Mathematics Test-11

Given:

\(\frac{\partial z}{\partial x} e^{y}=\frac{\partial z}{\partial y} e^{x}\)

In another form, the equation will be:

\(p e^{y}=q e^{x}\)

\(\Rightarrow \frac{p}{e^{x}}=\frac{q}{e^{y}}\)

Let \(p e^{-x}=q e^{-y}=a\)

\(\Rightarrow p=a e^{x} ; q=a e^{y} ;\)

The complete solution will be:

\(z=\int p d x+\int q d y\)

\(\Rightarrow z=\int a e^{x} d x+\int a e^{y} d y\)

\(\Rightarrow z=a e^{x}+a e^{y}+b\)

Where a, b are arbitrary constants.

Given lines are \(\frac{2 \mathrm{x}-2}{2 \mathrm{k}}=\frac{4-\mathrm{y}}{3}=\frac{\mathrm{z}+2}{-1}\) and \(\frac{\mathrm{x}-5}{1}=\frac{\mathrm{y}}{\mathrm{k}}=\frac{\mathrm{z}+6}{4}\)

Write the above equation of a line in the standard form of lines

\(\Rightarrow \frac{2(\mathrm{x}-1)}{2 \mathrm{k}}=\frac{-(\mathrm{y}-4)}{3}=\frac{\mathrm{z}+2}{-1} \Leftrightarrow \frac{(\mathrm{x}-1)}{\mathrm{k}}=\frac{\mathrm{y}-4}{-3}=\frac{\mathrm{z}+2}{-1}\)

So, the direction ratio of the first line is \((k,-3,-1)\)

\(\frac{\mathrm{x}-5}{1}=\frac{\mathrm{y}}{\mathrm{k}}=\frac{\mathrm{z}+6}{4}\)

So, direction ratio of second line is \((1, k, 4)\)

Lines are perpendicular,

\(\therefore(\mathrm{k} \times 1)+(-3 \times \mathrm{k})+(-1 \times 4)=0\)

\(\Rightarrow \mathrm{k}-3 \mathrm{k}-4=0\)

\(\Rightarrow-2 \mathrm{k}-4=0\)

\(\therefore \mathrm{k}=-2\)

Given \(A=\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right]\), and \(f(x)=x^{2}-2 x-3\)

The matrix A is also satisfying the above polynomial.

\(\Rightarrow f(A)=A^{2}-2 A-3\)

\(\Rightarrow f(A)=A A-2 A-3I\)

\(\Rightarrow f(A)=\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right]-2\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right]-3\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\)

\(\Rightarrow f(A)=\left[\begin{array}{ll}5 & 4 \\ 4 & 5\end{array}\right]-\left[\begin{array}{ll}2 & 4 \\ 4 & 2\end{array}\right]-\left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right]\)

\(\Rightarrow f(A)=\left[\begin{array}{ll}5 & 4 \\ 4 & 5\end{array}\right]-\left[\begin{array}{ll}5 & 4 \\ 4 & 5\end{array}\right]\)

\(\Rightarrow f(A)=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]\)

We have to find the value of \((\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})\)

\(\Rightarrow(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})=\vec{a} \times \vec{a}-\vec{a} \times \vec{b}+\vec{b} \times \vec{a}-\vec{b} \times \vec{b}\)

We know that,

\(\vec{a} \times \vec{b}=-\vec{b} \times \vec{a}\)

\(\Rightarrow(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})=0-\vec{a} \times \vec{b}-\vec{a} \times \vec{b}-0\)

\(\because(\vec{a} \times \vec{a}=\vec{b} \times \vec{b}=0)\)

\(\Rightarrow(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})=-2(\vec{a} \times \vec{b})\)

Given:

Mean of a set of observations \(x_{1}, x_{2}, x_{3}, \ldots, x_{10}\) is 50.

Here \(n=10\),

Mean \(=\frac{\sum x_{i}}{n}\)

\(=\frac{x_{1}+x_{2}+x_{3}+\ldots+x_{10}}{10}=50\)

\(\Rightarrow x_1+x_2+x_3+\ldots+x_{10}=500\) …. (1)

Now find out the mean of new observations

\(x_{1}+5, x_{2}+10, x_{3}+15, \ldots, x_{10}+50\)

Mean \(=\frac{\sum x_{i}}{n}\)

\(=\frac{\left(x_{1}+5\right)+\left(x_{2}+10\right)+\left(x_{3}+15\right)+\ldots+\left(x_{10}+50\right)}{10}\)

\(=\frac{\left(x_{1}+x_{2}+x_{3}+\ldots+x_{10}\right)+(5+10+15+\ldots+50)}{10}\)

From equation (1),

\(=\frac{500+5(1+2+3+\ldots+10)}{10}\)

\(=50+\frac{1}{2} \times \frac{10 \times 11}{2}\)

\(=50+27.5\)

\(=77.5\)

Let \(\mathrm{X}-\) axis intercept of line be \(\mathrm{A}(\mathrm{a}, 0)\) and \(\mathrm{Y}-\) axis intercept of line be \((0, \mathrm{b})\).

Hence equation of line will be,

\(\frac{x}{a}+\frac{y}{b}=1\)

According to question \(\mathrm{P}(1,2)\) is mid point of \(\mathrm{A}(\mathrm{a}, 0)\) and \(\mathrm{B}(0, \mathrm{b})\)

\(\Rightarrow \frac{a+0}{2}=1\) and \(\frac{0+b}{2}=2\)

\(\Rightarrow \mathrm{a}=2\) and \(\mathrm{b}=4\)

Hence equation of line will be:

\(\frac{x}{2}+\frac{y}{4}=1\)

Multiplying each term by 4 we get:

\(2 \mathrm{x}+\mathrm{y}=4\)

The sum of \(n\) terms \(=\frac{n}{2}(\) First term \(+n\)th term \()\)

Put \(n=1\), we get \(a_1=5(1)+1=6\)

Sum of \(n\) terms \(=\frac{n}{2}(\) First term \(+nth\) term)

\(\Rightarrow\) Sum of \(n\) terms \(=\frac{n}{2}(6+5 n+1)\)

\(\Rightarrow\) Sum of \(n\) terms \(=\frac{n}{2}(7+5 n)\)

\(I=\int \frac{d x}{x(x+4)}\)

\(=\frac{1}{4} \int \frac{4 \mathrm{dx}}{\mathrm{x}(\mathrm{x}+4)}\)

\(=\frac{1}{4} \int \frac{(\mathrm{x}+4-\mathrm{x}) \mathrm{dx}}{\mathrm{x}(\mathrm{x}+4)}\)

\(=\frac{1}{4}\left[\int \frac{x+4}{x(x+4)} d x-\int \frac{x}{x(x+4)} d x\right]\)

\(=\frac{1}{4}\left[\int \frac{\mathrm{dx}}{\mathrm{x}}-\int \frac{\mathrm{dx}}{(\mathrm{x}+4)}\right]\)

As we know, \(\int \frac{\mathrm{dx}}{\mathrm{x}}=\log |\mathrm{x}|+\mathrm{c}\) where \(\mathrm{c}\) is a constant

\(I=\frac{1}{4}[\log x-\log (x+4)]+c\)

\(=\frac{1}{4} \log \left|\frac{\mathrm{x}}{\mathrm{x}+4}\right|+\mathrm{c}\)\(\left(\because \log m-\log n=\log \frac{m}{n}\right)\)

Given:

\(\cos ^{2} \theta=\frac{(x+y)^{2}}{4 x y}\)

We know that,

Max value of \(\cos ^{2} \theta=1\)

Therefore,

\(1=\frac{(x+y)^{2}}{4 x y}\)

\(\Rightarrow 4 x y=(x+y)^{2}\)

\(\Rightarrow 4 x y=x^{2}+y^{2}+2 x y\)

\(\Rightarrow 0=x^{2}+y^{2}-2 x y\)

\(\Rightarrow 0=(x-y)^{2}\)

\(\Rightarrow 0=x-y\)

\(\Rightarrow x=y\)

Given:

\(\frac{x^{2}}{16}-\frac{y^{2}}{9}=1\)

Compare with standard equation, we get

\(a^{2}=16\) and \(b^{2}=9\)

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

\(=\sqrt{1+\frac{9}{16}}=\)

\(=\sqrt{\frac{16+9}{16}}=\sqrt{\frac{25}{16}}=\frac{5}{4}\)

Now, Equation of directrix

\(x=\pm \frac{a}{e}\)

\(=\pm \frac{4}{\left(\frac{5}{4}\right)}=\pm \frac{16}{5}\)

Let,

\(\mathrm{S}=(1+\mathrm{x})^{21}+(1+\mathrm{x})^{22}+(1+\mathrm{x})^{23}+\ldots(1+\mathrm{x})^{30} \ldots(\mathrm{i})\)

\((1+x) \mathrm{S}=(1+\mathrm{x})^{22}+(1+\mathrm{x})^{23} \ldots(1+\mathrm{x})^{30}+(1+\mathrm{x})^{31} \ldots(\mathrm{ii})\)

Subtracting (i) from (ii), we get

\(\mathrm{xS}=(1+\mathrm{x})^{31}-(1+\mathrm{x})^{21}\)

\(\mathrm{S}=\frac{(1+\mathrm{x})^{31}}{\mathrm{x}}-\frac{(1+\mathrm{x})^{21}}{\mathrm{x}}\)

Coefficient of \(\mathrm{x}^{\mathrm{r}}\)

\(={ }^{31} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{\mathrm{r}-1}-{ }^{21} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{\mathrm{r}-1}\)

\(=\left({ }^{31} \mathrm{C}_{\mathrm{r}}-{ }^{21} \mathrm{C}_{\mathrm{r}}\right) \mathrm{x}^{\mathrm{r}-1} \ldots(\mathrm{a})\)

Therefore, the coefficient of \(\mathrm{x}^{5}\) implies

\(\Rightarrow\mathrm{r}-1=5\)

\(\Rightarrow\mathrm{r}=6\)

Substituting in (a), we get coefficient of \(\mathrm{x}^{5}\)

\(=\left({ }^{31} \mathrm{C}_{6}-{ }^{21} \mathrm{C}_{6}\right)\)

If the \(1, \omega\) and \(\omega^{2}\) are the cube roots of unity, then\(1+\omega+\omega^{2}=0\)

If any row and column have all the elements as zero in the square matrix, then the determinant is zero.

\(\mathrm{D}=\left|\begin{array}{ccc}1 & \omega & \omega^{2} \\ \omega & \omega^{2} & 1 \\ \omega^{2} & 1 & \omega\end{array}\right|\)

\(\mathrm{R}_{3}=\mathrm{R}_{3}+\mathrm{R}_{1}+\mathrm{R}_{2}\)

\(\mathrm{D}=\left|\begin{array}{ccc}1 & \omega & 1+\omega+\omega^{2} \\ \omega & \omega^{2} & 1+\omega+\omega^{2} \\ \omega^{2} & 1 & 1+\omega+\omega^{2}\end{array}\right|\)

\(\mathrm{D}=\left|\begin{array}{ccc}1 & \omega & 0 \\ \omega & \omega^{2} & 0 \\ \omega^{2} & 1 & 0\end{array}\right|=0\)

Given:

The probability of getting plumbing contract \(=P(A)=\frac{2}{3}\)

Probability of getting not electric contract \(P(\bar{B})=\frac{5}{9}\)

Probability of getting atleast one contract \(P(A \cup B)=\frac{4}{5}\)

The probability of not getting electric contract \(=1-P(\bar{B})\)

\(=1-\frac{5}{9}=\frac{4}{9}\)

Probability of getting both the contracts \(P(A \cap B)=P(A)+P(B)-P(A \cup B)\)

\(\Rightarrow P(A \cap B)=\frac{2}{3}+\frac{4}{9}-\frac{4}{5}\)

\(= \frac{(90+60-108)}{135}\)

\(\therefore\) The probability that he will get both the contracts \(P(A \cap B)\) is \(\frac{14}{ 45}\).