Mathematics Test-30

We have given that,

\(A=\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right]\)

\(\Rightarrow A=2\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\)

\(\Rightarrow A=2 I\)

\(\Rightarrow A^5=2^5 I^5\)

\(\Rightarrow A^5=2^5 I\)

\(\Rightarrow A^5=2^4 \times 2 I\)

\(\Rightarrow A^5=2^4 \times 2\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\)

\(\Rightarrow A^5=16\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right]\)

\(\Rightarrow A^5=16 A\)

Two given line are,

\(\frac{x-2}{2}=\frac{y-1}{7}=\frac{z+3}{-3} \quad\dots\)(i)

\(\frac{x+2}{-1}=\frac{y-4}{2}=\frac{z-5}{4} \quad\ldots\)(ii)

Direction ratio of this two line \((2,7,-3)\) and \((-1,2,4)\) respectively.

we find the direction cosine \(\left(l_1, m_1, n_1\right)\) and \(\left(l_2, m_2, n_2\right)\),

\(l_1=\frac{2}{\sqrt{62}}, m_1=\frac{7}{\sqrt{62}}, n_1=\frac{-3}{\sqrt{62}}\)

And, \(l_2=\frac{-1}{\sqrt{21}}, m_2=\frac{2}{\sqrt{21}}, n_2=\frac{4}{\sqrt{21}}\)

If angle between the line (i) and (ii) is\(\theta\), then we know,

\(\cos \theta =l_1 l_2+m_1 m_2+n_1 n_2 \)

\(=\frac{-2}{\sqrt{62 \times 21}}+\frac{14}{\sqrt{62 \times 21}}-\frac{12}{\sqrt{62 \times 21}} \)

\(=\frac{14-14}{\sqrt{62 \times 21}}\)

\(=0=\cos \frac{\pi}{2}\)

So, \( \theta=\frac{\pi}{2}\)

The system of equations \(A X=0\) is said to be homogenous system of equations, then If \(|A| \neq 0\), then its solution \(X=0\), is called trivial solution.

If \(|A|=0\). Then \(A X=0\) has a non-trivial solution which means the system will have infinitely many solutions.

Given: \(2 x-3 y=0\) and \(2 x+a y=0\)

These equations can be written as: \(A X=B\) where,

\(A=\left[\begin{array}{cc}2 & -3 \\ 2 & \alpha\end{array}\right], X=\left[\begin{array}{l}x \\ y\end{array}\right]\) and \(B=\left[\begin{array}{l}0 \\ 0\end{array}\right]\)

As we know that, the given system is a homogenous system of equation. So, in order to say that the system has infinitely many solutions: \(|A|=0\).

\(| A |=2 \alpha+6=0\)

\(\Rightarrow \alpha=-3\)

\(\left|z_{1}^{2}-z_{2}^{2}\right|=\left|\bar{z}_{1}^{2}+\bar{z}_{2}^{2}-2 \bar{z}_{1} \bar{z}_{2}\right|\)

\(\left|z_{1}^{2}-z_{2}^{2}\right|=\left|z_{1}^{2}+z_{2}^{2}-2 z_{1} z_{2}\right|\)

\(\left|z_{1}+z_{2}\right|\left|z_{1}-z_{2}\right|=\left|z_{1}-z_{2}\right| \mid z_{1}-z_{2}\)

\(\Rightarrow\left|z_{1}+z_{2}\right|=\left|z_{1}-z_{2}\right|\)

\(\overline{z_{1}} \perp \overline{z_{2}}\)

\(\Rightarrow \arg \left(\frac{z_{1}}{z_{2}}\right)=2 n \pi \pm \frac{\pi}{2}\)

i.e. \(-\frac{\pi}{2}, \frac{\pi}{2}, \frac{3 \pi}{2}, \ldots \ldots \ldots\)

The minimum value of \(|a-b|=1\) (when \(a=1 \& b=2)\)

Given:

\(\left(1+3 x+3 x^{2}+x^{3}\right)^{2 n}\)

\(=(1+x)^{6 n}\)

Here, \(n\) is an even number.

So the middle terms is \((\frac{6 n}{2} { +1})=(3 n+1)^{\text {th }}\) term

So terms \((3 n+1)^{\text {th }}\) term is:

\(T_{3 n+1}={ }^{6 n} C_{3 n} x^{3 n} \)

\(=\frac{(6 n) !}{(3 n!)^{2}} x^{3 n}\)

Given,

\(r=1\) and angle \(=120^{\circ}\)

Length of chord \(=2 r \sin \left(\frac{\theta}{2}\right)\)

Where \(r\) is the radius of the circle and \(\theta\) is the angle from the center of the circle to the two points of the chord.

Now, Length of chord

\(=2(1) \sin \left(\frac{120^{\circ}}{2}\right) \)

\(=2 \sin \left(60^{\circ}\right) \)

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

\(=\sqrt{3} \text { units }\)