Mathematics Test-31

\(A B=A \cdots \cdots(I)\)

\(B A=B \cdots(I I)\)

From equation (I),

\(B \times(A B)=B\)

\(B^2 A=B\)

From equation (II),

\(B^2 A=B A\)

\(B^2=B\)

Given,

\(6\) letters word COLOUR is given.

There are \(6\) letters in the given word's out of \(6\) letters, two are identical.

The required number of word\(=\frac{6 !} 2\)

\(=\frac{(6 \times 5 \times 4 \times 3 \times 2)} 2\)

\(=30 \times 12\)

\(=360\)

With \(366\) days, the number of weeks in a year is

\(\frac{366}{7}=52(\frac{2}{7})\)

i.e., \(52\) complete weeks which contains \(52\) Mondays,

Now \(2\) days of the year are remaining.

These two days can be (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)

i.e., there are \(7\) pairs, in which Monday occurs in \(2\) pairs,

Probability \(= \frac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ outcomes}\)

So, probability is:

\(P(53\) Monday\()=\frac{2}{7}\)

First we shall find the point of intersection of the curve \(\mathrm{y}^{2}=2 \mathrm{x}\) and the straight line \(\mathrm{y}=\mathrm{x}\)

Put \(y=x\) in \(y^{2}=2 x,\) we get

\(\Rightarrow x^{2}=2 x\)

\(\Rightarrow x^{2}-2 x=0\)

\(\Rightarrow x(x-2)=0\)

\(\Rightarrow x=0\) or \(x=2\)

\(\Rightarrow y=0\) or \(y=2\)

Hence, (0,0) and (2,2) are the point of intersection of the curve \(\mathrm{y}^{2}=2 \mathrm{x}\) and the straight line \(y=x\).

The area under the curve \(\mathrm{y}=\mathrm{f}(\mathrm{x})\) between \(\mathrm{x}=\mathrm{a}\) and \(\mathrm{x}=\mathrm{b},\) is given by, Area \(=\) \(\int_{\mathrm{x}=\mathrm{a}}^{\mathrm{x}=\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{dx}\)

Here, \(f(x)=(\sqrt{2 x}-x)\)

Hence, Area \(=\int_{\mathrm{x}=0}^{\mathrm{x}=2}(\sqrt{2 \mathrm{x}}-\mathrm{x}) \mathrm{dx}\)

Area \(=\left[\frac{\mathrm{x}^{\frac{3}{2}}}{\frac{3}{2}}-\frac{\mathrm{x}^{2}}{2}\right]_{0}^{2}\)

Area \(=\left[\frac{2^{\frac{3}{2}}}{\frac{3}{2}}-\frac{2^{2}}{2}\right]-0\)

Area \(=\frac{2}{3}\)

Hence, the area of the region enclosed between the curve \(\mathrm{y}^{2}=2 \mathrm{x}\) and the straight line \(\mathrm{y}\) \(=x\) is \(\frac{2}{3}\)

As we know,

By definition,

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

\(n !\) is defined as:

\(n !=1 \times 2 \times 3 \times \ldots \times n\)

\(0 !=1\)

We have \({}^{\mathrm{n} }\mathrm{P}_{2}=30\)

\(\Rightarrow \frac{\mathrm{n} !}{(\mathrm{n}-2) !}=30\)

\(\Rightarrow \frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2) !}{(\mathrm{n}-2) !}=30\)

\(\Rightarrow \mathrm{n}(\mathrm{n}-1)=30\)

\(\Rightarrow \mathrm{n}^{2}-\mathrm{n}-30=0\)

\(\Rightarrow \mathrm{n}^{2}-6 \mathrm{n}+5 \mathrm{n}-30=0\)

\(\Rightarrow \mathrm{n}(\mathrm{n}-6)+5(\mathrm{n}-6)=0\)

\(\Rightarrow(\mathrm{n}+5)(\mathrm{n}-6)=0\)

\(\Rightarrow \mathrm{n}+5=0\) OR \(\mathrm{n}-6=0\)

\(\Rightarrow \mathrm{n}=-5\) OR \(\mathrm{n}=6\)

Since \(\mathrm{n}\) must be a positive whole number, \(\mathrm{n}=6\) is the correct answer.

Given:

\(\left(2 x^{2}-\frac{5}{x}\right)^{13}\)

Let \(a= 2x^2,~b=\frac{-5}{x}\)

The general term in the expansion of \((a+b)^n\) in:

\(T_{r+1}={ }^{n} C_{r}\left(a \right)^{n-r)} \cdot\left(b\right)^{r}\)

\(T_{r+1}={ }^{13} C_{r}\left(2 x^{2}\right)^{13-r} \cdot\left(-\frac{5}{x}\right)^{r}\)

\(=(-1)^{r} \cdot{ }^{13} C_{r} \cdot 2^{13-r} \cdot 5^{r} \cdot x^{26-3 r} ; \text{so}~ 13 \geq r \geq 0\)

Power of \(x\) in this term is \(26-3 r\).

Total power in all coefficients \(=\sum_{r=0}^{r-13}(26-3 r)\)

\(=26 \sum_{0}^{13} 1-3 \sum_{0}^{13} r \)

\(=26(14)-3\left(\frac{13.14}{2}\right) \)

\(=364-273 \)

\(=91\)

Given equation of circles are \(x^{2}+y^{2}-4=0\) and \(x^{2}+y^{2}-8 x+15=0\)

Now, the required line is the radical axis of the two circles are

\(\left(x^{2}+y^{2}-4\right)-\left(x^{2}+y^{2}-8 x+15\right)=0\)

\(\Rightarrow x^{2}+y^{2}-4-x^{2}-y^{2}+8 x-15=0\)

\(\Rightarrow 8 x-19=0\)

A quadratic equation has always two roots. Standard form of a quadratic equation.

\(a x^2+b x+c=0\) (where \(a\) can't be zero)

Given one root is \(3\).

Let first check the equation\(x^2-5 x+6=0\)

By putting \(x=3\) in the equaion\(x^2-5 x+6=0\)

\( (3)^2-5(3)+6=0\)

\(9-15+6=0\)

\(\therefore 0=0\)

It means LHS \(=\) RHS

So, \(x^2-5 x+6=0\) is a required equation which has root \(x=3\)

The perfect square numbers between 2 to 101 are: 1,4,9,16,25,36,49,64,81,100