Mathematics Test - 13

Using the definition of triple cross product:

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

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

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

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

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

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

\(=3 \vec{a}-\vec{a}\)

\(=2 \overrightarrow{\mathrm{a}}\)

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

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

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 \(\vec{a}=\hat{i}+2 \hat{k}\) and \(\vec{b}=3 \hat{i}+9 \hat{j}+6 \hat{k}\) respectively.

\(\therefore\) The vector equation of the required line is:

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

Given,

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

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

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

Therefore,

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

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

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

\(=2 \cos x\)

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

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

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

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

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

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

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

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

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

\( =74\)

Given:

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

For n = 1.

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

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

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

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

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

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

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

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

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

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

Given \(A^T=A\)

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

Both are equal if corresponding elements are equal,

\(\Rightarrow 2 x-3=x+2\)

\(\Rightarrow 2 x-x=3+2\)

\(\Rightarrow x=5\)

Given:

\(\lim _{x \rightarrow 2} \frac{\sin \left(e^{x-2}-1\right)}{\log (x-1)}\) .....(1)

We know that:

\(\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{+}} f(x)=l=\lim _{x \rightarrow a} f(x)\)

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

\(=\lim _{h \rightarrow 0} \frac{\sin \left(e^{h}-1\right)}{\log (1+h)}\)

This can be rearranged as:

\(=\lim _{h \rightarrow 0} \frac{\sin \left(e^{h}-1\right)}{e^{h}-1} \cdot \frac{e^{h}-1}{h} \cdot \frac{h}{\log (1+h)}\)

\(=1 \cdot 1 \cdot 1\)

\(=1\)

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, \(\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{3 x+\tan ^{2} x}{x}\left(\frac{0}{0}\right.\) form \()\)

\(=\lim _{x \rightarrow 0} \frac{3+2 \tan x \sec ^{2} x}{1}=3\)

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

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

Given,

The planes \(2 x-y-3 z-7=0\) and \(4 x-2 y+5 k z+9=0\) 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.,

\(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}} \neq \frac{d_{1}}{d_{2}}\)

Now,

If plane are parallel than ratio of coefficient of \(x, y\) and \(z\) are equal.

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

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

\(\Rightarrow 5 k=-6\)

So, \(k=\frac{-6}{5}\)

Now,

\(5 k+7\)

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

\(=(-6)+7\)

\(=1\)