Mathematics Test - 57

\(\cup \equiv+(O R)\)

\(\cap \equiv \cdot(A N D)\)

\(\left(A \cap B^{\prime}\right) \cup\left(A^{\prime} \cap B\right) \cup\left(A^{\prime} \cap B^{\prime}\right)\)

Or we can write it as,

\(=A B^{\prime}+A^{\prime} B+A^{\prime} B^{\prime}\) \(=A B^{\prime}+A^{\prime}\left(B+B^{\prime}\right)\) \(=A B^{\prime}+A^{\prime}\left[B+B^{\prime}=1\right]\) \(=A^{\prime}+B^{\prime} \quad\left[A s, A+A^{\prime} B=A+B\right] \quad-\) absorption law.

Again, write it in result from i.e. \(A^{\prime} \cup B^{\prime}\)

As we know,

Elementary row or column transformations do not change the value of the determinant of a matrix.

Given,

\(\left|\begin{array}{ccc}x & y & 3 \\ x^{2} & 5 y^{3} & 9 \\ x^{3} & 10 y^{3} & 27\end{array}\right|\)

Apply \(C_{1} \rightarrow C_{1}-C_{3}\)

\(=\left|\begin{array}{ccc}x-3 & y & 3 \\ x^{2}-9 & 5 y^{3} & 9 \\ x^{3}-27 & 10 y^{3} & 27\end{array}\right|\)

\(=\left|\begin{array}{ccc}x-3 & y & 3 \\ (\mathrm{x}-3)(x+3) & 5 y^{3} & 9 \\ (\mathrm{x}-3)\left(x^{2}+9+3 x\right) & 10 y^{3} & 27\end{array}\right|\)

\(=(x-3)\left|\begin{array}{ccc}1 & y & 3 \\ (x+3) & 5 y^{3} & 9 \\ \left(x^{2}+9+3 x\right) & 10 y^{3} & 27\end{array}\right|\)

\(\therefore(x-3)\) is a factors of the expansion of the determinant.

Given equation is \(y - b = a \sin x\)

There are two constants \(a\) and \(b\) so differentiate two times

Differentiating w.r.t \(x\)

\(y^{\prime}=a \cos x\)...(1)

Again differentiating w.r.t \(x\)

\(y^{\prime \prime}=-a \sin x\)

From equation (1), \(a =\frac{ y ^{\prime}}{\cos x }\)

\(y^{\prime \prime}=-\frac{y^{\prime}}{\cos x} \times \sin x \)

\(y^{\prime \prime}+y^{\prime} \tan x=0\)

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

\(\Rightarrow \frac{d y}{d x}=1+\frac{y}{x}+\left(\frac{y}{x}\right)^2\)

Substituting \(y=v x\) and \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)

\( \Rightarrow v + x \frac{ dv }{ dx }=1+ v + v ^2 \)

\( \Rightarrow x \frac{ dv }{ dx }=1+ v ^2\)

Integrating both sides we get,

\(\int \frac{ dx }{ x }=\int \frac{ dv }{1+ v ^2}\)

\(\Rightarrow \log x=\tan ^{-1} v + c , c =\) constant of integration

Putting the value of \(v\) we get,

\(\therefore \log x =\tan ^{-1} \frac{ y }{ x }+ c\)

Given:

1. \(\overline{\left( z ^{-1}\right)}=(\overline{ z })^{-1}\)

Let,\(z=a-i b\)

\(\Rightarrow z ^{-1}=\frac{1}{ z }=\frac{1}{ a - ib }=\frac{ a + ib }{ a ^{2}+ b ^{2}}\)

\(\Rightarrow \overline{ z ^{-1}}=\frac{ a - ib }{ a ^{2}+ b ^{2}} \quad \ldots\) (1)

\(\Rightarrow \overline{ Z }= a + ib\)

\(\Rightarrow(\overline{ z })^{-1}=\frac{1}{\overline{ z }}=\frac{1}{ a + ib }=\frac{ a - ib }{ a ^{2}+ b ^{2}} \quad \ldots\) (2)

\(\Rightarrow(\overline{ Z })^{-1}=\frac{1}{\bar{z}}=\frac{1}{a+ i b}=\frac{a- ib }{a^{2}+b^{2}}\)

from equation (1) and (2)

statement 1 is correct.

2. \(z z^{-1}=|z|^{2}\)

consider \(z=a-i b\)

\(\Rightarrow z^{-1}=\frac{a+i b}{a^{2}+b^{2}}\)

\(\Rightarrow z \cdot z^{-1}=(a-i b) \frac{a+i b}{a^{2}+b^{2}}=\frac{a^{2}+b^{2}}{a^{2}+b^{2}}=1\)

Statement 2 is not correct.

We have to find the sum of the series 2 + 6 + 18 + 54 +....+ 4374.

Here,

a = 2

r = 3

Let, a_n = 4374

As we know that, the the general term of a GP is given by

\(a_{n}=a r^{n-1}\)

\(\Rightarrow 4374=(2) \cdot(3)^{n-1}\)

\(\Rightarrow 3^{n-1}=2187\)

\(\Rightarrow 3^{n-1}=3^{7}\)

\(\therefore n-1=7\)

n = 8

As we know that, Sum of \(n\) terms of GP,

\(\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{a}\left(\mathrm{r}^{\mathrm{n}}-1\right)}{\mathrm{r}-1} \)

where \(r>1\)

\(\therefore S_{n}=\frac{2\left(3^{8}-1\right)}{3-1}\)

\(=3^{8}-1\)

= 6560

Here, we have to find the equation of the plane passing through the intersection of the planes \(\vec{r} \cdot(\hat{i}+3 \hat{j}-\hat{k})=0\) and \(\vec{r} \cdot(\hat{j}+2 \hat{k})=0\) and passing through the point \((2,1,-1) .\)

As we know that, the equation of the plane through the intersection of two planes \(\vec{r} \cdot \overrightarrow{n_{1}}=q_{1}\) and \(\vec{r} \cdot \overrightarrow{n_{2}}=q_{2}\) is given by \(\vec{r} \cdot\left(\overrightarrow{n_{1}}+\lambda \overrightarrow{n_{2}}\right)=q_{1}+\lambda q_{2}\)

Here, \(\overrightarrow{n_{1}}=\hat{i}+3 \hat{j}-\hat{k}, \overrightarrow{n_{2}}=\hat{j}+2 \hat{k}, q_{1}=0\) and \(q_{2}=0\)

So, the equation of the required plane is:

\(\vec{r} \cdot[\hat{i}+(3+\lambda) \hat{j}+(-1+2 \lambda) \hat{k}]=0 \cdot \cdot \cdot \cdot \cdot \cdot (1)\)

\(\because\) The plane represented by (1) passes through the point (2,1,-1) So, \(\vec{r}=2 \hat{i}+\hat{j}-\hat{k}\) will satisfy the equation (1) \(\Rightarrow(2 \hat{i}+\hat{j}-\hat{k}) \cdot(\hat{i}+(3+\lambda) \hat{j}+(-1+2 \lambda) \hat{k})=0\)

\(\Rightarrow 2+3+\lambda+1-2 \lambda=0\)

\(\Rightarrow \lambda=6\)

By substituting \(\lambda=6\) in equation (1) we get, \(\vec{r} \cdot(\hat{i}+9 \hat{j}+11 \hat{k})=0\)

Hence, the equation of the required plane is \(\vec{r} \cdot(\hat{i}+9 \hat{j}+11 \hat{k})=0\)

Given,

\(\operatorname{cosec}^{2} \cot ^{-1}\left(\frac{5}{12}\right)\)

Let \(\theta=\cot ^{-1}\left(\frac{5}{12}\right)\)

\(\Rightarrow \cot \theta=\frac{5}{12}\)

Therefore,

\(\operatorname{cosec}^{2} \cot ^{-1}\left(\frac{5}{12}\right)=\operatorname{cosec}^{2} \theta\)

\(=1+\cot ^{2} \theta\)\(\quad\quad(\because 1+\cot ^{2} \theta=\operatorname{cosec}^{2} \theta)\)

\(=1+\left(\frac{5}{12}\right)^{2}\)

\(=\frac{169}{144}\)

Given data is \(170,430,300,600\) and 470

Mean \(\bar{x}=\frac{\text { Sum of all the observations }}{\text { Total number of observations }}\)

\(=\frac{170+430+300+600+470}{5}\)

\(=394\)

For '\(n\)' observation \(x_{1}, x_{2} \ldots \ldots \ldots \ldots . . x_{n}\), the mean deviation about their mean \(\bar{x}\) is given by,

\(M . D=\frac{\sum_{i=1}^{n}\left|x_{i}-\bar{x}\right|}{N}\) where, \(N\) is the number of observations.

Mean deviation \(=\frac{|170-394|+|430-394|+|300-394|+|600-394|+|470-394|}{5}\)

\(=\frac{636}{5}\)

\(=127.2\)