Mathematics Test - 4

Given objective function is \(Z=-50 x+20 y\).

We have to minimize \(Z\) on given constraints

\(2 x-y \geq-5 \)

\(3 x+y \geq 3 \)

\(2 x-3 y \leq 12 \)

\(x \geq 0, y \geq 0\)

After plotting all the constraints we get the common region (Feasible region) as shown in the image.

There are four corner points \((0,5),(0,3),(1,0)\) and \((6,0)\)

Now, at corner points value of \(Z\) are as follows:

\(\begin{array}{|l|l|} \hline Corner~ Points & Value ~of ~Z=-50 x+20 y \\ \hline (0,5) & 100 \\ \hline (0,3) & 60 \\ \hline (1,0) & -50 \\ \hline (6,0) & -300 (minimum) \\ \hline \end{array}\)

Since common region is unbounded. So, value of \(Z\) may be minimum at \((6,0)\) and minimum value may be \(-300\).

Now to check if this minimum is correct or not, we have to draw region \(-50 x+20 y \leq-300\)

Since, there are some common region with feasible region(See image). So, \(-300\) will not be minimum value of \(Z\).

So, \(Z\) has no minimum value.

We know that:

The number of ways to select \(r\) things out of \(n\) things is given by \({ }^{{n}} C_{{r}}\).

\({ }^{{n}} {C}_{{r}}=\frac{{n} !}{({n}-{r}) ! \times({r}) !}=\frac{{n} \times({n}-1) \times \ldots({n}-{r}+1)}{{r} !}\)

Given that:

There are \(13\) points in a plane of which \(5\) are collinear. 

We know that:

To form a line we have to select two points out of \(13\) points.

\(\therefore\) Number of lines \(={ }^{13} {C}_{2}\)

\(=\frac{13 \times 12}{2 \times 1}\) 

\(=78\)

Also number of lines out of 5 points \(={ }^{5} {C}_{2}\)

\(=\frac{5 \times 4}{2 \times 1}\)

\(=10\)

But, these \(5\) points are collinear, and only one line can be formed out of these points.

\(\therefore\) The total number of straight lines obtained by joining these points in pairs.

\(=78-10+1\)\(\quad\)(We add \(1\), as one line can be obtained out of \(5\) collinear points).

\(=69\)

The sum of two rational numbers is a rational number.

Ex: Let two rational numbers are \(\frac12\) and \(\frac13\)

Now, \(\frac12+\frac13=\frac56\) which is a rational number.

Given,

\(\lim _{x \rightarrow 0} \frac{\tan ^{2} 3 x}{x^{2}}\)

Dividing by \(9\) in both numerator and denominator, we get:

\(=\lim _{x \rightarrow 0} \frac{\tan ^{2} 3 x}{x^{2}} \times \frac{9}{9}\)

\(=\lim _{x \rightarrow 0} \frac{9 \tan ^{2} 3 x}{(3 x)^{2}}\)

\(=9 \times \lim _{x \rightarrow 0} \frac{\tan 3 x}{3 x} \times \lim _{x \rightarrow 0} \frac{\tan 3 x}{3 x}\)

We know that,

\(\lim _{x \rightarrow 0} \frac{\tan x}{x}=1\)

Therefore,

\(=9 \times 1 \times 1\)

\(=9\)

Therefore, the value of \(\lim _{x \rightarrow 0} \frac{\tan ^{2} 3 x}{x^{2}}\) is \(9\).

We know that:

\({ }^{n} C_{r}=\frac{n !}{r !(n-r) !}\)

A particular player should not be included,

We have to select 5 players from (8 - 1) = 7 players.

Therefore, required number of ways,

\(={ }^{7} {C}_{5}\)

\(=\frac{7 !}{5 !(7-5) !}=21\)

Mean, \(\bar{x}=\frac{2+4+6+8+10}{5}=\frac{30}{5}=6\)

\(\therefore\) Variance \(=\frac{1}{n} \sum\left(x_{i}-\bar{x}\right)^{2}\)

\(=\frac{1}{5}\left\{(2-6)^{2}+(4-6)^{2}+(6-6)^{2}+(8-6)^{2}+(10-6)^{2}\right\}\)

\(=\frac{1}{5}\{16+4+0+4+16\}=\frac{1}{5} \times 40=8\)

Given:

\(\tan \theta+\cot \theta=\sqrt{3}\)

Formula:

\(a^{3}+b^{3}=(a+b)^{3}-3 a b(a+b)\)

\(a^{2}+b^{2}=(a+b)^{2}-2(a \times b)\)

\(\tan \theta \times \cot \theta=1\)

Calculation:

\(\tan \theta+\cot \theta=\sqrt{3}\)

Taking cube on both sides, we get

\((\tan \theta+\cot \theta)^{3}=(\sqrt{3})^{3}\)

\(\Rightarrow \tan ^{3} \theta+\cot ^{3} \theta+3 \times \tan \theta \times \cot \theta \times(\tan \theta+\cot \theta)=3 \sqrt{3}\)

\(\Rightarrow \tan ^{3} \theta+\cot ^{3} \theta+3 \sqrt{3}=3 \sqrt{3}\)

\(\Rightarrow \tan ^{3} \theta+\cot ^{3} \theta=0\)

Taking square on the both sides

\(\left(\tan ^{3} \theta+\cot ^{3} \theta\right)^{2}=0 \)

\(\Rightarrow \tan ^{6} \theta+\cot ^{6} \theta+2 \times \tan ^{3} \theta \times \cot ^{3} \theta=0 \)

\(\Rightarrow \tan ^{6} \theta+\cot ^{6} \theta+2=0 \)

\(\Rightarrow \tan ^{6} \theta+\cot ^{6} \theta=-2\)

\(\therefore\) The value of \(\tan ^{6} \theta+\cot ^{6} \theta\) is \(-2\).

Postulate is a statement that is assumed true without proof. listed below are some postulates related to geometry:

• Infinite lines can be drawn from a point.
• A line contains at least two points.
• Each line (i.e. straight line) and is a set of plane points.
• A plane contains at least three noncollinear points.
• A line drawn through two given points in a plane is also contained in the plane.
• Through any two points, there is exactly one line. If two planes intersect, their intersection is a straight line.
• Through any three noncollinear points, there is exactly one plane.
• If two planes intersect, then their intersection is a line.

So, we conclude that Infinite lines can be drawn from a point.

Given: \(A=\left[\begin{array}{ccc}1 & 0 & -2 \\ 2 & -3 & 4\end{array}\right]\)

We have, \(2 X+3 A=0\)

\(\Rightarrow 2 X=-3 A\)

\(\therefore X=(\frac{-3}{  2}) \times A\)

Now,

\(\mathrm{x}=\frac{-3}{2}\left[\begin{array}{ccc}1 & 0 & -2 \\ 2 & -3 & 4\end{array}\right]\)

\(\Rightarrow \mathrm{x}=\left[\begin{array}{ccc}1 \times \frac{-3}{2} & 0 \times \frac{-3}{2} & -2 \times \frac{-3}{2} \\ 2 \times \frac{-3}{2} & -3 \times \frac{-3}{2} & 4 \times \frac{-3}{2}\end{array}\right]\)

\(\therefore x=\left[\begin{array}{ccc}-\frac{3}{2} & 0 & 3 \\ -3 & \frac{9}{2} & -6\end{array}\right]\)

Given:

\(x^{e}=e^{x^{2}+y^{2}}\)

Taking log on both sides, we get:

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

\(\Rightarrow e \log x=\left(x^{2}+y^{2}\right) \log e\)

\(\left[\because \log m^{n}=n \log m\right]\)

\(\Rightarrow e \log x=x^{2}+y^{2} \quad [\because \log e=1]\)

Differentiating with respect to x, we get:

\(\Rightarrow {e}\left(\frac{1}{{x}}\right)=2 {x}+2 {y} \frac{{dy}}{{dx}}\)

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

\(\therefore \frac{{dy}}{{dx}}=\frac{{e}-2 {x}^{2}}{2 {xy}}\)