Mathematics Test - 56

Given: 

Equation of planes: \(x+y+z=6\) and \(2 x+3 y+4 z+5=0\)

The given equations can be re-written as: 

\(x+y+z-6=0\) and \(2 x+3 y+4 z+5=0\)

As we know that, the equation of a plane passing through the line of intersection of two planes \(a_{1} x+b_{1} y+c_{1} z+d_{1}=0\) and \(a_{2} x+b_{2} y+c_{2} z+d_{2}=0\) is given by:

\((a_{1} x+b_{1} y+c_{1} z+d_{1})+\lambda\left(a_{2} x+b_{2} y+c_{2} z+d_{2}\right)=0\) where \(\lambda \in R\)

\((x+y+z-6)+\lambda(2 x+3 y+4 z+5)=0\) ...(1)

\(\because\) It is given that plane passing through the line line of intersection of the plane \(x+y+\) \(z=6\) and \(2 x+3 y+4 z+5=0\) also passes through the point (1,1,1).

i.e The point (1,1,1) will satisfy the equation (1)

\((1+1+1-6)+\lambda(2+3+4+5)=0\)

\(-3+14 \lambda=0\)

\( \lambda=\frac{3}{14}\)

Now by substituting \(\lambda=\frac{3}{14}\) in equation (1) we get

\((x+y+z-6)+\frac{3}{14}.(2 x+3 y+4 z+5)=0\)

\( 20 x+23 y+26 z-69=0\)

\(\begin{aligned} & \text { Given, } P=\left[\begin{array}{ll}1 & 0 \\ \frac{1}{2} & 1\end{array}\right] \\ & \Rightarrow P^2=\left[\begin{array}{ll}1 & 0 \\ \frac{1}{2} & 1\end{array}\right]\left[\begin{array}{ll}1 & 0 \\ \frac{1}{2} & 1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right] \\ & \Rightarrow P^3=\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ \frac{1}{2} & 1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ \frac{3}{2} & 1\end{array}\right] \\ & \Rightarrow P^4=\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right]\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right] \\ & \vdots \\ & \Rightarrow P^n=\left[\begin{array}{cc}1 & 0 \\ \frac{n}{2} & 1\end{array}\right] \\ & \text { Hence, } P^{50}=\left[\begin{array}{cc}1 & 0 \\ 25 & 1\end{array}\right]\end{aligned}\)

Given,

\(P(n)=n^{2}+n\)

By putting \(n=1\), we get

\(P(1)=1^{2}+1\)

\(=2\)

As we can say that \(2\) is an even number.

Let \(P(k)\) is true for \(n=k\).

\(P(k) = \left(k^{2}+k\right)\) will be even.

\(\left(k^{2}+k\right)=2 m\) for some natural number \(m \ldots(i)\)

By putting \(n=k+1\), we get

\(P(k+1)=(k+1)^{2}+(k+1)\)

\(=k^{2}+3 k+2\)

\(=\left(k^{2}+k\right)+2(k+1)\)

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

\(P(k+1)=2 m+2(k+1)\)

\(=2[m+(k+1)]\), which is even.

So, \(P(k+1)\) is even.

\(\Rightarrow P(k+1)\) is true, whenever \(P(k)\) is true.

Thus, \(P(1)\) is true and \(P(k+1)\) is true, whenever \(P(k)\) is true.

Thus, by the Principle of Mathematical Induction, \(P(n)\) is true for all \(n \in N\) i.e., \(p(n)=\left(n^{2}+n\right)\) is even.

Given,

\(\cos \left(3015^{\circ}\right)\)

\(=\cos \left(360^{\circ} \times 8^{\circ}+135^{\circ}\right)\)\(\quad\quad(\because\cos (2 n \pi \pm \theta)=\cos \theta)\)

\(=\cos \left(135^{\circ}\right)\)

\(=\cos \left(90^{\circ}+45^{\circ}\right) \)\(\quad\quad(\because\cos (90+\theta)=-\sin \theta)\)

\(=-\sin 45^{\circ}\)

\(=\frac{-1}{\sqrt{2}}\)

Let \(S\) be the sample space

Number of all combinations of \(n\) things, taken \(r\) at a time, is given by \({ }^{n} C_{r}=\frac{n !}{(r) !(n-r) !}\)

Then, \(n(S)={ }^{52} C_{2}\)

\(=\frac{52 \times 51}{(2 \times 1)} \\\)

\(=1326\)

Let \(E=\) event of getting \(2\) kings out of \(4\)

\(\therefore n(E)={ }^{4} C_{2}=\frac{4 \times 3}{2 \times 1}=6\)

\(\therefore P(E)=\frac{n(E)}{n(S)}\)

\(=\frac{6}{1326}\)

\(=\frac{1}{221}\)

Given,

Slope of line = -2

Lines are:

\(2 x-y=1 \)....(i)

\(x+2 y=3\)...(ii)

On multiplying equation (i) by 2 we get,

\(4 x-2y=2 \)....(iii)

Adding equation (ii) and (iii) we get,

\(5 x=5\)

\(\Rightarrow x=1\)

On putting value of \(x\) in equation (i),

\(2(1)-y=1 \)

\(\Rightarrow y=1\)

Therefore,

The intersection point is \((1,1)\)

So line has the slope m = \(-2\) and passes through \((1,1)\)

Therefore,

Equation of the perpendicular line is:

\(\left(y-y_{1}\right)=m\left(x-x_{1}\right)\)

\(\Rightarrow y-1=-2(x-1) \)

\(\Rightarrow y+2 x-3=0\)

Let \(\Delta=\left|\begin{array}{lll}1-a & a-b-c & b+c \\ 1-b & b-c-a & c+a \\ 1-c & c-a-b & a+b\end{array}\right|\)

Applying \(\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}+\mathrm{C}_{3},\) we get:

\(\Delta=\left|\begin{array}{lll}1-a & a & b+c \\ 1-b & b & c+a \\ 1-c & c & a+b\end{array}\right|\)

Applying \(\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}, \mathrm{C}_{3} \rightarrow \mathrm{C}_{3}+\mathrm{C}_{2}\), we get:

\(\Delta=\left|\begin{array}{lll}1 & a & a+b+c \\ 1 & b & b+c+a \\ 1 & c & c+a+b\end{array}\right|\)

Taking common \(\mathrm{a}+\mathrm{b}+\mathrm{c}\) from \(\mathrm{C}_{3},\) we get:

\(\Delta=(a+b+c)\left|\begin{array}{lll}1 & a & 1 \\ 1 & b & 1 \\ 1 & c & 1\end{array}\right|\)

We know that if two columns of a determinant are identical, the value of the determinant is zero.

\(\therefore \Delta=0\)