Mathematics Test - 31

Given,

\(a_{n}=n \cdot(n+2)\)

As we know that,

\(S_{n}=\sum a_{n}\)

\(\Rightarrow S_{n}=\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n}[k \cdot(k+2)]\)

\(\Rightarrow S_{n}=2 \sum_{k=1}^{n} k+\sum_{k=1}^{n} k^{2}\)

As we know

The sum of first n natural numbers,

S=\(\frac{n(n+1)}2\)

The sum of first n² natural numbers,

S=\(\frac{n(n+1)(2n+1)}6\)

Then, we can write

\(\Rightarrow S_{n}=\frac{2n(n+1)}2 + \frac{n(n+1)(2n +1)}2\)

\(\Rightarrow S_{n}=\frac{n \cdot(n+1) \cdot(2 n+7)}{6}\)

Probability of two or more incidents happening when they are not related to each other \(=P_{1} \times P_{2} \times P_{3}\)

We know that:

\(P^{\prime}=1-P\)

\(P^{\prime}\) is the probability of a thing not happening and \(P\) is of that happening.

Probability of Ankit failing in atleast two subject = Probability of him failing in \(2\) subject + Probability of him failing in \(3\) subjects.

Probability of failing in Maths \(=1-\frac{5}{8}=\frac{3}{8}\)

Probability of failing in English \(=1-\frac{7}{9}=\frac{2}{9}\)

Probability of failing in Science \(=1-\frac{3}{5}=\frac{2}{5}\)

Probability of him failing in \(2\) subject

\(=\frac{3}{8} \times \frac{2}{9} \times \frac{3}{5}+\frac{5}{8} \times \frac{2}{9} \times \frac{2}{5}+\frac{3}{8} \times \frac{7}{9} \times \frac{2}{5}\)

\(=\frac{(18+20+42)}{360}\)

\(=\frac{80}{360}\)

Probability of him failing in three subjects \(=\frac{3}{8} \times \frac{2}{5} \times \frac{2}{9}=\frac{12}{360}\)

Required probability \(=\frac{80}{360}+\frac{12}{360}\) \(=\frac{92}{360}\)

Given:

Ina frequency distribution, the mean and median are 10 and 9 respectively

Relationship between Mean, Median and mode:

As we know that,Mode = 3 Median – 2 Mean

Mode = (3× 9) - (2× 10) = 7

Given:

\(\frac{1}{\log _{2} n}+\frac{1}{\log _{3} n}+\frac{1}{\log _{4} n}+\ldots . .+\frac{1}{\log _{100} n}\)

\(=\log _{n} 2+\log _{n} 3+\log _{n} 4+\ldots . .+\log _{n} 100\)

\(=\log _{n}[2 \times 3 \times 4 \times \ldots \times 100]\)

\(=\log _{n} 100 !\)

\(=\log _{100 !} 100 !\)

\(=1\)

\(\begin{aligned} & \because(a-b)^2+(b-c)^2+(c-a)^2>0 \\ & \Rightarrow 2\left(a^2+b^2+c^2-a b-b c-c a\right)>0 \\ & {\left[\because a^2+b^2+c^2=1\right]} \\ & \Rightarrow 2>2(a b+b c+c a) \Rightarrow a b+b c+c a<1\end{aligned}\)

Given,

Points are \(\mathrm{A}(-1,0)\) and \(\mathrm{B}(0,5)\).

The distance between tow points \(\mathrm{A}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) and \(\mathrm{B}\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)\) is given by:

\(\mathrm{AB}=\sqrt{\left(\mathrm{x}_{1}-\mathrm{x}_{2}\right)^{2}+\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)^{2}}\)

Let the point which is equidistant from the points \(\mathrm{A}\) and \(\mathrm{B}\) be \(\mathrm{C}(\mathrm{x},\mathrm{x})\).

As the point is equidistant we have \(\mathrm{A}\mathrm{C}=\mathrm{B}\mathrm{C}\).

Therefore, equating the right hand and left hand side using distance formula.

\(\mathrm{AC} =\mathrm{BC} \)

\(\sqrt{(-1-\mathrm{x})^{2}+(0-\mathrm{x})^{2}}=\sqrt{(0-\mathrm{x})^{2}+(5-\mathrm{x})^{2}}\)

\(\Rightarrow (-1-\mathrm{x})^{2}+(0-\mathrm{x})^{2} =(0-\mathrm{x})^{2}+(5-\mathrm{x})^{2}\)

\(\Rightarrow1+2 \mathrm{x}+\mathrm{x}^{2}+\mathrm{x}^{2} =\mathrm{x}^{2}+25-10 \mathrm{x}+\mathrm{x}^{2} \)

\(\Rightarrow1+2 \mathrm{x} =25-10 \mathrm{x}\)

\(\Rightarrow12 \mathrm{x} =24 \)

\(\Rightarrow\mathrm{x} =2\)

Thus, the point which is equidistant from both points is \((2,2)\).

\(f(x)=\left(\frac{3}{5}\right)^x+\left(\frac{4}{5}\right)^x-1 \\\)

\( \text { Putf }(x)=0 \\\)

\( \Rightarrow 0=\left(\frac{3}{5}\right)^x+\left(\frac{4}{5}\right)^x-1 \\\)

\( \Rightarrow\left(\frac{3}{5}\right)^x+\left(\frac{4}{5}\right)^x=1 \\\)

\( \Rightarrow 3^x+4^x=5^x \ldots \text { (i) }\)

For \(\mathrm{x}=1\)

\(3^1+4^1>5^1\)

For \(\mathrm{x}=3$\)

\(3^3+4^3=91<5^3\)

Only for \(\mathrm{x}=2\), equation (i) Satisfy

So, only one solution \((x=2)\)

\(\begin{aligned} & \mathrm{Q}=\mathrm{PAP}^{\mathrm{T}} \\ & \mathrm{P}^{\mathrm{T}} \cdot \mathrm{Q}^{2007} \cdot \mathrm{P}=\mathrm{P}^{\mathrm{T}} \cdot \mathrm{Q} \cdot \mathrm{Q} \ldots \mathrm{Q} \cdot \mathrm{P} \\ & =\mathrm{P}^{\mathrm{T}}\left(\mathrm{PAP} \mathrm{P}^{\mathrm{T}}\right)\left(\mathrm{P} \cdot \mathrm{AP} \mathrm{P}^{\mathrm{T}} \ldots\left(\mathrm{P}^{\mathrm{T}}\right) \mathrm{P} .\right. \\ & \Rightarrow\left(\mathrm{P}^{\mathrm{T}} \mathrm{P}\right) \mathrm{A}\left(\mathrm{P}^{\mathrm{T}} \mathrm{P}\right) \mathrm{A} \ldots \mathrm{A}\left(\mathrm{P}^{\mathrm{T}} \mathrm{P}\right) \\ & \mathrm{P}^{\mathrm{T}} \cdot \mathrm{P}=\left[\begin{array}{cc}\sqrt{3} / 2 & -1 / 2 \\ 1 / 2 & \sqrt{3} / 2\end{array}\right]\left[\begin{array}{cc}-\sqrt{3} / 2 & 1 / 2 \\ -1 / 2 & \sqrt{3} / 2\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\mathrm{I} \\ & \therefore \mathrm{P}^{\mathrm{T}} \cdot \mathrm{Q}^{2007} \cdot \mathrm{P}=\mathrm{A}^{2007} \\ & \mathrm{~A}^2=\left[\begin{array}{cc}1 & 1 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}1 & 1 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}1 & 2 \\ 0 & 1\end{array}\right] \\ & \therefore A^{2007}=\left[\begin{array}{cc}1 & 2007 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}\mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{d}\end{array}\right] \\ & \mathrm{a}=1, \mathrm{~b}=2007, \mathrm{c}=0, \mathrm{~d}=1 \\ & 2 \mathrm{a}+\mathrm{b}-3 \mathrm{c}-4 \mathrm{~d}=2+2007-4=2005\end{aligned}\)

Given that:

\(R\) is a relation on \(Z\) and is defined as:

\(R=\{(a, b): a-b\) is divisible by 7 where \(a, b \in Z\}\)

We know that:

A relation R on a set A is said to be an equivalence relation if R is reflexive, symmetric and transitive.

Therefore,

\(R\) is reflexive as:

\(a - a=0\) is divisible by 7 for all \(a \in Z\).

Suppose, if \((a, b) \in R\)

\( \Rightarrow 7\) divides \(a-b\) i.e.,

\(\Rightarrow a-b=7 m\), where \(m\in Z \)

\(\Rightarrow b-a=7 n\), where \(n=-m\)

\(\Rightarrow 7\) divides \(b-a\) too, which implies that:

\((b, a) \in R .\)

Therefore, \(R\) is symmetric.

Suppose, if \((a, b) \in R\) and \((b, c) \in R\), then:

\(a-b\) and \(b-c\) are divisible by 7.

\(\Rightarrow a-b=7 m\) and \(b-c=7 n\), where \(m, n \in Z\)

\(\Rightarrow a-c=7 q\), where \(q=m+n\)

\(\Rightarrow(a, c) \in R\)

Therefore, \(R\) is transitive.