Mathematics Test - 29

\(\text { Given, } \frac{d y}{d x}=2(y+2 \sin x-5) x-2 \cos x, y(0)=7 \\\)

\(\Rightarrow \frac{d y}{d x}+2 \cos x=2(y+2 \sin x-5) x \\\)

\(\text { Let } y+2 \sin x-5=t \\\)

\(\Rightarrow \frac{d y}{d x}+2 \cos x=\frac{d t}{d x}\)

Then, Eq. (i) becomes

\(\frac{\mathrm{dt}}{\mathrm{dx}}=2 \mathrm{tx} \\\)

\(\Rightarrow \frac{\mathrm{dt}}{\mathrm{t}}=2 \mathrm{xdx}\)

On integrating

\(\ln t=x^2+C \\\)

\(\Rightarrow \ln (y+2 \sin x-5)=x^2+C \ldots(\text { iii) } \\\)

\( \therefore y(0)=7 \\\)

\( \Rightarrow \ln (7+0-5)=0+C \\\)

\( \Rightarrow C=\ln 2 \\\)

\( \therefore \text { From Eq. (ii), } \\\)

\(\ln (y+2 \sin x-5)=x^2+\ln 2\)

Now, at \(\mathrm{x}=\pi\)

\(\ln (y(\pi)+2 \sin \pi-5)=\pi^2+\ln 2 \\\)

\( \Rightarrow \ln (y(\pi)-5)=\pi^2+\ln 2 \\\)

\( \Rightarrow y(\pi)-5=e^{\pi^2+\ln 2} \\\)

\( \Rightarrow y(\pi)=2 \mathrm{e}^{\pi^2}+5\)

Given that:

\({ }^{{n}} {P}_{{r}}=2760,{ }^{{n}} {C}_{{r}}=23\)

We know that,

\({ }^{{n}} {C}_{{r}}=\frac{{ }^{{n}} {P}_{{r}}}{{r} !}\)

\(\Rightarrow 23=\frac{2760}{r !}\)

\(\Rightarrow {r} !=\frac{2760}{23}=120\)

\(\Rightarrow {r} !=5 \times 24\)

\(\Rightarrow {r} !=5 \times 4 \times 6\)

\(\Rightarrow {r} !=5 \times 4 \times 3 \times 2 \times 1\)

\(\Rightarrow {r!}=5!\)

\(\therefore r=5\)

Given:

The mean of 10 observations \(x_{1}, x_{2}, x_{3} \ldots . x_{10}\) is 20.

Mean of \(x_{1}+2, x_{2}+4, x_{3}+6, \ldots x_{10}+20=\) Mean of \(\left(x_{1}, x_{2}, x_{3} \ldots x_{10}\right)+\) Mean of \((2,4)\) \((6, \ldots .20) \ldots(1)\)

Now mean of \((2,4,6, \ldots .20)\) is:

Mean \(= \frac{\text{Sum of total observations}}{\text{Total number of observation}}\)

\(\frac{(2+4+6+8+10+12+14+16+18+20)}{10}\)

\(=\frac{2 \times 55}{10}=11\)

From equation (1),

Mean of \(x_{1}+2, x_{2}+4, x_{3}+6, \ldots x_{10}+20=20+11=31\)

Given that the edge length \(( L )\) of the square changes at the rate \(=\frac{ dL }{ dt }=2\) cm/s

Area of the square \(A=L^{2}\)

Rate of change of area \(A =\frac{ dA }{ dt }=\frac{ dA }{ d L } \times \frac{ dL }{ dt }\)

\(\frac{ dA }{ dt }=\frac{ d }{ dL }\left[ L ^{2}\right] \times \frac{ d L }{ dt }\)

\(\frac{ dA }{ dt }=2 L \times 2=4 L\)

As, L \(=12\) cm

\(\frac{ dA }{ dt }=4 \times 12\)

\(\frac{ d A }{ dt }=48\) cm\(^{2} /\)s

\(\begin{aligned} & x=2 \sqrt{2} \cos t \sqrt{\sin 2 t}, y=2 \sqrt{2} \sin t \sqrt{\sin 2 t} \\ & \therefore \frac{d x}{d t}=\frac{2 \sqrt{2} \cos 3 t}{\sqrt{\sin 2 t}}, \frac{d y}{d t}=\frac{2 \sqrt{2} \sin 3 t}{\sqrt{\sin 2 t}} \\ & \therefore \frac{d y}{d x}=\tan 3 t,\left(a t t=\frac{\pi}{4}, \frac{d y}{d x}=-1\right) \\ & \text { and } \frac{d^2 y}{d x^2}=3 \sec ^2 3 t \cdot \frac{d t}{d x}=\frac{3 \sec ^2 3 t \cdot \sqrt{\sin 2 t}}{2 \sqrt{2} \cos 3 t} \\ & \left(. \text { At } \cdot t=\frac{\pi}{4}, \frac{d^2 y}{d x^2}=-3\right) \\ & \therefore \frac{1+\left(\frac{d y}{d x}\right)^2}{\frac{d^2 y}{d x^2}}=\frac{2}{-3}=\frac{-2}{3}\end{aligned}\)

Concept:

For a continuous random variable, \(f(x)\) is called probability density function if it satisfies-

\(\int_{-\infty}^{\infty} f(x) d x=1\)

Relation between Probability density function \(f(x)\) and Probability distribution function is:

\(f(x)=\frac{d}{d x}\{F(x)\}\)

Calculation:

Here Probability distribution function \(F ( x )\) is given

\(\because \int_{-\infty}^{\infty} f(x) d x=1\)

\(\Rightarrow \int_{-\infty}^{1} f(x) d x+\int_{1}^{3} f(x) d x+\int_{3}^{\infty} f(x) d x=1\)

\(\Rightarrow \int_{-\infty}^{1} \frac{d}{d x}\{F(x)\} d x+\int_{1}^{3} \frac{d}{d x}\{F(x)\} d x+\int_{3}^{\infty} \frac{d}{d x}\{F(x)\} d x=1\)

\(\Rightarrow \int_{-\infty}^{1} \frac{d}{d x}\{0\} d x+\int_{1}^{3} \frac{d}{d x}\left\{k(x-1)^{4}\right\} d x+\int_{3}^{\infty} \frac{d}{d x}\{1\} d x=1\)

\(\Rightarrow \int_{1}^{3} 4 k(x-1)^{3} d x=1\)

\(\Rightarrow 4 \frac{k}{4}\left[(x-1)^{4}\right]_{1}^{3}=1\)

\(\Rightarrow k\left[(3-1)^{4}\right]=1\)

\(\Rightarrow k=\frac{1}{16}\)

The objective function of a linear programming problem is a function to be optimized.It has either a maximum or minimum value or has no solution.The objective function in linear programming problems is the real-valued function whose value is to be either minimized or maximized subject to the constraints defined on the given LPP over the set of feasible solutions. The objective function of a LPP is a linear function of the form z = ax + by.

Given:

\(\mathrm{A}=\{2,4,6\}, \mathrm{B}=\{4,6,7\}\) and \(\mathrm{C}=\{2,7\}\)

We know that:

For any two non-empty sets \(A\) and \(B\), we have:

\(A \times B=\{(a, b) \mid a \in A\) and \(b \in B\}\)

Then,

\( \mathrm{A}-\mathrm{B}=\{2,4,6\}-\{4,6,7\}=\{2\}\)

Similarly,

\(B-C=\{4,6,7\}-\{2,7\}=\{4,6\}\)

Then,

\( (A-B) \times(B-C)=\{2\} \times\{4,6\}\)

\(=\{(2,4),(2,6)\}\)

Therefore,

\((A-B) \times(B-C)=\{(2,4),(2,6)\}\)

Given:

\(|\overrightarrow{\mathrm{a}}|=\alpha,|\overrightarrow{\mathrm{b}}|=\alpha\) and \(|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|=\alpha\)

As we know,

\(|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|=\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}+2 \mathrm{ab} \cos \theta}\)

\( \alpha=\sqrt{\alpha^{2}+\alpha^{2}+2(\alpha)(\alpha) \cos \theta}\)

\( \alpha^{2}=2 \alpha^{2}+2 \alpha^{2} \cos \theta\)

\(-1=2 \cos \theta\)

\( \cos \theta=-\frac{1}{2}\)

Now,

\(|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|=\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}-2 \mathrm{ab} \cos \theta}\)

\(|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|=\sqrt{\alpha^{2}+\alpha^{2}-2(\alpha)(\alpha) \cos \theta}\)

\(\because \cos \theta=-\frac{1}{2}\)

\(|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|=\sqrt{2 \alpha^{2}-2 \alpha^{2}\left(\frac{-1}{2}\right)}\)

\(|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|=\sqrt{2 \alpha^{2}+\alpha^{2}}\)

\(|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|=\sqrt{\mathrm{3}} {\alpha}\)

Total number of balls \(=4+5+6=15\)

Let \(S\) be the sample space.

Number of all combinations of \(n\) things, taken \(r\) at a time, is given by \({ }^{n} C_{r}=\frac{n !}{(r) !(n-r) !}\)

\(n(S)=\) Total number of ways of drawing \(3\) balls out of \(15={ }^{15} C_{3}\)

\(E=\) Event of drawing \(3\) balls, all of them are yellow.

\(n(E)=\) Number of ways of drawing \(3\) balls, all of them are yellow \(=\)Number of ways of drawing \(3\) balls from the total \(5={ }^{5} C_{3}\)( \(\because\) there are \(5\) yellow balls in the total balls)

\(P(E)=\frac{n(E)}{n(S)}=\frac{{ }^{5} C_{3}}{{ }^{15} C_{3}}\)

=\(\frac{\frac{5!}{3 ! 2 !}}{\frac{15!}{3 ! 12 !}}\)

=\(\frac{\left(\frac{5 \times 4}{2 \times 1}\right)}{\left(\frac{15 \times 14 \times 13}{3 \times 2 \times 1}\right)}\)

=\(\frac{2}{91}\)