Mathematics Test - 23

For quadratic equation \(a x^{2}+b x+c=0\), 

Sum of roots \(=\alpha+\beta=\frac{-b }{ a}\) 

Product of roots \(=\alpha \beta=\frac{c }{ a}\)

Given, \((5+\sqrt{2}) x^{2}-(4+\sqrt{5}) x+(8+2 \sqrt{5})=0\)

\(\therefore  \alpha \beta=\frac{(8+2 \sqrt{5}) }{(5+\sqrt{2})}\) and \(\alpha+\beta=\frac{(4+\sqrt{5}) }{(5+\sqrt{2})}\)

Now, \(\frac{2 \alpha \beta }{(a+\beta)}= \frac{\frac{2(8+2 \sqrt{5}) }{(5+\sqrt{2})} }{\frac{(4+\sqrt{5}) }{(5+\sqrt{2})}}\)

\(=\frac{2(8+2 \sqrt{5}) }{(5+\sqrt{2})} \times \frac{(5+\sqrt{2})}{(4+\sqrt{5})} \)

\(=\frac{2(8+2 \sqrt{5})}{(4+\sqrt{5})} \) \(=\frac{2×2(4+ \sqrt{5})}{(4+\sqrt{5})} \) \(=4\)

Any square matrix can be expressed as sum of symmetric and skew symmetric as.

\(A=\frac{1}{2}\left(A+A^T\right)+\frac{1}{2}\left(A-A^T\right)\)

Where the first term is symmetric and second term is skew-symmetric here \(S = P + Q\)

To find \(P\) i.e. the symmetric term is given by,

\(P=\frac{1}{2}\left(S+S^{ T }\right)\)

\(P=\frac{1}{2}\left(\left[\begin{array}{ll}1 & 6 \\ 7 & 2\end{array}\right]+\left[\begin{array}{ll}1 & 7 \\ 6 & 2\end{array}\right]\right)\)

\(P=\frac{1}{2}\left[\begin{array}{cc}2 & 13 \\ 13 & 4\end{array}\right]\)

\(P=\left[\begin{array}{cc}1 & \frac{13}{2} \\ \frac{13}{2} & 2\end{array}\right]\)

The right-angled triangle follow the Pythagoras theorem,

\(\frac{\sin 15}{0.5}=\frac{\sin 75}{R}\)

\(\Rightarrow \frac{\sqrt{3}-1}{(2 \sqrt{2}) 0.5}=\frac{\sqrt{6}+\sqrt{2}}{4 R}\Rightarrow \frac{2(\sqrt{3}-1)}{(2 \sqrt{2})}=\frac{\sqrt{6}+\sqrt{2}}{4 R}\)\(\Rightarrow \frac{(\sqrt{3}-1)}{(\sqrt{2})}=\frac{\sqrt{6}+\sqrt{2}}{4 R}\)

\(\Rightarrow R=\frac{\sqrt{6}+\sqrt{2}}{4} \times \frac{\sqrt{2}}{\sqrt{3}-1}\)

\(\Rightarrow D=2 R=\frac{\sqrt{6}+\sqrt{2}}{4} \times \frac{2 \sqrt{2}}{\sqrt{3}-1} \ldots \ldots . \text { (Diameter } {D}=2 {R})\)

\(\Rightarrow D=\frac{\sqrt{6}+\sqrt{2}}{2} \times \frac{\sqrt{2}}{\sqrt{3}-1} \Rightarrow \frac{\sqrt{6}+\sqrt{2}}{\sqrt{2} \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{3}-1} \)

\(\Rightarrow D=\frac{\sqrt{6}+\sqrt{2}}{\sqrt{2}} \times \frac{1}{\sqrt{3}-1} \)

\( \Rightarrow D=\frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}} \)

\(\Rightarrow D=\frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}} \times \frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}+\sqrt{2}} \)

\(\Rightarrow D=2+\sqrt{3} \)

Equation of line

\(\begin{aligned} & 3 y-2 z-1=0=3 x-z+4 \\ & \Rightarrow \frac{3 y-1}{2}=\frac{z-0}{1}=\frac{3 x+4}{1} \\ & \Rightarrow \frac{x+\frac{4}{3}}{1 / 3}=\frac{y-\frac{1}{3}}{2 / 3}=\frac{z-0}{1}\end{aligned}\)

\(\begin{aligned} & \mathrm{PR}=|\mathrm{PQ}| \cos \theta=|\mathrm{PQ}| \frac{\mathrm{PQ} \cdot \mathrm{P}}{|\mathrm{PQ}||\mathrm{P}|}=\frac{\mathrm{PQ} \cdot \mathrm{PQ}}{|\mathrm{PR}|} \\ & \mathrm{PR}=\left|\frac{\frac{1}{3}\left(2+\frac{4}{3}\right)+\frac{2}{3}\left(-1-\frac{1}{3}\right)+1(6-0)}{\sqrt{\frac{1}{9}+\frac{4}{9}+1}}\right|=4 \sqrt{\frac{14}{9}} \\ & \mathrm{OR}^2=\mathrm{PQ}^2-\mathrm{PR}^2 \\ & =\frac{100}{9}+\frac{16}{9}+36-\frac{224}{9} \\ & =\frac{100}{9}+\frac{16}{9}+36-\frac{224}{9} \\ & \mathrm{QR}=\sqrt{24}=2 \sqrt{6}\end{aligned}\)

If the function is continuous at \(x=0\), then

\(\underset{x \rightarrow 0}\lim f(x)\) will exist and \(f(0)=\underset{x \rightarrow 0}\lim f(x)\)

Now, \(\underset{x \rightarrow 0}\lim  f(x)=\underset{x \rightarrow 0}\lim \left(\frac{1}{x}-\frac{k-1}{e^{2 x}-1}\right)\)

\(=\underset{x \rightarrow 0}\lim \left(\frac{e^{2 x}-1-k x+x}{(x)\left(e^{2 x}-1\right)}\right) \)

\(=\underset{x \rightarrow 0}\lim \left[\frac{\left(1+2 x+\frac{(2 x)^2}{2!}+\frac{(2 x)^3}{3!}+\ldots .\right)-1-k x+x}{(x)\left(\left(1+2 x+\frac{(2 x)^2}{2!}+\frac{(2 x)^3}{3!}+\ldots . .\right)-1\right)}\right] \)

\( =\underset{x \rightarrow 0}\lim\left[\frac{(3-k) x+\frac{4 x^2}{2!}+\frac{8 x^3}{3!}+\ldots . .}{\left(2 x^2+\frac{4 x^3}{2!}+\frac{8 x^3}{3!}+\ldots .\right)}\right]\)

For the limit to exist, power of \(\mathrm{x}\) in the numerator should be greater than or equal to the power of \(\mathrm{x}\) in the denominator. Therefore, coefficient of \(\mathrm{x}\) in numerator is equal to zero

\( \Rightarrow 3-\mathrm{k}=0 \)

\(\Rightarrow \mathrm{k}=3\)

So the limit reduces to

\(\underset{x \rightarrow 0}\lim  \frac{\left(x^2\right)\left(\frac{4}{2!}+\frac{8 x}{3!}+\ldots\right)}{\left(x^2\right)\left(2+4 x 2!+\frac{8 x^2}{3!}+\ldots\right)} \)

\(=\underset{x \rightarrow 0}\lim  \frac{\frac{4}{2!}+\frac{8 x}{3!}+\ldots \cdot}{2+\frac{4 x}{2!}+\frac{8 x^2}{3!}+\ldots . .}=1\)

Hence \(f(0)=1\)

Given,

\(\sin ^{-1} {x}+\cos ^{-1} {y}=\frac{2 \pi}{5}\)

We know that, \(\sin ^{-1} {x}+\cos ^{-1} {x}=\frac{\pi}{2}\)

\(\sin ^{-1} x=\frac{\pi}{2}-\cos ^{-1} x\)....(i)

Similarly,

\(\sin ^{-1} {y}+\cos ^{-1} {y}=\frac{\pi}{2}\)

\(\cos ^{-1} {y}=\frac{\pi}{2}-\sin ^{-1} {y}\)...(i)

Now,

\(\sin ^{-1} {x}+\cos ^{-1} {y}=\frac{2 \pi}{5}\)

From equation (i) and (ii) we get,

\(\Rightarrow \frac{\pi}{2}-\cos ^{-1} {x}+\frac{\pi}{2}-\sin ^{-1} {y}=\frac{2 \pi}{5}\)

\(\Rightarrow \cos ^{-1} {x}+\sin ^{-1} {y}=\frac{3 \pi}{5}\)

Given,

\(\int_{1}^{3}|x-2| d x\)

Now,

\({f}({x})=|x-2| \)

\(\Rightarrow f(x)=\left\{\begin{array}{c}-(x-2), \quad 1 \leq x<2 \\ x-2, \quad 2 \leq x \leq 3\end{array}\right.\)

\(\int_{1}^{3}|x-2| d x=\int_{1}^{2}-({x}-2) d x+\int_{2}^{3}({x}-2) d x\)

\(\int_{1}^{3}|x-2| d x=\int_{1}^{2}(2-x) d x+\int_{2}^{3}({x}-2) d x\)

\(=\left[2 x-\frac{x^{2}}{2}\right]_{1}^{2}+\left[\frac{x^{2}}{2}-2 x\right]_{2}^{3}\)

\(=\left[\left(4-\frac{4}{2}\right)-\left(2-\frac{1}{2}\right)\right]+\left[\left(\frac{9}{2}-6\right)-\left(\frac{4}{2}-4\right)\right]\)

\(=2-\frac{3}{2}-\frac{3}{2}+2\)

\(=4-3=1\)

Let \(\alpha\) and \(\beta\) are roots of \(a x^{2}+b x+c=0\), then \(\alpha \times \beta=\frac{c}{a}\).

Given,

\(x^{2}+x+2=0\)

Comparing with \(a x^{2}+b x+c=0\), we get

\(a=1, b=1, c=2\)

Product of roots \(=\alpha \times \beta=\frac{c}{a}=\frac{2}{1}=2\)

Now,

\(\frac{\alpha^{10}+\beta^{10}}{\alpha^{-10}+\beta^{-10}}=\frac{\alpha^{10}+\beta^{10}}{\frac{1}{\alpha^{10}}+\frac{1}{\beta^{10}}} \)

\(=\frac{\alpha^{10}+\beta^{10}}{\frac{\beta^{10}+\alpha^{10}}{\alpha^{10} \cdot \beta^{10}}} \)

\(=\alpha^{10} \cdot \beta^{10} \)

\(=(\alpha \cdot \beta)^{10} \)

\(=(2)^{10} \)

\(=1024\)

Given,

Complex number is \(z=\frac{5}{2}+\frac{5 \sqrt{3}}{2} i\)

\(r \cos \theta=\frac{5}{2}, r \sin \theta=\frac{5 \sqrt{3}}{2}\)

By squaring and adding, we get -

\(\mathrm{r}^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=\frac{100}{4}=25\)

\(\therefore r=5\)

\(\Rightarrow \cos \theta=\frac{\frac{5}{2}}{r}=\frac{\frac{5}{2}}{5}=\frac{1}{2}\) and \(\sin \theta=\frac{\frac{5 \sqrt{3}}{2}}{r}=\frac{\frac{5 \sqrt{3}}{2}}{5}=\frac{\sqrt{3}}{2}\)

Since it is in first quadrant, \(\theta=\frac{\pi}{3}\)

So, on comparing with \(z=r(\cos \theta+i \sin \theta)\), we can write as:

\(5\left(\cos \left(\frac{\pi}{3}\right)+i \sin \left(\frac{\pi}{3}\right)\right)\)

Given,

\(6 \sin ^{2} x-2 \cos ^{2} x=4\)

\(\Rightarrow 6 \sin ^{2} x-2 \cos ^{2} x=4 \times 1\)

As we know that,

\(\sin ^{2} x+\cos ^{2} x=1\)

\(\Rightarrow 6 \sin ^{2} x-2 \cos ^{2} x=4\left(\sin ^{2} x+\cos ^{2} x\right)\)

\(\Rightarrow 6 \sin ^{2} x-2 \cos ^{2} x=4 \sin ^{2} x+4 \cos ^{2} x\)

\(\Rightarrow 6 \sin ^{2} x-4 \sin ^{2} x=4 \cos ^{2} x+2 \cos ^{2} x\)

\(\Rightarrow 2 \sin ^{2} x=6 \cos ^{2} x\)

\(\Rightarrow \tan ^{2} x=3\)

\(\therefore \tan x=\sqrt{3}\)