Mathematics Test - 2

Given, 

\(P(A)=\frac{1}{5}\), 

\(P(B)=\frac{1}{3}\), 

And, \(P(B \mid A)=\frac{1}{3}\)

We know that, 

\({P}({B} \mid {A})=\frac{{P}({A} \cap {B})}{{P}({A})}\)

\(\Rightarrow \frac{1}{3}=\frac{{P}({A} \cap {B})}{\frac{1}{5}}\)

\(\Rightarrow {P}({A} \cap {B})=\frac{1}{15}\)

Formula:

\(\mathrm{C}(\mathrm{n}, \mathrm{r})=\frac{n !}{r !(n-r) !}\)

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

Calculation:

In a cricket team,

The order of choosing a cricket player does not affect the team.

So, the number of ways that a cricket team of \(11\) players can be made out of \(15\) players is:

\(\mathrm{C}(15,11)=\frac{15 !}{11 !(15-11) !}\)

\(\Rightarrow \mathrm{C}(15,11)=\frac{15 !}{11 !(4) !}\)

\(\Rightarrow \mathrm{C}(15,11)=\frac{15 \times 14 \times 13 \times 12 \times 11 !}{11 !(4 \times 3 \times 2 \times 1)}\)

\(\Rightarrow \mathrm{C}(15,11)=15 \times 7 \times 13\)

\(\Rightarrow \mathrm{C}(15,11)=1365\)

Given:

\(y=x^{x}\)

Taking log both sides, we get,

\(\Rightarrow \log y=\log x^{x} \)

\(\Rightarrow \log y=x \log x\quad\left(\because \log m^{n}=n \log m\right)\)

Differentiating above with respect to x, we get:

\(\Rightarrow \frac{1}{y} \times \frac{d y}{d x}=x \times \frac{\log x}{d x}+\log x \times \frac{d x}{d x}\)

\(\Rightarrow \frac{1}{y} \times \frac{d y}{d x}=x \times \frac{1}{x}+\log x \times 1\)

\(\Rightarrow \frac{d y}{d x}=y \times[1+\log x]\)

\(\Rightarrow \frac{d y}{d x}=x^{x} \times[1+\log x]\)

Putting x = 1, we get:

\(\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=1^{1} \times[1+\log 1]\)

\(\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=1 \times[1+0] \quad(\because \log 1=0)\)

\(\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=1\)

A triangular region in the 3rd quadrant

Given inequalities x ≥ 0 , y ≥ 0 , 2x + y + 6 ≥ 0

Now take x = 0, y = 0 and 2x + y + 6 = 0

when x = 0, y = -6

when y = 0, x = -3

So, the points are A(0, 0), B(0, -6) and C(-3, 0)

So, the graph of the inequations x ≤ 0 , y ≤ 0 , and 2x + y + 6 ≥ 0 is a triangular region in the 3rd quadrant.

In vector, i component lies in X-axis,  j component lies in Y-axis and k component lies in Z-axis.

Given vector is \(\mathrm{A}=2 \hat{j}-\hat{\mathrm{k}}\).

The vector has \(j\) and \(k\) component.

In vector, j component lies in Y-axis and k component lies in Z-axis.

Given vector \(\mathrm{A}=2 \hat{\mathrm{j}}-\hat{\mathrm{k}}\) lies in \(\mathrm{YZ}\) - plane.

Given,

\(\mathrm{T}(\mathrm{k})\) \(=1+3+5+\ldots \ldots . .+(2 \mathrm{k-1)=k}^{2}+10\)

\(\mathrm{T}(\mathrm{k})=2 \mathrm{k}-1=\mathrm{k}^{2}+10\)

By putting \(\mathrm{k}=1\), we get

\(\mathrm{T}(1)=2 \times 1-1\)

\(=1^{2}+10\) 

\(=1 \neq 10\)

\(\mathrm{LHS} \neq \mathrm{RHS}\)

\(\therefore \mathrm{T}(1)\) is not true.

Let \(\mathrm{T}(\mathrm{k})\) is true, then

\(\mathrm{T(k)}=1+3+5+\ldots \ldots \ldots+(2 \mathrm{k-1)=k^{2}}+10.........(1)\)

\(= 1+3+5+\ldots \ldots . .+(2 \mathrm{k-1)+(2 k+1)=k^{2}+10+2 k+1}\)

Using equation \((1)\), we get

\(= 1+3+5+\ldots \ldots . .+(2 \mathrm{k}-1)+(2 \mathrm{k}+1)=(\mathrm{k}+1)^{2}+10\)

\(\therefore \mathrm{T}(\mathrm{k}+1)\) is true.

Thus, \(\mathrm{T(k)}\) is true \(\Rightarrow \mathrm{T(k+1)}\) is true.

Here, we have to find the distance of the line 3x - 4y + 12 = 0 from the point (4, 1)

Let P = (4, 1)

⇒ x₁ = 4 and y₁ = 1

Here, a = 3 and b = -4

Now substitute x₁ = 4 and y₁ = 1 in the equation 3x - 4y + 12 = 0, we get

⇒ |3⋅ x₁ - 4 ⋅ y₁ + 12| = |12 - 4 + 12| = 20

\(\Rightarrow \sqrt{a^{2}+b^{2}}=\sqrt{3^{2}+(-4)^{2}}=5\)

As we know that, the perpendicular distance \(\mathrm{d}\) from \(P\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) to the line \(\mathrm{ax}+\mathrm{by}+\mathrm{c}=0\) is given by \(d=\left|\frac{a x_{1}+b y_{1}+c}{\sqrt{a^{2}+b^{2}}}\right|\)

\(\Rightarrow d=\left|\frac{3 \cdot x_{1}-4 \cdot y_{1}+12}{\sqrt{3^{2}+(-4)^{2}}}\right|=\frac{20}{5}=4\)

Given:

a α $\frac{1}{b}$      ----(i)

b α $\frac{1}{c}$      ----(ii)

Calculation:

If b α $\frac{1}{c}$

⇒ b = $\frac{k}{c}$

Here k is a constant

Put this value of b in equation (i) we get,

a α $\frac{1}{\frac{k}{c}}$

⇒ a α $\frac{c}{k}$

⇒ a α c

∴ a is directly proportional to c.

Given lines are: 

\(x=a y+b\) and \(z=c y+d\)

\(\Rightarrow x-b=a y\) and \(z-d=c y\)

\(\therefore \frac{x -b}{a}=\frac{y}{1}=\frac{z-d}{c} \quad \quad \ldots\) (1)

Therefore, direction ration of line is \(( a , 1, c )\).

Again, lines are:

\(x=e y+f\) and \(z=g y+h\) 

\(\Rightarrow x-f=e y\) and \(z-h=g y\)

\(\therefore \frac{ x - f }{ e }=\frac{ y }{1}=\frac{ z - h }{g}\quad \quad \ldots\) (2)

Therefore, direction ration of line is \((e, 1, g)\).

Given both line are perpendicular:

We know that:

If two lines are perpendicular to each other’s then product of direction ratio of these lines is zero.

\(\therefore( a \times e )+(1 \times 1)+( c \times g )=0\)

\(\Rightarrow ae +1+ cg =0\)

or, \(a e+c g+1=0\)

By cramer’s rule:

If \(\Delta=0\) and \(\Delta_{1}=\Delta_{2}=\Delta_{3}=0\), then the system is consistent and has infinitely many solutions.

Given: \(x+y+z=3,2 x+2 y+2 z=6\) and \(a x+a y+a z=3 a\) As we know that,

\(\Delta=\left|\begin{array}{lll}a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3}\end{array}\right|, \Delta_{1}=\left|\begin{array}{lll}d_{1} & b_{1} & c_{1} \\ d_{2} & b_{2} & c_{2} \\ d_{3} & b_{3} & c_{3}\end{array}\right|, \Delta_{2}=\left|\begin{array}{lll}a_{1} & d_{1} & c_{1} \\ a_{2} & d_{2} & c_{2} \\ a_{3} & d_{3} & c_{3}\end{array}\right|\) and \(\Delta_{3}=\left|\begin{array}{ccc}a_{1} & b_{1} & d_{1} \\ a_{2} & b_{2} & d_{2} \\ a_{3} & b_{3} & d_{3}\end{array}\right|\)

\(\Rightarrow \Delta=\left|\begin{array}{ccc}1 & 1 & 1 \\ 2 & 4 & 1 \\ \alpha & \alpha & \alpha\end{array}\right|, \Delta_{1}=\left|\begin{array}{ccc}3 & 1 & 1 \\ 6 & 4 & 1 \\ 3 \alpha & \alpha & \alpha\end{array}\right|, \Delta_{2}=\left|\begin{array}{ccc}1 & 3 & 1 \\ 2 & 6 & 1 \\ \alpha & 3 \alpha & \alpha\end{array}\right|\) and \(\Delta_{3}=\left|\begin{array}{ccc}1 & 1 & 3 \\ 2 & 4 & 6 \\ \alpha & \alpha & 3 \alpha\end{array}\right|\)

\(\Rightarrow \Delta=0\) and \(\Delta_{1}=\Delta_{2}=\Delta_{3}=0\)

Thus, the given system has infinite solution for \(\alpha \in R\).