Mathematics Test - 8

Let the missing number be \(x\).

Given:

Mean = \(25\), Total elements = \(8\)

Mean of n elements \(=\frac{\text { Sum of all n elements }}{\text { Total number of elements (n) }}\)

\(\Rightarrow \frac{30+24+27+22+18+26+32+x}{8}=25\)

\(\Rightarrow 179+x=200\)

\(x=21\)

We know that,

Mean: \((\bar{x})=\frac{x_{1}+x_{2}+\ldots+x_{n}}{n}=\frac{\sum x_{i}}{n}\)

Mean of first 20 natural numbers \((\bar{x})=\frac{1+2+\ldots+20}{20}\)

\(=\frac{\sum_{x=1}^{20} x}{20} \)

\(=\frac{20(20+1)}{2 \times 20}=10.5\)

And variance is:

\(\sigma^{2}=\frac{\sum x_{i}^{2}}{n}-\bar{x}^{2} \)

\(=\frac{n(n+1)(2 n+1)}{6 n}-\bar{x}^{2} \)

\(=\frac{21 \times 41}{6}-10.5^{2} \)

\(=33.25=\frac{133}{4}\)

\(I=\int \frac{d x}{x\left(x^{7}-1\right)}\)

Let, \(x^{7}-1=t\)

Differentiate both sides w.r.t \(x\)

\(\Rightarrow 7 x^{6} d x=d t\)

\(\Rightarrow {dx}=\frac{{dt}}{7 {x}^{6}}\)

Substitute above values in given integration,

\(I=\frac{1}{7} \int \frac{{dt}}{{x}^{7} t}\)

\({I}=\frac{1}{7} \int \frac{{dt}}{{t}({t}+1)} \quad\left[\because {x}^{7}={t}+1\right]\)

\({I}=\frac{1}{7} \int \frac{{d} t}{{t}^{2}+{t}}\)

By completing square method,

\(I=\frac{1}{7} \int \frac{d t}{t^{2}+t+\frac{1}{4}-\frac{1}{4}}\)

\(I=\frac{1}{7} \int \frac{d t}{\left(t+\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}}\)

\({I}=\frac{1}{7} \log \left|\frac{{t}}{{t}+1}\right|+{C}\)

From eq. (i)

\(I=\frac{1}{7} \log \left|\frac{x^{7}-1}{x^{7}} \right| +C\)

To Find: Rate of change of \(\sqrt{x^{2}+48}\) with respect to \(x^{2}\)

Let \(u\) be \(\sqrt{x^{2}+48}\) and \(v\) be \(x^{2}\)

\(\frac{d u}{d x}=\frac{1 \times 2 x}{2 \sqrt{\left(x^{2}+48\right)}}=\frac{x}{\sqrt{\left(x^{2}+48\right)}}\)

\(\frac{d v}{d x}=2 x\)

\(\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{\frac{x}{\sqrt{\left(x^{2}+48\right)}}}{2 x}=\frac{x}{\sqrt{\left(x^{2}+48\right)}} \times \frac{1}{2 x}\)

\(\frac{d u}{d v}=\frac{1}{2 \sqrt{\left(x^{2}+48\right)}}\)

At \(x=4\)

\(\frac{d u}{d v}=\frac{1}{2 \sqrt{\left(4^{2}+48\right)}}\)

\(\frac{d u}{d v}=\frac{1}{2 \sqrt{64}}=\frac{1}{16}\)

\(\therefore \frac{d u}{d v}=\frac{1}{16}\)

Consider the third angle as A

Given, tan^-1 2, tan^-1 3 are two angles of a triangle.

We know that,

\(\tan ^{-1} {a}+\tan ^{-1} {b}=\tan ^{-1}\left(\frac{{a}+{b}}{1-{ab}}\right)\)

Consider

\(\tan ^{-1} 2+\tan ^{-1} 3=\tan ^{-1}\left(\frac{2+3}{1-2.3}\right)\)

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

\(\tan ^{-1} 2+\tan ^{-1} 3=\tan ^{-1}(-1)\)

\(\tan ^{-1} 2+\tan ^{-1} 3=\frac{3 \pi}{4}\)

Now, the sum of angles of the triangle is 180º i..e π.

\(\tan ^{-1} 2+\tan ^{-1} 3+{A}=\pi\)

\(\frac{3 \pi}{4}+A=\pi\)

\({A}=\frac{\pi}{4}\)

If tan^-1 2, tan^-1 3 are two angles of a triangle, then the third angle is \(\frac{\pi}{4}\).

Given,

\(2(\sec A-\cos A)^{2}+2(\operatorname{cosec} A-\sin A)^{2}-2(\cot A-\tan A)^{2}-2\)

\(=2\left[(\sec A-\cos A)^{2}+(\operatorname{cosec} A-\sin A)^{2}-(\cot A-\tan A)^{2}-1\right]\)

\(=2\left[\sec ^{2} A+\cos ^{2} A-2 \sec A \cos A+\operatorname{cosec}^{2} A+\sin ^{2} A-2 \operatorname{cosec} A \sin A-\cot ^{2} A-\tan ^{2} A+2 \cot A \tan A-1\right]\)

\(=2[\sec ^{2} A+\cos ^{2} A-2+\operatorname{cosec}^{2} A+\sin ^{2} A-2-\cot ^{2} A-\tan ^{2} =2[(\cos ^{2} A+\sin ^{2} A)+(\sec ^{2} A-\tan ^{2} A)+(\operatorname{cosec}^{2} A-\cot ^{2} A)-3]\)

\({\left[\because \cos ^{2} A+\sin ^{2} A=1, \sec ^{2} A-\tan ^{2} A=1 \text { and } \operatorname{cosec}^{2} A-\cot ^{2} A=1\right]}\)

\(=2[1+1+1-3]\)

\(=0\)

Given,

\(\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}} \)

\(\overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}\)

So,

\((\vec{a}+\vec{b})=(\hat{i}+3 \hat{i})+(\hat{2 j}-\hat{j})+(-\hat{3} k+2 \hat{k}) \)

\(\Rightarrow \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}=4 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathrm{k}} \)

\((\vec{a}-\vec{b})=(\hat{i}-3 \hat{i})+(\hat{2 j}-(-\hat{j}+(-\hat{3} k-2 \hat{k}) \)

\(\Rightarrow \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}=-2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}\)

Then,

\((\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=(4 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}) \cdot(-2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}) \)

\((\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=-8+3+5=0 \)

\(\text { If }(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=0\)

Then angle between two vectors is \(90^{\circ} .\)

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

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

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

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

The given equation is,

\(\frac{d y}{d x}=y\left(1-3 x^{2}\right)\)

\(\Rightarrow\frac{d y}{y}=\left(1-3 x^{2}\right) d x\)

Integrating both side, we get,

\(\int \frac{d y}{y}=\int\left(1-3 x^{2}\right) d x\)

We know that,

\(\int\frac{dx}{x}=\log x\) and \(\int x^n dx=\frac{x^{n+1}}{n+1}\)

\(\therefore\log y=x-x^{3}+c\)

We also know that,

If \(\log x = n\) then \(x=e^n\)

\(\therefore y=e^{x-x^{3}+c}\)

\(\Rightarrow y=e^{x-x^{3}}. e^{c}\)

\(\Rightarrow y=C e^{x\left(1-x^{2}\right)}\quad\) \([\)Let \(e^c=C]\)

Given,

Probability of success \(\frac{2}{3}\) and number of trials is \(10\).

Consider, the most probable number of successes in \(10\) trials with probability of success \(\frac{2}{3}\) is \(n\).

Now, \({n}=\frac{2}{3}+\frac{2}{3}+\frac{2}{3}+\ldots\) (\(10\) times)

\({n}=\frac{2}{3} \times 10 \)

\(\Rightarrow n=6.67 \)

\(\Rightarrow n=7\)

The most probable number of successes in \(10\) trials with probability of success \(\frac{2}{3}\) is \(7\).