Mathematics Test-27

Prime number: Prime number are those which are divisible by itself and 1.

Not a single five-digit prime number can be formed using the digits 1, 2, 3, 4, 5(without repetition).

This is because if one adds the digits, the result obtained will be = 1 + 2 + 3 + 4 + 5 = 15 which is divisible by 3.

So, any number obtained as a permutation of these 5 digits will be at least divisible by 3 and cannot be a prime number.

Here, \(m_{1}=m_{2}^{2} \Rightarrow m_{2}^{2}+m_{2}=\frac{-2 h}{b}\quad \ldots\) (i)

And,

\(m_{2}^{2} m_{2}=\frac{a}{b} \Rightarrow m_{2}=\left(\frac{a}{b}\right)^{\frac{1}{3}}\ldots\) (ii)

Putting this value of m² in equation (i), we get

\(\left\{\left(\frac{a}{b}\right)^{\frac{1}{3}}\right\}^{2}+\left(\frac{a}{b}\right)^{\frac{1}{3}}=\frac{-2 h}{b}\)

On cubing both sides, we get,

\(\left(\frac{a}{b}\right)^{2}+\frac{a}{b}+3\left(\frac{a}{b}\right)^{\frac{2}{3}} \cdot\left(\frac{a}{b}\right)^{\frac{1}{3}} \cdot\left\{\left(\frac{a}{b}\right)^{\frac{2}{3}}+\left(\frac{a}{b}\right)^{\frac{1}{3}}\right\}=\frac{-8 h^{3}}{b^{3}}\)

\(\left(\frac{a}{b}\right)^{2}+\frac{a}{b}-\frac{6 a h}{b^{2}}=\frac{-8 h^{3}}{b^{3}}\)

\(\left\{\because\left(\frac{a}{b}\right)^{\frac{2}{3}}+\left(\frac{a}{b}\right)^{\frac{1}{3}}=\frac{-2 h}{b}\right\} \mid\)

\(a b(a+b)-6 a b h+8 h^{3}=0\).

We can see from the table that we get '\(4\)' \(79\) times out of \(500\) trials.

Therefore, probability of getting 4 as outcome \(=\frac{\text { Event of occurence of getting } 4 }{\text { Total number of trials }}\)

\(=\frac{79}{500}\)

\(=0.158\).

\(\frac{{ }^{n} P_{r}}{{ }^{n} C_{r}}=\frac{3024}{126}\)

\({ }^{n} P_{r}=\frac{n !}{(n-r) !}\)

\({ }^{n} C_{r}=\frac{n !}{(n-r) ! \times r !} \)

\(\text { So, [ } \left.\frac{n !}{(n-r) !}\right] \div\left[\frac{n !}{(n-r) ! \times r !}\right]=24 \)

\(24=r ! \)

\(\text { So, } r=4 \)

\(\text { Now, }{ }^{n} P_{4}=3024\)

\(\frac{n !}{(n-4) !}=3024 \)

\(n(n-1)(n-2)(n-3)=9\times8 \times 7 \times6\)

\(n=9\)

Given:

\((x+a)^{47}-(x-a)^{47}\)

When we expand the above equation using binomial expansion

\((\mathrm{x}+\mathrm{y})^{\mathrm{n}}=\left(\sum_{\mathrm{k}=0}^{\mathrm{n}}{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{k}} \mathrm{x}^{\mathrm{k}} \mathrm{y}^{\mathrm{n}-\mathrm{k}}\right)\)

So the above equation becomes

\((\mathrm{x}+\mathrm{a})^{47}=\left(\sum_{\mathrm{k}=0}^{47}{ }^{47} \mathrm{C}_{\mathrm{k}} \mathrm{x}^{\mathrm{k}} \mathrm{a}^{47-\mathrm{k}}\right) \)

\((\mathrm{x}-\mathrm{a})^{47}=\left(\sum_{\mathrm{k}=0}^{47}{ }^{47} \mathrm{C}_{\mathrm{k}} \mathrm{x}^{\mathrm{k}}(-\mathrm{a})^{47-\mathrm{k}}\right)\)

\((x+a)^{47} \rightarrow\) There are 48 terms in the expansion and all are positive.

\((x-a)^{47} \rightarrow\) There are 48 terms in the expansion.

The terms with odd powers of a will be cancelled and those with even powers of a will add up.

24 terms will be positive and 24 negative in the expansion of \((x-a)^{47}\)

48 terms positive- [ 24 terms negative and 24 terms positive]

\(=48\) terms positive \(+24\) terms negative \(+24\) terms positive \(=24\) terms

Given equation is \(\left(2 x^{2}+5 x+5\right)\).

For any quadratic equation of form \(\left(a x^{2}+b x+c\right)\)

So, \(a=2>0\)

Also, \(b=5\) and \(c=5\)

If \(a>0\), then the minimum value of equation \(=c-\frac{b ^{2}}{4 a}\)

Minimum value of equation \(=5-\left(\frac{25}{8}\right)=\frac{15}{8}\)

Number of events of obtaining minimum one head \(=55+105\) \(=160\)

Now, \(P(E)=\) probability of event of obtaining minimum on head\(=\frac{\text { Number of events of obtaining minimum one head }}{\text { Number of total trials }}\)

\(=\frac{160}{200}\)

\(=0.8\)