Mathematics Test - 24

Given, \(z_{1}=2-i\) and \(z_{2}=1+{i}\)

\(\therefore\left|\frac{z_{1}+z_{2}+1}{z_{1}-z_{2}+1}\right|=\left|\frac{(2-i)+(1+i)+1}{(2-i)-(1+i)+1}\right|\)

\(=\left|\frac{4}{2-2 i}\right|=\left|\frac{4}{2(1-i)}\right|\)

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

\(=\left|\frac{2(1+i) }{1+1}\right| \quad\quad \left[\because i^{2}=-1\right]\)

\(=\left|\frac{2(1+i)}{2}\right|\)

\(=|1+i|=\sqrt{1^{2}+1^{2}}=\sqrt{2}\)

Given:

\(\sin 2 \theta=\cos 3 \theta, 0<\theta<\frac{\pi}{2}\)

As we know that, \(\sin 2 \theta=2 \sin \theta \cos \theta\) and \(\cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta\).

\( 2 \sin \theta \cos \theta=4 \cos ^{3} \theta-3 \cos \theta\)

\(2 \sin \theta=4 \cos ^{2} \theta-3\)

\(2 \sin \theta=4\left(1-\sin ^{2} \theta\right)-3=4-4 \sin ^{2} \theta-3\)

\(4 \sin ^{2} \theta+2 \sin \theta-1=0\)

Comparing the above equation with quadratic equation \(a x^{2}+b x+c=0, a=4, b=2\) and \(c=-1\)

Now substituting the values in the quadratic formula \(x=\frac{\left(-b \pm \sqrt{b^{2}-4 a c}\right)}{2 a}\) we get,

\(\sin \theta=\frac{-2 \pm \sqrt{2^{2}-4(4)(-1)}}{2(4)}=\frac{-2 \pm \sqrt{4+16}}{8}=\frac{-2 \pm \sqrt{20}}{8}=\frac{-1 \pm \sqrt{5}}{4}\)

Thus, \(\sin \theta=\frac{-1 \pm \sqrt{5}}{4}\)

Since, \(0<\theta<\frac{\pi }{2} = \theta\) lies between \(0^{\circ}\) to \(90^{\circ} =\) all ratios are positive.

\(\sin \theta=\frac{-1+\sqrt{5}}{4}\)

As we know that, \(\cos 2 \theta=\cos ^{2} \theta-\sin ^{2} \theta=1-2 \sin ^{2} \theta\)

\(\cos 2 \theta=1-2\left(\frac{-1+\sqrt{5}}{4}\right)^{2}=\frac{1+\sqrt{5}}{4}\)

For rectangle:

Let breadth \(={x}\)

Since breadth is \(3 \mathrm{~m}\) less than length,

Length is \(3 \mathrm{~m}\) more than breadth

So, Length \(=x+3 \)

Area \(=\) Length \( \times\) Breadth

\(=x(x+3)\)

For triangle:

Area of triangle:

\(=\frac{1}{2} \times\) Base \(\times\) Altitude

\(=\frac{1}{2} \times \mathrm{CD} \times\) Altitude

\(=\frac{1}{2} \times \mathrm{AB} \times\) Altitude \(\quad\quad\)(Opposite sides of rectangle are equal

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

\(=6 {x}\)

Given that area of a rectangular park is \(4 \mathrm{~m}^{2}\) more than the area of the isosceles park.

So, Area of rectangle \(=4+\) Area of triangle,

\(x(x+3)=4+6 x\)

\(x^{2}+3 x=4+6 x\)

\(x^{2}+3 x-6 x-4=0\)

\(x^{2}-3 x-4=0\)

Comparing equation with \(a x^{2}+b x+c=0\)

Here, \(a=1, b=-3, c=-4\)

We know that:

\(\ D =b^{2}-4 a c \)

\( =(-3)^{2}-4 \times 1 \times(-4) \)

\(=9+16 \)

\( =25\)

Thus, roots to equation are

\(\mathrm{x}=\frac{-b \pm \sqrt{D}}{2 a}\)

Putting values

\(x=\frac{-(-3) \pm \sqrt{25}}{2 \times 1} \)

\(x=\frac{+3 \pm \sqrt{5^{2}}}{2} \)

\(x=\frac{3 \pm 5}{2}\)

Solving

\(\begin{array}{l|l}{x}=\frac{3+5}{2} & {x}=\frac{3-5}{2} \\{x}=\frac{8}{2} & {x}=\frac{-2}{2} \\{x}=4 & {x}=-1\end{array}\)

So, \(x=4\) and \( x=-1\)

Since \({x}\) is the breadth of the rectangle, \({x}\) cannot be negative.

So, \(x=4\) is the solution.

Breadth \(=x=4 \mathrm{~m}\)

Length \(=x+3=4+3 m=7 \mathrm{~m}\)

Given: \(\int_{0}^{1} \frac{1-x}{1+x} d x\)

Let \(I=\int_{0}^{1} \frac{1-x}{1+x} d x\)

Adding and subtracting 1 to the numerator, we have 

\(\Rightarrow I=\int_{0}^{1} \frac{2-1-x}{1+x} d x\)

\(\Rightarrow I=\int_{0}^{1} \frac{2-(1+x)}{1+x} d x\)

Splitting the terms in numerator, we get 

\(\Rightarrow I=\int_{0}^{1}\left(\frac{2}{1+x}-\frac{1+x}{1+x}\right) d x\)

\(\Rightarrow I=\int_{0}^{1}\left(\frac{2}{1+x}-1\right) d x\)

\(\Rightarrow I=\int_{0}^{1} \frac{2}{1+x} d x-\int_{0}^{1} 1 d x\)

By integrating, we get

\(\Rightarrow I=[2 \log (1+x)]_{0}^{1}-[x]_{0}^{1}\)

\(\Rightarrow I=2[\log (1+x)]_{0}^{1}-[x]_{0}^{1}\)

Substituting the integral values, we get

\(\Rightarrow I=2[\log (1+1)-\log (1+0)]-[1-0]\)

\(\Rightarrow I=2[\log (2)-\log (1)]-[1-0]\)

\(\Rightarrow I=2[\log 2-0]-1\)

\(\Rightarrow I=2 \log 2-1\)

Therefore, the solution of definite integral \(\int_{0}^{1} \frac{1-x}{1+x} d x\) is \(2 \log 2-1\).

We have to select 2 persons out of 76 for handshakes.

\(\therefore\) Number of handshakes \(={ }^{76} \mathrm{C}_{2}=\frac{76 \times 75}{2 \times 1}\)

= 38 × 75

= 2850

According to the question,

We have to find the value of \(\left[\begin{array}{lll}\mathrm{x} & \mathrm{y} & \mathrm{z}\end{array}\right]\left[\begin{array}{lll}\mathrm{a} & \mathrm{h} & \mathrm{g} \\ \mathrm{h} & \mathrm{b} & \mathrm{f} \\ \mathrm{g} & \mathrm{f} & \mathrm{c}\end{array}\right]\).

As we know, the matrix \(\left[\begin{array}{lll}\mathrm{x} & \mathrm{y} & \mathrm{z}\end{array}\right]\) has 1 row and 3 columns, therefore the order of the matrix is \(1 \times\)3.

Also, since the matrix \(\left[\begin{array}{lll}\mathrm{a} & \mathrm{h} & \mathrm{g} \\ \mathrm{h} & \mathrm{ b} & \mathrm{f} \\ \mathrm{g} & \mathrm{f} & \mathrm{c}\end{array}\right]\) has 3 rows and 3 columns, therefore the order of the matrix is \(\mathrm{3} \times \mathrm{3}\).

We know that, if \(\mathrm{A}\) is a matrix with order \({m} \times {n}\) and \(\mathrm{B}\) is a matrix with order \({n} \times {p}\), then the matrix \(\mathrm{AB}\) has the order \({m} \times {p}\).

Now we have to compute the order of \(\left[\begin{array}{lll}\mathrm{x} & \mathrm{y} & \mathrm{z}\end{array}\right]\left[\begin{array}{lll}\mathrm{a} & \mathrm{h} & \mathrm{g} \\ \mathrm{h} & \mathrm{b} & \mathrm{f} \\ \mathrm{g} & \mathrm{f} & \mathrm{c}\end{array}\right]\), so the product of the orders of the matrices to get the result that is:

\((1 \times 3)(3 \times 3)\)

\(=(1 \times 3)\)

Therefore,

\(\left[\begin{array}{lll}\mathrm{x} & \mathrm{y} & \mathrm{z}\end{array}\right]\left[\begin{array}{lll}\mathrm{a} & \mathrm{h} & \mathrm{g} \\ \mathrm{h} & \mathrm{b} & \mathrm{f} \\ \mathrm{g} & \mathrm{f} & \mathrm{c}\end{array}\right]\)

\(=[\mathrm{ax}+\mathrm{h} \mathrm{y}+\mathrm{g} \mathrm{z} \quad \mathrm{hx}+\mathrm{by}+\mathrm{fz} \quad \mathrm{gx}+\mathrm{fy}+\mathrm{cz}]\)

The given differential equation is:

\(\left(e^{x}+e^{-x}\right) d y-\left(e^{x}-e^{-x}\right) d x=0 \)

\(\Rightarrow\left(e^{x}+e^{-x}\right) d y=\left(e^{x}-e^{-x}\right) d x \)

\(\Rightarrow d y=\left[\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\right] d x\)

Integrating both sides of this equation, we get:

\(\int d y=\int\left[\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\right] d x+\mathrm{C}\)

\(\Rightarrow y=\int\left[\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\right] d x+\mathrm{C}\quad\quad\)....(1)

Let \(\left(e^{x}+e^{-x}\right)=t\)

Differentiating both sides with respect to \(x\), we get:

\(\frac{d}{d x}\left(e^{x}+e^{-x}\right)=\frac{d t}{d x}\)

\(\Rightarrow e^{x}-e^{-x}=\frac{d t}{d t}\)

\(\Rightarrow\left(e^{x}-e^{-x}\right) d x=d t\)

Substituting this value in equation (1), we get:

\(y=\int \frac{1}{t} d t+\mathrm{C}\)

\(\Rightarrow y=\log (t)+\mathrm{C}\)

\(\Rightarrow y=\log \left(e^{x}+e^{-x}\right)+C\)

Now, we are given a complex number \({z}\), where \({z}={x}+{iy}\).

Now, we will apply the property of complex numbers. 

We will use the property: \(|{x}+{iy}|=\sqrt{{x}^{2}+({iy})^{2}}\).

We will put the value of \(z\) in the given condition \(\left|\frac{z-5 i}{z+5 i}\right|=1\) and also apply the above property.

Now, putting \(z=x+\) iy in the given condition, we get \(\left|\frac{{x}+{i} {y}-5 {i}}{{x}+{i} {y}+5 {i}}\right|=1\).

Now, separating the real part and imaginary part, we get \(\left|\frac{{x}+{i}({y}-5)}{{x}+{i}({y}+5)}\right|=1\).

Cross-multiplying both sides, we get:

\(|{x+i(y-5)|=|x+i(y+5)}|\)

Using the property \(|{x}+{iy}|=\sqrt{{x}^{2}+({iy})^{2}}\) in the above equation,

\(\sqrt{{x}^{2}+({i}({y}-5))^{2}}=\sqrt{{x}^{2}+({i}({y}+5))^{2}}\)

\({\sqrt{x^{2}-(y-5)^{2}}=\sqrt{x^{2}-(y+5)^{2}}}\) as \(i^{2}=-1\)

Squaring both sides, we get:

\({x^{2}-(y-5)^{2}=x^{2}-(y+5)^{2}}\)

Eliminating same terms from both the sides of the above equation, we get:

\(({y-5)^{2}=(y+5)^{2}}\)

\(({y-5)^{2}-(y+5)^{2}=0}\)

Using property \(a^{2}-b^{2}=(a-b)(a+b)\) in the above equation.

\(\Rightarrow({y-5-y+5)(y-5+y+5)=0}\)

\(\Rightarrow{-10 y=0}\) or \({y=0}\)

Now, \( y=0\) represents the \(x\)-axis. 

So, all the complex numbers satisfying \(\left|\frac{z-5 i}{z+5 i}\right|=1\) lies on the \(x\)-axis.

Clearly, there are 2 ways of answering each of the 5 questions i.e true or false.

\(\therefore\) Total number of different sequence of answers \(=2 \times 2 \times 2 \times 2 \times 2=32\)

There is only one correct sequence.

So, the maximum number of sequences except for the all correct answer sequence \(=32-1=31\)

\(\because\) Different students have given different sequence of answers, so the maximum possible number of students \(=31\)

Given: \(A=\left[\begin{array}{ccc}2 & -3 & 4 \\ 1 & 0 & -1 \\ 3 & 1 & -2\end{array}\right]\)

Here, we have to find the value of \(M_{13} \times C_{21}+C_{33} \times C_{31}\) for the given matrix

\(M_{13}=\left|\begin{array}{cc}1 & 0 \\ 3 & 1\end{array}\right|=1-0=1\)

\(C_{21}=(-1)^{3} \times\left|\begin{array}{cc}-3 & 4 \\ 1 & -2\end{array}\right|=-1 \times(6-4)=-2\)

\(C_{31}=(-1)^{4} \times\left|\begin{array}{cc}-3 & 4 \\ 0 & -1\end{array}\right|=1 \times(3-0)=3\)

\(C_{33}=(-1)^{6} \times\left|\begin{array}{cc}2 & -3 \\ 1 & 0\end{array}\right|=1 \times(0+3)=3\)

\(M_{13} \times C_{21}+C_{33} \times C_{31}=1 \times(-2)+3 \times 3=7\)