Mathematics Test - 38

Case 1:

It is a fundamental property.

Example

Consider a set of 5 data points

{1, 2, 3, 4, 5}

median = 3

Deviation from mean:

1 - 3 = -2

2 - 3 = -1

3 - 3 = 0

4 - 3 = 1

5 - 3 = 2

Mean of deviations (about mean) = \(\frac{0}{5}\) = 0

Case 2:

Consider data point 2

Deviations of data about 2

1 - 2 = - 1

2 - 2 = 0

3 - 2 = 1

4 - 2 = 2

5 - 2 = 3

Mean of deviation of data about 2 = \(\frac{(-1 + 0 + 1 + 2 + 3)}{5}\) = 1

Mean of deviation of data about median is minimum and this is true for all data.

If \(a_{1} x+b_{1} y+c_{1} z+d_{1}=0\) and \(a_{2} x+b_{2} y+c_{2} z+d_{2}=0\) are a plane equations, then angle between planes can be found using the following formula:

\(\cos \theta=\frac{ a _{1} a _{2}+ b _{1} b _{2}+ c _{1} c _{2}}{\sqrt{ a _{1}^{2}+ b _{1}^{2}+ c _{1}^{2}} \sqrt{ a _{2}^{2}+ b _{2}^{2}+ c _{2}^{2}}}\)

Here,

\(a_{1}=\sqrt{3}-1\)

\(a_{2}=-\sqrt{3}-1\)

\(b_{1}=-\sqrt{3}-1\)

\(b_{2}=\sqrt{3}-1\)

\(c_{1}=4\)

\(c_{2}=4\)

Therefore, required angle will be given by:

\(\cos \theta=\frac{ a _{1} a _{2}+ b _{1} b _{2}+ c _{1} c _{2}}{\sqrt{ a _{1}^{2}+ b _{1}^{2}+ c _{1}^{2}} \sqrt{a_{2}^{2}+ b _{2}^{2}+ c _{2}^{2}}}\)

\(\cos \theta=\frac{-2-2+16}{\sqrt{24} \sqrt{24}}\)

\(=\frac{12}{24}\)

\(=\frac{1}{2}\)

\(\cos \theta = \cos \frac{\pi}{3}\)

\(\Rightarrow \theta=\frac{\pi}{3}\)

Let \(\mathrm{ABCD}\) be a parallelogram such that \(\overline{\mathrm{AB}}=\overrightarrow{\mathrm{q}}, \overline{\mathrm{AD}}=\overrightarrow{\mathrm{p}}\) and \(\angle \mathrm{BAD}\) be an acute angle.

We have

\(\overline{\mathrm{AX}}=\left(\frac{\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}}}{|\overrightarrow{\mathrm{p}}|}\right)\left(\frac{\overrightarrow{\mathrm{p}}}{|\overrightarrow{\mathrm{p}}|}\right)=\frac{\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}}}{|\overrightarrow{\mathrm{p}}|^2} \vec{p}\)

From triangle law

Let \(\vec{r}=\overline{B X}=\overline{B A}+\overline{A X}=-\vec{q}+\frac{\vec{p} \cdot \vec{q}}{|\vec{p}|^2} \vec{p}\)

We know,

\(\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\)

Hence,

\(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}}{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}+\sqrt{\cos \left(\frac{\pi}{2}-x\right)}} d x\)

\(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x\)

If,

\(I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x\)\( \quad \ldots \ldots(i)\)

Then,

\(I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x\)\( \quad \ldots \ldots(ii)\)

Adding to \(eq^{n}(i)+(ii)\)

Hence,

\(2 I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}+\frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x \)

On changing the limit,

\(2 I=-\int_{\frac{\pi}{3}}^{\frac{\pi}{6}} 1 d x \)

\(2I=[-x]_{\frac{\pi}{3}}^{\frac{\pi}{6}}\)

Put the limit,

\(2I=-[\frac{\pi}{6}-\frac{\pi}{3}]\)

\(2 I=\frac{\pi}{6} \)

\(I=\frac{\pi}{12}\)

The solution of the set of constraints of a linear programming problem is a convex (open or closed) is called feasible region.A feasible region or solution space is the set of all possible points problem that satisfy the problem's constraints, potentially including inequalities, equalities, and integer constraints.

Given, \(\frac{\mathrm{dy}}{\mathrm{dx}}+(\tan \mathrm{x}) \mathrm{y}=\sin \mathrm{x}, \mathrm{x} \in\left[0, \frac{\pi}{3}\right]\)

which is a linear differential equation of the form of \(\frac{d y}{d x}+P y=Q\)

Here, \(P=\tan \mathrm{x}\)

\(\therefore \quad \text { IF }=\mathrm{e}^{\int \operatorname{Pd} x} \\\)

\(\Rightarrow \mathrm{e}^{\int \tan x d x}=\mathrm{e}^{\log (\sec x)}=\sec \mathrm{x}\)

Multiplying by \(\sec \mathrm{x}\) on both sides

\(\frac{d y}{d x}+(\tan x) y=\sin x \\\)

\(\sec x \frac{d y}{d x}+(\tan x \sec x) y=\sin x \sec x \\\)

\( \Rightarrow \frac{d}{d x}(y \sec x)=\tan x \Rightarrow y \sec x=\int \tan x d x \\\)

\( \Rightarrow y \sec x=\log (\sec x)+c \\\)

\( y=\cos x \log (\sec x)+c \cdot \cos x \\\)

\( y(0)=0 \\\)

\( \Rightarrow 0=1 \cdot 0+c \cdot 1 \Rightarrow c=0 \\\)

\( \therefore y=\cos x \cdot \log (\sec x) \\\)

\( \Rightarrow y\left(\frac{\pi}{4}\right)=\cos \left(\frac{\pi}{4}\right) \cdot \log \left(\sec \frac{\pi}{4}\right) \\\)

\( =\frac{1}{\sqrt{2}} \log (\sqrt{2}) \\\)

\(\left(\frac{1}{2 \sqrt{2}}\right) \log _e 2\)

We have,

\( \frac{3}{2} x^2+\frac{5}{3} x^3+\frac{7}{4} x^4+\ldots \\\)

= \( \left(2-\frac{1}{2}\right) x^2+\left(2-\frac{1}{3}\right) x^3+\left(2-\frac{1}{4}\right) x^4+\ldots \\\)

= \( 2\left(x^2+x^3+x^4+\ldots\right)-\left(\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\ldots\right) \\\)

= \( 2 \cdot \frac{x^2}{1-x}-\left[-\log _e(1-x)-x\right]\)

[ using sum of infinite GP \(=\frac{a}{1-r}\) and logarithmic series]

\(=\frac{2 x^2}{1-x}+x+\log _e(1-x) \\\)

\( =\frac{2 x^2+x-x^2}{1-x}+\log _e(1-x) \\\)

\( =\frac{x^2+x}{1-x}+\log _e(1-x) \\\)

\( =x\left(\frac{1+x}{1-x}\right)+\log _e(1-x)\)

We have,

\(\int_{0}^{\frac{\pi}{3}} \sqrt{1+\sin 2 x} d x\)

\(= \int_{0}^{\frac{\pi}{3}} \sqrt{\left(\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x\right)} d x\)

\(= \int_{0}^{\frac{\pi}{3}}(\cos x+\sin x) d x\)

\(=[\sin x-\cos x]_{0}^{\frac{\pi}{3}}\)

\(=\left[\sin \frac{\pi}{3}-\cos \frac{\pi}{3}\right]-(\sin 0-\cos 0)\)

\(=\left[\frac{\sqrt{3}}{2}-\frac{1}{2}\right]-(0-1)\)

\(=\left[\frac{\sqrt{3}-1}{2}\right]+1\)

\(= \frac{\sqrt{3}+1}{2}\)

Thus, \(\int_{0}^{\frac{\pi}{3}} \sqrt{1+\sin 2 x} d x=\frac{\sqrt{3}+1}{2}\)

Given inequalities x ≥ 0, y ≥ 0, 2x + y + 6 ≤ 0

Now take x = 0, y = 0 and 2x + y + 6 = 0

when x = 0, y = -6

when y = 0, x = -3

So, the points are A(0, 0), B(0, -6) and C(-3, 0)

Since region is outside from the line 2x + y + 6 = 0
So, it does not represent any figure.

Given,

The objective function,

\(Z=3 x+4 y\)

At O\( (0, 0) \),

\(Z=3 \times 0+4 \times 0=0\)

At A\( (52, 0) \),

\(Z=3 \times 52+4 \times 0=156\)

\(x+2 y=76\) .......(i)

\(2 x+y=104\).......(ii)

By multiplying equation (i) by 2, we get

\(2 x+4 y=152\)

\(2 x+y=104\)

\(3 y=48\)

\(y=16\)

By putting the value of \(y\) in equation (i), we get

\(x+2 \times 16=76\)

\(x=76-32\)

\(x=44\)

At E \( (44, 16) \),

\(Z=3 \times 44+4 \times 16=196\)

At D\( (0, 38) \),

\(Z=3 \times 0+4 \times 38=152\)

So,\(Z\) is maximum at \((44,16)\) and its maximum value is \(196 .\)