Mathematics Test - 16

Given:

\(u_{1}=1, u_{2}=1, u_{n+2}=u_{n+1}+u_{n} \cdot n \geq 1\)

\(u_{n}=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^{n}\right]\)

\(\therefore u_{2}=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{2}-\left(\frac{1-\sqrt{5}}{2}\right)^{2}\right]\)

\(=\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}+\frac{1-\sqrt{5}}{2}\right)\left(\frac{1+\sqrt{5}}{2}-\frac{1-\sqrt{5}}{2}\right)\)

\(=\frac{1}{\sqrt{5}}(1 \times \sqrt{5})=\) Thus, \(u_{n}\) is true for \(n=2 \quad\) (as given \(\left.u_{2}=1\right)\)

Let it be true for \(n=k>2\). Then \(u_{k}=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{k}-\left(\frac{1-\sqrt{5}}{2}\right)^{k}\right]\)

Consider that \(u_{k+1}=u_{k}+u_{k-1}\) [Using \(\left.u_{n+2}=u_{n+1}+u_{n}\right]\)

\(\therefore u_{k+1}=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{k}-\left(\frac{1-\sqrt{5}}{2}\right)^{k}\right]+\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{k-1}-\left(\frac{1-\sqrt{5}}{2}\right)^{k-1}\right] \)

\(=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{k}+\left(\frac{1+\sqrt{5}}{2}\right)^{k-1}\right] \)

\(=\left[\left(\frac{1-\sqrt{5}}{2}\right)^{k}+\left(\frac{1+\sqrt{5}}{2}\right)^{k-1}\right] \)

\(=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{k-1}\left\{\frac{1+\sqrt{5}}{2}+1\right\}\right] \)

\(=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{k-1}\left\{\frac{6+2 \sqrt{5}}{4}\right\}-\left(\frac{1-\sqrt{5}}{2}\right)^{k-1}\right. \)

\(=\left[\left\{\frac{6-2 \sqrt{5}}{4}\right\}\right] \)

\(=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{k-1}\left(\frac{\sqrt{5}+1}{2}\right)^{2}-\left(\frac{1-\sqrt{5}}{2}\right)^{k-1}\right.\)

\(=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{k-1}\left(\frac{\sqrt{5}+1}{2}\right)^{2}-\left(\frac{1-\sqrt{5}}{2}\right)^{k-1}\right. \)

\(=\left[\left(\frac{\sqrt{5}-1}{2}\right)^{2}\right]\)

So, \(u_{k+1}\) is also true.

So, by the principle of mathematical induction \(u_{n}\) is true \(\forall n>1\)

The slope of the tangent to the curve \(=\frac{d y}{d x}\)

The slope of normal to the curve \(=\frac{-1}{\left(\frac{ dy }{ dx }\right)}\)

Point-slope is the general form: \(y-y_{1}=m\left(x-x_{1}\right)\), Where \(m=\) slope

Here, \(y=3 x^{2}+1\)

\(\frac{ dy }{ dx }=6 x\)

\(\left.\frac{ dy }{ dx }\right|_{ x =2}=12\)

Slope of normal to the curve \(=\frac{-1}{\left(\frac{d y}{d x}\right)}=\frac{-1}{12}\)

Equation of normal to curve passing through \((2,13)\) is:

\(y-13=\frac{-1}{12}(x-2)\)

\(\Rightarrow 12 y-156=-x+2\)

\(\Rightarrow x+12 y-158=0\)

The vector equation of a line passing through a point with position vector \(\overrightarrow{r_{1}}\) and parallel to \(\vec{m}\) is given by:

\(\vec{r}=\overrightarrow{r_{1}}+\lambda \vec{m}\) where, \(\lambda\) is a scalar

Given here:

Equation of plane is \(\vec{r} \cdot(2 \hat{i}-3 \hat{j}-\hat{k})=0\) and the line is perpendicular to the given plane.

It is given that, the required line is perpendicular to the plane \(\vec{r} \cdot(2 \hat{i}-3 \hat{j}-\hat{k})=0\)

i.e., The required line is parallel to the vector \(\vec{m}=2 \hat{i}-3 \hat{j}-\hat{k}\)

The line passes through the point whose position vector is \(\hat{i}-2 \hat{j}+5 \hat{k}\) i.e., \(\overrightarrow{r_{1}}=\hat{i}-2 \hat{j}+5 \hat{k}\)

As we know that, the vector equation of a line passing through a point with position vector \(\overrightarrow{r_{1}}\) and parallel to \(\vec{m}\) is given by:

\(\vec{r}=\overrightarrow{r_{1}}+\lambda \vec{m}\)

So, the required equation of line is:

\(\vec{r}=(\hat{i}-2 \hat{j}+5 \hat{k})+\lambda \cdot(2 \hat{i}-3 \hat{j}-\hat{k})\)

Concept:

If \(\omega\) is the cube root of unity, i.e., \(\omega^3=1\). then \(1+\omega+\omega^2=0\)

\(\omega^4=\omega^3 \omega=\omega\left[\because \omega^3=1\right]\)

Given,

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

\(C_1^{\prime}=C_1+C_2+C_3\)

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

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

\(R_2^{\prime}=R_2-R_1\)

\(R_3^{\prime}=R_3-R_1\)

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

Expanding along first column:

\(x\left[\left(x+\omega^2-\omega\right)\left(x+\omega-\omega^2\right)-(1-\omega)\left(1-\omega^2\right)\right]\)

\(x\left[x^2+\omega x-\omega^2 x+\omega^2 x+\omega^3-\omega^4-\omega x-\omega^2+\omega^3-1+\omega^2+\omega-\omega^3\right]\)

\(x^3=0\left(\because \omega^3=1\right.\) and \(\left.\omega^4=\omega^3 \omega=\omega\right)\)

\(\therefore x=0\)

The given equation of the sphere is \(|2 \overrightarrow{\mathrm{r}}+(3 \overrightarrow{\mathrm{i}}-\overrightarrow{\mathrm{j}}+4 \overrightarrow{\mathrm{k}})|=4\)

Which can be re-written as \(\left|\overrightarrow{\mathrm{r}}-\left(-\frac{3}{2} \overrightarrow{\mathrm{i}}+\frac{1}{2} \overrightarrow{\mathrm{j}}-2 \overrightarrow{\mathrm{k}}\right)\right|=2\) (dividing the \(\mathrm{LHS}\) and RHS by 2).

Comparing this with the general equation \(|\overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{c}}|=\mathrm{a},\) we can say that \(\overrightarrow{\mathrm{c}}=-\frac{3}{2} \overrightarrow{\mathrm{i}}+\frac{1}{2} \overrightarrow{\mathrm{j}}-2 \overrightarrow{\mathrm{k}}\) and \(\mathrm{a}=2\)

It means that the center is \(\left(\frac{-3}{2}, \frac{1}{2},-2\right)\) and the radius is 2 .

Given:

\(p(P)=\frac{1}{4}\)

\(P\left(\frac{P}{Q}\right)=\frac{1}{2}, P\left(\frac{Q}{P}\right)=\frac{1}{3}\)

\(P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}\)

\(P\left(\frac{Q}{P}\right)=\frac{P(P \cap Q)}{P(P)},\)

\(\frac{1}{3}=\frac{P(P \cap Q)}{\frac{1}{4}},\)

\(P(P \cap Q)=\frac{1}{12}\)

\(\text { Also, } P\left(\frac{P}{Q}\right)=\frac{P(P \cap Q)}{P(Q)},\)

\(\frac{1}{2}=\frac{\frac{1}{12}}{P(Q)},\)

\(P(Q)=\frac{1}{6}\)

Required probability, \(P\left(\frac{P^{\prime}}{Q^{\prime}}\right)=\frac{P\left(P^{\prime} \cap Q^{\prime}\right)}{P\left(Q^{\prime}\right)}\)

\(=\frac{P(P \cup Q)^{\prime}}{1-P(Q)}=\frac{1-P(P \cup Q)}{1-P(Q)}\)

\(=\frac{1-(P(P)+P(Q)-P(P \cap Q))}{1-P(Q)}\)

\(=\frac{1-\left(\frac{1}{4}+\frac{1}{6}-\frac{1}{12}\right)}{1-\frac{1}{6}}\)

\(=\frac{\frac{8}{12}}{\frac{5}{6}}=\frac{4}{5}\)

Given Two vertices of Triangle \(A(2,4,6)\) and \(B(0,-2,-5)\) and if centroid \(\mathrm{G}(2,1,-1)\)

Let Third vertices be ( \(\mathrm{x}, \mathrm{y}, \mathrm{z}\) )

Now \(\frac{2+0+x}{3}=2, \frac{4-2+y}{3}=1, \frac{6-5+z}{3}=-1\)

\(\mathrm{x}=4, \mathrm{y}=1, \quad \mathrm{z}=-1\)

Third vertices \(C(4,1,-4)\\)

Now, Image of vertices \(\mathrm{C}(4,1,-4)\) in the given plane is \(\mathrm{D}(\alpha, \beta, \gamma)\)

Now

\(\frac{\alpha-4}{1}=\frac{\beta-1}{2}=\frac{\gamma+4}{4}=-2 \frac{(4+2-16-11)}{1+4+16} \\\)

\(\frac{\alpha-4}{1}=\frac{\beta-1}{2}=\frac{\gamma+4}{4}=\frac{42}{21} \Rightarrow 2 \\\)

\( \alpha=6, \beta=5, \gamma=4 \\\)

\( \text { Then } \alpha \beta+\beta \gamma+\gamma \alpha \\\)

\( =(6 \times 5)+(5 \times 4)+(4 \times 6) \\\)

\( =30+20+24 \\\)

\( =74\)

\(\lim _{n \rightarrow a}\left(\sqrt{n^2-n-1}+n \alpha+\beta\right)=0\)

[ This limit will be zero when \(\alpha<0\) as when \(\alpha>0\) then overall limit will be \(\infty\).]

\( \Rightarrow \lim _{n \rightarrow \alpha} \frac{\left(\sqrt{n^2-n-1}+n \alpha+\beta\right)\left(\sqrt{n^2-n-1}-(n \alpha+\beta)\right)}{\sqrt{n^2-n-1}-(n \alpha+\beta)}=0 \\\)

\( \Rightarrow \lim _{n \rightarrow \alpha} \frac{\left(n^2-n-1\right)-(n \alpha+\beta)^2}{\sqrt{n^2-n-1}-(n \alpha+\beta)}=0 \\\)

\( \Rightarrow \lim _{n \rightarrow \alpha} \frac{n^2-n-1-n^2 \alpha^2-2 n \alpha \beta-\beta^2}{\sqrt{n^2-n-1}-(n \alpha+\beta)}=0 \\\)

\(\Rightarrow \lim _{n \rightarrow \alpha} \frac{n^2\left(1-\alpha^2\right)-n(1+2 \alpha \beta)-\left(1+\beta^2\right)}{\sqrt{n^2-n-1}-(n \alpha+\beta)}\)

Here power of " \(\mathrm{n}\) " in the numerator is 2 and power of " \(\mathrm{n}\) " in the denominator is 1 .

To get the value of limit equal to zero power of " \(n\) " should be equal in both numerator and denominator, otherwise value of limit will be infinite \((\infty)\).

\(\therefore\) Coefficient of \(\mathrm{n}^2\) should be 0 in this case.

\( \therefore 1-\alpha^2=0 \\\)

\( \Rightarrow \alpha= \pm 1\)

But \(\alpha\) should be \(<0\)

\(\therefore \alpha=+1 \text { not possible } \\\)

\( \therefore \alpha=-1 \\\)

\( \Rightarrow \lim _{n \rightarrow \alpha} \frac{0-n(1+2 \alpha \beta)-(1+\beta)}{\left[\sqrt{1-\frac{1}{n}-\frac{1}{n^2}}-\alpha-\frac{\beta}{n}\right]}=0\)

Divide numerator and denominator by \(\mathrm{n}\ then we get,\)

\(\Rightarrow \lim _{n \rightarrow \alpha} \frac{-(1+2 \alpha \beta)-\frac{(1+\beta)}{n}}{\sqrt{1-\frac{1}{n}-\frac{1}{n^2}}-\alpha-\frac{\beta}{n}}=0 \\\)

\( \Rightarrow \frac{-(1+2 \alpha \beta)-0}{\sqrt{1-0-0}-\alpha-0}=0 \\\)

\( \Rightarrow \frac{-(1+2 \alpha \beta)}{1-\alpha}=0 \\\)

\( \Rightarrow-(1+2 \alpha \beta)=0 \\\)

\( \Rightarrow 1+2 \alpha \beta=0 \\\)

\( \Rightarrow 2 \alpha \beta=-1 \\\)

\( \Rightarrow \beta=-\frac{1}{2 \alpha}=-\frac{1}{2(-1)}=\frac{1}{2} \\\)

\( \therefore 8(\alpha+\beta) \\\)

\( =8\left(-1+\frac{1}{2}\right) \\\)

\( =8 \times-\frac{1}{2} \\\)

\( =-4\)