Mathematics Test - 23

Given: \({dy}=\left(4+{y}^{2}\right) {dx}\)

\(\Rightarrow \frac{{dy}}{4+{y}^{2}}={dx}\)

Integrating both sides, we get

\(\int \frac{{dy}}{2^{2}+{y}^{2}}=\int {dx}\)

\(\Rightarrow \frac{1}{2} \tan ^{-1} \frac{y}{2}=x+c\)

\(\Rightarrow \tan ^{-1} \frac{y}{2}=2 x+C\)

\(y=2 \tan (2 x+C)\)

The given differential equation is 

\(\frac{{dy}}{{dx}}=\frac{{x} \sqrt{1-{y}^{2}}}{{y} \sqrt{1-{x}^{2}}}\)

\(\frac{{ydy}}{\sqrt{1-{y}^{2}}}=\frac{{xdx}}{\sqrt{1-{x}^{2}}}\)

Integrating both the sides 

\(\int \frac{{ydy}}{\sqrt{1-{y}^{2}}}=\int \frac{{xdx}}{\sqrt{1-{x}^{2}}}\)

\(-\ln \sqrt{1-{y}^{2}}=-\ln \sqrt{1-{x}^{2}}+{c}\)

\(\ln \sqrt{1-{y}^{2}}-\ln \sqrt{1-{x}^{2}}+{c}=0\)

\(\ln \frac{\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}}=c \quad[\because \log m-\log n=\log (\frac{m}{n})]\)

\(\sqrt{\frac{1-y^{2}}{1-x^{2}}}=e^{c}=a\)

\(\frac{1-y^{2}}{1-x^{2}}=a^{2}=C\)

\(y^{2}-1=C\left(x^{2}-1\right)\)

Given: \(\sin ^{-1} \frac{4}{5}+\sec ^{-1} \frac{5}{4}-\frac{\pi}{2}\)

As we know that, sin^-1 (x) = cosec^-1 \((\frac{1}{x})\)

\(\Rightarrow \sin ^{-1} \frac{4}{5}+\sec ^{-1} \frac{5}{4}-\frac{\pi}{2}=\operatorname{cosec}^{-1}\left(\frac{5}{4}\right)+\sec ^{-1} \frac{5}{4}-\frac{\pi}{2}\)

As we know that, sec^-1 y + cosec^-1 y = \(\frac{\pi}{2}\), y ≤ - 1 or y ≥ 1

\(\Rightarrow \sin ^{-1} \frac{4}{5}+\sec ^{-1} \frac{5}{4}-\frac{\pi}{2}=\frac{\pi}{2}-\frac{\pi}{2}=0\)

As we know,

\(\sin x=y\) then \(x=\sin ^{-1} y\)

\(\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\)

Given,

\(\sin \left(\tan ^{-1} \frac{1}{10}+\cot ^{-1} {x}\right)=1\)

\(\Rightarrow \tan ^{-1} \frac{1}{10}+\cot ^{-1} {x}=\sin ^{-1}(1)\) \(\left(\because \sin ^{-1}(1)=\sin ^{-1} \left(\sin \left(\frac{\pi}{ 2}\right)\right)=\frac{\pi }{ 2}\right)\)

\(\Rightarrow \tan ^{-1} \frac{1}{10}+\cot ^{-1} x=\frac{\pi}{2}\)

Here, \(\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\)

Then \(x=\frac{1}{10}\)

Given:

\(\frac{d}{d x}\left(\frac{x^{4}+x^{2}+1}{x^{2}-x+1}\right)=a x+b\)__(1)

Now,

\(\left(x^{2}+x+1\right)\left(x^{2}-x+1\right)=\left(x^{2}+1\right)^{2}-x^{2} \quad\left[\because(a+b)(a-b)=a^{2}-b^{2}\right]\)

\(=x^{4}+2 x^{2}+1-x^{2}=x^{4}+x^{2}+1\)

So, factor of \(x^{4}+x^{2}+1\) is \(\left(x^{2}+x+1\right)\left(x^{2}-x+1\right)\)

\({LHS}=\frac{{d}}{{dx}}\left(\frac{{x}^{4}+{x}^{2}+1}{{x}^{2}-{x}+1}\right)\)

\(=\frac{\mathrm{d}}{\mathrm{dx}}\left[\frac{\left(\mathrm{x}^{2}+\mathrm{x}+1\right) \times\left(\mathrm{x}^{2}-\mathrm{x}+1\right)}{\left(\mathrm{x}^{2}-\mathrm{x}+1\right)}\right]=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}+\mathrm{x}+1\right)=2 \mathrm{x}+1\)

Compare with equation \(1^{\text {st }}\),

\(\therefore a=2\) and \(b=1\)

Let inclination of required line is \(\theta\),

So the coordinates of point \(B\) can be assumed as

\(\left(4-\frac{\sqrt{29}}{3} \cos \theta, 3-\frac{\sqrt{29}}{3} \sin \theta\right)\)

Which satisfices \(x-y-2=0\)

\(4-\frac{\sqrt{29}}{3} \cos \theta-3+\frac{\sqrt{29}}{3} \sin \theta-2=0 \)

\(\sin \theta-\cos \theta=\frac{3}{\sqrt{29}}\)

By squaring

\(\sin 2 \theta=\frac{20}{29}=\frac{2 \tan \theta}{1+\tan ^2 \theta} \)

\(\tan \theta=\frac{5}{2}\) only (because slope is greater than 1)

\(\sin \theta=\frac{5}{\sqrt{29}}, \cos \theta=\frac{2}{\sqrt{29}}\)

Point \(B:\left(\frac{10}{3}, \frac{4}{3}\right)\)

Which also satisfies \(x+2 y=6\)

Given: \(z^{2}+z+1=0\) means \(z=\omega\) or \(\omega^{2},\) where \(\omega\) represents complex cube root of \(1+i 0\)
From the properties of cube roots of \(1\), we know that: (\(1+\omega+\omega^{2}=0\)) And \(1 \times \omega \times \omega^{2}=1\)
\(\Rightarrow \omega \cdot \omega^{2}=1\) 
\(\Rightarrow\omega^{3}=1\) (\(\because\) roots of \(z^{2}+z+1=0\) are reciprocal of each other)
So, In \(\left(z+\frac{1}{z}\right)^{2}+\left(z^{2}+\frac{1}{z^{2}}\right)^{2}+\left(z^{3}+\frac{1}{z^{3}}\right)^{2}+\ldots .+\left(z^{6}+\frac{1}{z^{6}}\right)^{2}\) 
Putting \(z=\omega\) we get; \(\left(z+\frac{1}{z}\right)^{2}=\left(\omega+\frac{1}{\omega}\right)^{2}=\left(\omega+\omega^{2}\right)^{2}\)\(=(-1)^{2}=1\)
\(\left(z^{2}+\frac{1}{z^{2}}\right)^{2}=\left(\omega^{2}+\omega\right)^{2}\)\(=1\)
\(\left(z^{3}+\frac{1}{z^{3}}\right)^{2}=(1+1)^{2}\)\(=4\)
\(\left(z^{4}+\frac{1}{z^{4}}\right)^{2}=\left(\omega+\omega^{2}\right)^{2}\)\(=(-1)^{2}=1\)
\(\left(z^{3}+\frac{1}{z^{3}}\right)^{2}=\left(\omega^{2}+\omega\right)^{2}\)\(=1\)
\(\left(z^{6}+\frac{1}{z^{6}}\right)^{2}=(1+1)^{2}\)\(=4\)
Thus, adding them all, \(\left(z+\frac{1}{z}\right)^{2}+\left(z^{2}+\frac{1}{z^{2}}\right)^{2}+\left(z^{3}+\frac{1}{z^{3}}\right)^{2}+\ldots .+\left(z^{6}+\frac{1}{z^{6}}\right)^{2}\)\(=1+1+4+1+1+4\)\(=12\)

Given,

\(\angle B P C=\beta\)

Let \(A B=x\),

Then \(A P=2 A B=2 x\)

Now,

Triangle \({ABP}\) is right angled triangle with \({BP}\) as hypotenuse

Now, by Pythagoras theorem,

\(B P^2=A P^2+A B^2 \)

\(B P^2=(2 x)^2+x^2 \)

\(B P^2=5 x^2\)

Therefore,

\(B P=\sqrt{5} x\)

Now, according to question we are told that \({C}\) is mid-point of \({AB}\) therefore

\(A C=\frac{1}{2} A B\)

Thus \(A C=\frac{x}{2}\)

Now,

\(\tan \alpha=\frac{\left(\frac{x}{2}\right)}{(2 x)}=\frac{1}{4}\)

thus tan \(\alpha=\frac{1}{4}\)

Now, we know from figure that in triangle APB,

\(\tan (\alpha+\beta)=\frac{x}{2 x}=\frac{1}{2}\)

Therefore,

\(\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \times \tan \beta}=\frac{1}{2} \)

\(\Rightarrow 2(\tan \alpha+\tan \beta)=1-\tan \alpha \tan \beta \)

\(\Rightarrow 2\left(\frac{1}{4}+\tan \beta\right)=1-\frac{1}{4} \tan \beta \)

\(\Rightarrow \frac{9}{4} \tan \beta=\frac{1}{2} \)

\(\Rightarrow \tan \beta=\frac{2}{9}\)

Thus, the angle made by triangle CPB is nothing \(\tan \beta\) and is equal to \(\frac{2}{9}\).