Mathematics Test - 7

\(y=\sqrt{16-x^{2}}\) and \(x\)-axis

 At \(x\)-axis, \(y\) will be zero

\(y=\sqrt{16-x^{2}}\)

\(\Rightarrow 0=\sqrt{16-x^{2}}\)

\(\Rightarrow 16-x^{2}=0\)

\(\Rightarrow x^{2}=16\)

\(\therefore x=\pm 4\)

So, the intersection points are (4, 0) and (−4, 0)

Area of the curve, \(\mathrm{A}=\int_{-4}^{4} \sqrt{16-\mathrm{x}^{2}} \mathrm{dx} \) \(\because \int (\sqrt{a^2-x^2})= \frac{x} {2} \sqrt{a^2-x^2}+ \frac{a^2} {2}~ \text{sin}^{-1} \frac{x}{4}\)

\(=\left[\frac{x}{2} \sqrt{4^{2}-x^{2}}+\frac{4^{2}}{2} \sin ^{-1} \frac{x}{4}\right]_{-4}^{4}\)

\(=\left[\frac{4}{2} \sqrt{4^{2}-4^{2}}+\frac{4^{2}}{2} \sin ^{-1} \frac{4}{4}\right]-\left[\frac{-4}{2} \sqrt{4^{2}-(-4)^{2}}+\frac{4^{2}}{2} \sin ^{-1} \frac{-4}{4}\right]\)

\(=8 \frac{\pi}{2}+8 \frac{\pi}{2}\) \(=8 \pi\) sq. units

Given,

\(|\frac2{(x-4)}|>1\)

\(\Rightarrow \frac2{|x-4|}>1\)

\(\Rightarrow 2>|x-4|\)

\(\Rightarrow|x-4|<2\)

\(\Rightarrow-2

\(\Rightarrow-2+4

\(\Rightarrow 2

\(\Rightarrow x \in(2,6)\), where \(x \neq 4\)

\(\Rightarrow x \in(2,4) \cup(4,6)\)

Given: Vector \(\vec{r}=a \hat{i}+b \hat{j}\) is equally inclined to both \(x\) and \(y\) axes and the magnitude of the vector is \(2\) units.

i.e.,\(|\vec{r}|=2\)

\(\Rightarrow \sqrt{a^{2}+b^{2}}=2\)

\(\Rightarrow a^{2}+b^{2}=4 \quad \ldots\) (1)

\(\because\) The vector \(\vec{r}=a \hat{i}+b \hat{j}\) is equally inclined to both \(x\) and \(y\) axes

Let \(\theta\) be the angle between the vector \(\vec{r}=a \hat{i}+b \hat{j}\) and both the \(x\) and \(y\) axes.

\(\Rightarrow \cos \theta=\frac{a}{2}=\frac{b}{2} \)

\(\Rightarrow a = b\)

So, by substituting \(a=b\) in equation (1), we get

\(\Rightarrow 2 b^{2}=4 \Rightarrow b=\sqrt{2}\)

So, \(a=b=\sqrt{2}\)

Given,

The constraints \(-x_{1}+x_{2} \leq 1, -x_{1}+3 x_{2} \leq 9\) and \(x_{1}, x_{2} \geq 0\)

By drawing these equations, we get 

Thus we can say that the constraints \(-x_{1}+x_{2} \leq 1, -x_{1}+3 x_{2} \leq 9\) and \(x_{1}, x_{2} \geq 0\) defines on unbounded feasible space because in the graph shaded area is not enclosed by these given lines.

Given that:

\(R=\left\{(a, b): a^{2} \geq b\right\}\)

As we know:

A relation R on a set A is said to be an equivalence relation if R is reflexive, symmetric and transitive.

We know that:

\(\mathrm{a}^{2} \geq \mathrm{a}\)

Therefore, \((\mathrm{a}, \mathrm{a}) \in \mathrm{R}\), for all \(\mathrm{a} \in \mathrm{Z}\).

Thus, relation \(\mathrm{R}\) is reflexive.

Let \((\mathrm{a}, \mathrm{b}) \in \mathrm{R}\)

\(\Rightarrow a^{2} \geq b\) but \(b^{2} \ngeq a\) for all \(a, b \in Z\).

So, if \((a, b) \in R\), then it does not implies that \((b, a)\) also belongs to \(\mathrm{R}\).

Thus, relation \(\mathrm{R}\) is not symmetric.

Now, let \((a, b) \in R\) and \((b, c) \in R\).

\(\Rightarrow a^{2} \geq b\) and \(b^{2} \geq c\)

This does not implies that \(\mathrm{a}^{2} \geq \mathrm{c}\), therefore \((\mathrm{a}, \mathrm{c})\) does not belong to \(\mathrm{R}\) for all \(\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{Z}\).

Thus, relation \(\mathrm{R}\) is not transitive.

Concept:

Poisson distribution formula,

\(P(x)=\frac{e^{-\lambda} \lambda^{x}}{x !}\)

where \(\lambda=\) mean value of occurrence within an interval

\(P(x)=\) probability of \(x\) occurrence within an interval

Calculation:

Mean \((\lambda)=3, x=3\)

We need to find \(P\{|x-3|<1\}\),

After simplifying we get \(P(2

So between \(2\) and \(4\), only one integer value possible is \(3\).

\(\Rightarrow P\{|x-3|<1\}=P(3)\)

\(P(x)=\frac{e^{-\lambda} \lambda^{x}}{x !}\)

\(\Rightarrow P(3)=\frac{e^{-3} 3^{3}}{3 !}\)

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

\(=\left(\frac{9}{2}\right) e^{-3}\)

To find the distance of the point (2,1,0) from the plane \(2 x+y+2 z+5=0\)

The distance between the point and the plane is given by \(\left|\frac{A x_{1}+B y_{1}+C z_{1}-d}{\sqrt{A^{2}+B^{2}+C^{2}}}\right|\)

Here, \(x_{1}=2, y_{1}=1\) and \(z_{1}=0\)

So, the distance of the given point form the given plane \(=\left|\frac{2 \times 2+1 \times 1+2 \times 0+5}{\sqrt{2^{2}+(1)^{2}+(2)^{2}}}\right|\)

\(=\left|\frac{10}{\sqrt{9}}\right|\)

So, the distance between the plane and the point is \(\frac{10}{3}\) units.

Concept:

Solution of Linear Differential equation:

If the D.E. has a form of\(\frac{d x}{d y}+P x=Q\)then, where P and Q are functions of y.

The solution is given as,\(\mathrm{x} \times \mathrm{I} . \mathrm{F} .=\int \mathrm{I} . \mathrm{F} . \times \mathrm{Qdy}+\mathrm{c}\)

where, I.F. is integrating factor which is given as,

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

Calculation:

Given:\(y d x-\left(x+2 y^{2}\right) d y=0\)

\(\mathrm{y} \frac{\mathrm{dx}}{\mathrm{dy}}=\mathrm{x}+2 \mathrm{y}^{2}\)

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

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

Differential equation is in form of,\(\frac{d x}{d y}+P x=Q\)

Integrating factor,I. F. \(=e^{\int P d y}\)

I.F. \(=e^{\int-\frac{1}{y} d y}\)

\(\Rightarrow\) I. F. \(=e^{-\ln y}\)

\(\Rightarrow\) I.F. \(=\frac{1}{y}\)

Differential equation is given as,

\(x \times \frac{1}{y}=\int \frac{1}{y} \times(2 y) d y+c\)

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

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