Mathematics Test - 12

We have given three complex numbers \(3+4 {i}, 8-6 {i}\) and \(13+9 {i}\).

Let \(A, B, C\) represent complex number \(3+4 i, 8-6 i\) and \(13+9 i\).

\({AB}=|(3+4 {i})-(8-6 {i})|\)

\(A B=|3+4 i-8+6 i|\)

\({AB}=|-5+10 i|=\sqrt{(-5)^{2}+(10)^{2}}\)

\({AB}=\sqrt{25+100}=\sqrt{125}\)

\(B C=|(8-6 i)-(13+9 i)|\)

\({BC}=|8-6 {i}-13-9 {i}|=|-5-15 {i}|\)

\({BC}=\sqrt{(-5)^{2}+(-15)^{2}}\)

\({BC}=\sqrt{25+225}=\sqrt{250}\)

\({CA}=|(13+9 {i})-(3+4 {i})|\)

\({CA}=|13+9 {i}-3-4 {i}|=|10+5 {i}|\)

\({CA}=\sqrt{(10)^{2}+(5)^{2}}\)

\({CA}=\sqrt{100+25}=\sqrt{125}\)

\(\Rightarrow {BC}^{2}={AB}^{2}+{CA}^{2}\)

\(\Rightarrow\) One of the angle is \(90^{\circ}\).

Thus, nature of the triangle formed by the points representing complex number \(3+4 {i}, 8-6 {i}\) and \(13+9i\) is right angled triangle.

We are given a total of 7 men and a total of 6 women. Now, we have to select five persons to form a committee out of which at least 3 should be men, which means the minimum number of men is 3, therefore we get the following three cases in which the committee can be formed.

Let's take a look at the three cases:

Case I: Here, in the committee of 5 persons we have 3 men and 2 women, therefore the total number of ways that can be done is by selecting 3 men from 7 men and 2 women from 6 women. Now, for this we will apply the combination formula for selecting objects from a total of \(n\) objects: \({ }^{n} C_{r}=\frac{n !}{r !(n-r) !}\)

So, the ways in which we select 3 men from 7 men and 2 women from 6 women are as follows:

\(\Rightarrow{ }^{7} C_{3} \times{ }^{6} C_{2}=\left(\frac{7 !}{3 !(7-3) !}\right) \times\left(\frac{6 !}{2 !(6-2) !}\right)=35 \times 15=525 \quad\quad\ldots \ldots(1)\)

Case II: Here, in the committee of 5 persons we have 4 men and 1 woman, therefore the total number of ways that can be done is by selecting 4 men from 7 men and 1 woman from 6 women. Now, again we will apply the combination formula: \({ }^{n} C_{r}=\frac{n !}{r !(n-r) !}\)

So, the ways in which we select 3 men from 7 men and 1 woman from 6 women are as follows:

\(\Rightarrow{ }^{7} C_{4} \times{ }^{6} C_{1}=\left(\frac{7 !}{4 !(7-4) !}\right) \times\left(\frac{6 !}{1 !(6-1) !}\right)=35 \times 6=210 \quad\quad\ldots \ldots(2)\)

Case III: Here, in the committee of 5 persons we have 5 men and 0 women, therefore the total number of ways that can be done is by selecting 5 men from 7 men and 0 women from 6 women. Now, again we will apply the combination formula: \({ }^{n} C_{r}=\frac{n !}{r !(n-r) !}\)

So, the ways in which we select 5 men from 7 men is as follows:

\(\Rightarrow{ }^{7} C_{5}=\left(\frac{7 !}{5 !(7-5) !}\right)=21 \quad\quad\ldots \ldots(3)\)

Now, we will be adding all these cases to get the total number of ways in which five persons are selected to form a

committee so that at least three men are in the committee, therefore from equation 1,2, and \(3:\)

\(525+210+21=756\)

The last digit of the four whole numbers can be:

\(0, 1, 2, 3, 4, 5, 6, 7, 8, 9\)

The chance that any of the four numbers is divisible by 2 or \(5=\frac{6}{10}=\frac{3}{5}\)

Therefore,

The chance that any of the four numbers is not divisible by \(2\) or \(5=1-\frac{3}{5}\)

\(=\frac{2}{5}\)

So, the chance that all of the four numbers are divisible by \(2\) or \(5=\left(\frac{2}{5}\right) \times\left(\frac{2}{5}\right) \times\left(\frac{2}{5}\right) \times\left(\frac{2}{5}\right)\)

\(=\frac{16}{625}\)

This is the chance that the last digit in the product will not be \(0,2,4,5,6,8\) and this is also the chance that the last digit in the product is \(1,3,7\) or \(9\).

Given: \({i} z^{3}+{z}^{2}-{z}+{i}=0\)

\(\Rightarrow i z^{3}+z^{2}+\left(i^{2}\right) z+i=0 \quad\left(i^{2}=-1\right)\)

\(\Rightarrow {i} z^{3}+{i}^{2} {z}+{z}^{2}+{i}=0\)

\(\Rightarrow {iz}\left({z}^{2}+{i}\right)+1\left({z}^{2}+{i}\right)=0\)

\(\left(z^{2}+i\right)(i z+1)=0\)

\(\left(z^{2}+i\right) i(z-i)=0 \quad\left(-i^{2}=1\right)\)

\(z^{2}=-i\) or \(z=i\)

If \(z=i\) then \(|z|=|i|=1\)

If \(z^{2}=-i\) then \(\left|z^{2}\right|=|-i|=1 \)

\(\Rightarrow\left|z^{2}\right|=1 \)

\(\Rightarrow|z|=1\)

Therefore, if \({iz}^{3}+{z}^{2}-{z}+{i}=0\), then the value of \(|{z}|\) is \(1\).

Given: Mean of 100 observations is 50 and standard deviation is 10.

If 5 is added to each observation

∵ There are total 100 observations

500 is added to sum of the observations and now dividing it by the no. of observations, we get that the mean is increased by 5.

As we know that, the standard deviation of N observations is given by:

\(\sigma=\sqrt{\frac{1}{N} \times \sum_{i=1}^{N}\left(x_{i}-\mu\right)^{2}}\) where, \(\mu\) is the arithmetic mean

∵ Every observation increased by 5 and mean also increased by 5 and standard deviation is the square root of the difference of mean and each observation divided by the number of observations.

So, standard deviation will remain same.

Thus, new mean is 55 and new standard deviation is still 10.

The general term in the expansion of \(\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)^{11}\) is given by,

\(\mathrm{T}_{\mathrm{r}+1}=(-1)^{\mathrm{r}} \times{ }^{11} \mathrm{C}_{\mathrm{r}} \times \mathrm{x}^{11-\mathrm{r}} \times\left(\frac{1}{\mathrm{x}}\right)^{\mathrm{r}}\)

\(=(-1)^{\mathrm{r}} \times{ }^{11} \mathrm{C}_{\mathrm{r}} \times \mathrm{x}^{11-\mathrm{r}} \times \mathrm{x}^{-\mathrm{r}}\)

\(=(-1)^{\mathrm{r}} \times{ }^{11} \mathrm{C}_{\mathrm{r}} \times \mathrm{x}^{11-2 \mathrm{r}}\)

Here, \(n\) is 11 (i.e., odd), so there will be two middle terms.

\(\frac{1}{2}(\mathrm{n}+1)\) th term \(=6\) th term and \(\frac{1}{2}(\mathrm{n}+3)\) th term \(=7\) th term

\(\therefore \mathrm{T}_{6}=\mathrm{T}_{5+1}=(-1)^{5} \times{ }^{11} \mathrm{C}_{5} \times \mathrm{x}^{11-10}\)

\(=-\frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} \times \mathrm{x}\)

\(=-462 \mathrm{x}\)

And, \(\mathrm{T}_{7}=\mathrm{T}_{6+1}=(-1)^{6} \times{ }^{11} \mathrm{C}_{6} \times \mathrm{x}^{11-12}\)

\(=\frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} \times \mathrm{x}^{-1}\)

\(=\frac{462}{\mathrm{x}}\)

Given:

In 2008 A = Rs. 3.2, B = Rs. 5.3

Each year A is increased by Rs. 0.5 and B is increased by Rs. 0.25

Let after n year A will be Rs. 0.4 more than B.

\(\Rightarrow A + n(0.5) = B + n(0.25) + 0.4\)

\(\Rightarrow 3.2 + 0.5n = 5.3 + 0.25n + 0.4\)

\(\Rightarrow 0.5n - 0.25n = 5.7 - 3.2\)

\(\Rightarrow 0.25n = 2.5\)

\(\Rightarrow n = \frac{2.5}{0.25} = 10\) years

So, after \(10\) years from \(2008\) the year will be \(2018\).

It is given that: 

\(y=e^{x+e^{x+e^{x+\cdots \infty}}}\)

\(\therefore y=e^{x+\left(e^{x+e^{x+} \cdots \infty}\right)}=e^{x+y}\)

Differentiating both sides with respect to x and using the chain rule, we get:

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

\(\Rightarrow \frac{d y}{d x}=e^{x+y} \frac{d}{d x}(x+y)\)

\(\Rightarrow \frac{d y}{d x}=y\left(1+\frac{d y}{d x}\right)\)

\(\Rightarrow \frac{d y}{d x}=y+y \frac{d y}{d x}\)

\(\Rightarrow(1-y) \frac{d y}{d x}=y\)

\(\Rightarrow \frac{d y}{d x}=\frac{y}{1-y}\)

As we know,

The standard equation of a hyperbola:

\(\frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1\)

where \(2 a\) and \(2 b\) are the length of the transverse axis and conjugate axis respectively and centre \((h, k)\).

The center is the midpoint of the 2 foci.

The eccentricity \(=\frac{\sqrt{a^{2}+b^{2}}}{a}\)

Length of latus recta \(=\frac{2 b ^{2}}{ a }\)

Distance from center to focus \(=\sqrt{a^{2}+b^{2}}\)

Given,

Foci are \((5,0)\) and \((-3,0)\).

Center \(=\left(\frac{5+(-3)}{2}, \frac{0+0}{2}\right)=(1,0)\)

Now distance of focus from the center \(=\sqrt{a^{2}+b^{2}}\)

\(\Rightarrow \sqrt{(5-1)^{2}+(0-0)^{2}}=\sqrt{ a ^{2}+ b ^{2}} \)

\(\Rightarrow 16=a ^{2}+ b ^{2}\)...(i)

Eccentricity \(=2\)

\(\therefore \frac{\sqrt{a ^{2}+ b^{2}}}{ a }=2\)

\(\Rightarrow \frac{\sqrt{16}}{ a }=2\)

\(\Rightarrow \frac{4}{ a }=2\)

\(\Rightarrow a =2\)

\(\Rightarrow a^{2}=4\)

Putting the value of \(a\) in (i), we get

\( 4+b^{2}=16 \)

\(\Rightarrow b^{2}=12\)

The equation of the hyperbola is given by:

\(\frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1 \)

\(\Rightarrow \frac{(x-1)^{2}}{4}-\frac{(y-0)^{2}}{12}=1 \)

\(\Rightarrow \frac{(x-1)^{2}}{4}-\frac{y^{2}}{12}=1\)