Mathematics Test-20

Given,

\(n^{\text {th }}\) terms of two A.P's are \(3 n+8\) and \(7 n+15\).

So, \(12^{\text {th }}\) term in \(1^{\text {st }} \mathrm{AP}\):

a_n= 3 n + 8

For n = 12,

\(a _{12}= 3 \times 12+8\)

\(=36+8=44\)

Similarly, \(12^{\text {th }}\) term of \(2^{\text {nd }}\) AP:

a_n = 7n + 15

For n = 12,

\(a_{12}=7 \times 12+15\)

\(=84+15=99\)

\(\therefore\) The required ratio \(=\frac{44 }{ 99}\)

\(=\frac{4 } 9=4: 9\)

If \({z}=\frac{{z}_{1}}{{z}_{2}}\) so, modulus of \(|{z}|=\frac{\left|{z}_{1}\right|}{\left|{z}_{2}\right|}\)

\(z=\frac{1+i}{1+\sqrt{3} \mathrm{i}}\)

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

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

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

Given, \(XM= 3\) units, \(MB= 3\) units and\(XM \perp AB\)

As we know that, perpendicular line on a chord from the centre bisect it.

Therefore, \(AM=MB= 3\) units

Given \(Y=\left[\begin{array}{ll}3 & 2 \\ 1 & 4\end{array}\right]\)

And \(2 \mathrm{X}+\mathrm{Y}=\left[\begin{array}{cc}1 & 0 \\ -3 & 2\end{array}\right]\)

\(\Rightarrow 2 \mathrm{X}+\left[\begin{array}{ll}3 & 2 \\ 1 & 4\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ -3 & 2\end{array}\right]\)

\(\Rightarrow 2 \mathrm{X}=\left[\begin{array}{cc}1 & 0 \\ -3 & 2\end{array}\right]-\left[\begin{array}{cc}3 & 2 \\ 1 & 4\end{array}\right]\)

\(\Rightarrow 2 X=\left[\begin{array}{ll}-2 & -2 \\ -4 & -2\end{array}\right]\)

\(\Rightarrow X=\frac{1}{2}\left[\begin{array}{ll}-2 & -2 \\ -4 & -2\end{array}\right]\)

\(\therefore X=\left[\begin{array}{ll}-1 & -1 \\ -2 & -1\end{array}\right]\)

\(1+i\left[(1-1)-(1+i)^2\right]-(1-i)\left[(1-i)^2-i(1+i)\right]+\)

\(1\left[(1-i)(1+i)-i^2\right]\)

\(=(1+i)\left[1-1^2-2 i\right]-(1-i)\left[1+1^2-2 i-i-1^2\right]+\left[1-1^2-1^2\right]\)

\(=(1+i)\left[-2 i^2-2 i\right]-(1-i)(1-3 i)+\left[1-2 i^2\right]\)

\(=(1+i)[+2-2 i]-(1-i)(1-3 i)-1\)

\(2\left[1-1^2\right]-(1-i)(1-3 i)-1\)

\(4-\left(1-3 i-i+3 i^2\right)-1\)

\(3-(1-4 i-3)\)

\(3-(-2-4 i)\)

\(=5+4 i\)

Given,

\(\vec{b}_1=\hat{i}+2 \hat{j}+2 \hat{k}\)

\(\vec{b}_2=3 \hat{i}+2 \hat{j}+6 \hat{k}\)

The angle \(\theta\) between the two lines is given by,

\(\cos \theta=\left|\frac{\vec{b}_1 \cdot \vec{b}_2}{\left|\overrightarrow{b_1}\right|\left|\vec{b}_2\right|}\right|\)

\(=\left|\frac{(\hat{i}+2 \hat{j}+2 \hat{k}) \cdot(3 \hat{i}+2 \hat{j}+6 \hat{k})}{\sqrt{1+4+4} \sqrt{9+4+36}}\right|\)

\(=\left|\frac{3+4+12}{3 \times 7}\right|=\frac{19}{21} \)

\( \theta =\cos ^{-1}\left(\frac{19}{21}\right)\)

Given:

\(\cos \alpha=\frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}\)

Using the half-angle formula \(\cos 2 \theta=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\) and the fact that \(c=2 \mathrm{~b}-\mathrm{a},\) we get;

\(\Rightarrow \frac{1-\tan ^{2} \frac{\alpha}{2}}{1+\tan ^{2} \frac{\alpha}{2}}=\frac{\mathrm{a}}{3 \mathrm{~b}-\mathrm{a}}\)

\(\Rightarrow \frac{2 \tan ^{2} \frac{\alpha}{2}}{2}=\frac{(3 \mathrm{~b}-\mathrm{a})-\mathrm{a}}{(3 \mathrm{~b}-\mathrm{a})+\mathrm{a}}\)

\(\Rightarrow \tan ^{2} \frac{\alpha}{2}=\frac{3 \mathrm{~b}-2 \mathrm{a}}{3 \mathrm{~b}}\)

Similarly, \(\tan ^{2} \frac{\gamma}{2}=\frac{2 \mathrm{a}-\mathrm{b}}{3 \mathrm{~b}}\)

Now, \(\tan ^{2} \frac{\alpha}{2}+\tan ^{2} \frac{\gamma}{2}\)

\(=\frac{3 \mathrm{~b}-2 \mathrm{a}}{3 \mathrm{~b}}+\frac{2 \mathrm{a}-\mathrm{b}}{3 \mathrm{~b}}\)

\(=\frac{2 \mathrm{~b}}{3 \mathrm{~b}}\)

\(=\frac{2}{3}\)

Given:

(2 - i) (x - iy) = 3 + 4i

⇒ 2x - 2iy - ix + i²y = 3 + 4i

⇒ 2x - 2iy - ix - y = 3 + 4i (∵ i² = -1)

⇒ (2x – y) + i(-x - 2y) = 3 + 4i

Equating real and imaginary parts,

2x - y = 3 ----(1)

-x - 2y = 4 ----(2)

Solving equation (1) and (2), we get

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

Now, the value of 5x can be calculated as:

5x = 5 × \(\frac{2}{5}\) = 2

We know that,

\(P(E)=\frac{n(E)}{n(S)}\)

Where \(n(E)=\) Number of a favorable outcome

\({n}({S})=\) Total number of outcome

Now,

We know that in a simultaneous throw of two dice,

\(n(S)=6 \times 6=36\)

Let \(E=\) event of getting a total of \(7\)

\(\Rightarrow\{(1,6),(2,5),(3,4),(5,2),(6,1)\}\)

\(P(E)=\frac{n(E)}{n(S)}\)

\(\Rightarrow \frac6{36}=\frac1 6\)

\(\therefore\) The probability of getting a total of \(7\) is \(\frac16\).

Given:\(x^{2} d y+y^{2} d x=0\)

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

\(\Rightarrow \frac{\mathrm{dy}}{\mathrm{y}^{2}}=-\frac{\mathrm{dx}}{\mathrm{x}^{2}}\)

Integrating both sides, we get

\(\int \frac{\mathrm{dy}}{\mathrm{y}^{2}}=-\int \frac{\mathrm{dx}}{\mathrm{x}^{2}}\)

\(\frac{-1}{\mathrm{y}}=-\frac{-1}{\mathrm{x}}+\mathrm{c}\)

\(\frac{-1}{\mathrm{y}}=\frac{1}{\mathrm{x}}+\mathrm{c}\)

\(\frac{1}{x}+\frac{1}{y}=-c\)

\(\mathrm{x}+\mathrm{y}=-\mathrm{cxy}\)

Here \(c\) is differential constant, take \(-c=\frac{1}{c}\) (Because \(\frac{1}{c}\) is also a constant)

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