Mathematics Test - 21

\(\mathrm{x}=(7+4 \sqrt{3})^{2 \mathrm{n}} \)

\(\mathrm{x}=3+\mathrm{f}\)

\(\mathrm{f}\) is fractional part of \(\mathrm{x}\):

\( 0<\mathrm{f}<1\)

Let \(\mathrm{f}^{\prime}=(7-4 \sqrt{3})^{2 \mathrm{n}}\)

\(0<7-4 \sqrt{3}<1 \)

\(0<(7-4 \sqrt{3})^{\mathrm{n}}<1 \)

\(0<\mathrm{f}^{\prime}<1\)

\(\mathrm{I}+\mathrm{f}+\mathrm{f}^{\prime}=2\left[{ }^{2 \mathrm{n}} \mathrm{C}_{0}(7)^{2 \mathrm{n}}+{ }^{2 \mathrm{n}} \mathrm{C}_{2}(7)^{2 \mathrm{n}-2}+\ldots\right] \)

\(=\text { even }\)

\(0+\mathrm{f}+\mathrm{f}^{\prime}<2\) is integer

\({ }^{\prime} \mathrm{I}^{\prime}\) is integer and \(\mathrm{f}+\mathrm{f}^{\prime}\) is also integer.

\(\Rightarrow \mathrm{f}+\mathrm{f}^{\prime}=1\)

\(\mathrm{f}^{\prime}=1-\mathrm{f} \)

\(\mathrm{x}(1-\mathrm{f})=\mathrm{xf}^{\prime}=(7+4 \sqrt{3})^{2 \mathrm{n}}(7-4 \sqrt{3})^{2 \mathrm{n}} \)

\(=1\)

The word ALLAHABAD contains 9 letters, in which A occur 4 times, L occurs twice and the rest of the letters occur only once.

Number of Permutations of ‘\(n\)’ objects where there are \(n_1\) repeated items, \(n_2\) repeated items, \(n_k\) repeated items taken ‘\(r\)’ at a time:

\(p(n, r)=\frac{n !}{n_{1} ! n_{2} ! n_{3} ! \ldots n_{k} !}\) 

Therefore,  Number of different words formed by the word ALLAHABAD using all the letters \(=\frac{9 !}{4 ! \times 2 !}\)

\(=\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 !}{4 ! \times 2}\) \(=7560\)

Now, let us take both L together and consider (LL) as 1 letter.

Then, we will have to arrange 8 letters, in which A occurs 4 times and the rest of the letters occur only once.

So, the number of words having both \(L\) together will be \(=\frac{8 !}{4 !}\)

\(=\frac{8 \times 7 \times 6 \times 5 \times 4 !}{4 !}\) \(=1680\)

Therefore, the number of words with both L not occurring together will be \(= 7560 - 1680\) \(= 5880\)

Given, A box contains \(4\) tennis balls, \(6\) season balls and \(8\) dues balls

We know that, Probability \(=\frac{\text { Favourable outcomes }}{\text { Total outcomes }}\)

Let us assume that all balls are unique.

There are a total of \(18\) balls.

Number of all combinations of \(n\) things, taken \(r\) at a time, is given by \({ }^{n} C_{r}=\frac{n !}{(r) !(n-r) !}\)

Total ways \(=3\) balls can be chosen in \({ }^{18} \mathrm{C}_{3}\) ways

\(=\frac{18 !}{3 ! \times 15 !}\)

\(=\frac{18 \times 17 \times 16}{3 \times 2 \times 1}\) \(=816\)

There are \(4\) tennis balls, \(6\) season balls and \(8\) dues balls, \(1\) tennis ball, \(1\) season ball and \(1\) dues Ball drawn.

Therefore, favorable ways \(=4 \times 6 \times 8\) \(=192\)

Probability \(=\frac{192}{816}\) \(=\frac{4}{17}\)

As we know,

Identification of curves represented by the general equation of the second degree:

Let a general equation of second degree in \(x\) and \(y\) be \(a x^{2}+2 h x y+b y^{2}+2 g x+2 f y+c=0\)...(1).

Then this equation represents,

1. A parabola if \(\Delta \neq 0\) and \(h^{2}=a b\)

2. An ellipse if \(\Delta \neq 0\) and \(h^{2}>a b\)

3. A hyperbola if \(Δ ≠ 0\) and \(h^2 > ab\)

4. A pair of straight line or empty set if \(\Delta=0\) and \(h^{2} \geq a b\)

5. A unique point if \(\Delta=0\) and \(h^2 < ab\)

6. A circle if \(a = b ≠ 0, h = 0\) and \(g^2 + f^2 -ac > 0\)

The given equation is \(2 x^{2}+2 y^{2}-5 x-7 y-3=0\).

Comparing it wth the equation (1) we get,

\(a=2, h=0, b=2, g=-(\frac{5 }{ 2}), f=-(\frac{7 }{ 2}), c=-3\)

As, \(a=b=2(>0), h=0\)

As we know,

\(g^2 + f^2 -ac > 0\)

\(\therefore (-\frac{5 }{ 2})^2+ (-\frac{7 }{ 2})^2-(2 \times -3)>0\)

\(\Rightarrow \frac{25 }{4}+ \frac{49}{4}+6>0\)

\(\Rightarrow \frac{25 +49+24}{4}>0\)

\(\Rightarrow \frac{98}{4}>0\)

\(\Rightarrow 24.5>0\)

So, the given equation represents a circle.

\(\begin{aligned}& \rightarrow P_x=\frac{2 \times 5+1 \times 0}{1+2}=\frac{10}{3} \\& P_y=\frac{0 \times 2+9 \times 1}{1+2}=3 \\& P=\left(\frac{10}{3}, 3\right)\end{aligned}\)

Similarly \(\rightarrow \mathrm{Q}_{\mathrm{x}}=\frac{1 \times 5+2 \times 0}{1+2}=\frac{5}{3}\)

\(\begin{aligned}& Q_y=\frac{1 \times 0+2 \times 9}{1+2}=6 \\& Q:\left(\frac{5}{3}, 6\right)\end{aligned}\)

Now \(\mathrm{m}_1=\frac{3-0}{\frac{10}{3}-0}=\frac{9}{10}\)

\(\begin{aligned}& \mathrm{m}_2=\frac{6-0}{\frac{5}{3}-0}=\frac{18}{5} \\& \text { from }(1) \&(2) \\& 9 x+5 y=45 \\& 9 x-2 y=0 \\& \frac{-+-}{7 y=45} \Rightarrow y=\frac{45}{7} \\& \Rightarrow x=\frac{10}{7}\end{aligned}\)

which satisfy \(\mathrm{y}-\mathrm{x}=5\)

Consider the equation

\(\begin{aligned} & x^2+2 x+2=0 \\ & x=\frac{-2 \pm \sqrt{4-8}}{2}=-1 \pm i \\ & \text { Let } \alpha=-1+i, \beta=-1-i \\ & \alpha^{15}+\beta^{15}=(-1+i)^{15}+(-1-i)^{15} \\ & =\left(\sqrt{2} e^{i \frac{3 \pi}{4}}\right)^{15}+\left(\sqrt{2} e^{-i \frac{3 \pi}{4}}\right)^{15} \\ & =(\sqrt{2})^{15}\left[e^{\frac{i 45 \pi}{4}}+e^{\frac{-i 45 \pi}{4}}\right] \\ & =(\sqrt{2})^{15} \cdot 2 \cos \frac{45 \pi}{4}=(\sqrt{2})^{15} \cdot 2 \cos \frac{3 \pi}{4} \\ & =\frac{-2}{\sqrt{2}}(\sqrt{2})^{15} \\ & =-2(\sqrt{2})^{14}=-256\end{aligned}\)

Given that,

\(A=\left[\begin{array}{ll}x & 2 \\ 4 & x\end{array}\right]\) and \(\operatorname{det}\left\{A^{2}\right\}=64\)

\(\Rightarrow|\mathrm{A}|={x}^{2}-8 \quad \quad \ldots(1)\)

Given \(\left|\mathrm{A}^{2}\right|=64\)

\(\Rightarrow|\mathrm{A}|^{2}=64 \quad \quad\) \(\left[\because\left|\mathrm{A}^{\mathrm{n}}\right|=|\mathrm{A}|^{\mathrm{n}}\right]\)

\(\Rightarrow|\mathrm{A}|=(64)^{\frac{1}{2}}=8\quad \quad \ldots(2)\)

From equation (1) and (2),

\(x^{2}-8=8\)

\(\Rightarrow x^{2}=16\)

\(\Rightarrow x=\pm 4\)

We know that,

\((i) ^{2}=-1\)

\((-i)^{4}=1\)

\(\left(\frac{1-i}{1+i}\right)^{n^{2}}=1\)

On rationalizing,

\(\left(\frac{1-i}{1+i} \times \frac{1-i}{1-i}\right)^{n^{2}}=1 \)

\(\Rightarrow \left(\frac{(1-i)^{2}}{1-i^{2}}\right)^{n^{2}}=1 \)

\(\Rightarrow\left(\frac{1+i^{2}-2 i}{1-(-1)}\right)^{n^{2}}=1 \)

\(\Rightarrow\left(\frac{1-1-2 i}{1+1}\right)^{n^{2}}=1 \)

\(\Rightarrow\left(\frac{-2 i}{2}\right)^{n^{2}}=1 \)

\(\Rightarrow(-i)^{n^{2}}=1\)

If we put \(n =2\) then,

\( (-i)^{2^{2}}=1\)

\( (-i)^{4}=1\)

Satisfy the equation.

Given,

A room contains \(3\) red, \(5\) green and \(4\) blue chairs

Total number of chairs \(=3+5+4\)

\(=12\)

Number of all combinations of \(n\) things, taken \(r\) at a time, is given by \({ }^{n} C_{r}=\frac{n !}{(r) !(n-r) !}\)

Let \(\mathrm{S}\) be the sample space.

Then, \(n(s)=\) Number of ways of picking \(2\) chairs out of \(12\)

\(={ }^{12} C_{2}\)

\(=\frac{12 !}{2 !10 !}\)

\(=\frac{12 \times 11}{2 \times 1}\)

\(=66\)

Let \(n(E)=\) number of events of selecting \(2\) chairs for selecting no blue chairs.

\(={ }^{8} \mathrm{C}_{2}\)

\(=\frac{8 !}{2 !6 !}\)

\(=\frac{8 \times 7}{2 \times 1}\)

\(=28\)

Therefore, required probability \(\mathrm{P}(\mathrm{E})=\frac{n(E)}{n(S)}\)

\(=\frac{28}{66}\)

\(=\frac{14}{33}\)

Given, 2x + 1 > 3

⇒ 2x > 3 – 1

⇒ 2x > 2

⇒ x > 1

⇒ x ∈ (1, ∞)