Mathematics Test - 24

Given:

\(|\overrightarrow{\mathrm{a}}|=\alpha,|\overrightarrow{\mathrm{b}}|=\alpha\) and \(|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|=\alpha\)

As we know,

\(|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|=\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}+2 \mathrm{ab} \cos \theta}\)

\( \alpha=\sqrt{\alpha^{2}+\alpha^{2}+2(\alpha)(\alpha) \cos \theta}\)

\( \alpha^{2}=2 \alpha^{2}+2 \alpha^{2} \cos \theta\)

\(-1=2 \cos \theta\)

\( \cos \theta=-\frac{1}{2}\)

Now,

\(|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|=\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}-2 \mathrm{ab} \cos \theta}\)

\(|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|=\sqrt{\alpha^{2}+\alpha^{2}-2(\alpha)(\alpha) \cos \theta}\)

\(\because \cos \theta=-\frac{1}{2}\)

\(|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|=\sqrt{2 \alpha^{2}-2 \alpha^{2}\left(\frac{-1}{2}\right)}\)

\(|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|=\sqrt{2 \alpha^{2}+\alpha^{2}}\)

\(|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|=\sqrt{\mathrm{3}} {\alpha}\)

\(\begin{aligned}& x^2-70 x+\lambda=0 \\& \alpha+\beta=70 \\& \alpha \beta=\lambda \\& \therefore \alpha(70-\alpha)=\lambda\end{aligned}\)

Since, 2 and 3 does not divide \(\lambda\)

\(\therefore \alpha=5, \beta=65, \lambda=325\)

By putting value of \(\alpha, \beta, \lambda\) we get the required value 60 .

Given determinant is \(\left|\begin{array}{ccc}\mathrm{i} & \mathrm{i}^{2} & \mathrm{i}^{3} \\ \mathrm{i}^{4} & \mathrm{i}^{6} & \mathrm{i}^{8} \\ \mathrm{i}^{9} & \mathrm{i}^{12} & \mathrm{i}^{15}\end{array}\right|\)

Since, we have,

\(\mathrm{i}=\sqrt{-1}\)

\(\mathrm{i}^{2}=-1, \mathrm{i}^{3}=-\mathrm{i}, \mathrm{i}^{4}=1, \mathrm{i}^{6}=-1, \mathrm{i}^{8}=1, \mathrm{i}^{9}=\mathrm{i}, \mathrm{i}^{12}=1\), and \(\mathrm{i}^{15}=-\mathrm{i}\)

\(=\left|\begin{array}{ccc}\mathrm{i} & -1 & -\mathrm{i} \\ 1 & -1 & 1 \\ \mathrm{i} & 1 & -\mathrm{i}\end{array}\right|\)

\(=\mathrm{i}(\mathrm{i}-1)+1(-\mathrm{i}-\mathrm{i})-\mathrm{i}(1+\mathrm{i})\)

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

\(=-4 i\)

Given \(-2<2 x-1<2\)

\(\Rightarrow-2+1<2 x<2+1\)

\(\Rightarrow-1<2 x<3\)

\(\Rightarrow\frac{-1}2

\(\Rightarrow x \in(\frac{-1}2,\frac32)\)

Intersection:

Let A and B be two sets. The intersection of A and B is the set of all those elements which are present in both sets A and B.

The intersection of A and B is denoted by A ∩ B

i.e., A ∩ B = {x : x ∈ A and x ∈ B}

The Venn diagram for intersection is as shown below:

Union:

Let A and B be two sets. The union of A and B is the set of all those elements which belong to either A or B or both A and B.

The union of A and B is denoted by A ∪ B.

i.e., A ∪ B = {x : x ∈ A or x ∈ B}

The Venn diagram for the union of any two sets is shown below:

A ∪ B = A + B - A ∩ B

As we know, 

P ∪ Q = P + Q - P ∩ Q

Putting the values given in the question,

P = P + Q -  P ∩ Q

P ∩ Q = Q

Given:

The mean of 10 observations \(x_{1}, x_{2}, x_{3} \ldots . x_{10}\) is 20.

Mean of \(x_{1}+2, x_{2}+4, x_{3}+6, \ldots x_{10}+20=\) Mean of \(\left(x_{1}, x_{2}, x_{3} \ldots x_{10}\right)+\) Mean of \((2,4)\) \((6, \ldots .20) \ldots(1)\)

Now mean of \((2,4,6, \ldots .20)\) is:

Mean \(= \frac{\text{Sum of total observations}}{\text{Total number of observation}}\)

\(\frac{(2+4+6+8+10+12+14+16+18+20)}{10}\)

\(=\frac{2 \times 55}{10}=11\)

From equation (1),

Mean of \(x_{1}+2, x_{2}+4, x_{3}+6, \ldots x_{10}+20=20+11=31\)

\(\left|\begin{array}{ccc}1+x & 1 & 1 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z\end{array}\right|=0\)

\(R_{1}=R_{1}-R_{2}\)

\(\left|\begin{array}{ccc}x & -y & 0 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z\end{array}\right|=0\)

\(R_{2}=R_{2}-R_{3}\)

\(\left|\begin{array}{ccc}x & -y & 0 \\ 0 & y & -z \\ 1 & 1 & 1+z\end{array}\right|=0\)

Now, Expanding along \(R_{1}\), we get:

\(x[y(1+z)-(-z)]-(-y)[0-(-z)]+0=0\)

\(\Rightarrow x[y+y z+z]+y[z]=0\)

\(\Rightarrow x y+x y z+x z+y z=0\)

Dividing by \({xyz}\),

\(\frac{1}{\mathrm{z}}+1+\frac{1}{\mathrm{y}}+\frac{1}{\mathrm{x}}=0\)

\(\Rightarrow {x}^{-1}+{y}^{-1}+{z}^{-1}=-{1}\)

Given:

\(\underset{{{{x} \rightarrow \infty}}}{\lim}\left(\frac{{x}^{2}+{x}+1}{{x}+1}-{px}-{q}\right)=-3\)

On simplifying, we get:

\(\Rightarrow \underset{{{x \rightarrow \infty}}}{\lim}\left(\frac{x^{2}+x+1-p x^{2}-p x-q x-q}{x+1}\right)=-3\)

\(\Rightarrow \underset{{{x \rightarrow \infty}}}{\lim}\left(\frac{x^{2}(1-p)+x(1-p-q)+1-q}{x+1}\right)=-3\)

As we can see limit gives finite value. So, this is possible only when the coefficient of higher degree term will be zero.

Therefore,

\((1-p)=0\)

Or, \( {p}=1\)

\(\Rightarrow \underset{{{x \rightarrow \infty}}}{\lim}\left(\frac{{x}(1-{p}-{q})+1-{q}}{{x}+1}\right)=-3\) ....(1)

Dividing and multiplying by\(x\) in both numerator and denominator, we get:

\(\Rightarrow \underset{{{x \rightarrow \infty}}}{\lim}\left(\frac{{x}\left[(1-{p}-{q})+\frac{(1-{q})}{{x}}\right]}{{x}\left[1+\frac{1}{{x}}\right]}\right)=-3\)

\(\Rightarrow \underset{{{x \rightarrow \infty}}}{\lim}\left(\frac{\left[(1-{p}-{q})+\frac{(1-{q})}{{x}}\right]}{\left[1+\frac{1}{{x}}\right]}\right)=-3\)

\(\Rightarrow\left(\frac{[(1-p-q)+0]}{[1+0]}\right)=-3\)

\(\Rightarrow(1-p-q)=-3\)

\(\Rightarrow(1-1-q)=-3 \quad(\because p=1)\)

\(\therefore q=3\)

Given,

25, -125, 625, -3125, …….

Here,

First term, a = 25

Common ratio, r =\(\frac{-125}{25}\) = -5

As we know that, if a₁, a₂, …., a_n is a GP then the general term is given by:

a_n = a × r^{n - 1}

Where a is the first term and r is the common ratio.

The general term is:

\(a_{n}=25 \times(-5)^{n-1}\)

\(=(-1)^{n-1} 5^{n+1}\)