Mathematics Test - 4

\begin{equation}\begin{array}{l}
\text { We have } \frac{3}{4}-\frac{5}{4^{2}}+\frac{3}{4^{3}}-\frac{5}{4^{4}}+\frac{3}{4^{5}}-\frac{5}{4^{6}}+\ldots \\
\text { Sum }=\left(\frac{3}{4}+\frac{3}{4^{3}}+\frac{3}{4^{5}}+\ldots . to \times 0\right)-\left(\frac{5}{4^{2}}+\frac{5}{4^{4}}+\frac{5}{4^{6}}+\ldots t 0 \infty\right) \\
=\frac{3}{1-\left(\frac{1}{4}\right)^{2}-\frac{5}{4^{2}}}{1-\left(\frac{1}{4}\right)^{2}} \\
=\frac{3}{4} \times \frac{16}{15}-\frac{5}{16} \times \frac{16}{15}=\frac{7}{15}
\end{array}\end{equation}

Let any point \(P=(a, \beta)\). Any tangent to the parabola is \(y=m x+1 / m\) and if it passes through \((a, \beta)\) Then, \(\beta=m \alpha+\frac{1}{m}\)
\(\Rightarrow m ^{2} \alpha- m \beta+1=0\)
Its roots are \(m _{1}, Sm _{1}\). So \(m _{1+} Sm _{1}= B / a\)
and \(m _{1} \cdot Sm _{1}=1 / a\)
\(\Rightarrow 5\left(\frac{\beta}{b \alpha}\right)^{2}=\frac{1}{\alpha}\)
\(\Rightarrow 5 \beta^{2}=36 \alpha\)
Thus the locus is \(5 y^{2}=36 x,\) which is a parabola.

Sides are in the ratio \(1: \sqrt{3}: 2\),
Let a \(=k, b=\sqrt{3} k, c=2 k\)
\(\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{3 k^{2}+4 k^{2}-k^{2}}{2 \cdot \sqrt{3} k \cdot 2 k}=\frac{\sqrt{3}}{2}\)
\(A=\frac{\pi}{6}\)
\(\cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c}=\frac{k^{2}+4 k^{2}-3 k^{2}}{2 \cdot k \cdot 2 k}=\frac{1}{2}\)
\(B=\frac{\pi}{3}\)
\(C=\pi-(A+B)=\frac{\pi}{2}\)

Limit is of the form \(1^{\infty},\) so we have
\(\lim _{x \rightarrow 0}\left(\frac{a^{x}+b^{x}+c^{x}}{3}\right)^{\frac{2}{x}}\)
\(=\lim _{e^{x} \rightarrow 0}\left(\frac{a^{x}+b^{x}+c^{x}}{3}-1\right) \frac{2}{x}\)
Now, \(\lim _{x \rightarrow 0}\left(\frac{a^{x}+b^{x}+c^{x}}{3}-1\right) \frac{2}{x}\)
\(=\lim _{x \rightarrow 0}\left(\frac{a^{x}+b^{x}+c^{x}-3}{3 x}\right)^{2}\)
\(=\lim _{x \rightarrow 0}\left(\frac{a^{x}-1}{x}+\frac{b^{x}-1}{x}+\frac{c^{x}-1}{x}\right) \frac{2}{3}\)
\(=\left\{\lim _{x \rightarrow 0} \frac{a^{x}-1}{x}+\lim _{x \rightarrow 0} \frac{b^{x}-1}{x}+\lim _{x \rightarrow 0} \frac{c^{x}-1}{x}\right\} \frac{2}{3}\)
\(=(\ln a+\ln b+\ln c) 2 / 3\)
\(=2 / 3(\ln abc )\)
\(\therefore\) Desired limit \(=e^{\frac{2}{3} \ln a b c}=e^{\ln (a b c)^{2 / 3}}=(a b c)^{2 / 3}\)

\(y=x^{2}, y=\left|x^{2}-2\right|\)

Shaded area \(=\int_{1 / 2}^{1}\left(2-x^{2}-x^{2}\right) d x\)
\(=\int_{1 / 2}^{1}\left(2-2 x^{2}\right) d x\)
\(=2 \int_{1 / 2}^{1}\left(1-x^{2}\right) d x\)
\(=2\left[x-\frac{x^{3}}{3}\right]_{1 / 2}^{1}\)
\(=\frac{5}{12}\) Square units

Hence option D is correct.

\(\frac{d y}{d x}=\frac{y}{2 y \ln y+y-x}\)
\(\Rightarrow \frac{d x}{d y}=2 \ln y+1-\frac{x}{y}\)
\(\Rightarrow \frac{d x}{d y}+\frac{x}{y}=2 \ln y+1\)
\(I F=e^{\int \frac{1}{y} d y}=e^{\ln y}=y\)
\(\Rightarrow d(x y)=(2 y \ln y+y) d y\)
Integrating both sides
\(\Rightarrow \int d(x y)=\int(2 y \ln y+y) d y\)
\(\Rightarrow x y=y^{2} \ln y-\frac{y^{2}}{2}+\frac{y^{2}}{2}+C\)
\(\Rightarrow x y=y^{2} \ln y+C\)
\(\Rightarrow\) Putting \(x =3\) and \(y =4,\) we get
\(\Rightarrow 12=16 \ln 4+C\)
\(\Rightarrow C=12-32 \ln 2\)
\(\Rightarrow\) Thus we get the solution as
\(\Rightarrow x y=y^{2} \ln y+12-32 \ln 2\)

\begin{equation}\begin{array}{l}
\text { Let } y=\sum_{r=1}^{89} \log _{10} \tan \frac{\pi r}{180} \\
\Rightarrow y=\log _{10}\left[\prod_{r=1}^{89} \tan \frac{\pi r}{180}\right] \\
\Rightarrow y=\log _{10}\left[\tan \frac{\pi}{180} \tan \frac{2 \pi}{180} \tan \frac{3 \pi}{180} \ldots \ldots \tan \frac{87 \pi}{180} \tan \frac{88 \pi}{180} \tan \frac{89 \pi}{180}\right] \\
\Rightarrow y=\log _{10}\left[\tan \frac{\pi}{180} \tan \frac{2 \pi}{180} \tan \frac{3 \pi}{180} \ldots \ldots \cot \left(\frac{\pi}{2}-\frac{87 \pi}{180}\right) \cot \left(\frac{\pi}{2}-\frac{88 \pi}{180}\right) \cot \left(\frac{\pi}{2}-\frac{89 \pi}{180}\right)\right] \\
\Rightarrow y=\log _{10}\left[\tan \frac{\pi}{180} \tan \frac{2 \pi}{180} \tan \frac{3 \pi}{180} \ldots \ldots \cot \frac{3 \pi}{180} \cot \frac{2 \pi}{180} \cot \frac{\pi}{180}\right] \\
\Rightarrow y=\log _{10} 1=0 \\\end{array}\end{equation}

\((4)^{\log _{9} 3}+(9)^{\log _{2} 4}=(10)^{\log _{x} 83}\)
Taking L.H.S.
\((4)^{\log _{9} 3}+(9)^{\log _{2} 4}\)
\(=2^{2 \log _{9} 3}+3^{2 \log _{2} 4}\)
\(=2^{\log _{9} 9}+3^{2 \log _{2} 2^{2}}\)
\(=2^{\log _{9} 9}+3^{4 \log _{2} 2}\)
\(=2^{1}+3^{4}\left(\because \log _{a} a=1\right)\)
\(=2+81=83\)
Taking \(L . H . S=R . H . S\)
\(83=10^{\log _{x} 83}=83^{\log _{x} 10},\left(\because a^{\log _{x} y}=y^{\log _{x} a}\right)\)
\(\Rightarrow \log _{x} 10=1\)
\(\therefore x=10^{1}=10\)

\(\sqrt{\log _{2} x}=0.5+\log _{2} \sqrt{x}\)
where \(x>0\) and \(\log _{2} x>0\)
\(\Rightarrow x>1\)
We have, \(\sqrt{\log _{2} x}=0.5+\log _{2} \sqrt{x}\)
\(\Rightarrow \sqrt{\log _{2} x}=\frac{1}{2}+\frac{\log _{2} x}{2},\left[\because \log a^{n}=n \log a\right]\)
On squaring we get, \(\log _{2} x=\frac{1+2 \log _{2} x+\left(\log _{2} x\right)^{2}}{4}\)
\(\Rightarrow 4 \log _{2} x=1+2 \log _{2} x+\left(\log _{2} x\right)^{2}\)
\(\Rightarrow\left(\log _{2} x\right)^{2}-2 \log _{2} x+1=0\)
\(\Rightarrow\left(\log _{2} x-1\right)^{2}=0\)
\(\Rightarrow \log _{2} x=1\)
\(\Rightarrow x=2\)
Hence, \(x\) is a prime number.
Hence, option 'B' is correct.