Mathematics Test-33

We have,

\({i} {z}^{3}+{z}^{2}-{z}+{i}=0\)

On dividing by \({i}\), we get

\(\Rightarrow {z}^{3}+\frac{{z}^{2}}{{i}}-\frac{{z}}{{i}}+1=0\)

\(\Rightarrow {z}^{3}-{i} {z}^{2}+{i} {z}+1=0 \quad\left[\because \frac{1}{{i}}=-{i}\right]\)

\(\Rightarrow {z}^{3}-{i} {z}^{2}+{iz}-{i}^{2}=0 \quad\left[\because {i}^{2}=-1\right]\)

\(\Rightarrow {z}^{2}({z}-{i})+{i}({z}-{i})=0\)

\(\Rightarrow\left({z}^{2}+{i}\right)({z}-{i})=0\)

So, \(z=i\) or \(z^{2}=-i\).

Now, \(z=i \)

\(\Rightarrow|z|=|i|\)

\(\Rightarrow|z|=1\)

And, \(z^{2}=-i\)

\(\Rightarrow\left|z^{2}\right|=|-i|=1\)

\(\Rightarrow|z|=1\)

\(A(\vec{a}), B(\vec{b}), C(\vec{c})\) are the vertices of the triangle \(A B C\).

\(D\) is the midpoint of \(BC\)

\(\Rightarrow D =\left(\frac{\overrightarrow{ b }+\overrightarrow{ c }}{2}\right)\)

\(E\) is the midpoint of \(CA\)

\(\Rightarrow E=\left(\frac{\vec{a}+\vec{c}}{2}\right)\)

\(F\) is the midpoint of \(A B\)

\(\Rightarrow F =\left(\frac{\overrightarrow{ a }+\overrightarrow{ b }}{2}\right)\)

Now, \(AD , BE\) and \(CF\) are the medians of triangle \(ABC\)

All three medians intersects at point \(P\).

\(\because P\) divides \(AD\) in the ratio \(2: 1\) and other medians also intersect at \(P\).

\(\Rightarrow P\) divides all three medians in the ratio \(2: 1\)

\(\Rightarrow P\) is the centroid of the triangle \(ABC\).

The position vector of the centroid of a triangle \(ABC\) with position vectors of the vertices being \(\vec{a}, \vec{b}\) and \(\vec{c}\) respectively is given by:

\(P=\frac{\vec{a}+\vec{b}+\vec{c}}{3}\)

Clearly \(f(A)=I+A+A^2+\ldots \ldots+A^{16}\)

\(A^2=A A=\left[\begin{array}{ll}0 & 5 \\ 0 & 0\end{array}\right]\left[\begin{array}{ll}0 & 5 \\ 0 & 0\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]=O\)

\(A^3=0, A^4=0, \ldots \ldots A^{16}=0\)

\(\therefore f(A)=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]+\left[\begin{array}{ll}0 & 5 \\ 0 & 0\end{array}\right]+0+0+\ldots \ldots+0\)

\(=\left[\begin{array}{ll}1 & 5 \\ 0 & 1\end{array}\right]\)

Let the first element of A.P. is 'a' and common difference is 'd'.

T₂₅ = T₁₅ + 70

According to the question,

T₂₅ = T₁₅ + 70

As we know,

\(a_{n}=a+(n-1) d\)

⇒ a + (25 – 1)d = a + ( 15 – 1)d + 70

⇒ 24d – 14d = 70

⇒ 10d = 70

⇒ d = 7

∴ The common difference is 7.

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 \(C _2 \rightarrow C _2+ C _3\), 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 \(C _1 \rightarrow C _1+ C _2, C _3 \rightarrow C _3+ C _2\) 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 \(a + b + c\) from \(C _3\), 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.

\(\therefore \Delta=0\)

Let \(\vec{c}=x \hat{i}+y \hat{j}\), then

\(\vec{b} \perp \vec{c} \Rightarrow \vec{b} \cdot \vec{c}=4 x+3 y=0\)

\(\Rightarrow \frac{x}{3}=\frac{y}{-4}=\lambda\)

\(\Rightarrow x=3 \lambda\)

\(y=-4 \lambda\)

\(\therefore \vec{c}=\lambda(3 \hat{i}-4 \hat{j})\)

Let the required vector be \(\vec{a}=a_1 \hat{i}+a_2 \hat{j}\), then the projections of \(\vec{a}\) on \(\vec{b}\) and \(\vec{c}\) are \(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\) and \(\frac{\vec{a} \cdot \vec{c}}{|\vec{c}|}\) respectively

\(\therefore \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}=1\) and \(\frac{\vec{a} \cdot \vec{c}}{|\vec{c}|}=2\) (given)

\(\Rightarrow 4 a_1+3 a_2=5\)

and \(3 a_1-4 a_2=10\)

\(\Rightarrow a_1=2, a_2=-1\)

Therefore, the required vector \(=2 \hat{i}-\hat{j}\)

A circle is a simple closed curve all of whose points are at the same distance from a fixed point.

A circle is a figure consisting of all points in a plane that lie at a given point, a certain distance from the center; uniformly it is a curve formed by a point that moves in a plane so that its distance from a given point is constant. The distance between any point on the circle and the center is called the radius. Radius is a positive number.