Mathematics Test - 19

The number of ways to select \(r\) things out of \(n\) things is given by \({ }^{n} C_{r}\)

\({ }^{{n}} {C}_{{r}}=\frac{{n} !}{({n}-{r}) ! \times({r}) !}=\frac{{n} \times({n}-1) \times \ldots({n}-{r}+1)}{{r} !}\)

\({ }^{{n}} {C}_{{r}}={ }^{{n}} {C}_{{n}-{r}}\)

Here, the number of points \(=12\)

Octagon is formed by 8 points.

\(\therefore\) Number of octagon \(=\) Number of ways of selecting 8 points out of 12

\(={ }^{12} {C}_{8}\)

\(={ }^{12} {C}_{4} \quad\left(\because{ }^{{n}} {C}_{{r}}={ }^{{n}} {C}_{{n}-{r}}\right)\)

\(=\frac{12 !}{4 ! 8 !}\) \(=495\)

Given, \(\alpha\) and \(\beta\) are the roots of the equation \(x^{2}-2 x+4=0\).

\(\Rightarrow \alpha+\beta=2\) and \(\alpha \beta=4\)

Consider, \(\alpha^{3}+\beta^{3}=(\alpha+\beta)\left(\alpha^{2}+\beta^{2}-\alpha \beta\right)\)

\(\Rightarrow \alpha^{3}+\beta^{3}=(\alpha+\beta)\left(\alpha^{2}+\beta^{2}+2 \alpha \beta-3 a \beta\right)\)

\(\left.\Rightarrow \alpha^{3}+\beta^{3}=(\alpha+\beta)\left[(\alpha+\beta)^{2}-3 a \beta\right)\right]\)

\(\Rightarrow \alpha^{3}+\beta^{3}=(2)\left[(2)^{2}-3(4)\right]\)

\(\Rightarrow \alpha^{3}+\beta^{3}=(2)(-8)\)

\(\Rightarrow \alpha^{3}+\beta^{3}=-16\)

So, if \(\alpha\) and \(\beta\) are the roots of the equation \(x^{2}-2 x+4=0\), then the value of \(\alpha^{3}+\beta^{3}=-16\).

The given equation is,

$\frac{{}^{15}{\mathrm{C}}_r}{{}^{15}{\mathrm{C}}_{r-1}}=\frac{11}{5}$

$\Rightarrow \frac{\frac{15!}{r!(15-r)!}}{\frac{15!}{(r-1)!(15-r+1)!}}=\frac{11}{5}$

$\Rightarrow \frac{(r-1)!(16-r)!}{r!(15-r)!}=\frac{11}{5}$

$\Rightarrow \frac{(r-1)!(16-r)!(15-r)!}{r(r-1)!(15-r)!}=\frac{11}{5}$

$\Rightarrow \frac{16-r}{r}=\frac{11}{5}$

$\begin{array}{rl}\Rightarrow 86-5r& =11r\end{array}$

$\Rightarrow 16r=80$

$\Rightarrow r=\frac{80}{16}=5$

Given: \(|\overrightarrow{\mathrm{a}}|=10,|\overrightarrow{\mathrm{b}}|=2\) and \(\overrightarrow{\mathrm{a}}. \overrightarrow{\mathrm{b}}=12\)

As we know, \(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=|\overrightarrow{\mathrm{a}}| \cdot|\overrightarrow{\mathrm{b}}| \cos \theta\)

\(\Rightarrow 12=10 \times 2 \times \cos \theta\)

\(\Rightarrow \cos \theta=\frac{12}{20}=\frac{3}{5}\)

As we know, \(\sin \theta=\sqrt{1-\cos ^{2} \theta}=\sqrt{1-\left(\frac{3}{5}\right)^{2}}=\frac{4}{5}\)

\(|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{a}}| \cdot|\overrightarrow{\mathrm{b}}| \sin \theta\)

\(\Rightarrow|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=10 \times 2 \times \frac{4}{5}=16\)

We know that,

\(\cot (A-B)=\frac{\cot A \times \cot B+1}{\cot B-\cot A}\)

It is given that,

\(\tan A-\tan B=x\) and \(\cot B-\cot A=y\)

\(\because \cot B-\cot A=y\)

\(\Rightarrow \cot B-\cot A=\frac{1}{\tan B}-\frac{1}{\tan A}=\frac{\tan A-\tan B}{\tan B \times \tan A}=\frac{x}{\tan B \times \tan A}=y\)

\(\Rightarrow \frac{x}{y}=\tan A \times \tan B\)

\(\Rightarrow \frac{y}{x}=\cot A \times \cot B\)

As we know that, \(\cot (A-B)=\frac{\cot A \times \cot B+1}{\cot B-\cot A}\)

\(\Rightarrow \cot (A-B)=\frac{\frac{y}{x}+1}{y}=\frac{1}{x}+\frac{1}{y}\)

We know that:

The number of ways to select \(r\) things out of \(n\) things is given by \({ }^{{n}} C_{{r}}\).

\({ }^{{n}} {C}_{{r}}=\frac{{n} !}{({n}-{r}) ! \times({r}) !}=\frac{{n} \times({n}-1) \times \ldots({n}-{r}+1)}{{r} !}\)

Given that:

There are 13 points in a plane of which 5 are collinear.

We know that:

To form a line we have to select two points out of 13 points.

\(\therefore\) Number of lines\(={ }^{13} {C}_{2}\)

\(=\frac{13 \times 12}{2 \times 1}\)

\(=78\)

Also number of lines out of 5 points\(={ }^{5} {C}_{2}\)

\(=\frac{5 \times 4}{2 \times 1}\)

\(=10\)

But, these 5 points are collinear, and only one line can be formed out of these points.

∴ The total number of straight lines obtained by joining these points in pairs.

\(=78-10+1\)\(\quad\)(We add 1, as one line can be obtained out of 5 collinear points).

\(=69\)

Let \(A=x_{1}+i y_{1}\) and \(B=x_{2}+i y_{2}\)

If \(A=B\) then \(x_{1}=x_{2}\) and \(y_{1}=y_{2}\)

Given: \(\mathrm{A}+\mathrm{iB}=\frac{4+2 \mathrm{i}}{1-2 \mathrm{i}}\)

\(\Rightarrow \mathrm{A}+\mathrm{iB}=\frac{4+2 \mathrm{i}}{1-2 \mathrm{i}} \times \frac{1+2 \mathrm{i}}{1+2 \mathrm{i}}\)

\(\Rightarrow \mathrm{A}+\mathrm{iB}=\frac{4+10 \mathrm{i}+4 \mathrm{i}^{2}}{1-4 \mathrm{i}^{2}}\)

We know i² = -1

\(\Rightarrow \mathrm{A}+\mathrm{i} \mathrm{B}=\frac{4+10 \mathrm{i}-4}{1+4}\)

\(\Rightarrow \mathrm{A}+\mathrm{iB}=\frac{10 \mathrm{i}}{5}\)

\(\Rightarrow \mathrm{A}+\mathrm{iB}=2 \mathrm{i}\)

\(\Rightarrow \mathrm{A}+\mathrm{iB}=0+2 \mathrm{i}\)

Comparing real and imaginary parts, we get,

⇒ A = 0

Let, \(\overrightarrow{\mathrm{OA}}=-\hat{\mathrm{i}}+\frac{1}{2} \hat{\mathrm{j}}+4 \hat{\mathrm{k}}, \overrightarrow{\mathrm{OB}}=\hat{\mathrm{i}}+\frac{1}{2} \hat{\mathrm{j}}+4 \hat{\mathrm{k}}\)

\(\overrightarrow{\mathrm{OC}}=\hat{\mathrm{i}}-\frac{1}{2} \hat{\mathrm{j}}+4 \hat{\mathrm{k}}\) and \(\overrightarrow{\mathrm{OD}}=-\hat{\mathrm{i}}-\frac{1}{2} \hat{\mathrm{j}}+4 \hat{\mathrm{k}}\)

\(\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=\left(\hat{i}+\frac{1}{2} \hat{\jmath}+4 \hat{k}\right)-\left(-\hat{i}+\frac{1}{2} \hat{j}+4 \hat{k}\right)\)

\(=2 \hat{i}\)

\(|\overrightarrow{A B}|=\sqrt{2^{2}}=2\)

\(\overrightarrow{B C}=\overrightarrow{O C}-\overrightarrow{O B}=\left(\hat{i}-\frac{1}{2} \hat{\jmath}+4 \hat{k}\right)-\left(-\hat{i}-\frac{1}{2} \hat{j}+4 \hat{k}\right)\)

\(=2 \hat{i}\)

\(|\overrightarrow{B C}|=\sqrt{2^{2}}=2\)

Area of rectangle, \(\mathrm{ABCD}=|\mathrm{AB}| \times|\mathrm{BC}|=2 \times 2\)

\(=4\) square units

\(\mathrm{R}=\{(a, b): a=b-2, b>6\}\)
Now, Since \(b>6,(2,4) \notin \mathrm{R}\).
Also, as \(3 \neq 8-2,\)
\( \therefore(3,8) \notin \mathrm{R}\) and, as \(8 \neq 7-2, \)
\(\therefore(8,7) \notin \mathrm{R}\)
Now, consider \((6,8) .\)We have \(8>6\) and also, \(6=8-2, \)\(\therefore(6,8) \in \mathrm{R}\)

Given curve is:

\(3 x^{2}-y^{2}=8\)

On differentiating, we get \(6 x-2 y \frac{d y}{d x}=0\)

\(\Rightarrow \frac{d y}{d x}=\frac{3 x}{y}\)

Slope of normal \(=\frac{-y}{3 x}\)

Slope of given line \(=\frac{-1}{3}\)

A.T.Q.

\(\frac{-1}{3}=\frac{-y}{3 x}\)

\(\Rightarrow y=x\)

Putting the value of y in above equation, we get \(3 x^{2}-y^{2}=8\)

We get \(x=2,-2\) and \(y=2,-2\)

Equation of normal

\(y-y_{1}=\frac{-1}{3}\left(x-x_{1}\right)\)

On solving, we get \(x+3 y+8=0\) and \(x+3 y-8=0\)