Mathematics Test-11

Given:

5 letter word MUNCH is given.

Since U and N are always included so first we select 2 letters from remaining 3 letters M, C and H, which can be done in \({ }^{3} \mathrm{C}_{2}=3\) ways.

Now these 4 letters can be arrange in \(4 !=24\) ways.

So, the required number is 72 ways \((3 \times 24)\).

Given,

\(x=y+8\)

\(\Rightarrow y=x-8\)

\(x+y=5\)

\(\Rightarrow y=-x+5\)

As we know,

Equation of straight line is \(y=m x+c\).

\(\therefore\) Slope of line \(y=x-8\) is \(m_{1}=1\)

Also, Slope of line \(y=-x+5\) is \(m_{2}=-1\).

Now, \(\theta=\tan ^{-1}\left|\frac{m_{2}-m_{1}}{1+m_{1} m_{2}}\right|\)

\(\Rightarrow \theta=\tan ^{-1}\left|\frac{-1-1}{1+(-1)(1)}\right|\)

\(\Rightarrow \theta=\tan ^{-1}\left|\frac{-2}{1-1}\right|\)

\(\Rightarrow \theta=\tan ^{-1}(\infty)=90^{\circ} \quad\left(\because \tan 90^{\circ}=\infty \Rightarrow \tan ^{-1} \infty=90^{\circ}\right)\)

Given: \(I_1=\int_e^{e^2} \frac{d x}{\log x}\) and \(I_2=\int_1^2 \frac{e^x}{x} d x\)

\(\Rightarrow I_1=\int_e^{e^2} \frac{d x}{\log x} \text { put } \log \mathrm{x}=\mathrm{z}\)

Such that \(x=e^z\)

Such that \(d x=e^z d z\)

When \(x=e, z=\) loge

\(x=e^2, z=\log e^2=2 \log e=z\)

Such that \(I_1=\int_1^2\left(e^z d z\right) / z=\int_1^2\left(e^x / z\right) d x=I_2\)

Such that \(I_1=I_2\)

\(\Rightarrow I_1-I_2=0\)

In complex number theory \(i\) represent 'imaginary unit' and satisfy the equation, \(\mathrm{i}^{2}=-1\).

Factor out \(\mathrm{i}^{\mathrm{n}}\) from given expression and simplify.

\(\mathrm{i}^{\mathrm{n}}+\mathrm{i}^{\mathrm{n}+1}+\mathrm{i}^{n+2}+\mathrm{i}^{\mathrm{n}+3}\)

\(=\mathrm{i}^{\mathrm{n}}\left[1+\mathrm{i}+\mathrm{i}^{2}+\mathrm{i}^{3}\right]\)

\(=i^{n}[1+i-1-i]\)

\(=\mathrm{i}^{\mathrm{n}} \times 0\)

\(=0\)

The \((\mathrm{r}+1)\) th term in the expansion of \(\left(\sqrt{\frac{\mathrm{x}}{3}}-\frac{\sqrt{3}}{\mathrm{x}^{2}}\right)^{10}\) is given by:

\(\mathrm{T}_{\mathrm{r}+1}=(-1)^{\mathrm{r}} \cdot{ }^{10} \mathrm{C}_{\mathrm{r}}\left(\sqrt{\frac{\mathrm{x}}{3}}\right)^{10-\mathrm{r}} \cdot\left(\frac{\sqrt{3}}{\mathrm{x}^{2}}\right)^{\mathrm{r}}\)

\(=(-1)^{\mathrm{r}}{ }^{10} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{5-\frac{5 \mathrm{r}}{2}} 3^{\mathrm{r}-5}\).....(1)

For term to be independent of \(\mathrm{x}\),

\(5-\frac{5 \mathrm{r}}{2}=0\)

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

Substituting the values in (1) we get:

\(\mathrm{T}_{3}=(-1)^{2}{ }^{10} \mathrm{C}_{2} 3^{-3}\)

\(=\frac{10(9)}{2(27)}\)

\(=\frac{5}{3}\)

\(A B C D\) is a parallelogram so, \(A B \| C D\) and \(A D \| B C\)

\(\overrightarrow{A C}=\overrightarrow{A B}+\overrightarrow{B C}\) and

\(\overrightarrow{A D}=\overrightarrow{A B}+\overrightarrow{B D}\)

\(\therefore \overrightarrow{B D}=\overrightarrow{A D}-\overrightarrow{A B}\)

\(\therefore \overrightarrow{A C}-\overrightarrow{B D}=\overrightarrow{A B}+\overrightarrow{B C}-(\overrightarrow{A D}-\overrightarrow {A B})\quad\{\because \overrightarrow{A D}=\overrightarrow{B C}\}\)

\(=\overrightarrow{A B}+\overrightarrow{B C}-\overrightarrow{A D}+\overrightarrow{A B}\)

\(=2 \overrightarrow{A B}\)

\(R=\{(a, b]: a \leq b\}\) \(\Rightarrow( a, a] \in R\)
\(\therefore R\) is reflexive. \(( 2,4) \in \mathrm{R}[\) as \(2<4]\), \((4,2) \notin R\) as \(4>2\)
\(\therefore R\) is not symmetric. \((a, b),(b, c) \in R\), \(a \leq b\) and \(b \leq c\)
\(\Rightarrow a \leq c\) \(\Rightarrow(a, c) \in R\) \(\therefore R\) is transitive.
\({R}\) is reflexive and transitive but not symmetric.

Given \(\left[\begin{array}{cc}3 x+7 & 5 \\ y+1 & 2-3 x\end{array}\right]=\left[\begin{array}{cc}0 & y-2 \\ 8 & 4\end{array}\right]\)

Comparing the corresponding elements we get,

\(3 x+7=0\)

\(\Rightarrow x=-\frac{7}{3}\)

\(y-2=5\)

\(\Rightarrow y=7\)

\(y+1=8\)

\(\Rightarrow y=7\)

\(2-3 x=4\)

\(\Rightarrow x=-\frac{2}{3}\)

Since we get two different values of \({x}\), which is not possible. It is not possible to find the values of \({x}\) and \({y}\) for which the given matrices are equal.

Given the exprission is \(\left(2 \mathrm{x}+\frac{1}{\mathrm{x}}\right)^{8}\).

Here \(n=8(\mathrm{n}\) is even number)

\(\therefore\) Middle term \(=\left(\frac{\mathrm{n}}{2}+1\right)=\left(\frac{8}{2}+1\right)=5\) th term

General term in the expansion of \((x+y)^{n}\) is given by

\(\mathrm{T}_{(\mathrm{r}+1)}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \times \mathrm{x}^{\mathrm{n}-\mathrm{r}} \times \mathrm{y}^{\mathrm{r}}\)

\(\Rightarrow T_{5}=T_{(4+1)}={ }^{8} \mathrm{C}_{4} \times(2 \mathrm{x})^{(8-4) \times}\left(\frac{1}{\mathrm{x}}\right)^{4}\)

\(\Rightarrow\mathrm{~T}_{5}={ }^{8} \mathrm{C}_{4} \times 2^{4}\)

Given,

P\( (-2,4,-5) \) and Q\( (1,2,3) \)

We know the direction cosines of the line passing through two points \(\mathrm{P}\left(x_1, y_1, z_1\right)\) and \(\mathrm{Q}\left(x_2, y_2, z_2\right)\) are given by,

\(\frac{x_2-x_1}{\mathrm{PQ}}, \frac{y_2-y_1}{\mathrm{PQ}}, \frac{z_2-z_1}{\mathrm{PQ}}\)

where \(\mathrm{PQ}=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2}\)

Here \(P\) is \( (-2,4,-5) \) and \(Q\) is \( (1,2,3) \).

So, \(\mathrm{PQ}=\sqrt{(1-(-2^2+(2-4)^2+(3-(-5^2} ) )=\sqrt{77}\)

Thus, the direction cosines of the line joining two points is,

\((\frac{3}{\sqrt{77}}, \frac{-2}{\sqrt{77}}, \frac{8}{\sqrt{77}})\)