Mathematics Test - 7

Let \(x\) and \(y\) respectively be the number of machines \(A\) and \(B\) which the factory owner should buy.

Now, according to the given information, the linear programming problem is:

Maximise \(Z=60 x+40 y\)

Subject to the constraints:

\(1000 x+1200 y \leq 9000\)

\(\Rightarrow 5 x+6 y \leq 45 \ldots(1)\)

\(12 x+8 y \leq 72 \)

\(\Rightarrow 3 x+2 y \leq 18 \ldots(2) \)

\(x\geq 0, y \geq 0 \ldots \text { (3) }\)

The inequalities (1), (2), (3) can be graphed as:

The shaded portion OABC is the feasible region.

The value of Z at the corner points are given in the following table.

Given: \(|\overrightarrow{\mathrm{a}}|=\frac{1}{\sqrt{3}},|\overrightarrow{\mathrm{b}}|=2\) and \(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}\) is a unit vector

Let the angle between the vectors \(\vec{a}\) and \(\vec{b}\) is \(\theta\)

We know that, \(|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \sin \theta\)

\(\Rightarrow 1=\frac{1}{\sqrt{3}} \times 2 \times \sin \theta\)

\(\Rightarrow \sqrt{3}=2 \times \sin \theta\)

\(\Rightarrow \frac{\sqrt{3}}{2}=\sin \theta\)

⇒ θ​ = 60°

Lami's theorem:

It states that if three forces acting at a point are in equilibrium, each force is proportional to the sine of the angle between the other two forces. Consider three forces \(F_{A}, F_{B}, F_{C}\) acting on a particle or rigid body making angles \(\alpha, \beta\) and \(\gamma\) with each other.

Therefore,

\(\frac{F_{A}}{\sin \alpha}=\frac{F_{B}}{\sin \beta}=\frac{F_{C}}{\sin \gamma}\)

Given,

\(\angle B A C=180^{\circ}-\left(30^{\circ}+60^{\circ}\right)\)

\(\angle B A C=90^{\circ}\)

By Lami’s Theorem,

\(\frac{800}{\sin 90^{\circ}}=\frac{T_{A C}}{\sin 150^{\circ}}\)

\(\mathrm{T}_{\mathrm{AC}}=400 \mathrm{~N}\)

Horizontal Range \((\mathrm{R})=3 \mathrm{~km}\)

Angle of projection \((\theta)=30^{\circ}\)

Acceleration due to gravity \((\mathrm{g)=9.8 \mathrm{~m/s}^2}\)

Horizontal range for the projection velocity \(\mathrm{u}_{0}\), is given by the relation:

\(\mathrm{R=\frac{u_{0}^{2} \sin 2 \theta}{g}} \)

\(3=\frac{\mathrm u_{0}^{2} \sin 60^{\circ}}{\mathrm g} \)

\(\frac{\mathrm u_{0}^{2}}{\mathrm g}=2 \sqrt{3} \quad \ldots \ldots .(\mathrm{i})\)

The maximum range \(\mathrm{R}_{\max }\) is achieved by the bullet when it is fired at an angle of \(45^{\circ}\) with the horizontal, that is

\(\mathrm{R}_{\max }=\frac{\mathrm u_{0}^{2}}{\mathrm g} \quad \ldots \ldots . (\mathrm{ii})\)

On comparing equations (i) and (ii), we get:

\(\mathrm{R}_{\max }=2 \times 1.732 \)

\(\mathrm{R}_{\max }=3.464 \mathrm{~km}\)

Given, \(1+\sin x \cdot \sin ^{2} \frac{x}{2}=0\)

\(\Rightarrow 2+2 \sin x \cdot \sin ^{2} \frac{x}{2}=0\)

\(\Rightarrow 2+\sin x(1-\cos x)=0\)

\(\Rightarrow 4+2 \sin x(1-\cos x)=0\)

\(\Rightarrow 4+2 \sin x-\sin 2 x=0\)

\(\Rightarrow \sin 2 x=2 \sin x+4\)

Above is not possible for any value of \(x\) as \(L H S\) has maximum value 1 and \(RHS\) has minimum value 2 .

Here, probability of medicine to cure patient \(=\frac{1}{2}\)

And, probability of none cured \(=1-\frac{1}{2}=\frac{1}{2}\)

The probability that at least one patient is cured by this medicine = 1 - none patient cured by this medicine

\(=1-(\frac{1}{2}) \times(\frac{1}{2}) \times(\frac{1}{2}) \times(\frac{1}{2})\)

\(=1-(\frac{1}{2})^{4}\)

\(=\frac{15}{16}\)

Equation of circle having centre \(({h}, {k})\) and radius \({r}\) is \(({x}-{h})^{2}+({y}-{k})^{2}=0\)

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

\(x^{2}+y^{2}-2 h x-2 h y+h^{2}+k^{2}-r^{2}=0\)

Comparing the above equation with \(x^{2}+y^{2}+2 g x+2 f y+c=0\), we get,

\(h=-g, k=-f\) and \(h 2+k 2-r 2=c\)

Now, \(\left(x^{2}+2 g x+g^{2}\right)+\left(y^{2}+2 f y+f^{2}\right)=g^{2}+f^{2}-c\)

\((x+g)^{2}+(y+f)^{2}=\sqrt{g^{2}+f^{2}-c}\)

\(\{x-(-g)\}^{2}+\{y-(-f)\}^{2}=\sqrt{g^{2}+f^{2}-c}\)

The equation \(x^{2}+y^{2}+2 g x+2 f y+c=0\) always represents a circle whose centre is \((-g,-f)\), that is \(\left(-\frac{1}{2}\right.\) coefficient of \(x,-\frac{1}{2}\) coefficient of \(\left.y\right)\) and radius is \(\sqrt{g^{2}+f^{2}-c}\).

If \(g^{2}+f^{2}-c=0\) then the radius of the circle becomes zero (degenerate circle). In this case, the circle reduces to the point \((-g,-f) .\) Such a circle is known as a point circle. In other words, the equation \(x^{2}+\) \(y^{2}+2 g x+2 f y+c=0\) represents a point circle.

Here, we have to find the equation of the plane passing through the points A (1, 0, 2), B (2, 1, 1) and C (-1, 2, 1)

Here, x₁ = 1, y₁ = 0, z₁ = 2, x₂ = 2, y₂ = 1, z₂ = 1, x₃ = - 1, y₃ = 2 and z₃ = 1.

As we know, equation of the plane in Cartesian form passing through three non-collinear points (x₁, y₁, z₁), (x₂, y₂, z₂) and (x₃, y₃, z₃) is given by:

\(\left|\begin{array}{ccc}x-x_{1} & y-y_{1} & z-z_{1} \\ x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1}\end{array}\right|=0\)

\(\Rightarrow\left|\begin{array}{ccc}x-1 & y-0 & z-2 \\ 1 & 1 & -1 \\ -2 & 2 & -1\end{array}\right|=0\)

⇒ (x - 1) × (-1 + 2) - (y - 0) × (-1 - 2) + (z - 2) × (2 + 2) = 0

⇒ x - 1 - 3y + 4z - 8 = 0

⇒ x - 3y + 4z - 9 = 0

So, the equation of the required plane is x - 3y + 4z - 9 = 0

\(\tan 30^\circ=\frac{x}{b}\)

\(b=\frac{x}{\tan 30^\circ}\)

Taking moment about \(Q\)

\(\Rightarrow F \times X=V_{R} \times 2.732 x\)

\(\therefore V_{R}=0.366 \mathrm{~F}\)

\(\Rightarrow V_{Q}=F-0.366 \mathrm{~F}\)

\(=0.634 \mathrm{~F}\)

Let force in the member \(\mathrm{PQ}\) is \(\mathrm{F}_{1}\)

\(\therefore F_{1} \sin 45^\circ=V_{Q}\)

\(\Rightarrow F_{1} \sin 45^\circ=0.634 \mathrm{~F}\)

Force in member \(QR\)

\(=\mathrm{F}_{1} \cos 45^\circ\)

\(=0.634 {F}\)

Given:

\({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}+2 \times\left({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}\right)+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-2}\)

\(={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}+\left({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}\right)+\left({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}\right)+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-2}\)

\(={ }^{(\mathrm{n}+1)} \mathrm{C}_{\mathrm{r}}+{ }^{(\mathrm{n}+1)} \mathrm{C}_{\mathrm{r}-1} \quad\left(\because{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}={ }^{(\mathrm{n}+1)} \mathrm{C}_{\mathrm{r}}\right)\)

\(={ }^{(n+2)} C_{\mathrm{r}}\)