Mathematics Test-6

Given: \(p(x)=(4 e)^{2 x}\)

Let \(I=\int \mathrm{p}(\mathrm{x}) \mathrm{dx}\)

\(=\int(4 e)^{2 x} d x\)

Let \(2 x=t\)

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

\(\Rightarrow 2 \mathrm{dx}=\mathrm{dt}\)

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

Now,

\(\int \mathrm{p}(\mathrm{x}) \mathrm{dx}\)

\(=\frac{1}{2} \int(4 \mathrm{e})^{\mathrm{t}} \mathrm{dt}\)

\(=\frac{1}{2} \frac{(4 e)^{t}}{\ln 4 \mathrm{e}}+\mathrm{c}\)

\(=\frac{1}{2} \frac{(4 \mathrm{e})^{t}}{(\ln 4+\ln \mathrm{e})}+\mathrm{c} \quad(\because \log \mathrm{mn}=\log \mathrm{m}+\log \mathrm{n})\)

\(=\frac{1}{2} \frac{(4 \mathrm{e})^{t}}{(1+2 \ln 2)}+\mathrm{c}\)

Any square matrix can be expressed as sum of symmetric and skew-symmetric as

\(A=\frac{1}{2}\left(A+A^T\right)+\frac{1}{2}\left(A-A^T\right)\)

Where the first term is symmetric and second term is skew-symmetric Here \(S = P + Q\)

To find Q i.e. the skew-symmetric matrix is given by,

\(Q=\frac{1}{2}\left(S-S^T\right)\)

\(Q =\frac{1}{2}\left(\left[\begin{array}{cc}6 & -8 \\ 2 & 10\end{array}\right]-\left[\begin{array}{cc}6 & 2 \\ -8 & 10\end{array}\right]\right)\)

\(Q =\frac{1}{2}\left[\begin{array}{cc}0 & -10 \\ 10 & 0\end{array}\right]\)

\(Q=\left[\begin{array}{cc}0 & -5 \\ 5 & 0\end{array}\right]\)

Concept:

Properties of Determinants:

1) The value of the determinant remains unchanged if both rows and columns are interchanged.

2) If any two rows (or columns) of a determinant are interchanged, then sign of determinant changes.

3) If any two rows (or columns) of a determinant are identical (all corresponding elements are same), then the value of the determinant is zero.

4) If some or all elements of a row or column of a determinant are expressed as the sum of two (or more) terms, then the determinant can be expressed as the sum of two (or more) determinants.

5) det (A^T) = det (A)

6) det (kA) = kⁿdet(A)

7) det(A.B) = det(A).det(B)

Explanation:

Given,

Determinant (Δ) = 15

Two columns C₂ and C₃ of determinant (Δ) are interchanged

From the properties of the determinants:

If any two rows (or columns) of a determinant are interchanged, then sign of determinant changes.

Hence the value of determinant will be Δ = -15

\(3 \cos \theta=5 \sin \theta \Rightarrow \frac{\sin \theta}{\cos \theta}=\frac{3}{5}\) \(\Rightarrow \tan \theta=\frac{3}{5}=\frac{\text { Altitude }}{\text { Base }}\)

By Pythagoras Theorem,

\((\text { Hypotenuse})^2=(\text { Base })^2+(\text {Altitude})^2\)

\(\quad\quad\quad\quad\quad\quad\quad=(5)^2\quad+\quad\quad(3)^2\quad\quad\)

\(\quad \quad \quad \quad \quad\quad\quad =25+9=34\)

\(\therefore\) Hyp. \(=\sqrt{34}\)

Now \(\sin \theta=\frac{\text { Altitude }}{\text { Hypotenuse }}=\frac{3}{\sqrt{34}}\)

\(\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}=\frac{5}{\sqrt{34}}\)

\(\sec \theta=\frac{1}{\cos \theta}=\frac{\sqrt{34}}{5}\)

Now, \(\frac{5 \sin \theta-2 \sec ^3 \theta+2 \cos \theta}{5 \sin \theta+2 \sec ^3 \theta-2 \cos \theta}\)

\(=\frac{3 \cos \theta+2 \cos \theta-\frac{2}{\cos ^3 \theta}}{3 \cos \theta-2 \cos \theta+\frac{2}{\cos ^3 \theta}}\)

\(=\frac{5 \cos \theta-\frac{2}{\cos ^3 \theta}}{\cos \theta+\frac{2}{\cos ^3 \theta}} \quad\{3 \cos \theta=5 \sin \theta\}\)

\(=\frac{\frac{5 \cos ^4 \theta-2}{\cos ^3}}{\frac{\cos ^4 \theta+2}{\cos ^3 \theta}}=\frac{5 \cos ^4 \theta-2}{\cos ^3 \theta} \times \frac{\cos ^3 \theta}{\cos ^4 \theta+2}\)

\(=\frac{5 \cos ^4 \theta-2}{\cos ^4 \theta+2}\)

\(=\frac{5\left(\frac{5}{\sqrt{34}}\right)^4-2}{\left(\frac{5}{\sqrt{34}}\right)^4+2}=\frac{\frac{5 \times 625}{1156}-2}{\frac{625}{1156}+2}\)

\(=\frac{\frac{3125-2372}{625+2312}}{\frac{1156}{156}}\)

\(=\frac{813}{1156} \times \frac{1156}{2937}=\frac{271}{979}\)

We have \(|\vec{r}|=3 \sqrt{3}\)

Since, \(\vec{r}\) is equally inclined to the three axes, direction cosines of the unit vector \(\vec{r}\) will be same.

i.e., \(l=m=n\)

Now, we know that,

\(l^{2}+ m ^{2}+ n ^{2}=1 \)

\(\Rightarrow l^{2}+l^{2}+l^{2}=1 \)

\(\Rightarrow 3l^{2}=1 \)

\(\Rightarrow l^{2}=\frac{1}{3} \)

\(\Rightarrow l=\pm \frac{1}{\sqrt{3}} \)

So, \(\hat{ r }=\pm \frac{1}{\sqrt{3}} \hat{ i } \pm \frac{1}{\sqrt{3}} \hat{ j } \pm \frac{1}{\sqrt{3}} \hat{ k } \)

\(\therefore \vec{ r }=|\vec{ r }| \hat{ r } \)

\(=2 \sqrt{3}\left[\pm \frac{1}{\sqrt{3}} \hat{ i } \pm \frac{1}{\sqrt{3}} \hat{ j } \pm \frac{1}{\sqrt{3}} \hat{ k }\right] \)

\(=\pm 2[\hat{ i }+\hat{ j }+\hat{ k }]\)

As we know,

Form of a linear differential equation:

\(\frac{d y}{d x}+P y=Q\), where \(P\) and \(Q\) are the functions of \(x\)

Integrating factor \(=e^{\int P d x}\)

General solution is given by:

\(y \times e^{\int P d x}=\int Q \times e^{\int P d x} d x+C\)

Given:

Differential equation is \(\left(1-x^2\right) \frac{d y}{d x}-x y=1\)

Divide by \(\left(1-x^2\right)\) on both sides, we get

\(\Rightarrow \frac{{dy}}{{dx}}-\frac{{xy}}{\left(1-{x}^2\right)}=\frac{1}{\left(1-{x}^2\right)}\)

Now, by comparing the above equation with \(\frac{{dy}}{{dx}}+{Py}={Q}\)

So, \({P}=-\frac{{x}}{\left(1-{x}^2\right)}\)

Now,

Integrating factor \(={e}^{\int {P}} {dx}\)

\( ={e}^{\int \frac{-{x}}{\left(1-{x}^2\right)} d {x}}\)

\(={e}^{\frac{1}{2} \int \frac{-2 {x}}{\left(1-{x}^2\right)} d {x}}\)

Let \(1-x^2=t\)

Differentiating with respect to \(x\), we get

\(\Rightarrow(0-2 {x}) {dx}={dt} \)

\(\Rightarrow-2 {xdx}={dt}\)

Integrating factor \(={e}^{\frac{1}{2} \int \frac{{dt}}{t}}\)

\(={e}^{\frac{1}{2} \log \mathrm{t}}\)

\(={e}^{\log t^{\frac{1}{2}}}\)

\(={t}^{\frac{1}{2}}\)

\(=\left(1-{x}^2\right)^{\frac{1}{2}}\)

\(=\sqrt{1-{x}^2}\)

\(\int \frac{1}{ x } dx =\log x + c\)

Let \(I=\int \frac{1}{1+e^{ x }} dx =\int \frac{ e ^{- x }}{ e ^{- x }+1} dx\) Assume \(e^{-x}+1=t\)

Differenatiang with respect to \(x\), we get

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

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

\(=\int \frac{- dt }{ t }=-\log t + c =-\log \left( e ^{- x }+1\right)+ c =\log \left(\frac{1}{ e ^{-x}+1}\right)+ c\)

\( =\log \left(\frac{ e ^{ x }}{1+ e ^{ x }}\right)+ c\)

There are \(2\) diagonals with \(8\) squares each on the chess board and we have to choose \(6\) consecutive square.

We will assume block of \(6\) consecutive squares as one.

So, we are left with \(3\) places on the chess board where it can be arranged by selecting any one of the place out of three i. e., \({ }^{3} \mathrm{C}_{1}=3\).

Similarly, for other diagonal too we will get \(3\) different ways

So, in total we have \(6\) different ways of choosing \(6\) consecutive squares.

I. 15x² – 11x + 2 = 0

15x² – 5x – 6x + 2 = 0

5x (3x – 1) – 2 (3x – 1) = 0

(3x – 1) (5x – 2) = 0

x =$\frac{1}{3},\frac{2}{5}$

II. 10y² – 9y + 2 = 0

10y² – 5y – 4y + 2 = 0

5y (2y – 1) –2 (2y – 1) = 0

(2y – 1) (5y – 2) = 0

y =$\frac{1}{2},\frac{2}{5}$

Therefore,

x ≤ y

As we know,

The equation of a circle is given by:

\(\left(x-h\right)^{2}+\left(y-k\right)^{2}=r^{2}\)

Where \(\left( h , k \right)\) is the center of the circle and \(r\) is the radius.

Given,

\(x=2+3 \cos \theta \)

\(\Rightarrow 3 \cos \theta=x-2 \)..(i)

Also \( y=1-3 \sin \theta \)

\(\Rightarrow 3 \sin \theta=-y+1 \)...(ii)

Squaring adding (i) & (ii), we get

\(\left( x -2\right)^{2}+\left(-( y -1)\right)^{2}=3^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)\)

\(\Rightarrow( x -2)^{2}+\left(-( y -1)\right)^{2}=3^{2}\)

Comparing it with standard equation of a circle, we get

Centre \(=\left(h, k\right)=\left(2,1\right)\) and radius \(=3\)