Mathematics Test - 13

Using the definition of triple cross product:

$\overrightarrow{\mathrm{i}}\times (\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{i}})=(\overrightarrow{\mathrm{i}}\cdot \overrightarrow{\mathrm{i}})\overrightarrow{\mathrm{a}}-(\overrightarrow{\mathrm{i}}\cdot \overrightarrow{\mathrm{a}})\overrightarrow{\mathrm{i}}=\overrightarrow{\mathrm{a}}-\mathrm{a}\overrightarrow{\mathrm{i}}$

$\overrightarrow{\mathrm{j}}\times (\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{j}})=(\overrightarrow{\mathrm{j}}\cdot \overrightarrow{\mathrm{j}})\overrightarrow{\mathrm{a}}-(\overrightarrow{\mathrm{j}}\cdot \overrightarrow{\mathrm{a}})\overrightarrow{\mathrm{j}}=\overrightarrow{\mathrm{a}}-\mathrm{a}\overrightarrow{\mathrm{j}}$

$\overrightarrow{\mathrm{k}}\times (\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{k}})=(\overrightarrow{\mathrm{k}}\cdot \overrightarrow{\mathrm{k}})\overrightarrow{\mathrm{a}}-(\overrightarrow{\mathrm{k}}\cdot \overrightarrow{\mathrm{a}})\overrightarrow{\mathrm{k}}=\overrightarrow{\mathrm{a}}-\mathrm{ak}$

$\therefore \overrightarrow{\mathrm{i}}\times (\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{i}})+\overrightarrow{\mathrm{j}}\times (\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{j}})+\overrightarrow{\mathrm{k}}\times (\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{k}})$

$=(\overrightarrow{a}-a\overrightarrow{i})+(\overrightarrow{a}-a\overrightarrow{j})+(\overrightarrow{a}-a\overrightarrow{k})$

$=3\overrightarrow{\mathrm{a}}-(\mathrm{a}\overrightarrow{\mathrm{i}}+\mathrm{a}\overrightarrow{\mathrm{j}}+\mathrm{a}\overrightarrow{\mathrm{k}})$

$=3\overrightarrow{a}-\overrightarrow{a}$

$=2\overrightarrow{\mathrm{a}}$

Let, $\overrightarrow{a}$ and $\overrightarrow{b}$ be the position vector of the points $A$ and $B$ respectively, then vector equation of line passing through $A$ and $B$ is given by: 

$\overrightarrow{r}=\overrightarrow{a}+\lambda (\overrightarrow{b}-\overrightarrow{a})$

Given here: 

The required line passes through the points $A(1,0,2)$ and $B(3,9,6)$.

So, the position vector of the points $A(1,0,2)$ and $B(3,9,6)$ are $\overrightarrow{a}=\hat{i}+2\hat{k}$ and $\overrightarrow{b}=3\hat{i}+9\hat{j}+6\hat{k}$ respectively.

$\therefore$ The vector equation of the required line is:

$\overrightarrow{r}=\hat{i}+2\hat{k}+\lambda (2\hat{i}+9\hat{j}+4\hat{k})$

Given,

$\frac{\sin 7x+6\sin 5x+17\sin 3x+12\sin x}{\sin 6x+5\sin 4x+12\sin 2x}$

$=\frac{\sin 7\mathrm{x}+\sin 5\mathrm{x}+5\sin 5\mathrm{x}+5\sin 3\mathrm{x}+12\sin 3\mathrm{x}+12\sin \mathrm{x}}{\sin 6\mathrm{x}+5\sin 4\mathrm{x}+12\sin 2\mathrm{x}}$

By using, $\sin x+\sin y=2\sin \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)$

Therefore,

$=\frac{(2\sin (\frac{7x+5x}{2})\cos (\frac{7x-5x}{2})+5\times (2\sin (\frac{5x+3x}{2})\cos (\frac{5x-3x}{2}))+12\times (2\sin (\frac{3x+x}{2})\cos (\frac{3x-x}{2}))}{(\sin 6x+5\sin 4x+12\sin 2x)}$

$=\frac{2\sin 6x\cos x+10\sin 4x\cos x+24\sin 2x\cos x}{(\sin 6x+5\sin 4x+12\sin 2x)}$

$=\frac{2\cos x(\sin 6x+5\sin 4x+12\sin 2x)}{(\sin 6x+5\sin 4x+12\sin 2x)}$

$=2\cos x$

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

$\mathrm{x}=4,\mathrm{y}=1,\mathrm{z}=-1$

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

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

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

$\alpha =6,\beta =5,\gamma =4$

$\text{ Then }\alpha \beta +\beta \gamma +\gamma \alpha$

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

$=30+20+24$

$=74$

Given:

$p(n)=x(x^{n-1}-n\cdot a^{n-1}+a^n(n-1))$ is divisible by $(x-a{)}^2$

For n = 1.

$p(1)=x(x^{1-1}-1\cdot a^{1-1}+a^1(1-1))$

$\mathrm{p}(1)=0$

It is not divisible by $(x-a{)}^2$.

$\mathrm{p}(2)=\mathrm{x}(\mathrm{x}-2\mathrm{a}+{\mathrm{a}}^2)$

$={\mathrm{x}}^2-2\mathrm{ax}+{\mathrm{a}}^2\mathrm{x}$

$=(\mathrm{x}-\mathrm{a}{)}^2-{\mathrm{a}}^2+{\mathrm{a}}^2(\mathrm{x})$

$=(\mathrm{x}-\mathrm{a}{)}^2+{\mathrm{a}}^2(\mathrm{x}-1)$

$p(2)$ is also not divisible by $(x-a{)}^2$

$\{1,2\}$ falls in the set of first n natural numbers.

$A=\left[\begin{array}{cc}4& x+2\\ 2x-3& x+1\end{array}\right]$

Given $A^T=A$

$\Rightarrow \left[\begin{array}{cc}4& 2x-3\\ x+2& x+1\end{array}\right]=\left[\begin{array}{cc}4& x+2\\ 2x-3& x+1\end{array}\right]$

Both are equal if corresponding elements are equal,

$\Rightarrow 2x-3=x+2$

$\Rightarrow 2x-x=3+2$

$\Rightarrow x=5$

Given:

$\underset{x\to 2}{lim}\frac{\sin (e^{x-2}-1)}{\log (x-1)}$ .....(1)

We know that:

$\underset{x\to a^{-}}{lim}f(x)=\underset{x\to a^{+}}{lim}f(x)=l=\underset{x\to a}{lim}f(x)$

On substituting, $\mathrm{h}=\mathrm{x}-2$ in (1), we get:

$=\underset{h\to 0}{lim}\frac{\sin (e^h-1)}{\log (1+h)}$

This can be rearranged as:

$=\underset{h\to 0}{lim}\frac{\sin (e^h-1)}{e^h-1}\cdot \frac{e^h-1}{h}\cdot \frac{h}{\log (1+h)}$

$=1\cdot 1\cdot 1$

$=1$

And, $f(2)=k$

∴ For the function to be continuous the value of the function f(x) at x = 2 must equal the limiting value of 1, i.e., k = 1

Now, $\underset{x\to 0}{lim}f(x)=\underset{x\to 0}{lim}\frac{3x+{\tan }^{2}x}{x}(\frac{0}{0}$ form $)$

$=\underset{x\to 0}{lim}\frac{3+2\tan x{\sec }^{2}x}{1}=3$

Since, $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=0$

$\therefore \mathrm{f}(0)=3$

Given,

The planes $2x-y-3z-7=0$ and $4x-2y+5kz+9=0$ are parallel

We know that,

If plane $a_{1}x+b_{1}y+c_{1}z+d_1=0$ and

$a_{2}x+b_{2}y+c_{2}z+d_2=0$ are a parallel i.e.,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\ne \frac{d_1}{d_2}$

Now,

If plane are parallel than ratio of coefficient of $x,y$ and $z$ are equal.

$\frac{2}{4}=\frac{-1}{-2}=\frac{-3}{5k}$

$\Rightarrow \frac{1}{2}=\frac{-3}{5k}$

$\Rightarrow 5k=-6$

So, $k=\frac{-6}{5}$

Now,

$5k+7$

$=5\times \left(\frac{-6}{5}\right)+7$

$=(-6)+7$

$=1$