Mathematics Test-21

\(\mathrm{dy}=\left(1+\mathrm{y}^2\right) \mathrm{dx}\)

\(\Rightarrow \frac{d y}{1+y^2}=\mathrm{dx}\)

Integrating both sides, we get,

\(\Rightarrow \tan ^{-1} y=x+c\)

\(\Rightarrow y=\tan (x+c)\)

\(\therefore\) The solution of the differential equation \(d y=\left(1+y^2\right) d x\) is \(y=\tan (x+c)\).

\(\left|\frac{\beta-\alpha}{1-\alpha \beta}\right|^{2}=\left(\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right)\left(\frac{\beta-\alpha}{1-\alpha \beta}\right)\)

\(\Rightarrow\left|\frac{\beta-\alpha}{1-\alpha \beta}\right|^{2}=\frac{\beta \beta-\beta \alpha-\alpha \beta+\alpha \alpha}{(1-\alpha \beta)(1-\alpha \beta)}\)

\(\Rightarrow\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|^{2}=\frac{|\beta|^{2}-\beta \alpha-\alpha \beta+|\alpha|^{2}}{1-\alpha \beta-\alpha \beta+|\alpha|^{2}|\beta|^{2}}\)

\(\Rightarrow\left|\frac{\beta-\alpha}{1-\alpha \beta}\right|^{2}=\frac{|\alpha|^{2}-\beta \bar{\alpha}-\alpha \bar{\beta}+1}{1-\alpha \beta-\alpha \beta+|\alpha|^{2}}\), \(\quad[\because|\beta|=1]\)

\(\Rightarrow\left|\frac{\beta-\alpha}{1-\alpha \beta}\right|=1\)

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

\(A^2=\left[\begin{array}{ll} i & 0 \\ 0 & i \end{array}\right]\left[\begin{array}{ll} i & 0 \\ 0 & i \end{array}\right]=\left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right]\)

\(A ^4= A ^2 \times A ^2=\left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right]\left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\)

\(A ^4= I\)

\(A ^{4 n }= I ^{ n }= I =\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\)

x² – 5(k – 1)x + (8k + 1) = 0 has equal roots.

\(b^{2}-4 a c=0\)

\(\Rightarrow\{5(k-1)\}^{2}-4(8 k+1)=0\)

\(\Rightarrow\{5(k-1)\}^{2}=4(8 k+1)\)

\(\Rightarrow 25\left(k^{2}-2 k+1\right)=32 k+4\)

\(\Rightarrow 25k^{2}-82k+21=0\)

\(\Rightarrow(k-3)(25 k-7)=0\)

\(\Rightarrow k=3\) is the integer value.

As two dices are thrown simultaneously then combination of total cases(outcomes) \(=6\times6\)

Now, Total cases in which sum of \(9\) can be obtained are:

\((5,4),(4,5),(6,3),(3,6)\)

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

\(\therefore P(9)=\frac{4}{36}\)

\(=\frac{1}{9}\)

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 \(\alpha \in R\).

\(\frac{x \operatorname{cosec}^2 30^{\circ} \sec ^2 45^{\circ}}{8 \cos ^2 45^{\circ} \sin ^2 60^{\circ}}=\tan ^2 60^{\circ}-\tan ^2 30^{\circ}\)

\(\Rightarrow \frac{x(2)^2(\sqrt{2})^2}{8\left(\frac{1}{\sqrt{2}}\right)^2\left(\frac{\sqrt{3}}{2}\right)^2}=(\sqrt{3})^2-\left(\frac{1}{\sqrt{3}}\right)^2\)

\(\Rightarrow \frac{x \times 4 \times 2}{8 \times \frac{1}{2} \times \frac{3}{4}}=3-\frac{1}{3} \Rightarrow \frac{8 x}{3}=\frac{8}{3}\)

\(\Rightarrow x=\frac{8}{3} \times \frac{3}{8}=1\)

A vector \(\vec{v}\) in the plane of \(\vec{a}\) and \(\vec{b}\) is \(\vec{v}=\vec{a}+\lambda \vec{b}\) \(\Rightarrow \vec{v}=(\hat{i}+\hat{j}+\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})\)

Projection of \(\vec{v}\) on \(\vec{c}=\frac{1}{\sqrt{3}}\)

\(\Rightarrow \frac{\vec{v} \cdot \vec{c}}{|\vec{c}|}=\frac{1}{\sqrt{3}}\)

\(\Rightarrow \frac{(1+\lambda)-(1-\lambda)-(1+\lambda)}{\sqrt{3}}=\frac{1}{\sqrt{3}}\)

\(\Rightarrow(1+\lambda)-(1-\lambda)-(1+\lambda)=1\)

\(\lambda=2\)

So, \(\vec{v}=3 \hat{i}-\hat{j}+3 \hat{k}\)

General term: General term in the expansion of \((x+y)^{n}\) is given by \(\mathrm{T}_{({r}+1)}={ }^{\mathrm{n}} {C}_{{r}} \times {x}^{{n}-{r}} \times{y}^{{r}}\).

Given: Sum of the \(5^{\text {th }}\) and \(6^{\text {th }}\) terms is zero

\(\Rightarrow \mathrm{T}_{5}+\mathrm{T}_{6}=0\)

\(\Rightarrow{ }^{\mathrm{n}} \mathrm{C}_{4} \times \mathrm{a}^{\mathrm{n}-4} \times(-\mathrm{b})^{4}+{ }^{\mathrm{n}} \mathrm{C}_{5} \times \mathrm{a}^{\mathrm{n}-5} \times(-\mathrm{b})^{5}=0\)

\(\Rightarrow{ }^{\mathrm{n}} \mathrm{C}_{4} \times \mathrm{a}^{\mathrm{n}-4} \times \mathrm{b}^{4}-{ }^{\mathrm{n}} \mathrm{C}_{5} \times \mathrm{a}^{\mathrm{n}-5} \times \mathrm{b}^{5}=0\)

\(\Rightarrow{ }^{\mathrm{n}} \mathrm{C}_{4} \times \mathrm{a}^{\mathrm{n}-4} \times \mathrm{b}^{4}={ }^{\mathrm{n}} \mathrm{C}_{5} \times \mathrm{a}^{\mathrm{n}-5} \times \mathrm{b}^{5}\)

\(\Rightarrow \frac{\mathrm{n} !}{(\mathrm{n}-4) ! 4 !} \times \frac{\mathrm{a}^{\mathrm{n}-4}}{\mathrm{a}^{\mathrm{n}-5}}=\frac{\mathrm{n} !}{(\mathrm{n}-5) ! 5 !} \times \frac{\mathrm{b}^{5}}{\mathrm{~b}^{4}}\)

\(\Rightarrow \frac{\mathrm{a}}{(\mathrm{n}-4)}=\frac{\mathrm{b}}{5}\)

\(\therefore \frac{\mathrm{a}}{\mathrm{b}}=\frac{(\mathrm{n}-4)}{5}\)