem if three students \((A, B\) and \(C)\), try it independently, with respective probabilities \(\frac{4}{7}, \frac{3}{8}\) and \(\frac{1}{2}\)?

A
\(\frac{97}{112}\)

B
\(\frac{95}{112}\)

C
\(\frac{97}{111}\)

D
\(\frac{97}{125}\)

Solution

Given:

\(P(A)=\frac{4}{7}, P(B)=\frac{3}{8} \text { and } P(C)=\frac{1}{2}\)

We know that,

\(P(A \cup B \cup C)=P(A)+P(B)+P(C)-P(A \cap B)-P(B \cap C)-P(C \cap A)+P(A \cap B \cap C)\)

\(\Rightarrow P(A \cap B)=P(A) \times P(B)\)

So, the probability of the question getting solved is

\(P(A \cup B \cup C)=\frac{4}{7}+\frac{3}{8}+\frac{1}{2}-\left(\frac{4}{7} \times \frac{3}{8}\right)-\left(\frac{3}{8} \times \frac{1}{2}\right)-\left(\frac{1}{2} \times \frac{4}{7}\right)+\left(\frac{4}{7} \times \frac{3}{8} \times \frac{1}{2}\right)\) \(\Rightarrow P(A \cup B \cup C)=\frac{4}{7}+\frac{3}{8}+\frac{1}{2}-\left(\frac{3}{14}\right)-\left(\frac{3}{16}\right)-\left(\frac{2}{7}\right)+\left(\frac{3}{28}\right)\)

\(\Rightarrow P(A \cup B \cup C)=\frac{81}{56}-\left(\frac{3}{14}\right)-\left(\frac{3}{16}\right)-\left(\frac{2}{7}\right)+\left(\frac{3}{28}\right)\)

\(\Rightarrow P(A \cup B \cup C)=\frac{81}{56}+\frac{3}{28}-\left(\frac{24+21+32}{112}\right)\)

\(\Rightarrow P(A \cup B \cup C)=\frac{87}{56}-\frac{77}{112}\)

\(\Rightarrow P(A \cup B \cup C)=\frac{174-77}{112}\)

\(\Rightarrow P(A \cup B \cup C)=\frac{97}{112}\)
Question 3
1 / -0

If \(A=\left[\begin{array}{ccc}1 & 2 & x \\ 3 & -1 & 2\end{array}\right]\) and \(B=\left[\begin{array}{l}y \\ x \\ 1\end{array}\right]\) be such that \(A B=\left[\begin{array}{l}6 \\ 8\end{array}\right]\),then:

A
y=2x

B
y=−2x

C
y=x

D
y=−x

Solution

Let \(A=\left[\begin{array}{ccc}1 & 2 & x \\ 3 & -1 & 2\end{array}\right]\) and \(B=\left[\begin{array}{l}y \\ x \\ 1\end{array}\right]\)

\(A B=\left[\begin{array}{ccc}1 & 2 & x \\ 3 & -1 & 2\end{array}\right]\left[\begin{array}{l}y \\ x \\ 1\end{array}\right]\)

\(\Rightarrow\left[\begin{array}{l}6 \\ 8\end{array}\right]=\left[\begin{array}{c}y+2 x+x \\ 3 y-x+2\end{array}\right]\)

\(\Rightarrow\left[\begin{array}{l}6 \\ 8\end{array}\right]=\left[\begin{array}{c}y+3 x \\ 3 y-x+2\end{array}\right]\)

\(\Rightarrow \mathrm{y}+3 \mathrm{x}=6 \text { and } 3 \mathrm{y}-\mathrm{x}=6\)

On solving, we get

\(\begin{aligned} & x=\frac{6}{5} \text { and } y=\frac{12}{5} \\ & \Rightarrow y=2 x\end{aligned}\)
Question 4
1 / -0

If \(A=\left[\begin{array}{cc}\sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha\end{array}\right]\), then for what value of \(\alpha, A\) is an identity matrix?

A
\(0^{\circ}\)

B
\(45^{\circ}\)

C
\(90^{\circ}\)

D
\(60^{\circ}\)

Solution

Given, \(A=\left[\begin{array}{cc}\sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha\end{array}\right]\)
As we know, Any square matrix in which all the elements are zero except those in the principal diagonal is called a diagonal matrix. A diagonal matrix in which all the principal diagonal elements are equal to \(1\) is called an identity matrix. It is also known as a unit matrix whereas an identity matrix of order \(n\) is denoted by \(I\).
Value of \(\alpha\) such that \(A\) is an identity matrix.
i.e., \(A=I\)
\(\begin{aligned}
&\left[\begin{array}{cc}
\sin \alpha & -\cos \alpha \\
\cos \alpha & \sin \alpha
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\end{aligned}\)
\(\therefore \sin \alpha=1\)
\(\Rightarrow \sin \alpha=\sin 90^\circ\)\(\Rightarrow \alpha= 90^\circ\)
And \(\cos \alpha=0 \)
\(\Rightarrow \cos \alpha=\cos 90^{\circ}\)\(\Rightarrow \alpha= 90^{\circ}\)
\(\therefore\) The required value of \(\alpha\) is \(90^{\circ}\).
Question 5
1 / -0

The differential equation of the family of curves \(y=c_1 e^x+c_2 e^{-x}\) is:

A
\(\frac{d^2 y}{d x^2}+y=0\)

B
\(\frac{d^2 y}{d x^2}-y=0\)

C
\(\frac{d^3 y}{d x^3}+y=0\)

D
\(\frac{d^4 y}{d x^4}-1=0\)

Solution

\(\begin{array}{l}y=c_1 e^x+c_2 e^{-x} \\ \Rightarrow \frac{d y}{d x}=\frac{d}{d x} c_1 e^x+\frac{d}{d x} c_2 e^{-x} \\ \Rightarrow \frac{d y}{d x}=c_1 e^x-c_2 e^{-x} \\ \Rightarrow \frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}\left(c_1 e^x-c_2 e^{-x}\right) \\ \Rightarrow \frac{d^2 y}{d x^2}=c_1 e^x+c_2 e^{-x}=y \\ \Rightarrow \frac{d^2 y}{d x^2}-y=0\end{array}\)
Question 6
1 / -0

Find the value of \(k\) if \(\underset{{{x \rightarrow 7}}}{\lim} g(x)=k\) where \(g(x)=\sqrt{8 x-7}\)

A
3

B
6

C
7

D
None of these

Solution

Given:

\(\underset{{{x \rightarrow 7}}}{\lim} g(x)=k\)

where, \(g(x)=\sqrt{8 x-7}\)

As we know that, if \(\lim _{x \rightarrow a} f(x)\) does not result into indeterminate form, then we use direct substitution in order to find the limits.

Here, also we can see that \(\underset{{{x \rightarrow 7} }}{\lim} g(x)\) does not result into any indeterminate form.

So, we can substitute \(x=7\) in the expression \(g(x)=\sqrt{8 x-7}\) in order to find the value of \({k}\).

\(\underset{{{x \rightarrow 7}}}{\lim} g(x)=k\), where \(g(x)=\sqrt{8 x-7}\), i.e.,

\(\Rightarrow \underset{{{x \rightarrow 7}}}{\lim} \sqrt{8 x-7}=k\)

\(\Rightarrow \sqrt{(8\times7)-7}=k\)

\(\Rightarrow \sqrt{49}=k\)

\(\Rightarrow 7=k\)

or, \(k=7\)
Question 7
1 / -0

Find the multiplicative inverse of 4 - 3i ?

A

\(\frac{4}{25}\)-i \(\frac{3}{25}\)

B

\(\frac{3}{25}\)-i \(\frac{4}{25}\)

C

\(\frac{3}{25}\)+i \(\frac{4}{25}\)

D

\(\frac{4}{25}\)+i \(\frac{3}{25}\)

Solution

CONCEPT:

If \(\mathrm{z}=\mathrm{x}+\) i\(\mathrm{y}\) then conjugate of \(\mathrm{z}\) is given by \(\mathrm{\bar{z}=x}\)-i \(\mathrm{y}\)

If \(\mathrm{z}=\mathrm{x}+\) i\(\mathrm{y}\) then \(|\mathrm{z}|=\sqrt{x^{2}+y^{2}}\)

\(\mathrm{z=x}+\) i\(\mathrm{y}\) then multiplicative inverse of \(\mathrm{z}\) is given by \(\mathrm{z^{-1}=\frac{\bar{z}}{|z|^{2}}}\)

CALCULATION:

Let \(\mathrm{z}=4-3\)i

In order to find out the multiplicative of \(z\) first we need to find out \(\bar{\mathrm{z}}\) and \(\mathrm{|z|}\)

As we know that,

If \(\mathrm{z=x}+\) i\(\mathrm{y}\) then \(\mathrm{\bar{z}=x}\)-i \(\mathrm{y}\) and \(\mathrm{|z|=\sqrt{x^{2}+y^{2}}}\)

\(\Rightarrow \mathrm{\bar{z}}=4+3\)i and \(\mathrm{|z|}^{2}=25\)

As we know that, \(\mathrm{z}^{-1}=\mathrm{\frac{\bar{z}}{|z|^{2}}}\)

\(\Rightarrow \mathrm{z}^{-1}=\frac{4+3 i}{25}=\frac{4}{25}\) + i\(\frac{3}{25}\)
Question 8
1 / -0

The total number of subsets of a finite set A has 56 more elements than the total number of subsets of another finite set B. What is the number of elements in set A?

A
5

B
6

C
7

D
8

Solution

Let set A and B have m and n elements, respectively.

2^m−2ⁿ = 56

2ⁿ(2^m−n−1) = 56 = 8×7 = 2³×7

Comparing both sides, we get

2ⁿ = 2³ and 2^m−n = 7

⇒ n = 3 and 2^m−n = 8

⇒ 2^m−n = 2³ ⇒ m-n = 3

⇒ m−3 = 3 ⇒ m = 6

Number of the elements in A is 6.

Question 9

1 / -0

Find the mode of the following data

Class	0 -10	10 - 20	20 - 30	30 - 40	40 - 50
frequency	6	10	14	12	4

25

26.66

36.66

20 Solution

From the given data it is cleared data the highest frequency is 14 and it belongs to 20 - 30 class interval so 20 - 30 is the modal class.

L = 20, \(f_1=14, f_2=12, f_0=10,\) I = 10

Mode \(=L+\left[\frac{f_1-f_0}{2 f_1-f_0-f_2}\right] \times h\)

⇒ 20 + [\(\frac{(14 - 10)}{(2 × 14 - 10 - 12)}\)] × 10 = 26.66

Question 10
1 / -0

Find the angle between the line \(\frac{x-1}{3}=\frac{y+1}{-1}=\frac{z-3}{2}\) and the plane \(3 x+4 y+z+5=0\).

A
\(\sin ^{-1}\left(\sqrt{\frac{7}{52}}\right)\)

B
\(\cos ^{-1}\left(\sqrt{\frac{7}{52}}\right)\)

C
\(\sin ^{-1}\left(\sqrt{\frac{5}{52}}\right)\)

D
\(\cos ^{-1}\left(\sqrt{\frac{5}{52}}\right)\)

Solution

Given,

Equation of line: \(\frac{x-1}{3}=\frac{y+1}{-1}=\frac{z-3}{2}\)

Equation of plane: \(3 x+4 y+z+5=0\)

As we know,

The angle between the line \(\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}\) and the plane \(a_{2} x+b_{2} y+c_{2} z+d=0\) is given by,

\(\sin \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\left(\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\right)\left(\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}\right)}\)

On comparing given equation of line and plane with above equation, we get

\(a_{1}=3, b_{1}=-1, c_{1}=2, a_{2}=3, b_{2}=4\) and \(c_{2}=1\)

\(\sin \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\left(\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\right)\left(\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}\right)}\)

\(\Rightarrow \sin \theta=\frac{3\times 3 +(-1)\times 4+2 \times 1}{\left(\sqrt{(3)^{2}+(-1)^{2}+(2)^{2}}\right)\left(\sqrt{(3)^{2}+(4)^{2}+(1)^{2}}\right)}\)

\(\Rightarrow \sin \theta=\frac{9 -4+2 }{\left(\sqrt{9+1+4}\right)\left(\sqrt{9+16+1}\right)}\)

\(\Rightarrow \sin \theta=\frac{7}{\sqrt{14} \cdot \sqrt{26}}\)

\(\Rightarrow \sin \theta=\sqrt{\frac{7}{52}} \)

\(\Rightarrow \theta=\sin ^{-1}\left(\sqrt{\frac{7}{52}}\right)\)

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Re-Attempt Weekly Quiz Competition

Mathematics Test - 69

Result

Mathematics Test - 69

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Question 1

\((a-b)(b-c)(c-a)(a+b+c)\)

\((a-b)(b-c)(c-a)(a-b-c)\)

\((a+b)(b-c)(c-a)(a-b+c)\)

\((a+b)(b+c)(c+a)(a+b+c)\)

Solution

Question 2

\(\frac{97}{112}\)

\(\frac{95}{112}\)

\(\frac{97}{111}\)

\(\frac{97}{125}\)

Solution

Question 3

y=2x

y=−2x

y=x

y=−x

Solution

Question 4

\(0^{\circ}\)

\(45^{\circ}\)

\(90^{\circ}\)

\(60^{\circ}\)

Solution

Question 5

\(\frac{d^2 y}{d x^2}+y=0\)

\(\frac{d^2 y}{d x^2}-y=0\)

\(\frac{d^3 y}{d x^3}+y=0\)

\(\frac{d^4 y}{d x^4}-1=0\)

Solution

Question 6

3

6

7

None of these

Solution

Question 7

\(\frac{4}{25}\)-i \(\frac{3}{25}\)

\(\frac{3}{25}\)-i \(\frac{4}{25}\)

\(\frac{3}{25}\)+i \(\frac{4}{25}\)

\(\frac{4}{25}\)+i \(\frac{3}{25}\)

Solution

Question 8

5

6

7

8

Solution

Question 9

25

26.66

36.66

20

Solution

Question 10

\(\sin ^{-1}\left(\sqrt{\frac{7}{52}}\right)\)

\(\cos ^{-1}\left(\sqrt{\frac{7}{52}}\right)\)

\(\sin ^{-1}\left(\sqrt{\frac{5}{52}}\right)\)

\(\cos ^{-1}\left(\sqrt{\frac{5}{52}}\right)\)

Solution

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