Mathematics Test - 45

Let x package nuts and y package bolts are produced.

Let \(\mathrm{z}\) be the profit function, which we have to maximize.

Here, \(\mathrm{z}=17.50 \mathrm{x}+7 \mathrm{y} \ldots . .(1)\) is objective function.

And constraints are,

\(x+3 y \leq 12 \ldots .(2)\)

\(3 x+y \leq 12 \ldots .(3)\)

\(x \geq 0 \ldots(4)\)

\(y \geq 0 \ldots(5)\)

On plotting graph of above constraints or inequalities (2),(3),(4) and (5) we get shaded region as feasible region having corner points A, O, B and C.

For coordinate of 'C' two equations,

\(x+3 y=12 \ldots(6)\)

\(3 x+y=12 \ldots .(7)\)

On solving we get, \(x=3\) and \(y=3\)

So coordinate of \(\mathrm{C}\) are \((3,3)\)

Now value of \(z\) is evaluated at corner point as shown in the graph.

At point 'O' (0, 0),

The value of \(z\) \(= 17.50 \times 0 + 7 \times 0 =0\)

At point 'A' (0, 4),

The value of \(z\) \(= 17.50 \times 0 + 7 \times 4 =28\)

At point 'B' (4, 0),

The value of \(z\) \(= 17.50 \times 4 + 7 \times 0 =70\)

At point 'C' (3, 3),

The value of \(z\) \(= 17.50 \times 3 + 7 \times 3 =73.5\)

\(z\), is maximum at C. So \(x=3,y=3\)

Therefore, maximum profit is Rs. \(73.5\) when 3 package nuts and 3 package bolt are produced.

Let \(\Delta=\left|\begin{array}{ccc}\sec ^{2} x & \tan ^{2} x & 1 \\ 2 & 1 & 1 \\ 10 & 8 & 2\end{array}\right|\)

Apply \({C}_{1} \rightarrow {C}_{1}-{C}_{2}\)

\(\Delta=\left|\begin{array}{ccc}\sec ^{2} x-\tan ^{2} x & \tan ^{2} x & 1 \\ 2-1 & 1 & 1 \\ 10-8 & 8 & 2\end{array}\right|\)

\(\Delta=\left|\begin{array}{ccc}1 & \tan ^{2} x & 1 \\ 1 & 1 & 1 \\ 2 & 8 & 2\end{array}\right| \;\left(\because \sec ^{2} x-\tan ^{2} x=1\right)\)

As we know,

If two rows or two columns of a determinant are identical the value of the determinant is zero.

Here \({C}_{1}\) and \({C}_{3}\) are identical.

So, \(\Delta=0\)

Let, Probability of \(X=\) \(P(X)=\frac{1}{5}\)

Probability of \(Y=P(Y)=\frac{1}{4}\)

Probability of \(Z=P(Z)=\frac{1}{3}\)

The probability of success of at least two

\(=[P(A) P(B)\{1-P(C)\}]+[\{1-P(A)\} P(B) P(C)]+[P(A) P(C)\{1-P(B)\}]+[P(A) P(B) P(C)]\)

\(=\left[\frac{1}{4} \times \frac{1}{3} \times \frac{4}{5}\right]+\left[\frac{3}{4} \times \frac{1}{3} \times \frac{1}{5}\right]+\left[\frac{2}{3} \times \frac{1}{4} \times \frac{1}{5}\right]+\left[\frac{1}{4} \times \frac{1}{3} \times \frac{1}{5}\right]\)

\(=\frac{4}{60}+\frac{3}{60}+\frac{2}{60}+\frac{1}{60}\)

\(=\frac{10}{60}\)

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

\(\begin{aligned} & \operatorname{Lt}_{n \rightarrow \infty} \sqrt{\frac{d}{n}}\left(\frac{\sqrt{a_1}-\sqrt{a_2}}{a_1-a_2}+\frac{\sqrt{a_2}-\sqrt{a_3}}{a_2-a_3}+\ldots . .+\frac{\sqrt{a_{n-1}}-\sqrt{a_n}}{a_{n-1}-a_n}\right) \\ & =\operatorname{Lt}_{n \rightarrow \infty} \sqrt{\frac{d}{n}}\left(\frac{\sqrt{a_1}-\sqrt{a_2}+\sqrt{a_2}+\sqrt{a_3}+\ldots \ldots+\sqrt{a_{n-1}}-\sqrt{a_n}}{-d}\right) \\ & =\operatorname{Lt}_{n \rightarrow \infty} \sqrt{\frac{d}{n}}\left(\frac{\sqrt{a_n}-\sqrt{a_1}}{d}\right) \\ & =\operatorname{Lt}_{n \rightarrow \infty} \frac{1}{\sqrt{n}}\left(\frac{\sqrt{a_1+(n-1) d}-\sqrt{a_1}}{\sqrt{d}}\right) \\ & =\operatorname{Lt}_{n \rightarrow \infty} \frac{1}{\sqrt{d}}\left(\sqrt{\frac{a_1}{n}+d-\frac{d}{n}-\frac{\sqrt{a_1}}{n}}\right) \\ & =1\end{aligned}\)

\( -1 \leq \frac{2 \sin ^{-1}\left(\frac{1}{4 x^2-1}\right)}{\pi} \leq 1 \\\)

\( \Rightarrow-\frac{\pi}{2} \leq \sin ^{-1}\left(\frac{1}{4 x^2-1}\right) \leq \frac{\pi}{2} \\\)

\( \Rightarrow-1 \leq \frac{1}{4 x^2-1} \leq 1 \\\)

\( \therefore \frac{1}{4 x^2-1}+1 \geq 0 \\\)

\( \Rightarrow \frac{1+4 x^2-1}{4 x^2-1} \geq 0 \\\)

\( \Rightarrow \frac{4 x^2}{4 x^2-1} \geq 0 \\\)

\( \Rightarrow \frac{4 x^2}{(2 x+1)(2 x-1)} \geq 0 \ldots \ldots ;(1) \\\)

\( \therefore x \in\left(-\alpha,-\frac{1}{2}\right) \cup\{0\} \cup\left(\frac{1}{2}, \alpha\right) \\\)

\( \text { And } \frac{1}{4 x^2-1}-1 \leq 0 \\\)

\( \Rightarrow \frac{1-4 x^2+1}{4 x^2-1} \leq 0 \\\)

\( \Rightarrow \frac{2-4 x^2}{4 x^2-1} \leq 0 \\\)

\( \Rightarrow \frac{2 x^2-1}{4 x^2-1} \geq 0 \\\)

\( \Rightarrow \frac{(\sqrt{2} x+1)(\sqrt{2} x-1)}{(2 x+1)(2 x-1)} \geq 0 \\\)

\( x \in\left(-\alpha,-\frac{1}{\sqrt{2}}\right) \cup\left(-\frac{1}{2} ; \frac{1}{2}\right) \cup\left(\frac{1}{\sqrt{2}}, \alpha\right)\)

From (3) and (4), we get

\(\therefore \mathrm{x} \in\left[-\alpha,-\frac{1}{\sqrt{2}}\right) \cup\left[\frac{1}{\sqrt{2}}, \alpha\right) \cup\{0\}\)

Given,

\(d y=x \sec \frac{y}{x} d x+\frac{y}{x} d x\)

\(\frac{d y}{d x}=x \sec \frac{y}{x}+\frac{y}{x}\)

Now,

By putting \(y=v x\), and \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)

\(v+x \frac{d v}{d x}=x \sec v+v\)

\(\Rightarrow \frac{d v}{d x}=\sec v\)

\(\Rightarrow \cos v d v=d x\)

By integrating both sides

We get,

\(\int \cos v d v=\int d x \sin v=x+c\)

Put \(v=\frac{y}{x}\) in the above equation we get,

sin \(\frac{y}{x}=x+c\)

Put \(y(1)=\frac{\pi}{2}\) in the above equation we get,

\(c=0\)

\(y=x \sin ^{-1} x\)

Put \(x=\frac{1}{2}\) in the above equation we get,

\(y=\frac{1}{2} \sin ^{-1} \frac{1}{2}\)

\(\Rightarrow y=\frac{\pi}{12}\)

Given here:

\(\lim _{x \rightarrow 0} \frac{\sqrt{4+x}-2}{3 x}\)

On substituting the limit value in the given function, we get:

\(\lim _{x \rightarrow 0} \frac{\sqrt{4+x}-2}{3 x}\)

\(=\frac{\sqrt{(4+0)}-2}{3(0)}\)

\(=\frac{0}{0}\)

Conjugate of \(\sqrt{4+x}-2\) is \(\sqrt{4+x}+2\)

Multiply and divide by \(\sqrt{4+x}+2\) in the given expression, we get:

\(\therefore \lim _{x \rightarrow 0} \frac{\sqrt{4+x}-2}{3 x} \times \frac{\sqrt{4+x}+2}{\sqrt{4+x}+2}\)

\(=\lim _{x \rightarrow 0} \frac{4+x-4}{3 x \sqrt{4+x}+2} \quad\left(\because(a+b)(a-b)=\left(a^{2}-b^{2}\right)\right)\)

\(= \lim _{x \rightarrow 0} \frac{x}{3 x(\sqrt{4+x}+2)}\)

\(=\lim _{x \rightarrow 0} \frac{1}{3(\sqrt{4+x}+2)}\)

Now, putting the value of x in the limit, we get:

\(=\frac{1}{3(2+2)}\)

\(=\frac{1}{12}\)

A non-empty subset \({H}\) of a group \(({G}, *)\) is a group of \({G}\) iff,

\(\Rightarrow \mathbf{a}, \mathbf{b} \in \mathbf{H} \Rightarrow \mathbf{a} * \mathbf{b} \in \mathbf{H}\)

\(\Rightarrow {a} \in {H} \Rightarrow \frac{1}{ {a}} \in {H} \Rightarrow {a}^{-1}=\frac{1 }{{a}}\)

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 For Poisson Distribution we have

Mean \(=\) Variance \(=\) (Standard Deviation)\(^{2}\)

\(\therefore\) Standard Deviation \(=\sqrt{\text { Mean }}=\sqrt{\mu}\)

Given,

\(3 x+y+3 z=8\) and \(9 x+3 y+9 z=15\)

Now,

Divide \(9 x+3 y+9 z=15\) by \(3\)

We get,

\(3 x+y+3 z=5\)

Now,

We know that,

Distance between two parallel plane \(a x+b y+c z+d_{1}=0\) and \(a x+b y+c z+d_{2}=0\) is \(\left|\frac{{d}_{1}-{d}_{2}}{\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}}\right|\)

So,

Distance between \(3 x+y+3 z=8\) and \(3 x+y+3 z=5\)

\(=\left|\frac{8-5}{\sqrt{3^{2}+1^{2}+3^{2}}}\right| \)

\(=\frac{3}{\sqrt{19}}\)