Mathematics Test-34

From I:

\(\Rightarrow\)\(x^{2}+5 x-36=0\)

\(\Rightarrow\)\(x^{2}+9 x-4 x-36=0\)

\(\Rightarrow\)\(x(x+9)-4(x+9)=0\)

\(\Rightarrow\)\((x+9)(x-4)=0\)

\(\Rightarrow\)\(x=-9,4\)

From II:

\(\Rightarrow\)\(y^{2}+24 y+135=0\)

\(\Rightarrow\)\(y^{2}+9 y+15 y+135=0\)

\(\Rightarrow\)\(y(y+9)+15(y+9)=0\)

\(\Rightarrow\)\((y+9)(y+15)=0\)

\(\Rightarrow\)\(y=-9,-15\)

Therefore, \(x \geq y\)

Given,

n^th term of the sequence is \(a_{n}=\frac{n^{2}}{2^{n}}\)

To find the 8th term for the same, substitute n = 8

We obtain

\(a_{8}=\frac{8^{2}}{2^{8}}\)

\(=\frac{64}{256}\)

Let \(\Delta=\left|\begin{array}{ccc}\sin ^2 x & \cos ^2 x & 1 \\ \cos ^2 x & \sin ^2 x & 1 \\ -10 & 12 & 2\end{array}\right|\)

Applying \(C _1 \rightarrow C _1+ C _2\), we get:

\(\Delta=\left|\begin{array}{ccc}\sin ^2 x+\cos ^2 x & \cos ^2 x & 1 \\ \cos ^2 x+\sin ^2 x & \sin ^2 x & 1 \\ -10+12 & 12 & 2\end{array}\right|\)

\(\Delta=\left|\begin{array}{ccc}1 & \cos ^2 x & 1 \\ 1 & \sin ^2 x & 1 \\ 2 & 12 & 2\end{array}\right| \quad\left(\because \sin ^2 x+\cos ^2 x=1\right)\)

As we know, if two rows or two columns of a determinant are identical the value of the determinant is zero.

Here \(C_1\) and \(C_3\) are identical.

So, \(\Delta=0\)

There are \(8\) chair on each side of the table.

Let the sides be represented by \({A}\) and \({B}\).

Let four persons sit on side \({A}\), then number of ways of arranging \(4\) persons on \(8\) chairs on side \({A}={ }^{8} {P}_{4}\)

And two persons sit on side \({B}\).

The number of ways of arranging \(2\) persons on \(8\) chairs on side \({B}={ }^{8} {P}_{2}\)

The remaining \(10\) persons can be arranged in remaining \(10\) chairs in \(10 !\) ways.

Hence, the total number of ways in which the persons can be arranged is \({ }^{8} {P}_{4} \times{ }^{8} {P}_{2} \times 10 !=\frac{8 ! 8 ! 10 !}{4 ! 6 !}\)

As we know,

\({ }^{\mathrm{n}} C_{\mathrm{r}}=\frac{\mathrm{n} !}{(\mathrm{r} !(\mathrm{n}-\mathrm{r}) !)}\)

\((1+x)^{\mathrm{n}}={ }^{\mathrm{n}} C_{0} \times 1^{(n-0)} \times x^{0}+{ }^{\mathrm{n}} C_{1} \times 1^{(\mathrm{n}-1)} \times x^{1}+{ }^{\mathrm{n}} C_{2} \times 1^{(n-2)} \times x^{2}+\ldots .+{ }^{n} C_{n} \times 1^{(n \cdot n)} \times x^{\mathrm{n}}\)

Given expansion is \((1+x)^{2 n}\)

\(\Rightarrow{ }^{2 n} C_{0} \times 1^{(2 n-0)} \times x^{0}+{ }^{2 n} C_{1} \times 1^{(2 n-1)} \times x^{1}+\ldots+{ }^{2 n} C_{2 n} \times 1^{(2 n-2 n)} \times x^{2 n}\)

First term \(={ }^{2 n} C_{0} \times 1 \times 1=1\)

Last term \(={ }^{2 n} C_{2 n} \times 1 \times x^{2 n}=1 \times x^{2 n}=x^{2 n}\)

\(\Rightarrow\) Sum \(=1+x^{2 n}\)

Coefficient of \(1=1\), coefficient of \(x^{2 n}=1\)

\(\therefore\) sum of the coefficients \(=1+1=2\).

Total outcomes are HHH, HHT, HTH, THH, HTT, THT, TTH, TTT

Favourable outcomes for losing game are HHT, HTH, THH, HTT, THT

Probability \(= \frac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ outcomes}\)

Therefore probability of losing the game is:

\(P(\)Losing the game\()=\frac{6}{8}\)

\(\Rightarrow P(\)Losing the game\()=\frac{3}{4}\)

Equation of \(AM\): \(y=\frac{-1}{3(x+3)}\)

or \(x+3 y+3=0\), Slope of line \(=\frac{-1}{3}\)

So, \(y-1=\frac{-1}{3(x-0)}\)

\(\Rightarrow 3 y+x-3=0\)

\(\Rightarrow\) Equation of \(AM\) is \(3 y+x-3=0\), which is chord of auxiliary circle, \(x^{2}+9 y^{2}=9\).

Coordinates of point \(A(3,0)\) and \(B(0,1) .\)

Distance of \(AM\) from the origin \(=\left|\frac{(0+0-3)}{(\sqrt{(} 9+1)}\right|=\frac{3}{\sqrt{10}}\)

The equation of an auxiliary circle of an ellipse is \(x^{2}+y^{2}=9\)

From figure, line \(AM\) cuts auxiliary circle at \(M\).

By solving these equations, we have \(M(\frac{-12}{5},\frac{9}{5})\) co-ordinates of point \(M\).

Area of triangle \(AOM=\frac{1}{2}|0(0-\frac{9}{5})+3(\frac{9}{5}-0)-\frac{12}{5}(0-0)|\) 

\(=\frac{27}{10}\)

Given,

\(\int^{\pi}_{-\pi} \tan x \sec ^{2} x d x\)

this is of form \(\int_{-a}^{a} f(x) d x\)

\(f(x)=\tan x \sec ^{2} x\)

\(f(-x)=\tan (-x) \sec ^{2}(-x)=-\tan x \sec ^{2} x\)

Thus, \(f(-x)=-f(x)\)

Using the Property,

if \(f\) is an odd function, if \(f(-x)=-f(x)\)

then, \(\int_{-a}^{a} f(x) d x=0\)

\(\therefore \int_{-\pi}^{\pi} \tan x \sec ^{2} x d x=0\)