Mathematics Test-14

Given:

\(\mathrm{I=\int \frac{x e^{x}}{\sqrt{1+e^{x}}} d x}\) .....(A)

Take \(1+\mathrm{e}^{\mathrm{x}}=\mathrm{t}^{2} \quad \ldots .(1)\)

Differentiating with respect to \(\mathrm{x}\), we get

\( \mathrm{e}^{\mathrm{x}} \mathrm{dx}=2 \mathrm{tdt}\)

From equation (1), we get

\(\mathrm{e^{x}=t^{2}-1}\)

So, \(\mathrm{x}=\log \left(\mathrm{t}^{2}-1\right)\)

Now, put the above values in equation (A)

\(\mathrm{I}=\int \frac{\log \left(\mathrm{t}^{2}-1\right)}{\sqrt{\mathrm{t}^{2}}} 2 \mathrm{tdt}\)

\(=2 \times \int \frac{\log \left(\mathrm{t}^{2}-1\right)}{\mathrm{t}} \times \mathrm{tdt}\)

\(=2 \int \log \left(\mathrm{t}^{2}-1\right) \mathrm{d t}\)

Using integration by parts rule, we get

Integration by parts:

\( \int \mathrm{uvdx}=\mathrm{u} \int \mathrm{vdx}-\int\left(\frac{\mathrm{du}}{\mathrm{dx}} \times \int \mathrm{v} \mathrm{dx}\right) \mathrm{dx}+\mathrm{c}\)

\(=2\left[\mathrm{\log \left(t^{2}-1\right) \times t-2 \int \frac{t^{2}}{t^{2}-1} d t}\right]\)

\(=\mathrm{2 t \log \left(t^{2}-1\right)-4 \int\left[1+\frac{1}{t^{2}-1}\right] d t}\)

\(=\mathrm{2 t \log \left(t^{2}-1\right)-4 t-4 \times \frac{1}{2} \log \left(\frac{t-1}{t+1}\right)+c}\)

\(=\mathrm{2 t \log \left(t^{2}-1\right)-4 t-2 \log \left(\frac{t-1}{t+1}\right)+c}\)

\(=\mathrm{2 t\left(\log \left(t^{2}-1\right)-2\right)-2 \log \left(\frac{t-1}{t+1}\right)+c}\)

Resubstitute the value of \(\mathrm{t}\), we get

\(=\mathrm{2(x-2) \sqrt{1+e^{x}}-2 \log \frac{\sqrt{1+e^{x}}-1}{\sqrt{1+e^{x}}+1}+c}\)

\(=(2 \mathrm{x}-4) \sqrt{1+\mathrm{e}^{\mathrm{x}}}-2 \log \frac{\sqrt{1+\mathrm{e}^{\mathrm{x}}}-1}{\sqrt{1+\mathrm{e^{x}}}+1}+\mathrm{c}\).......(2)

On comparingequation(2) with question, we get

\(\mathrm{f(x) = 2 x-4}\)

Given sequence,

1, 5, 9, 13, 17, ...

Difference between the first term and second term = 4

Difference between the second term and third term = 4

Thus, clearly it is an AP with common difference 4.

\(\mathrm{n}^{\text {th }}\) term of the \(\mathrm{AP}=a_{n}\)

\(a_{n}= a_{1}+(n-1) d\)

\(\Rightarrow a_{n}=1+(n-1) 4\)

\(\Rightarrow a_{n}=1+4 n-4\)

\(\Rightarrow a_{n}=4 n-3\)

Given,

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

Direction ratio \(P Q\),

\(P Q =(1-(-5)) \hat{i}+(5-3) \hat{j}+(-2-1) \hat{k} \)

\(=6 \hat{i}+2 \hat{j}-3 \hat{k}\)

For direction cosine \(\Rightarrow \frac{P Q}{|P Q|}\)

\(|P Q|= \sqrt{6^2+2^2+3^2}\)

\(=\sqrt{36+4+9}=\sqrt{49} \)

\(|P Q| =7\)

\(\therefore\) direction cosine \(=\left(\frac{6}{7}, \frac{2}{7}, \frac{-3}{7}\right)\)

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

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

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

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

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

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:

1. \(\overline{\left( z ^{-1}\right)}=(\overline{ z })^{-1}\)

Let, \(z=a-i b\)

\(\Rightarrow z ^{-1}=\frac{1}{ z }=\frac{1}{ a - ib }=\frac{ a + ib }{ a ^{2}+ b ^{2}}\) 

\(\Rightarrow \overline{ z ^{-1}}=\frac{ a - ib }{ a ^{2}+ b ^{2}} \)     .....(i)

\(\Rightarrow \overline{ z }= a + ib\) 

\(\Rightarrow(\overline{ z })^{-1}=\frac{1}{\overline{ z }}=\frac{1}{ a + ib }=\frac{ a - ib }{ a ^{2}+ b ^{2}}\)     .....(ii)

\(\Rightarrow(\overline{ z })^{-1}=\frac{1}{\overline{z}}=\frac{1}{a+ i b}=\frac{a- ib }{a^{2}+b^{2}}\)

From equation (i) and (ii)

Statement 1 is correct.

2. \(z z^{-1}=|z|^{2}\)

Consider \(z=a-i b\)

\(\Rightarrow z^{-1}=\frac{a+i b}{a^{2}+b^{2}}\) 

\(\Rightarrow z \cdot z^{-1}=(a-i b) \frac{a+i b}{a^{2}+b^{2}}=\frac{a^{2}+b^{2}}{a^{2}+b^{2}}=1\)

Statement 2 is not correct.

Given, \(AL=3.2\) units

Since the angular bisector divides the chord into equal parts.

\(\therefore AL=LB\)

\(LB=3.2\) units

\(AB= AL+LB\)

\(\Rightarrow 3.2+3.2\)

\(=6.4\) units

Let \(A\) and \(B\) be the positions of the two flag posts and \(P(x, y)\) be the position of the man.

Accordingly, \(PA+PB=10\)

We know that if a point moves in a plane in such a way that the sum of its distances from two fixed points is constant, then the path is an ellipse and this constant value is equal to the length of the major axis of the ellipse.

Therefore, the path described by the man is an ellipse where the length of the major axis is \(10~m\), while points \(A\) and \(B\) are the foci.

Taking the origin of the coordinate plane as the center of the ellipse, while taking the major axis along the \(x\)-axis, the ellipse can be diagrammatically represented as

The equation of the ellipse will be of the form \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where \(a\) is the semi-major axis

Accordingly, \(2 a=10\)

\(\Rightarrow a=5\)

Distance between the foci \((2 c)=8\)

\(\Rightarrow c=4\)

On using the relation \(c=\sqrt{a^{2}-b^{2}}\), we obtain

\(4=\sqrt{25-b^{2}}\)

\(\Rightarrow 16=25-b^{2}\)

\(\Rightarrow b^{2}=25-16=9\)

Thus, the equation of the path traced by the man is \(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\)