Mathematics Test-39

Given that A

\(\mathrm{A}=\left[\begin{array}{ccc}\frac{2}{3} & 1 & \frac{5}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3}\end{array}\right]\) and \(\mathrm{B}=\left[\begin{array}{ccc}\frac{2}{5} & \frac{3}{5} & 1 \\ \frac{1}{5} & \frac{2}{5} & \frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5}\end{array}\right]\)

\(\therefore\) \(3 \mathrm{~A}-5 \mathrm{~B}=3\left[\begin{array}{ccc}\frac{2}{3} & 1 & \frac{5}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3}\end{array}\right]-5\left[\begin{array}{ccc}\frac{2}{5} & \frac{3}{5} & 1 \\ \frac{1}{5} & \frac{2}{5} & \frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5}\end{array}\right]\)

\(\Rightarrow 3 \mathrm{~A}-5 \mathrm{~B}=\left[\begin{array}{lll}2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2\end{array}\right]-\left[\begin{array}{lll}2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2\end{array}\right]\)

\(\Rightarrow 3 \mathrm{~A}-5 \mathrm{~B}=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]\)

The word "RUMOUR" has two repetitive \(R\) and \(U\).

Therefore, number of arrangements \(=\frac{6 !}{2 ! \times 2 !}\)

\(=\frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1 \times 2 \times 1} \)

\(=180\)

Let \(P(h, k)\) be any point on the locus. Then

Given: \(PA=P B\)

\(\Rightarrow P A^{2}=P B^{2}\)

\(\Rightarrow(h-2)^{2}+(k-3)^{2}=(h+1)^{2}+(k-2)^{2}\)

\(\Rightarrow h^{2}-4 h+4+k^{2}-6 k+9=h^{2}+2 h+1+k^{2}-4 k+4\)

\(\Rightarrow-4 h-6 k+9=2 h-4 k+1\)

\(\Rightarrow 6 h+2 k=8\)

The equation of the locus of a point equidistant from the point \(A(2,3)\) and \(B(-1,2)\) is \(6 x+2 y=8\)

Let the number of white marbles be \(x\).

Since only two colour marbles are present, and total probability we know of all the events is equal to \(1\).

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

\(P(\text {white})=1-P(\text {red})\)

\(\frac{x}{24}=1-(\frac{2}{3})\)

\(\Rightarrow \frac{x}{24}=\frac{1}{3}\)

\(\Rightarrow {x}=8\)

So there are \(8\) white marbles.

Given:

\(y=3 x^{2}+2\)

\(x\) changes from 10 to 10.1

So, \(\Delta x=10.1-10=0.1\)

Now, \(y=3 x^{2}+2\)

Differentiating with respect to \(x,\) we get

\(\Rightarrow \frac{dy}{dx}=6 x\)

As we know, \(\Delta y=\frac{y}{dx} \Delta x\)

\(\Rightarrow \Delta y=6 x \Delta x\)

Put \(x=10\) and \(\Delta x=0.1\)

\(\therefore \Delta y=6 \times 10 \times 0.1=6 \approx 6.03\)

Given,

A code consists of two distinct letters followed by two distinct numbers by using digits from \(1\) to \(9\).

As we know that, there \(26\) English alphabets.

So, the number of ways in which \(2\) letters can be chosen from \(26\) letters \(=\) \({ }^{26} P_{2}=26 \times 25=650\)

Similarly, the number of ways in which \(2\) digits can be chosen from \(9\) digits \(=\) \({ }^{9} P_{2}=9 \times 8=72\)

\(\therefore\) Total number of such codes that can be generated \(=650 \times 72=46800\)

Given:

\(\left( y ^{2}+ \frac{c }{ y} \right)^{5}\)

\(\left( y ^{2}+ \frac{c }{ y} \right)^{5}={ }^{5}C _{0}\left(\frac{ c }{ y }\right)^{0}\left( y ^{2}\right)^{5-0}+{ }^{5}C_{1}\left(\frac{ c }{ y }\right)^{1}\left( y ^{2}\right)^{5-1}+\ldots+{ }^{5} C _{5}\left(\frac{ c }{ y }\right)^{5}\left( y ^{2}\right)^{5-5}\)

\(=\sum_{r=0}^{5} {}^{5}C_{r}\left(\frac{c}{y}\right)^{r}\left(y^{2}\right)^{5-r}\) .....(i)

We need coefficient of \(y \Rightarrow 2(5-r)-r=1\)

\(\Rightarrow 10-3 r=1\)

\(\Rightarrow r=3\)

Put \(r=3\) in (i),

\(={}^{5}C_{3}\left(\frac{c}{y}\right)^{3}\left(y^{2}\right)^{2}\)

\(={}^{5}C_{3}c^{3}y\)

So, coefficient of \(y ={ }^{5} C _{3} \cdot c ^{3}\)

\(=10 c ^{3}\)

As we know,

If the point lies inside the circle no tangent can be drawn.

If the point lies on the circle then only one tangent can be drawn.

If the point lies outside the circle then a maximum of two tangent lines can be drawn on the circle.

Given,

Point \(=(2,6)\)

Equation of circle is \(x^{2}+y^{2}=40\)

\(x^{2}+y^{2}-40=0\)

Substituting \((2,6)\) in this equation, we get

\(2^{2}+6^{2}-40=0\)

\(\Rightarrow 4+36-40=0\)

\(\Rightarrow40-40=0\)

So, the point \((2,6)\) lies on the circle.

Therefore, only one tangent can be drawn.

Given,

\(2 x^{2}+x+4=0\)

\(2{x}^{2}+{x}=-4\)

Dividing the equation by \(2\), we get

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

\({x}^{2}+2 \times {x} \times (\frac {1} {4})=-2\)

By adding \((\frac {1} {4})^2\) to both sides of the equation, we get

\((x)^{2}+2 \times x \times \frac{1} {4}+ (\frac{1} {4})^{2}\) \(=(\frac {1} {4})^2-2\)

\((x+\frac{1} {4})^{2}=\frac{1} {16}-2\)

\((x+\frac{1} {4})^ 2\)\(=\frac{-31}{16}\)

The square root of a negative number is imaginary, therefore, there is no real root for the given equation.

We have \(5\) numbers \(-2,-1,0,1,2\)

Whose squares are \(4, 1,0,1,4\)

So, square of \(3\) numbers is less than \(2\)

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

Therefore Probability is:

\(P\left(x^2<2\right)=\frac{3}{5}\)