Mathematics Test - 5

\begin{equation}\begin{aligned}
&S . D . \sigma=\sqrt{n p q} \geq 0\\
&\text { Mean, } np =25 \text { and } q<1\\
&\therefore \sigma=\sqrt{n p q}<\sqrt{n p}\\
&\sigma<5\\
&\Rightarrow 0 \leq \sigma<5
\end{aligned}\end{equation}

\begin{equation}\begin{array}{l}
\sin \left(\sin ^{-1} \frac{1}{7}+\cos ^{-1} x\right)=1 \\
\Rightarrow \sin ^{-1} \frac{1}{7}+\cos ^{-1} x=\frac{\pi}{2} \\
\Rightarrow \sin ^{-1} \frac{1}{7}=\frac{\pi}{2}-\cos ^{-1} x \\
\Rightarrow \sin ^{-1} \frac{1}{7}=\sin ^{-1} x \\
\Rightarrow x=\frac{1}{7}
\end{array}\end{equation}

\((\sqrt{8}+i)^{50}=3^{49}(a+i b)\)
Taking magnitude and squaring on both sides, we get \((8+1)^{50}=3^{98}\left(a^{2}+b^{2}\right)\)
\(9^{50}=3^{98}\left(a^{2}+b^{2}\right)\)
\(3^{100}=3^{98}\left(a^{2}+b^{2}\right)\)
\(\Rightarrow a^{2}+b^{2}=9\)

We have
\(\frac{k+1}{k}+\frac{k+2}{k+1}=\frac{-b}{a} \ldots(1)\)
and \(\frac{k+1}{k} \cdot \frac{k+2}{k+1}=\frac{c}{a}\)
\(\Rightarrow \frac{k+2}{k}=\frac{c}{a}\)
or \(\frac{2}{k}=\frac{c}{a}-1=\frac{c-a}{a}\)
or \(k=\frac{2 a}{c-a}\)\(\ldots\)(2)
Now eliminate k putting the value of \(k\) in \(1^{\text {st }}\) relation, we get \(\frac{c+a}{2 a}+\frac{2 c}{c+a}=\frac{-b}{a}\)
\(\Rightarrow(c+a)^{2}+4 a c=-2 b(a+c)\)
\(\Rightarrow(a+c)^{2}+2 b(a+c)=-4 a c\)
Adding \(b^{2}\) on both sides, \((a+b+c)^{2}=b^{2}-4 a c\)

Doing \(R _{3} \rightarrow R _{3}- xR _{2}\) and
\(R _{2} \rightarrow R _{2}- x R _{1}\)
we get\(\quad,\) \(\begin{aligned} f(x) &=\left|\begin{array}{ccc}a & -1 & 0 \\ 0 & a+x & -1 \\ 0 & 0 & a+x\end{array}\right| \\ &=a(a+x)^{2} \end{aligned}\)
\(So\) \(f(2 x)-f(x)\)
\(=a\left[(a+2 x)^{2}-(a+x)^{2}\right]\)
\(=a(a+2 x-a-x)(a+2 x+a+x)\)
\(=a x(2 a+3 x)\)
So, \(f(2 x)-f(x)=\operatorname{ax}(2 a+3 x)\)
Thus, \(f(2 x)-f(x)\) is divisible by \(a, x\) and \((2 a+3 x)\).

\(\left(1+2 x+3 x^{2}+\ldots \ldots \ldots\right)^{-3 / 2}\)
\(=\left((1-x)^{-2}\right)^{-3 / 2}\)
\(=(1-x)^{3}=1-3 x+3 x^{2}-x^{3}\)
\(\therefore\) coefficient of \(x^{5}=0\)

The equation of the normal at \(\left(x_{1}, y_{1}\right)\) to the given ellipse is \(\frac{ a ^{2} x }{ x _{1}}-\frac{ b ^{2} y }{ y _{1}}= a ^{2}- b ^{2}\)
Here \(x _{1}=\) ae and \(y _{1}= b ^{2} / a\)
So the equation of the normal at positive end of the latus rectum is \(\frac{ a ^{2} x }{ ae }-\frac{ b ^{2} y }{\frac{ b ^{2}}{ a }}= a ^{2} e ^{2}\)
\(\left(\because b^{2}=a^{2}\left(1-e^{2}\right)\right)\)
\(\Rightarrow \frac{ ax }{ e }- ay = a ^{2} e ^{2}\)
\(\Rightarrow x-e y-e^{3} a=0\)

\begin{equation}\begin{array}{l}
B=\left[\begin{array}{cc}
0 & -1 \\
1 & 0
\end{array}\right], B^{2}=\left[\begin{array}{cc}
-1 & 0 \\
0 & -1
\end{array}\right] \text { and } \\
B^{4}=\left[\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}\right]=I \\
B^{1027}=\left(B^{4}\right)^{256} \cdot B^{3} \\
=(I)^{256} \cdot B^{3} \\
=I \cdot B^{3} \\
=B^{3} \\
=B \cdot B^{2} \\
=\left[\begin{array}{cc}
0 & -1 \\
1 & 0
\end{array}\right]\left[\begin{array}{cc}
-1 & 0 \\
0 & -1
\end{array}\right]=\left[\begin{array}{cc}
0 & 1 \\
-1 & 0
\end{array}\right]
\end{array}\end{equation}