Mathematics Test - 12

We have to find find probability when, P-hits, Q-hits

R-does not hits=\(PQ ^{\prime} R ^{\prime}+ P ^{\prime} QR R ^{\prime}+ PQR\)

\(=\frac{3}{4} \times \frac{1}{2} \times \frac{3}{8}+\frac{1}{4} \times \frac{1}{2} \times \frac{3}{8}+\frac{3}{4} \times \frac{1}{2} \times \frac{3}{8}\)

\(=\frac{9+3+9}{64}=\frac{21}{64}\).

The given equation \(x^{4}-2 x^{2} \sin ^{2} \frac{\pi x}{2}+1=0\)

On solving, we get-

\(\left(x^{2}-\sin ^{2} \frac{\pi x}{2}\right)^{2}+1-\sin ^{4} \frac{\pi x}{2}=0\)

So, \(\left(x^{2}-\sin ^{2} \frac{\pi x}{2}\right)^{2}=0\) and \(1-\sin ^{4} \frac{\pi x}{2}=0\)

So, \(1-\sin ^{4} \frac{\pi x}{2}=0 \Rightarrow \frac{\pi x}{2}=(2 n+1) \frac{\pi}{2}\)

So, \(x=2 n+1\)

\(\left(x^{2}-\sin ^{2} \frac{\pi x}{2}\right)^{2}=0\)

\((2 n+1)^{2}-1=0\)

\(n=0,-1\)

\(x=1,-1\)

We know that,\(|\vec{a}-\vec{b}|^{2}=(\vec{a}-\vec{b}) \cdot(\vec{a}-\vec{b})\)

\(=\vec{a} \cdot \vec{a}-\vec{a} \cdot \vec{b}-\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}\)

\(=|\vec{a}|^{2}-2(\vec{a} \cdot \vec{b})+|\vec{b}|^{2}\)

\(=(2)^{2}-2(4)+(3)^{2}\)

So, \(|\vec{a}-\vec{b}|=\sqrt{5}\)

From figure, we have

\(P=5, O Q=6\)

and \(O M=\frac{5}{2}, C M=3\)

\(\therefore\) ln \(\Delta O M C, O C^{2}=O M^{2}+M C^{2}\)

\(\Rightarrow  O C^{2}=\left(\frac{5}{2}\right)^{2}+(3)^{2} \Rightarrow O C=\frac{\sqrt{61}}{2}\)

Thus, the required circle has its centre \(\left(\frac{5}{2}, 3\right)\)

and radius \(\frac{\sqrt{61}}{2}\).

So, its equation is \(\left(x-\frac{5}{2}\right)+(y-3)^{2}=\left(\frac{61}{4}\right)\).

Thus, \(\lambda=\frac{61}{4}\)

The system of linear non-homogeneous equations in matrix form is written as:

\(A X=B\)

Let, A be a \(3 \times 3\) matrix:

\(A =\left[\begin{array}{lll} a _{11} & a _{12} & a _{13} \\ a _{21} & a _{22} & a _{23} \\ a _{31} & a _{32} & a _{33}\end{array}\right], X =\left[\begin{array}{l} x _{1} \\ x _{2} \\ x _{3}\end{array}\right], B =\left[\begin{array}{l} b _{1} \\ b _{2} \\ b _{3}\end{array}\right]\)

\(a_{11} x_{1}+a_{12} x_{2}+a_{13} x_{3}=b_{1}\)

\(a_{21} x_{1}+a_{22} x_{2}+a_{23} x_{3}=b_{2}\)

\(a_{31} x_{1}+a_{32} x_{2}+a_{33} x_{3}=b_{3}\)

The set of values \(x_{1}, x_{2}, x_{3}\) which satisfies the above equations are called the solutions of the system.

\(\left[\begin{array}{ccc}2 & -1 & 3 \\ 1 & 1 & 1 \\ 1 & -1 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}9 \\ 6 \\ 2\end{array}\right]\)

\(2 x-y+3 z=9 \quad \quad \ldots\)(1)

\(x+y+z=6 \quad \quad \ldots\)(2)

\(x-y+z=2 \quad \quad \ldots\)(3)

On solving equation (1), (2), and (3) using the elimination method we will get the values

\(x = 1, y = 2, z = 3\)

Given:

\(\frac{dx}{dy}+\frac{y}{x}=0\)

On solving, we get-

\(\Rightarrow xdx=ydy\)

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

\(x^{2}+y^{2}=2 c\) or

\(x^{2}+y^{2}=k\)

The given problem can be written as,

\({ }^{n} C_{r}+{ }^{n} C_{r+1}+{ }^{n} C_{r+1}+{ }^{n} C_{r+2}\)

We know that:

\({ }^{n} C_{r}=\frac{n !}{r !(n-r) !}\)

Therefore,

\({ }^{n} C_{r}+{ }^{n} C_{r+1}+{ }^{n} C_{r+1}+{ }^{n} C_{r+2}\)

\(=\frac{{n} !}{{r} !({n}-{r})!}+\frac{{n} !}{({r}+1) !({n}-{r}-1) ! }+\frac{{n} !}{({r}+1) !({n}-{r}-1) ! }+\frac{{n} !}{({r}+2) !({n}-{r}-2) ! }\)

\(=\frac{{n} !}{{r} !({n}-{r}-1) ! }\left[\frac{1}{({n}-{r})}+\frac{1}{({r}+1)}\right]+\frac{{n} !}{({r}+1) !({n}-{r}-2) ! }\left[\frac{1}{({n}-{r}-1)}+\frac{1}{({r}+2)}\right]\)

\(=\frac{{n} !}{{r} !({n}-{r}-1) ! }\left[\frac{{r}+1+{n}-{r}}{({n}-{r})({r}+1)}\right]+\frac{{n} !}{({r}+1) !({n}-{r}-2) ! }\left[\frac{{r}+2+{n}-{r}-1}{({n}-{r}-1)({r}+2)}\right]\)

\(=\frac{{n} ! \times({n}+1)}{{r} ! \times({r}+1) \times({n}-{r}-1) \times({n}-{r})}+\frac{{n} !}{({r}+1) !({n}-{r}-2) !} \times \frac{({n}+1)}{({n}-{r}-1)({r}+2)}\)

\(=\frac{({n}+1) !}{({r}+1) !({n}-{r}) !}+\frac{{n} ! \times(\mathbf{n}+1)}{({r}+2)({r}+1) \times({n}-{r}-1)({n}-{r}-2)}\)

\(=\frac{({n}+1) !}{({r}+1) !({n}-{r}) !}+\frac{({n}+1) !}{({r}+2) !({n}-{r}-1) !}\)

\(=\frac{({n}+1) !}{({r}+1) !({n}-{r}-1) !}\left[\frac{1}{{n}-{r}}+\frac{1}{{r}+2}\right]\)

\(=\frac{({n}+1) !}{({r}+1) !({n}-{r}-1) !}\left[\frac{{n}-{r}+{r}+2}{({r}+2)({n}-{r})}\right]\)

\(=\frac{({n}+2)({n}+1) !}{({r}+2)({r}+1) !({n}-{r})({n}-{r}-1) !}\)

\(=\frac{({n}+2) !}{({r}+2) !({n}-{r}) !}\)

\(=^{{n}+2} {C}_{{r}+2}\)

\(=C(n+2, r+2)\)

mean \(\mu=n p=8\) ...(1)

variance \(\sigma^{2}=\mathrm{npq}=4\) ..(2)

Dividing equation (2) by (1), we get

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

As we know, \(p+q=1\)

\(\Rightarrow p=1-q=\frac{1}{2}\)

Put the value of \(n\) in equation (1), we get

\(n=16\)

Now,

\(P(x=1)={ }^{16} C_{1}\left(\frac{1}{2}\right)^{1} \times\left(\frac{1}{2}\right)^{16-1}\)

\(\Rightarrow 16 \times \frac{1}{2^{16}}\)

\(\Rightarrow 2^{4} \times \frac{1}{2^{16}}\)

\(\therefore \frac{1}{2^{12}}\)

The given equation is,

$=(\cos 2p\pi +i\sin 2p\pi )(\cos 2q\pi +i\sin 2q\pi )$

$=\cos 2(p+q)\pi +i\sin 2(p+q)\pi$

$=(\cos \pi +i\sin \pi {)}^{2(p+q)}$

$=(-1+0{)}^{2(p+q)}$

$=(-1{)}^{2(p+q)}=1$

Given:

\(\cos ^{4} x+\sin ^{4} x=2 \cos (2 x+\pi) \cos (2 x-\pi)\)

On solving, we get-

\(\Rightarrow\left(\cos ^{2} x+\sin ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x\)

\(=\cos 4 x+\cos 2 \pi\)

\(\Rightarrow 1-\frac{1}{2} \sin ^{2} 2 x=\cos 4 x+1\)

\(\Rightarrow-\frac{1}{2}\left(\frac{1-\cos 4 x}{2}\right)=\cos 4 x\)

\(\Rightarrow-\frac{1}{4}=\frac{3}{4} \cos 4 x\)

\(\Rightarrow \cos 4 x=-\frac{1}{3}\)

\(\Rightarrow \sin 4 x=\frac{2 \sqrt{2}}{3}\)

Again \(\cos 4 x=-\frac{1}{3}\)

\(\Rightarrow 1-2 \sin ^{2} 2 x=-\frac{1}{3}\)

\(\Rightarrow \sin 2 x=\sqrt{\frac{2}{3}}\)