Mathematics Test - 7

If \(\overrightarrow{O A}\) is the position vector of \(A\) and \(\overrightarrow{O B}\) is position vector of \(B\) then,

\(\overrightarrow{A B}=\) Position Vector of \(B\) - Position Vector of \(A\)...(I)

\(\overrightarrow{B A}=\) Position Vector of \(A\) - Position Vector of \(B\)...(II)

Given,

\(\overrightarrow{O A}=\vec{a}\) and \(\overrightarrow{O B}=\vec{b}\)

Then, by equation (II)

\(\overrightarrow{B A}=\text { position vector of } A-\text { position vector of } B \)

\(\overrightarrow{B A}=\overrightarrow{O A}-\overrightarrow{O B} \)

\(\overrightarrow{B A}=\vec{a}-\vec{b}\)

Given,

\(z=\frac{3-2 i \sin \theta}{2-i \sin \theta}\)

Multiplying the numerator and denominator by \((2+i \sin \theta)\), we det

\(z=\frac{3-2 i \sin \theta}{2-i \sin \theta} \times \frac{2+i \sin \theta}{2+i \sin \theta}\)

\(\Rightarrow z=\frac{6+3 i \sin \theta-4 i \sin \theta-2 i^{2} \sin ^{2} \theta}{(2)^{2}-(i \sin \theta)^{2}}\)

\(\Rightarrow z=\frac{6-i \sin \theta+2 i^{2} \sin ^{2} \theta}{4-i^{2} \sin ^{2} \theta}\)

\(\Rightarrow z=\frac{6-i \sin \theta+2 \times(-1) \sin ^{2} \theta}{4-(-1) \sin ^{2} \theta} \quad\left[\because i^{2}=-1\right]\)

\(\Rightarrow z=\frac{6-i \sin \theta+2 \times(-1) \sin ^{2} \theta}{4-(-1) \sin ^{2} \theta}\)

\(\Rightarrow z=\frac{6-2 \sin ^{2} \theta}{4+\sin ^{2} \theta}+i\left(\frac{-\sin \theta}{41 \sin ^{2} \theta}\right)\)

Imaginary part of \(z\),

\(\operatorname{Im}(z)=\frac{-\sin \theta}{4+\sin ^{2} \theta}\)

As we know,

For \(z\) to be purely real, \(\operatorname{Im}(z)=0\)

\(\therefore \frac{-\sin \theta}{4 \sin ^{2} \theta}=0 \)

\(\sin \theta=0 \)

\(\theta=n \pi\) where \(n\) belongs to an integer

Given: \(\mathrm{y}=\mathrm{x}^{3}\) and \(\mathrm{y}=\mathrm{x}^{2}\)

Finding a point of intersection:

\(\Rightarrow \mathrm{x}^{3}-\mathrm{x}^{2}=0 \Rightarrow \mathrm{x}^{2}(\mathrm{x}-1)=0 \Rightarrow \mathrm{x}=0,1\)

Draw the graph of the curve \(\mathrm{y}=\mathrm{x}^{2}\) and \(\mathrm{y}=\mathrm{x}^{3}\)

Let the required area be A.

Using the formula of the area under the curve, \(\mathrm{A}=\left|\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x})-\mathrm{g}(\mathrm{x}) \mathrm{dx}\right|\)

\(\Rightarrow \mathrm{A}=\left|\int_{0}^{1}\left(\mathrm{x}^{3}-\mathrm{x}^{2}\right) \mathrm{dx}\right|=\left|\left[\frac{\mathrm{x}^{4}}{4}-\frac{\mathrm{x}^{3}}{3}\right]_{0}^{1}\right|\)

Substitute the limit to evaluate the area.

\(\Rightarrow \mathrm{A}=\left|\frac{1}{4}-\frac{1}{3}-0+0\right|=\frac{1}{12}\)

Let f (x) = x⁵ sin4 x

f (-x) = (-x)⁵ sin⁴ (-x) = -x⁵ (-sin x )⁴ = -x⁵ sin⁴ x

⇒ f(-x) = - f(x)

So, f(x) is an odd function.

We know that, when \(\mathrm{f}(\mathrm{x})\) is odd \(\Leftrightarrow \int_{-\mathrm{a}}^{\mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=0\)

So, \(\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \mathrm{x}^{5} \sin ^{4} \mathrm{xdx}=0\)

We know that,

The highest order derivative present in the differential equation is the order of the differential equation.

The degree is the highest power of the highest order derivative in the differential equation after the equation has been cleared from fractions and the radicals as for as the derivatives are concerned.

Given, the differential equation is,

${\left[1+{\left(\frac{dy}{dx}\right)}^3\right]}^{\frac{7}{3}}=7\frac{d^{2}y}{d{x}^2}$

The order and degree of a differential equation are the integers not in fraction form.

To find the order and degree of a differential equation, take the cube on both side.

${\left({\left[1+{\left(\frac{dy}{dx}\right)}^3\right]}^{\frac{7}{3}}\right)}^3={\left[7\frac{d^{2}y}{d{x}^2}\right]}^3$

${\left[1+{\left(\frac{dy}{dx}\right)}^3\right]}^7={\left[7\frac{d^{2}y}{d{x}^2}\right]}^3$

The highest order derivative present in the differential equation$\frac{d^{2}y}{d{x}^2}$Its order is 2.

So, the order of the given differential equation is 2

The highest degree of the derivative present in the differential equation${\left[\frac{d^{2}y}{d{x}^2}\right]}^3$. Its degree is 3.

So, the Order and degree of the differential equation

${\left[1+{\left(\frac{dy}{dx}\right)}^3\right]}^{\frac{7}{3}}=7\frac{d^{2}y}{d{x}^2}$are 2 and 3 respectively

Equality of complex numbers:

Two complex numbers \(z_{1}=x_{1}+i y_{1}\) and \(z_{2}=x_{2}+i y_{2}\) are equal if and only if \(x_{1}=x_{2}\) and \(y_{1}=y_{2}\).

or \(\operatorname{Re}\left(z_{1}\right)=\operatorname{Re}\left(z_{2}\right)\) and \(\operatorname{Im}\left(z_{1}\right)=\operatorname{lm}\left(z_{2}\right)\)

Given: \({x}+{iy}=\frac{3+4 {i}}{2-{i}}\)

\(\Rightarrow {x}+{iy}=\frac{3+4 {i}}{2-{i}} \times \frac{2+{i}}{2+{i}}\)

\(\Rightarrow {x}+{iy}=\frac{6+11 {i}+4 {i}^{2}}{4-{i}^{2}}\)

As we know \({i}^{2}=-1\)

\(\Rightarrow {x}+{iy}=\frac{6+11 {i}-4}{4+1}\)

\(\Rightarrow {x}+{iy}=\frac{2+11 {i}}{5}=\frac{2}{5}+{i} \frac{11}{5}\)

Comparing real and imaginary parts, we get,

\({x}=\frac{2}{5}\) and \({y}=\frac{11}{5}\)

Given,

A box contains \(3\) white and \(2\) black balls.

Total balls \(= 3 + 2 = 5\)

So probability of first ball to be black \( = \frac{2}{5}\) and

Probability of second ball to be also black \(=\frac{1}{4}\) (∵second time we have \(2 -1 = 1\) black and \(5 - 1 = 4\) total balls)

∴ The probability that both the balls are black \(=\frac{ 2}{5} × \frac{1}{4}\)

\(= \frac{1}{10}\)

Let, \(z=\frac{1-\mathrm{i}}{1+\mathrm{i}}\)

or \(\mathrm{z}=\frac{1-\mathrm{i}}{1+\mathrm{i}} \times \frac{1-\mathrm{i}}{1-\mathrm{i}}=\frac{1+\mathrm{i}^{2}-2 \mathrm{i}}{1-\mathrm{i}^{2}}\)

⇒ z = -\(\frac{2i}{2}\) = -i

x + iy = -i

so, x = 0 and y = -1

We know that, \(\operatorname{Arg}(z)=\tan ^{-1}\left(\frac{y}{x}\right)\)

\(\operatorname{Arg}(z)=\tan ^{-1}\left(\frac{-1}{0}\right)=-\tan ^{-1}\left(\frac{1}{0}\right)\)

\(\operatorname{Arg}(z)=-\tan ^{-1}(\infty)=-\frac{\pi}{2}\)

Given,

\(\frac{(x-3)}{1}=\frac{(y-2)}{-1}=\frac{(z+\lambda)}{-2}=k\)

Let \(P\) be any point on the plane,

\(P=(k+3,-k-2,-2 k-\lambda)\)

\(P\) lie on the plane, \(2 x-4 y+3 z=2,\) Which means \(P\) satisfy the equationn of plane.

So, \(2(k+3)-4(-k-2)+3(-2 k-\lambda)=2\)

Solving above equation, we have \(\lambda=4\)

Now, the shortest distance between the below lines:

\(\frac{(x-3)}{1}=\frac{(y-2)}{-1}=\frac{(z+4)}{-2}\) and \(\frac{(x-1)}{12}=\frac{y}{9}=\frac{z}{4}\) is:

\(\begin{aligned} \mathrm{d}^{2} &=\left|\begin{array}{ccc}x_{1}-x_{2} & y_{1}-y_{2} & z_{1}-z_{2} \\ l_{1} & m_{1} & n_{1} \\ l_{2} & m_{2} & n_{2}\end{array}\right| \\ &=\left|\begin{array}{ccc}3-1 & -2-0 & -4-0 \\ 1 & -1 & -2 \\ 12 & 9 & 4\end{array}\right| \end{aligned}\)

\(=\mid2(14)+2(28)-4(21) \mid\)

\(=0\)

\(d^{2}=0\) or \({d}=0\)

Given,

\(7,13,15,11,4\)

Formula used:

\(\mathrm{S} . \mathrm{D}=\sqrt{\frac{\sum|x-m|^{2}}{n}}\)

Mean \((m)=\frac{\text { Total of observations }}{\text { number of observations }}\)

\(\mathrm{S.D}=\) standard deviation

\(\sum=\) summation

\(x=\) observation

\(m=\) mean of the observations

\(n=\) number of observation

Mean of \(7,13,15,11,4\)

\(=\frac{50}{5}=10\)

\(\text { S.D }=\sqrt{\frac{(7-10)^{2}+(13-10)^{2}+(15-10)^{2}+(11-10)^{2}+(4-10)^{2}}{5}}\)

\(=\sqrt{\frac{9+9+25+1+36}{5}}\)

\(=\sqrt{\frac{80}{5}} \)

\(=\sqrt{16}=4\)

\(\therefore\) The standard deviation is \(4\).