Mathematics Test - 11

Probability of Solving the problem by first person \(=\frac{1}{3}\)

Probability of Solving the problem by second person \(=\frac{1}{4}\)

Probability of Solving the problem by third person \(=\frac{1}{5}\)

We know that, probability of occurrence of an event \(= 1-\) Probability of not happening of the same event

Then, probability of Not Solving the problem by first person \(=1-\left(\frac{1}{3}\right)=\frac{2}{3}\)

Probability of not Solving the problem by second person \(=1-\left(\frac{1}{4}\right)=\frac{3}{4}\)

Probability of not Solving the problem by third person \(=1-\left(\frac{1}{5}\right)=\frac{4}{5}\)

So, the probability of none solving the problem \(=\frac{2}{3} \times \frac{3}{4} \times \frac{4}{5}=\frac{2}{5}\)

Given:

Curve \(1: y=|x-1|\)

\(y=1-x\) for \(x<1\)

\( y=x-1\) for \(x \geq 1\)

Curve \(2: y=3-|x|\)

\( y=3+x\) for \(x<0\)

\(y=3-x\) for \(x \geq 0\)

Area enclosed\((A)\) \(=\int_{x_{1}}^{x_{2}}\left(y_{1}-y_{2}\right) d x\)

\(A=\int_{-1}^{2}\left( 3-|x|-|x-1|\right) d x\)

\(A=\left|\int_{-1}^{0} \left(3+x-(1-x)\right) d x\right|+\left|\int_{0}^{1} \left(3-x-(1-x)\right) d x\right|+\left|\int_{1}^{2}\left(3-x-(x-1)\right) d x\right|\)

\(A=\left|\int_{-1}^{0}(2+2 x) d x\right|+\left|\int_{0}^{1} 2 d x\right|+\left|\int_{1}^{2}(4-2 x) d x\right|\)

\(A=\left|\left[2 x+x^{2}\right]_{-1}^{0}\right|+\left|[2 x]_{0}^{1}\right|+\left|\left[4 x-x^{2}\right]_{1}^{2}\right|\)

\(A=1+2+1=4\) sq. units

As we know,

If \(z=x+i y \ldots(1)\) be any complex number, then its modulus is given by,

\(|z|=\sqrt{x^{2}+y^{2}}\)

Let,

\(z=\frac{1+7 i}{(2-i)^{2}}\)

\(=\frac{1+7 i}{2^{2}+i^{2}-4 i} \)

\(=\frac{1+7 i}{4-1-4 i} \)

\(=\frac{1+7 i}{3-4 i} \)

\(=\frac{1+7 i}{3-4 i} \times \frac{3+4 i}{3+4 i} \)

\(=\frac{3+4 i+21 i+28 i^{2}}{3^{2}-(4 i)^{2}}\)

\(=\frac{3+25 i-28}{9+16} \)

\(=\frac{-25+25 i}{25} \)

\(\therefore z=-1+i \ldots(2)\)

Comparing equation (1) and (2), we get

\(x=-1 \text { or } y=1 \)

\(\text { As, }|z|=\sqrt{x^{2}+y^{2}} \)

\(\therefore|z|=\sqrt{(-1)^{2}+1^{2}} \)

\(\Rightarrow|z|=\sqrt{1+1} \)

\(\Rightarrow|z|=\sqrt{2}\)

So, the modulus of \(\frac{1+7 i}{(2-i)^{2}}\) is \(\sqrt{2}\).

Given,

\(\sqrt{3}-3 \sqrt{3} \tan ^{2} \mathrm{~A}=3 \tan \mathrm{A}-\tan ^{3} \mathrm{~A} \)

\(\Rightarrow \sqrt{3}\left(1-3 \tan ^{2} A\right)=3 \tan A-\tan ^{3} A \)

\(\Rightarrow \sqrt{3}=\frac{\left(3 \tan \mathrm{A}-\tan ^{3} \mathrm{~A}\right)}{\left(1-3 \tan ^{2} \mathrm{~A}\right)} \)

\(\Rightarrow \sqrt{3}=\tan 3 \mathrm{~A}\)

We know that,

\(\tan 60^{\circ}=\sqrt{3}\)

Therefore,

\(\tan 60^{\circ}=\tan 3 A \)

\(3 A=60^{\circ} \)

\(A=\frac{60^{\circ}}{3}=20^{\circ} \)

\(\therefore A=20^{\circ}\)

Recorder data are \(10,5,8,16,18,20,8,10,16,20,18,11,16,14\) and 12

Let us arrange the following data in ascending order to find out the median and mode, \(5,8,8,10,10,11,12,14,16,16,16,18,18,20,20\)

Total number of observation \(=15\)

\(\operatorname{Mean}=\frac{\sum \mathrm{x}}{\mathrm{n}}=\frac{5+8+8+10+10+11+12+14+16+16+16+18+18+20+20}{15}=\frac{202}{15}\)

Median \(=\) value of \(\left(\frac{n+1}{2}\right)^{\text {th }}\) observation

\(\therefore\) Median \(=8^{\text {th }}\) observation \(=14\)

From above observation 16 occurs maximum number of times.

\(\therefore\) Mode \(=16\)

Given: One of the values is wrongly written as \(16\) instead of \(18\)

\(5,8,8,10,10,11,12,14,16,16,18,18,18,20,20\)

Mean \(=\frac{\sum \mathrm{x}}{\mathrm{n}}=\frac{5+8+8+10+10+11+12+14+16+16+18+18+18+20+20}{15}=\frac{204}{15}\)

Median \(=\) value of \(\left(\frac{\mathrm{n}+1}{2}\right)^{\text {th }}\) observation

\(\therefore\) Median \(=8^{\text {th }}\) observation \(=14\)

From above observation 18 occurs maximum number of times.

\(\therefore \text { Mode }=18\)

We can see that Mean and mode are changed.

Mean \(\mu=n p=8\) ...(1)

Variance \(\sigma^{2}=\mathrm{npq}=4\) ..(2)

Dividing equation (2) by (1), we get

\(q=\frac{1}{2}\)

As we know, \(p+q=1\)

\(p=1-q=\frac{1}{2}\)

Put the value of \(n\) in equation (1), we get

\(n=16\)

Now,

\(P(x=1)={ }^{16} C_{1}\left(\frac{1}{2}\right)^{1} \times\left(\frac{1}{2}\right)^{16-1}\)

\(=16 \times \frac{1}{2^{16}}\)

\(=2^{4} \times \frac{1}{2^{16}}\)

\(P(x=1)\) is \(\frac{1}{2^{12}}\).

Given,

\(\int \frac{3 a x}{b^{2}+c^{2} x^{2}} d x\)

Let \(I=\int \frac{3 a x}{b^{2}+c^{2} x^{2}} d x\).

Substituting \(b^{2}+c^{2} x^{2}=t\)

\(\Rightarrow\left(2 \mathrm{c}^{2} \mathrm{x}\right) \mathrm{dx}=\mathrm{dt}\)

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

\(\Rightarrow \mathrm{I}=\frac{3 \mathrm{a}}{2 \mathrm{c}^{2}} \int \frac{1}{\mathrm{t}} \mathrm{dt}\)

\(\Rightarrow I=\frac{3 \mathrm{a}}{2 \mathrm{c}^{2}} \log |t|+C\)

\(\Rightarrow \mathrm{I}=\frac{3 \mathrm{a}}{2 \mathrm{c}^{2}} \log \left|\mathbf{b}^{2}+\mathrm{c}^{2} \mathrm{x}^{2}\right|+\mathrm{C}\)

\(\therefore \int \frac{3 a x}{b^{2}+c^{2} x^{2}} d x=\frac{3 \mathrm{a}}{2 \mathrm{c}^{2}} \log \left|\mathrm{b}^{2}+\mathrm{c}^{2} \mathrm{x}^{2}\right|+\mathrm{C}\)

If \(\vec{a}\) and \(\vec{b}\) are two non-zero vectors then scalar product will be:

\(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta \quad \ldots\) (1)

Where, \(\theta\) is angle between \(\vec{a}\) and \(\vec{b}\) and, \(|\vec{a}|=\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\) and \(|\vec{b}|=\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}} \quad \ldots\) (2)

Given:

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

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

Then,

\(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta \)

\(\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} \)

\(\cos \theta=\frac{(2 \hat{i}+\hat{j}+3 \hat{k}) \cdot(3 \hat{i}-2 \hat{j}+\hat{k})}{\sqrt{2^{2}+1^{2}+3^{2}} \cdot \sqrt{3^{2}+(-2)^{2}+1^{2}}} \)

\(\cos \theta=\frac{6-2+3}{\sqrt{14} \sqrt{14}} \)

\(\cos \theta=\frac{7}{14}=\frac{1}{2} \)

\(\cos \theta=\cos 60^{\circ} \)

\(\theta=60^{\circ}\)