Mathematics Test - 6

Given, \(P(A)=2 P(B)=6 P(C)\)

We know that when events are mutually exclusive and exhaustive,

\(P(A)+P(B)+P(C)=1\)

Let, \(P(A)\) = \(k\)

Then, \(P(B)= \frac{P(A)}{2}=\frac{k}{2}\)

\(P(B)=\frac{k}{2}\)

\(\Rightarrow P(C)= \frac{k}{6}\)

So, according to the concept \(k+\frac{k}{2}+\frac{k}{6}=1\)

\(\Rightarrow \frac{10 k}{6}=1\)

\(\Rightarrow k=\frac{3}{5}\)

Therefore, \(P(B)\) = \(\frac{k}{2}=\frac{3}{5 \times 2}\) \(=0.3\)

Given,

\(\lim _{x \rightarrow \infty} x \sin \left(\frac{\pi}{x}\right)\)

Dividing and multiplying by \(x\) in above, we get,

\(=\lim _{x \rightarrow \infty} \frac{\sin \left(\frac{\pi}{x}\right)}{\left(x\right)}\)

Dividing and multiplying by \(\pi\) in above, we get,

\(=\lim _{x \rightarrow \infty} \frac{\sin \left(\frac{\pi}{x}\right)}{\left(\frac{\pi}{x}\right)} \times \pi\)

If \(x \rightarrow \infty\), then \(t \rightarrow 0\)

\(=\lim _{t \rightarrow 0} \frac{\sin t}{t} \times \pi\)

We know that,

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

Therefore, it will be,

\(=1 \times \pi\)

\(=\pi\)

Given:

Maximize \(Z=3 x+5 y\) 

Subject to:

\(x+2 y \leq 20 \)

\(x+y \leq 15 \)

\(y \leq 5 \)

\(x, y \geq 0\)

We need to maximize \(\mathrm{Z}=3 x+5 y\)

First, we will convert the given inequations into equations, we obtain the following equations:

\(x+2 y=20, x+y=15, y=5, x=0\) and \(y=0\)

The line \(x+2 y=20\) meets the coordinate axis at \(A(20,0)\) and \(B(0,10)\). Join these points to obtain the line \(x+2 y=20\), \((0,0)\) satisfies the inequation \(x+2 y \leq 20\).

So, the region in \(x y\)-plane that contains the origin represents the solution set of the given equation.

The line \(x+y=15\) meets the coordinate axis at \(C(15,0)\) and \(D(0,15)\). Join these points to obtain the line \(x+y=15\). \((0,0)\) satisfies the inequation \(x+y \leq 15 .\) So, the region in \(x y\)-plane that contains the origin represents the solution set of the given equation.

\(y=5\) is the line passing through \((0,5)\) and parallel to the \(X\) axis.The region below the line \(y=5\) will satisfy the given inequation.

Region represented by \(x \geq 0\) and \(y \geq 0\) :

Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations.

These lines are drawn using a suitable scale.

The corner points of the feasible region are \(O(0,0), C(15,0), E(10,5)\) and \(F(0,5)\) The values of \(Z\) at these corner points are as follows.

\(\begin{array}{|l|l|} \hline \text{Corner point} & Z=3 x+5 y \\ \hline O(0,0) & 3 \times 0+5 \times 0=0 \\ \hline C(15,0) & 3 \times 15+5 \times 0=45 \\ \hline E(10,5) & 3 \times 10+5 \times 5=55 \\ \hline F(0,5) & 3 \times 0+5 \times 5=25 \\ \hline \end{array}\)

We see that the maximum value of the objective function \(Z\) is 55 which is at \(E(10,5)\).

Thus, the optimal value of \(Z\) is 55.

R on the set {1, 2, 3} be defined by R={(1, 2)}

It is clear that R is transitive.

A homogeneous relation R over a set X is transitive if for all elements a, b, c in X , whenever R relates a to b and b to c, then R also relates a to c.

Given,

\(\lim _{x \rightarrow 0} \frac{\tan 6 x+4 x}{4 x+\tan x}\)

We know that,

\(\lim _{x \rightarrow a}\left[\frac{f(x)}{g(x)}\right]=\frac{\lim _{x \rightarrow a} f(x)}{\lim _{x \rightarrow a} g(x)}\), provided \(\lim _{x \rightarrow a} g(x) \neq 0\)

Now,

Dividing by \(6 x\) in numerator and denominator,

We get,

\(=\lim _{x \rightarrow 0} \frac{\frac{\tan 6 x}{6 x}+\frac{4 x}{6 x}}{\frac{4 x}{6 x}+\left(\frac{1}{6}\right) \frac{\tan x}{x}} \quad\left[\because \lim _{x \rightarrow 0} \frac{\tan x}{x}=1, \lim _{x \rightarrow 0} \frac{\tan 6 x}{6 x}=1\right] \)

\(=\lim _{x \rightarrow 0} \frac{1+\frac{4}{6}}{\frac{4}{6}+\frac{1}{6}}\)

\(=\frac{1+\frac{4}{6}}{\frac{4}{6}+\frac{1}{6}}\)

\(=\frac{\frac{6+4}{6}}{\frac{4+1}{6}}\)

\(=\frac{10}{5}\)

\(=2\)

Given:

\(\left(x^{2}+\frac{1}{x^{3}}\right)^{n}\)

General term in the expansion of \((a+b)^{n}\) is given by:

\(T_{(r+1)}={ }^{n} C_r \times a^{n-7} \times b^{r}\)

So,

\(T_{r+1}={ }^{n} C_{r}\left(x^{2}\right)^{n-7}\left(\frac{1}{x^{3}}\right)^{r}\)

\(\Rightarrow T_{r+1}={ }^{n} C_{r} x^{2 n-5 r}\)

So, \(2 n-5 r=1\).....(1)

According to question:

\({ }^{n} C_{r}={ }^{n} C_{23}\)

so, \(r=23\) or \(n-r=23 \quad \ldots\)(2)

From (1) and (2),

Minimum value is \(n=38\)

Given curve are \(y=\sqrt{x}\) and \(2 y=x\)

\(\Rightarrow y=\sqrt{x}=\frac{x}{2}\)

Squaring both sides, we get

\(\Rightarrow x=\left(\frac{x^{2}}{4}\right)\)

\(\Rightarrow x^{2}-4 x=0\)

\(\Rightarrow x(x-4)=0\)

\(\therefore x=0\) and \(x=4\)

Now, \(y=\sqrt{x}\)

If \(x=0\) than \(y=0\)

If \(x=4\) then \(y=2\)

The area in the given case can be evaluated by horizontal component, thus the area is given by:

\(A=\int_{0}^{2}\left(2 y-y^{2}\right) d y\)

\(=2\left[\frac{y^{2}}{2}\right]_{0}^{2}-\left[\frac{y^{3}}{3}\right]_{0}^{2}\)

\(=4-\frac{8}{3}\)

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

Therefore, the required area is \(\frac{4}{3}\) units.

Let the integer be \(x\).

\(=x+ \frac{1 }{ x}\)

\(\Rightarrow\) This can also be written as \(\frac{\left(x^{2}+1\right)}{  x}\)

So, from the options we can find that which one will satisfy this expression And we can see that option \(2\) will satisfy the expression.

\(\Rightarrow \frac{\left(5^{2}+1\right) }{ 5}=\frac{26 }{ 5}\)

Let \(z=\frac{(2-\mathrm{i})(1+2 \mathrm{i})}{(3+\mathrm{i})(2-3 \mathrm{i})}\)

\(\Rightarrow z=\frac{2+4 i-i-2 i^{2}}{6-9 i+2 i-3 i^{2}}=\frac{2+4 i-i+2}{6-9 i+2 i+3}\)

\(\Rightarrow z=\frac{4+3 i}{9-7 i}\)

\(\Rightarrow z=\frac{4+3 \mathrm{i}}{9-7 i} \times \frac{9+7 i}{9+7 i}\)

\(\Rightarrow z=\frac{36+28 \mathrm{i}+27 \mathrm{i}+21 \mathrm{i}^{2}}{81-49 \mathrm{i}^{2}}=\frac{36+28 \mathrm{i}+27 \mathrm{i}-21}{81+49}\)

\(\Rightarrow z=\frac{15+55 \mathrm{i}}{130}\)

\(\Rightarrow z=\frac{15}{130}+\mathrm{i} \frac{55}{130}\)

We know that, Conjugate of \(\mathrm{z}=\overline{\mathrm{z}}=\mathrm{x}-\mathrm{iy}\)

\(\Rightarrow \overline{\mathrm{z}}=\frac{15}{130}-\mathrm{i} \frac{55}{130}\)

Range: It is simply the difference between the maximum value and the minimum value given in a data set.

Given data are 35, 12, 21, 24, 15, 7, 16, 12, 30, 32, 13, and 17

From the data the minimum value = 7

From the data the maximum value = 35

Range = maximum value – minimum value

= 35 – 7

= 28