Mathematics Test - 37

Let A be the \(k^{t h}\) A.M., then \(\mathrm{H}\) will be the \(k^{t h} \mathrm{H.M}\)., \(\mathrm{Now}, A=2+k d=2+k\left[\frac{3-2}{10}\right]=\frac{20+k}{10}\)
\(\frac{1}{H}=\frac{1}{2}+k d^{\prime}=\frac{1}{2}+k\left[\frac{\frac{1}{3}-\frac{1}{2}}{10}\right]=\frac{30-k}{60}\)
\(\therefore A+\frac{6}{H}=5 \Longrightarrow H(5-A)=6\)
Hence, option 'B' is correct.

Let the digits of a three-digit number be \(x, y\) and \(z\) and the number be \(100 z+10 y+x,\) where \(x, y, z\) are in G.P. \(\therefore y^{2}=x z\)
\(\Rightarrow x z\) must be a square number If \(x z=4,\) then \(y=2 \Rightarrow 421\) is the number, but \(421-400=21\) which is not a three digit number. \(\therefore x z=9,\) i.e., \(x=1, z=9\)
\(\therefore y=3,\) so that \(x, y, z\) are in \(A . P .\) \(\therefore\) The number is \(931 .\)
The other number will be \(531,\) so that 1,3 and 5 are in \(A\). P. Their sum \(=931+531=1462\)

Let, \(S_{\infty}=1+\left(1+\frac{1}{2}\right)\left(\frac{1}{3}\right)+\left(1+\frac{1}{2}+\frac{1}{2^{2}}\right)\left(\frac{1}{3^{2}}\right)+\ldots \ldots \infty\)
Now, \(t_{r}=\left(1+\frac{1}{2}+\frac{1}{2^{2}}+\ldots+\frac{1}{2^{r-1}}\right)\left(\frac{1}{3^{r-1}}\right)=\left(\frac{1-(1 / 2)^{r}}{1-(1 / 2)}\right)\left(\frac{1}{3^{r-1}}\right)\)
-.. Sum of G.P series ]
\(\Rightarrow t_{r}=6\left(\frac{1}{3^{r}}-\frac{1}{6^{r}}\right)\)
\(S_{\infty}=\sum_{r=1}^{\infty} t_{r}=6 \sum_{r=1}^{\infty}\left(\frac{1}{3^{r}}-\frac{1}{6^{r}}\right)\)
\(\Rightarrow S_{\infty}=6\left(\left[\frac{1 / 3}{1-(1 / 3)}-\frac{1 / 6}{1-(1 / 6)}\right]\right) \quad \ldots .\) [Sum of Infinite G.P series \(]\)
\(\Rightarrow S_{\infty}=\frac{9}{5}\)

\(\frac{1}{2}+\frac{2}{3} \operatorname{cosec}^{2} \theta+\frac{3}{8} \sec ^{2} \theta\)
\(=\frac{1}{2}+\frac{2}{3}+\frac{2}{3} \cot ^{2} \theta+\frac{3}{8}+\frac{3}{8} \tan ^{2} \theta\)
\(=\frac{37}{24}+\frac{2}{3} \cot ^{2} \theta+\frac{3}{8} \tan ^{2} \theta\)
Now, \(A M \geq G M\) \(\frac{\frac{2}{3} \cot ^{2} \theta+\frac{3}{8} \tan ^{2} \theta}{2} \geq \sqrt{\frac{2}{3} \cot ^{2} \theta \times \frac{3}{8} \tan ^{2} \theta}\)
\(\frac{2}{3} \cot ^{2} \theta+\frac{3}{8} \tan ^{2} \theta \geq 1\)
Therefore, \(\frac{37}{24}+\frac{2}{3} \cot ^{2} \theta+\frac{3}{8} \tan ^{2} \theta \geq \frac{37}{24}+1=\frac{61}{24}\)

\(\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{3}\right) b^{2}-2\left(a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{4}\right) b+\left(a_{2}^{2}+a_{3}^{3}+a_{4}^{2}\right) \leq 0\)
where, \(a_{1}, a_{2}, a_{3}, a_{4}, b \in R\)
or, \(\left(a_{1}^{2} b^{2}-2 a_{1} a_{2} b+a_{2}^{2}\right)+\left(a_{2}^{2} b^{2}-2 a_{2} a_{3} b+a_{3}^{2}\right)+\left(a_{3}^{2} b^{2}-2 a_{3} a_{4} b+a_{4}^{2}\right) \leq 0\)
or, \(\left(a_{1} b-a_{2}\right)^{2}+\left(a_{2} b-a_{3}\right)^{2}+\left(a_{3} b-a_{4}\right)^{2} \leq 0\)
since all individual terms are positive so, they all are equal to zero separately
or, \(\left(a_{1} b-a_{2}\right)^{2}=0\) or \(\left(a_{2} b-a_{3}\right)^{2}=0\) or \(\left(a_{3} b-a_{4}\right)^{2}=0\)
or, \(\left(a_{1} b-a_{2}\right)=0\) or \(\left(a_{2} b-a_{3}\right)=0\) or \(\left(a_{3} b-a_{4}\right)=0\)
or, \(b=\frac{a_{2}}{a_{1}}\) or \(b=\frac{a_{3}}{a_{2}}\) or \(b=\frac{a_{4}}{a_{3}}\)
or, \(\frac{a_{2}}{a_{1}}=\frac{a_{3}}{a_{2}}=\frac{a_{4}}{a_{3}}=b\)
\(\mathrm{So}, a_{1}, a_{2}, a_{3}\) and \(a_{4}\) are in G.P. Hence, \(A\) is the correct option.

Let an arbitrary point be \(P(x, y)\), Then it is given that \(P A=P B\)
or \(P A^{2}=P B^{2}\)
\((x+2)^{2}+(y-3)^{2}=(x-6)^{2}+(y+5)^{2}\)
\((x+2)^{2}-(x-6)^{2}=(y+5)^{2}-(y-3)^{2}\)
\((2 x-4)(8)=(2 y+2)(8)\)
\(2 x-4=2 y+2\)
\(x-2=y+1\)
\(x=y+3\)
or
\(x-y=3\)

Converting from polar to rectangular form. \(A \equiv\left(2, \frac{\pi}{2}\right)\)
Here, \(r=2\) and \(\theta_{1}=\frac{\pi}{2}\)
Hence, \(x_{1}=r \cos \theta_{1}\) and \(y_{1}=r \sin \theta_{1}\)
\(\therefore x_{1}=0\) and \(y_{1}=2\)
so, \(\left(2, \frac{\pi}{2}\right) \equiv(0,2)\)
\(B \equiv\left(5, \tan ^{-1} \frac{4}{3}\right)\)
Here, \(r=5\) and \(\tan \theta_{2}=\frac{4}{3}\)
\(\therefore \sin \theta_{2}=\frac{4}{5}\) and \(\cos \theta_{2}=\frac{3}{5}\)
Hence, \(x_{2}=r \cos \theta_{2}\) and \(y_{2}=r \sin \theta_{2}\)
\(\therefore x_{2}=3\) and \(y_{2}=4\)
So, \(\left(5, \tan ^{-1} \frac{4}{3}\right) \equiv(3,4)\)
By distance formula, \(A B=\sqrt{(3-0)^{2}+(4-2)^{2}}\)
\(A B=\sqrt{13}\)

Distance between the points in polar form is \(d=\sqrt{r_{1}^{2}+r_{2}^{2}-2 r_{1} r_{2} \cos \left(\theta_{1}-\theta_{2}\right)}\)
\(\Rightarrow d=\sqrt{a^{2}+9 a^{2}-6 a^{2} \cos \frac{\pi}{3}}\)
\(\Rightarrow d=a \sqrt{7}\)
Alternate Method:
Converting polar coordinates \(\left(a, \frac{\pi}{2}\right)\) to cartesian form \(x_{1}=a \cos \frac{\pi}{2}\) and \(y_{1}=a \sin \frac{\pi}{2}\)
So, the cartesian form of \(\left(a, \frac{\pi}{2}\right)\) is \((0, a)\) Converting polar coordinates \(\left(3 a, \frac{\pi}{6}\right)\) to cartesian form \(x_{2}=3 a \cos \frac{\pi}{6}\) and \(y_{2}=3 a \sin \frac{\pi}{6}\)
\(x_{2}=\frac{3 a \sqrt{3}}{2}, y_{2}=\frac{3 a}{2}\)
So, the cartesian form of \(\left(3 a, \frac{\pi}{6}\right)\) is \(\left(\frac{3 a \sqrt{3}}{2}, \frac{3 a}{2}\right)\) Distance \(d=\sqrt{\frac{27 a^{2}}{4}+\frac{a^{2}}{4}}\)
\(\Rightarrow d=\sqrt{7 a^{2}}\)
\(\Rightarrow d=a \sqrt{7}\)

\(x=r \cos \theta\) and \(y=r \sin \theta\) where \(r=\sqrt{x^{2}+y^{2}}\)
Hence
\(x \cos \alpha+y \sin \alpha=p\)
\(r \cos \theta \cos \alpha+r \sin \theta \sin \alpha=p\)
\(r(\cos \theta \cos \alpha+\sin \theta \sin \alpha)=p\)
\(r \cos (\theta-\alpha)=p\)
This is the required polar equation of the given line.

Let the roots be \(\alpha, \beta\) Then the equation of the line will be \(\frac{x}{\alpha}+\frac{y}{\beta}=1\) where \(\alpha\) is the \(x\) intercept and \(\beta\) is the \(y\) intercept. Also, the area of the right triangle formed by the line and the coordinate axes is \(\frac{\alpha \cdot \beta}{2}=A\)
Or \(\alpha \cdot \beta=2 A \ldots(i)\)
since the line passes through ( 1,1 ) hence \(\frac{1}{\alpha}+\frac{1}{\beta}=1\)
\(\Rightarrow \frac{\alpha+\beta}{\alpha \cdot \beta}=1\)
\(\Rightarrow \alpha+\beta=\alpha \cdot \beta\)
\(\Rightarrow \alpha+\beta=2 A \ldots\) from \(\mathrm{i}\)
Hence, sum of roots is \(=\alpha+\beta=2 A\)
And, Product of roots \(\alpha \beta=2 A\) Thus the equation with the roots \(\alpha\) and \(\beta\) is \((x-\alpha)(x-\beta)=0\)
\(\Rightarrow x^{2}-(\alpha+\beta) x+\alpha \beta=0\)
\(\Rightarrow x^{2}-2 A x+2 A=0\)