Mathematics Test - 47

Given,

\(\cot \left[\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{8}\right]=\)

\(=\cot [\tan ^{-1}\left(\frac{\frac{1}{2}+\frac{1}{8}}{1-\frac{1}{2} \times \frac{1}{8}}\right)] \quad\left(\because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{\tan x+\tan y}{1-\tan x \tan y}\right)\right)\)

\(=\cot [\tan ^{-1} \frac{\left(\frac{2+8}{16}\right)}{\left(\frac{16-1}{16}\right)}]\)

\(=\cot [\tan ^{-1} \frac{\left(\frac{10}{16}\right)}{\left(\frac{15}{16}\right)} ]\)

\(=\cot [\tan ^{-1} \frac{10}{15}]\)\(=\cot^{-1} \frac{3}{2}\)

Now,

\(\cot \left[\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{8}\right] \)

\(=\cot [\cot ^{-1} \frac{3}{2}]\)

\(=\frac{3}{2}\)

Here, we have to find the area of the region bounded by the curves \(y=x^{3}+4 x+2\), the line \(x=0, x=4\) and the \(x\)-axis.

So, the area enclosed by the given curves is given by: 

\(\int_{0}^{4}\left(x^{3}+4 x+2\right) d x\)

As we know that, 

\(\int \mathrm{x}^{\mathrm{n}} \mathrm{dx}=\frac{\mathrm{x}^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{C}\)

\(\Rightarrow \int_{0}^{4}\left(x^{3}+4 x+2\right) d x\)

\(=\left[\frac{x^{4}}{4}+\frac{4 x^{2}}{2}+2 x\right]_{0}^{4}\)

\(= \frac{1}{4}(256)+\frac{64}{2}+(2 \times 4)\)

\(=104\) square units

Given,

\(\sin 1^{\circ}+\sin 2^{\circ}+\ldots+\sin 359^{\circ}\)

We know that,

\(\sin (180-\theta)=\sin \theta\).....(i)

\(\sin (180+\theta)=-\sin \theta\)....(ii)

Therefore, Rewrite the given expression in terms of \((180-\theta)\) and \((180+\theta)\) we get,

\(\sin 1^{\circ}+\sin 2^{\circ}+\cdots+\sin \left(180^{\circ}\right)+\sin \left(180^{\circ}+1^{\circ}\right)+\sin \left(180^{\circ}+2^{\circ}\right)+\cdots+\sin \left(180^{\circ}+179^{\circ}\right)\)

\(=\sin 1^{\circ}+\sin 2^{\circ}+\cdots+\sin \left(179^{\circ}\right)+\sin \left(180^{\circ}\right)-\sin \left(1^{\circ}\right)-\sin \left(2^{\circ}\right)+\cdots-\sin \left(179^{\circ}\right)\)

\(=\sin 180^{\circ}\)

\(=0\)

Given,

nth term of a series is T_n = 3n + 2

Here, we have to find the value of\(S_{n}=\sum_{k=1}^{n} T_{k}=?\)

Sum of series\(=\mathrm{S}_{\mathrm{n}}=\sum(3 \mathrm{n}+2)=\sum 3 \mathrm{n}+\sum 2\)

As we know that,

\(1+2+3+4+\ldots+\mathrm{n}=\sum \mathrm{n}=\frac{\mathrm{n}(\mathrm{n}+1)}{2}\)

\(\Rightarrow \mathrm{S}_{\mathrm{n}}=3 \sum \mathrm{n}+\sum 2\)

\(\Rightarrow \mathrm{S}_{\mathrm{n}}=\frac{3 \mathrm{n}(\mathrm{n}+1)}{2}+2 \mathrm{n}=\frac{\mathrm{n}(3 \mathrm{n}+7)}{2}\)
 

Given, \(\mathrm{a}_{\mathrm{ij}}=2^{\mathrm{j}-\mathrm{i}}\)

Now, \(A=\left[\begin{array}{ccc}2^0 & 2^1 & 2^2 \\ 2^{-1} & 2^0 & 2^1 \\ 2^{-2} & 2^{-1} & 2^0\end{array}\right]\)

\(=\left[\begin{array}{ccc}1 & 2 & 4 \\ \frac{1}{2} & 1 & 2 \\ \frac{1}{4} & \frac{1}{2} & 1\end{array}\right]\)

\(A^2=\left[\begin{array}{ccc}1 & 2 & 4 \\ \frac{1}{2} & 1 & 2 \\ \frac{1}{4} & \frac{1}{2} & 1\end{array}\right]\left[\begin{array}{ccc}1 & 2 & 4 \\ \frac{1}{2} & 1 & 2 \\ \frac{1}{4} & \frac{1}{2} & 1\end{array}\right]\)

\(=\left[\begin{array}{ccc}1+1+1 & 2+2+2 & 4+4+4 \\ \frac{1}{2}+\frac{1}{2}+\frac{1}{2} & 1+1+1 & 2+2+2 \\ \frac{1}{4}+\frac{1}{4}+\frac{1}{4} & \frac{1}{2}+\frac{1}{2}+\frac{1}{2} & 1+1+1\end{array}\right]\)

\(=\left[\begin{array}{ccc}3 & 6 & 12 \\ \frac{3}{2} & 3 & 6 \\ \frac{3}{4} & \frac{3}{2} & 3\end{array}\right]\)

\(=3\left[\begin{array}{ccc}1 & 2 & 4 \\ \frac{1}{2} & 1 & 2 \\ \frac{1}{4} & \frac{1}{2} & 1\end{array}\right]\)

\(=3 \mathrm{~A}\)

Similarly, \(\mathrm{A}^3=3^2 \mathrm{~A}\)

\( A^4=3^3 A \\\)

\( \therefore A^2+A^3+\ldots \ldots+A^{10} \\\)

\( =3 A+3^2 A+3^3 A+\ldots \ldots+3^9 A \\\)

\( =A\left(3+3^2+3^3+\ldots \ldots+3^9\right) \\\)

\( =A\left(\frac{3\left(3^9-1\right)}{3-1}\right)=\frac{3\left(3^9-1\right)}{2} A=\left(\frac{3^{10}-3}{2}\right) A\)

Given two lines are \(x-6 y=0\) and \(x-y=0\).

We know \(\frac{0+x_1+x_2}{3}=1\)

\(\Rightarrow x_1+x_2=3 \).....(i)

and \( y_1+y_2=0\).........(ii)

Also, \(x_1-6 y_1=0 \).............(iii)

\(x_2-y_2=0\)......(iv)

[Since the points \(\left(x_1, y_1\right)\) and \(\left(x_2, y_2\right)\) lie on the lines \(A B\) and \(A C\) respectively]

Now on solving, the co-ordinates of \(B\) and \(C\) are \(\left(\frac{18}{5}, \frac{3}{5}\right)\) and \(\left(\frac{-3}{5}, \frac{-3}{5}\right)\) respectively.

Hence, the equation of third side i.e., \(B C\) is \(2 x-7 y-3=0\).

Given:

No. of Students in class \(=30\) students

\(30\) students have to sit on \(40\) seats

Total seats \(=(5\) row \() \times(8\) seats \()=40\) seats

Total no. of ways \(={ }^{40} C_{30}\)

For \(6^{\text {th }}\) seat of \(5^{\text {th }}\) row to be empty, all \(30\) students have to sit on remaining \(39\) seats.

No. of favourable ways \(={ }^{39} C_{30}\)

Probability (sixth seat in the fifth row will be empty)\(=\frac{\text { No. of favourable ways }}{\text { Total no. of ways }}\)

\(=\frac{{ }^{39} C_{30}}{{ }^{40} C_{30}}\)

\(=\frac{1}{4}\)

\( \because a_1, a_2, a_3 \ldots \ldots . a_n \text { are in H.P. } \\\)

\( \therefore \frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3} \cdots \frac{1}{a_n} \text { are in A.P. } \\\)

\( \therefore \frac{1}{a_2}-\frac{1}{a_1}=\frac{1}{a_3}-\frac{1}{a_2}=\ldots \ldots \ldots \ldots i=\frac{1}{a_n}-\frac{1}{a_{n-1}}=d \text { (say) } \\\)

\( \text { Then } a_1 a_2=\frac{a_1-a_2}{d}, a_2 a_3=\frac{a_2-a_3}{d} \\\)

\( \because \ldots \ldots \ldots+a_{n-1} a_n=\frac{a_{n-1}-a_n}{d}\)

Adding all equations, we get

\(\therefore a_1 a_2+a_2 a_3+\ldots \ldots \ldots+a_{n-1} a_n \\\)

\( =\frac{a_1-a_2}{d}+\frac{a_2-a_3}{d}+\ldots .+\frac{a_{n-1}-a_n}{d} \\\)

\( =\frac{1}{d}\left[a_1-a_2+a_2-a_3+\ldots .+a_{n-1}-a_n\right]=\frac{a_1-a_n}{d}\)

Also, \(\frac{1}{a_n}=\frac{1}{a_1}+(n-1) d\)

\( \Rightarrow \frac{a_1-a_n}{a_1 a_n}=(n-1) d \\\)

\( \Rightarrow \frac{a_1-a_n}{d}=(n-1) a_1 a_n\)

\( x d y-2 y d x=0 \)

\( \Rightarrow x d y=2 y d x \)

\( \Rightarrow \frac{d y}{y}=2 \frac{d x}{x}\)

Integrating both sides, we get

\( \Rightarrow \int \frac{ dy }{ y }=2 \int \frac{ dx }{ x } \)

\( \Rightarrow \ln y =2 \ln x +\ln c \)

\(\Rightarrow \ln y =\ln x ^2+\ln c \)

\( \Rightarrow \ln y -\ln x ^2=\ln c \)

\( \Rightarrow \ln \frac{ y }{ x ^2}=\ln c \)

\(\therefore y = x ^2 c\)

Given: \(|\overrightarrow{\mathrm{a}}|=10,|\overrightarrow{\mathrm{b}}|=2\) and \(\overrightarrow{\mathrm{a}}. \overrightarrow{\mathrm{b}}=12\)

As we know, \(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=|\overrightarrow{\mathrm{a}}| \cdot|\overrightarrow{\mathrm{b}}| \cos \theta\)

\(\Rightarrow 12=10 \times 2 \times \cos \theta\)

\(\Rightarrow \cos \theta=\frac{12}{20}=\frac{3}{5}\)

As we know, \(\sin \theta=\sqrt{1-\cos ^{2} \theta}=\sqrt{1-\left(\frac{3}{5}\right)^{2}}=\frac{4}{5}\)

\(|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{a}}| \cdot|\overrightarrow{\mathrm{b}}| \sin \theta\)

\(|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=10 \times 2 \times \frac{4}{5}=16\)