Mathematics Test - 26

Let \(\vec{u}\) and \(\vec{v}\) be two non-zero vectors.

It is given that the magnitude of the sum of two non-zero vectors is equal to the magnitude of their difference.

\(\Rightarrow|\vec{u}+\vec{v}|=|\vec{u}-\vec{v}|\)

\(\Rightarrow|\vec{u}+\vec{v}|^{2}=|\vec{u}-\vec{v}|^{2}\)

As we know that, if \(\vec{u}\) be a vector then

\(|\vec{u}|^{2}=\vec{u} \cdot \vec{u}\)

\(\Rightarrow(\vec{u}+\vec{v}) \cdot(\vec{u}+\vec{v})=(\vec{u}-\vec{v}) \cdot(\vec{u}-\vec{v})\)

\(\Rightarrow|\vec{u}|^{2}+|\vec{v}|^{2}+2|\vec{u}| \times|\vec{v}| \times \cos \theta=|\vec{u}|^{2}+|\vec{v}|^{2}-2|\vec{u}| \times|\vec{v}| \times \cos \theta\)

\(\Rightarrow 4 \times|\vec{u}| \times|\vec{v}| \times \cos \theta=0\)

\(\because \vec{u}\) and \(\vec{v}\) be two non-zero vectors \(\Rightarrow|\vec{u}| \times|\vec{v}| \neq 0\)

\(\Rightarrow \cos \theta=0 \Rightarrow \theta=90^{\circ}\)

So, \(\vec{u}\) and \(\vec{v}\) are perpendicular to each other.

1. Both variance and standard deviation are measures of variability in the population.

This statement is correct.

2.The standard deviation is the square root of the variance.

So, the given statement is not correct.

The given differential equation is:

\(\left(e^{x}+e^{-x}\right) d y-\left(e^{x}-e^{-x}\right) d x=0 \)

\(\Rightarrow\left(e^{x}+e^{-x}\right) d y=\left(e^{x}-e^{-x}\right) d x \)

\(\Rightarrow d y=\left[\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\right] d x\)

Integrating both sides of this equation, we get:

\(\int d y=\int\left[\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\right] d x+\mathrm{C}\)

\(\Rightarrow y=\int\left[\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\right] d x+\mathrm{C}\quad\quad\)....(1)

Let \(\left(e^{x}+e^{-x}\right)=t\)

Differentiating both sides with respect to \(x\), we get:

\(\frac{d}{d x}\left(e^{x}+e^{-x}\right)=\frac{d t}{d x}\)

\(\Rightarrow e^{x}-e^{-x}=\frac{d t}{d t}\)

\(\Rightarrow\left(e^{x}-e^{-x}\right) d x=d t\)

Substituting this value in equation (1), we get:

\(y=\int \frac{1}{t} d t+\mathrm{C}\)

\(\Rightarrow y=\log (t)+\mathrm{C}\)

\(\Rightarrow y=\log \left(e^{x}+e^{-x}\right)+C\)

We know that,

Equation of ellipse:$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Eccentricity (e) =$\sqrt{1-\frac{b^2}{a^2}}$

Where, vertices =(± a, 0) and focus =(± ae, 0)

Now, according to the equation,vertices of the ellipse(± 5, 0) and foci (±4, 0)

So, a = ±5

⇒a²=25

⇒ae = 4

⇒ e =$\frac{4}{5}$

Now,$\frac{4}{5}=\sqrt{1-\frac{b^2}{5^2}}$

$\begin{array}{l}\Rightarrow \frac{16}{25}=\frac{25-{\mathrm{b}}^2}{25}\\ \Rightarrow 16=25-{\mathrm{b}}^2\\ \Rightarrow {\mathrm{b}}^2=9\end{array}$

∴ Equation of ellipse =$\frac{x^2}{25}+\frac{y^2}{9}=1$

Distance from the point (0,0) of the line 3x + 4y + 15 = 0

$\Rightarrow p_1=\frac{3\times 0+4\times 0+15}{\sqrt{9+16}}$

$\Rightarrow \frac{15}{5}=3$

$\therefore p_1=3$

Distance from the point (0,0) of the line 6x + 8y-13 = 0

$\begin{array}{rl}\Rightarrow p_2& =\frac{6\times 0+8\times 0-13}{\sqrt{36+64}}\\ & =\frac{-13}{10}\\ & =-1.3=1.3\end{array}$

Distance between these lines$=p_1+p_2$

$=3+1.3=4.3$units

On solving the given equation-

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

\(\Rightarrow\left(D^{2}+\mu\right) x=0\)

\(\Rightarrow m^{2}+\mu=0\)

\(m^{2}=-\mu\)

\(\Rightarrow \mathrm{m}=\pm \sqrt{u} i\)

\(\Rightarrow x(t)=c_{1} \cos \sqrt{u} t+c_{2} \sin \sqrt{u} t\)

The functions \(\sin \theta \& \cos \theta\) are always periodic

\(\therefore \mathrm{x}(\mathrm{t})\) is periodic function.

Given:

\(\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}\) and \(\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}\)

It is known that the shortest distance between the two lines.

\(\frac{\mathrm{x}-\mathrm{x}_{1}}{\mathrm{a}}=\frac{\mathrm{y}-\mathrm{y}_{1}}{\mathrm{~b}}=\frac{\mathrm{z}-\mathrm{z}_{1}}{\mathrm{c}}\) and \(\frac{\mathrm{x}-\mathrm{x}_{2}}{\mathrm{a}}=\frac{\mathrm{y}-\mathrm{y}_{2}}{\mathrm{~b}}=\frac{\mathrm{z}-\mathrm{z}_{2}}{\mathrm{c}}\) is given by,

\(\mathrm{d}=\frac{\left|\begin{array}{ccc} \mathrm{x}_{2}-\mathrm{x}_{1} & \mathrm{y}_{2}-\mathrm{y}_{1} & \mathrm{z}_{2}-\mathrm{z}_{1} \\ \mathrm{a}_{1} & \mathrm{~b}_{1} & \mathrm{c}_{1} \\ \mathrm{a}_{2} & \mathrm{~b}_{2} & \mathrm{c}_{2} \end{array}\right|}{\sqrt{\left(\mathrm{b}_{1} \mathrm{c}_{2}-\mathrm{b}_{2} \mathrm{c}_{1}\right)^{2}+\left(\mathrm{c}_{1} \mathrm{a}_{2}-\mathrm{c}_{2} \mathrm{a}_{1}\right)^{2}+\left(\mathrm{a}_{1} \mathrm{~b}_{2}-\mathrm{a}_{2} \mathrm{~b}_{1}\right)^{2}}} \quad\quad\)...(1)

Comparing the given equations, we obtain

\(\mathrm{x}_{1}=-1, \mathrm{y}_{1}=-1, \mathrm{z}_{1}=-1 \)

\(\mathrm{a}_{1}=7, \mathrm{~b}_{1}=-6, \mathrm{c}_{1}=1\)

\(\mathrm{x}_{2}=3, \mathrm{y}_{2}=5, \mathrm{z}_{2}=7\)

\(\mathrm{a}_{2}=1, \mathrm{~b}_{2}=-2, \mathrm{c}_{2}=1\)

Then, \(\left|\begin{array}{ccc}\mathrm{x}_{2}-\mathrm{x}_{1} & \mathrm{y}_{2}-\mathrm{y}_{1} & \mathrm{z}_{2}-\mathrm{z}_{1} \\ \mathrm{a}_{1} & \mathrm{~b}_{1} & \mathrm{c}_{1} \\ \mathrm{a}_{2} & \mathrm{~b}_{2} & \mathrm{c}_{2}\end{array}\right|=\left|\begin{array}{ccc}4 & 6 & 8 \\ 7 & -6 & 1 \\ 1 & -2 & 1\end{array}\right|\)

\(=4(-6+2)-6(7-1)+8(-14+6)\)

\(=-16-36-64=-116\)

Now \(\sqrt{\left(\mathrm{b}_{1} \mathrm{c}_{2}-\mathrm{b}_{2} \mathrm{c}_{1}\right)^{2}+\left(\mathrm{c}_{1} \mathrm{a}_{2}-\mathrm{c}_{2} \mathrm{a}_{1}\right)^{2}+\left(\mathrm{a}_{1} \mathrm{~b}_{2}-\mathrm{a}_{2} \mathrm{~b}_{1}\right)^{2}}\)

\(=\sqrt{(-6+2)^{2}+(1+7)^{2}+(-14+6)^{2}}\)

\(=\sqrt{16+36+64}\)

\(=\sqrt{116}\)

\(=2 \sqrt{29}\)

Substituting all the values in equation (1), we obtain

\(\mathrm{d}\) \(=\frac{116}{2 \sqrt{29} }\)

\(=\frac{-58}{\sqrt{29} }\)

\(=\frac{-2 \times 29}{\sqrt{29} }\)

\(=-2 \sqrt{29}\)

Since distance is always non-negative, the distance between the given lines is \(2 \sqrt{2} 9\) units.

Let \(U=\) No. of students in higher secondary group, \(M=\) No. of higher secondary students who qualified NDA exam and \(J=\) No. of higher secondary students who qualified JEE exam.
Given: \(n(U)=100, n(M)=50, n(J)=40\) and \(n(M \cap J)=30\)
As we know that, if \(A\) and \(B\) are two finite sets then, \(n(A \cup B)=n \)
\(n(A)+n(B)-n(A \cap B) \)
\(\Rightarrow n(M \cup J)=n(M)+n(J)-n(M \cap J)=50+40-30 \) \( =60\)
\(\therefore\) There are 60 higher secondary students who have qualified either NDA or JEE exam.
As, we know that the number of higher secondary students who have neither qualified NDA nor JEE exam is given by: \(n[(M \cup J)]\) \(\Rightarrow n[(M \cup J)]=n(U)-n(M \cup J)=100-60=40\)

It is given that,

First angle,${\theta }_1={75}^{\circ }=75\times \frac{\pi }{180}=\frac{5\pi }{12}$

Second angle,${\theta }_2={60}^{\circ }=60\times \frac{\pi }{180}=\frac{\pi }{3}$

∴Radius obtained from the first angle,$r_1=l\times \frac{12}{5\pi }$

Radius obtained from the second angle,$r_2=l\times \frac{3}{\pi }$

So,$\frac{r_1}{r_2}=\frac{12}{5\times 3}=\frac{4}{5}$

Therefore,$r_1:r_2=4:5$