Mathematics Tes...

Let A be a 3×3 real matrix such that \(\mathrm{A}\left(\begin{array}{l}1 \\ 1 \\ 0\end{array}\right)=\left(\begin{array}{l}1 \\ 1 \\ 1\end{array}\right) ; \mathrm{A}\left(\begin{array}{l}1 \\ 0 \\ 1\end{array}\right)=\left(\begin{array}{c}-1 \\ 0 \\ 1\end{array}\right)\) and \(\mathrm{A}\left(\begin{array}{l}0 \\ 0 \\ 1\end{array}\right)=\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right)\) \(\text { If } \mathrm{X}=\left(\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3\right)^{\mathrm{T}} \text { and } \mathrm{I} \text { is an identity matrix of order } 3 \text {, then the system }\)

\((A-2 I) X=\left(\begin{array}{l}4 \\ 1 \\ 1\end{array}\right)\)

If \(u_{1}=u_{2}=1\) and \(u_{n+2}=u_{n+1}+u_{n} \cdot n \geq 1\). Then use mathematical induction to show that \(u_{n}=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^{n}\right]\) for all \(n>1\).

Find the equation of the normal to the curve \(y=3 x^{2}+1\), which passes through \((2,13)\).

Find the vector equation of the line passing through the point with position vector \(\hat{i}-2 \hat{j}+5 \hat{k}\) and perpendicular to the plane \(\vec{r} \cdot(2 \hat{i}-3 \hat{j}-\hat{k})=0\).

If \(f(x)=\left|\begin{array}{ccc}x^3 & 2 x^2+1 & 1+3 x \\ 3 x^2+2 & 2 x & x^3+6 \\ x^3-x & 4 & x^2-2\end{array}\right|\) for all \(x \in \mathbb{R}\), then \(2 f(0)+f^{\prime}(0)\) is equal to:

If \(\omega\) is a cube root of unity, then a root of the equation is:

\(\left|\begin{array}{ccc}x+1 & \omega & \omega^2 \\ \omega & x+\omega^2 & 1 \\ \omega^2 & 1 & x+\omega\end{array}\right|=0\)

\(P\) and \(Q\) are considering to apply for a job. The probability that \(P\) applies for the job is \(\frac{1}{4}\), the probability that \(P\) applies for the job given that \(Q\) applies for the job is \(\frac{1}{2}\), and the probability that \(Q\) applies for the job given that \(P\) applies for the job is \(\frac{1}{3}\). Then the probability that \(P\) does not apply for the job given that \(Q\) does not apply for the job is:

Let two vertices of a triangle \(\mathrm{ABC}\) be \((2,4,6)\) and \((0,-2,-5)\), and its centroid be \((2,1,-1)\). If the image of the third vertex in the plane \(x+2 y+4 z=11\) is \((\alpha, \beta, \gamma)\), then \(\alpha \beta+\beta \gamma+\gamma \alpha\) is equal to :

\(\text { If } \lim _{n \rightarrow \infty}\left(\sqrt{n^2-n-1}+n \alpha+\beta\right)=0 \text {, then } 8(\alpha+\beta) \text { is equal to: }\)