Mathematics Test - 20

We are given the differential equation as \(x \frac{d y}{d x}+2 y=x^{2}\).

First, we will convert our differential equation into the linear form.

We have:

\(x \frac{d y}{d x}+2 y=x^{2}\)

Dividing both the sides by \(x\), we get,

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

So, comparing with \(\frac{d y}{d x}+P y=Q\), we get,

\(P=\frac{2}{x}\) and \(Q=x\).

Now, to solve further we will find the integrating factor.

The integrating factor is given as

\(I.F.=e^{\int P\ d x}\)

As, \(P=\frac{2}{x}\), so we get,

\(\Rightarrow I . F.=e^{\int \frac{2}{x} d x}\)

As, \(\int \frac{1}{x} d x=\log x\), so we get,

\(\Rightarrow I. F.=e^{2 \log x}\)

As, \(n \log x=\log x^{n}\), so we get,

\(\Rightarrow I. F.=e^{\log x^{2}}\)

Now, we know that,

\(e^{\log x}=x, e^{\log x^{2}}=x^{2}\), so we get,

\(\Rightarrow I .F.=x^{2}\)

Now, we know the solution is given as,

\(y \times I. F.=\int(Q \times I. F.) d x+C\)

As, \(I.F.=x^{2}, Q=x\), so we get,

\(\Rightarrow y x^{2}=\int\left(x \times x^{2}\right) d x\)

\(\Rightarrow y x^{2}=\int x^{3} d x\)

As, \(\int x^{n} d x=\frac{x^{n+1}}{n}\), so,

\(\Rightarrow y x^{2}=\frac{x^{4}}{4}+C\)

Dividing both the sides by \(x^{2}\), we get,

\(\Rightarrow y=\frac{x^{2}}{4}+\frac{C}{x^{2}}\)

Now, we have \(y(1)\) as 1, i.e., \(y(1)=1\).

So, putting \(x=1\) and \(y=1\), we get,

\(\Rightarrow 1=\frac{1^{2}}{4}+\frac{C}{1^{2}}\)

\(\Rightarrow 1=\frac{1}{4}+\frac{C}{1}\)

\(\Rightarrow 1=\frac{1}{4}+C\)

Solving for \(C\), we get,

\(\Rightarrow C=1-\frac{1}{4}=\frac{3}{4}\)

\(\Rightarrow C=\frac{3}{4}\)

So, putting it in the value of \(y\), we get,

\(\Rightarrow y=\frac{x^{2}}{4}+\frac{3}{4 x^{2}}\)

Given,

\(A=\left[\begin{array}{lll}1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{array}\right]\)

\( \mathrm{A}^{4}=\mathrm{A}^{2}+\mathrm{A}^{2}-\mathrm{I}=2 \mathrm{~A}^{2}-\mathrm{I}\)

\(= \mathrm{A}^{6}=\mathrm{A}^{4}+\mathrm{A}^{2}-\mathrm{I}=3 \mathrm{~A}^{2}-2 \mathrm{I}\)

Similarly,

\(\mathrm {A^{50}=25 A^{2}-24 I}\)

\( \mathrm{A}^{2}=\left[\begin{array}{lll}1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{array}\right]\left[\begin{array}{lll}1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{array}\right]\)

\(\therefore \mathrm{A}^{50}=\left[\begin{array}{ccc}25 & 0 & 0 \\ 25 & 25 & 0 \\ 25 & 0 & 25\end{array}\right]-\left[\begin{array}{ccc}24 & 0 & 0 \\ 0 & 24 & 0 \\ 0 & 0 & 24\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0 \\ 25 & 1 & 0 \\ 25 & 0 & 1\end{array}\right]\)

\(\Rightarrow \mathrm{A}^{50}=\left[\begin{array}{ccc}1 & 0 & 0 \\ 25 & 1 & 0 \\ 25 & 0 & 1\end{array}\right]\)

\(\therefore\left|\mathrm{A}^{50}\right|=1\)

Let \(y\) be an arbitrary element of Range \(f\).Then, there exists \(x \in R-\left\{-\frac{4}{3}\right\}\) such that \(y=f(x)\)
\(\Rightarrow y=\frac{4 x}{3 x+4}\)\(\Rightarrow 3 x y+4 y=4 x\)\(\Rightarrow x(4-3 y)=4 y\)\(\Rightarrow x=\frac{4 y}{4-3 y}\)
Let us define g: Range \(\mathrm{f} \rightarrow \mathrm{R}-\left\{-\frac{4}{3}\right\}\) as \({g}({y})=\frac{4 {y}}{4-3 {y}}\)
Now,\(g o f(x)=g(f(x))=g\left(\frac{4 x}{3 x+4}\right)=\frac{4\left(\frac{4 x}{3 x+4}\right)}{4-3\left(\frac{4 x}{3 x+4}\right)}=\frac{16 x}{12 x+16-12 x}=\frac{16 x}{16}=x\) and \(f o g(y)=f(g(y))=f\left(\frac{4 y}{4-3 y}\right)=\frac{4\left(\frac{4 y}{4-3 y}\right)}{3\left(\frac{4 y}{4-3 y}\right)+4}=\frac{16 y}{12 y+16-12 y}=\frac{16 y}{16}=y\) \(\therefore g o f=I_{R-\left\{-\frac{4}{3}\right\}}\) and fog \(=I_{\text {Range } f}\)
Thus, \(g\) is the inverse of \(f\) i.e. \(f^{-1}=g\).
Thus, the inverse of \(f\) is the map \(g:\) Range \(f \rightarrow R-\left\{-\frac{4}{3}\right\},\) which is given by \(g(y)=\frac{4 y}{4-3 y}\).

We know that,

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

It is given that,

\(\tan \left\{2 \tan ^{-1}\left(\frac{1}{3}\right)\right\},\) here \(x=\frac{1} {3}\) and \(-1

\(\Rightarrow 2 \times \tan ^{-1}\left(\frac{1}{3}\right)=\tan ^{-1}\left(\frac{2 \times \frac{1}{3}}{1-\left(\frac{1}{3}\right)^{2}}\right)=\tan ^{-1}\left(\frac{3}{4}\right)\)

\(\Rightarrow \tan \left\{2 \tan ^{-1}\left(\frac{1}{3}\right)\right\}=\tan \left(\tan ^{-1}\left(\frac{3}{4}\right)\right)=\frac{3}{4}\)

Given:

A bag contains 5 black and 3 white.

So, Total number of balls =5+3=8

Now, P(both the balls are white)$=\frac{3}{8}\times \frac{2}{7}=\frac{3}{28}$

We know that:

Suppose a set of \(n\) objects has \(n_{1}\) of one kind of object, \(n_{2}\) of a second kind, \(n_{3}\) of a third kind, and,

So, on with \(n=n_{1}+n_{2}+n_{3}+\ldots+n_{k}\) then the number of distinguishable permutations of the \(n\) objects is:

\(=\frac{n !}{n_{1} ! \times n_{2} ! \times n_{3} ! \ldots \ldots . n_{k} !}\)

Given: Three a’s and two b’s

Given: \(\int(x-1) e^{-x} d x\)

Let \(I=\int(x-1) e^{-x} d x\)

On separating the terms, we have

\(I=\int x e^{-x} d x-\int e^{-x} d x\)

For first term, using product rule

\(I=-x e^{-x}-\int 1 \cdot(-) e^{-x} d x-\int e^{-x} d x+C\)

On rewriting the terms, we have

\(I=-x e^{-x}+\int e^{-x}-\int e^{-x} d x+C\)

Here, two integrals get cancelled out and we are only left with algebraic terms which do not requires any further integration.

\(\Rightarrow I=-x e^{-x}+C\), where \(C\) is the constant of integration.

Given,

\(A=\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right]\)

\(A^{2}=A \cdot A\)

\(=\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right] \times\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right]\)

\(=\left[\begin{array}{cc}1+1 & -1-1 \\ -1-1 & 1+1\end{array}\right]\)

\(=\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right]\)

\(A^{3}=A^{2} \cdot A\)

\(=\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right] \times\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right]\)

\(=\left[\begin{array}{cc}2+2 & -2-2 \\ -2-2 & 2+2\end{array}\right]\)

\(=\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right]\)

\(=\left[\begin{array}{cc}4 & -4 \\ -4 & 4\end{array}\right]\)

Now,

\(A^{3}=\left[\begin{array}{cc}4 & -4 \\ -4 & 4\end{array}\right]\) and

\(2 A^{2}=2 \times\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right]\)

\(=\left[\begin{array}{cc}4 & -4 \\ -4 & 4\end{array}\right]\)

\(\therefore A^{3}-2 A^{2}=\left[\begin{array}{cc}4 & -4 \\ -4 & 4\end{array}\right]-\left[\begin{array}{cc}4 & -4 \\ -4 & 4\end{array}\right]\)

\(=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]\)

So, the expression \( A^{3}-2 A^{2}\) is a null matrix.

Given: \(3 x+2 y+1=0\)

\(\Rightarrow 2 y=-3 x-1\)

\(\Rightarrow y=\frac{-3}{2} x-\frac{1}{2}\)

Compare this with the slope-intercept form to find the slope of the line: \(y=m x+c\)

So, \(m=\frac{-3}{2}\)

Now the slope of the given line is \(m=\frac{-3}{2}\), since the given line is parallel to the required line

we know that the slope of the parallel is the same, i.e. under the condition of parallel lines \(m_{1}=m_{2}\)

Now the slope of the required line is also equals to \(m=\frac{-3}{2}\)

Equation of line:

\(y-y_{1}=m\left(x-x_{1}\right)\)

\(\Rightarrow y+2=\frac{-3}{2}(x+1)\)

\(\Rightarrow 2 \times(y+2)=-3 \times(x+1)\)

\(\Rightarrow 2 y+4=-3 x-3\)

\(\Rightarrow 3 x+2 y+4+3=0\)

\(\Rightarrow 3 x+2 y+7=0\)

Given:

\(y=x \frac{d y}{d x}+\left(\frac{d y}{d x}\right)^{-1}\)

The given differential equation can be written as:

\(\Rightarrow y=x \frac{d y}{d x}+\frac{1}{\frac{d y}{d x}}\)

\(\Rightarrow y \frac{d y}{d x}=x\left(\frac{d y}{d x}\right)^{2}+1\)

As we know that, the highest order derivative occurring in a differential equation is called the order of a differential equation

So, for the given differential equation order is 1.

As we know that, the power of the highest order derivative which occurs in it, after it is made free from radicals and fractions is called the degree of a differential equation.

So, for the given differential equation degree is 2.