Mathematics Test - 18

Cube root of unity \(=1, \omega, \omega^{2}\)

Properties of the cube root of unity:

(i) \(1, \omega, \omega^{2}\) are in G.P. \(\quad \quad \ldots (1)\)

(ii) \(1+\omega+\omega^{2}=0 \quad \quad \quad \ldots (2)\)

(iii) \(\omega^{3}=1 \quad\quad \quad \quad \quad \quad \ \ldots (3)\)

Given, \(\omega\) is a cube root of unity.

And, \(\left(1-\omega+\omega^{2}\right)\left(1+\omega-\omega^{2}\right)\)

\(\Rightarrow\left(1+\omega^{2}-\omega\right)\left(1+\omega-\omega^{2}\right) \quad\quad \ldots . (\mathrm{A})\)

From equation (2),

\(1+\omega^{2}=-\omega\) and \(1+\omega=-\omega^{2}\)

Put in equation \((\mathrm{A})\),

\(=(-\omega-\omega)\left(-\omega^{2}-\omega^{2}\right)\)

\(=(-2 \omega)\left(-2 \omega^{2}\right)\)

\(= 4 \omega^{3}\)

Using equation (3),

\(\Rightarrow 4 \times 1=4\)

\(\sin ^{-1} x+\sin ^{-1} y=\frac{5 \pi}{6}\)

We know that:

\(\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\)

\(\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\)

Let, \(\cos ^{-1} x+\cos ^{-1} y=a\)

Adding the two equation we get,

\(\left(\sin ^{-1} x+\cos ^{-1} x\right)+\left(\sin ^{-1} y+\cos ^{-1} y\right)=\frac{5 \pi}{6}+a\)

\(\Rightarrow \frac{\pi}{2}+\frac{\pi}{2}=\frac{5 \pi}{6}+a\)

\(\Rightarrow \pi-\frac{5 \pi}{6}=a \Rightarrow a=\frac{\pi}{6}\)

So, \(\cos ^{-1} x+\cos ^{-1} y=\frac{\pi}{6}\)

|A|=48
|B|=25
|C|=18
|A∪B∪C|=60 [Total] 
|A∩B∩C|=5

Let \(e^{x}=t\)\(\quad \dots (1)\)

Differentiating w.r.t. \(x\), we get

\(e^{x}=\frac{d t}{dx}\)

\(e^{x} d x=d t\)

New limit to \(eq^{n}(1)\)

Now, \(x=0 \Rightarrow t=1\)

\(x=1 \Rightarrow t=e\)

\(\therefore \int_{0}^{1} \frac{e^{x}}{1+e^{2 x}} d x =\int_{1}^{e} \frac{d t}{1+t^{2}} ~~~~~\quad\quad{\left[\because \int \frac{d t}{1+t^{2}}=\tan ^{-1} t\right] } \)

\(=\left[\tan ^{-1} t\right]_{1}^{e} ~~~~~~~~~~{\left[\because \tan \frac{\pi}{4}=1\right] } \)

Put the limit,

\(=\left[\tan ^{-1} e-\tan ^{-1} 1\right] \)

\(={\left[\tan ^{-1} e-\frac{\pi}{4}\right.}]\)

\(\therefore \int_{0}^{1} \frac{e^{x}}{1+e^{2 x}} d x=\tan ^{-1} e-\frac{\pi}{4}\)

The objective function, \(5 x+4 y \geq 2, x \leq 6, y \leq 7\)

\(\Rightarrow \frac{x}{\frac{2}{5}}+\frac{y}{\frac{1}{2}}=1\)

At point \(\mathrm{A}\left(\frac{2}{5}, 0\right), \mathrm{z}=\frac{2}{5}+0=\frac{2}{5}\)

At point \(\mathrm{B}(6,0), \mathrm{z}=6+0=6\)

At point \(C(6,7), z=6+14=20\)

At point \(\mathrm{D}(0,7), z=0+2(7)=14\)

At point \(\mathrm{E}\left(0, \frac{1}{2}\right), \mathrm{z}=0+2\left(\frac{1}{2}\right)=1\)

\(\therefore\) The maximum value of \(\mathrm{z}\) is 20 .

\(\begin{aligned} & \vec{a}=2 \hat{i}-7 \hat{j}+5 \hat{k} \\ & \vec{b}=\hat{i}+\hat{k} \\ & \vec{c}=\hat{i}+2 \hat{j}-3 \hat{k} \\ & \vec{r} \times \vec{a}=\vec{c} \times \vec{a} \Rightarrow(\vec{r}-\vec{c}) \times \vec{a}=0 \\ & \therefore \vec{r}=\vec{c}+\lambda \vec{a} \\ & \vec{r} \cdot \vec{b}=0 \Rightarrow \vec{c} \cdot \vec{b}+\lambda \vec{b} \cdot \vec{a}=0 \\ & -2+\lambda(7)=0 \Rightarrow \lambda=\frac{2}{7} \\ & \therefore \vec{r}=\vec{c}+2 \frac{\vec{a}}{7}=\frac{1}{7}(11 \hat{i}-11 \hat{k}) \\ & |\vec{r}|=\frac{11 \sqrt{2}}{7}\end{aligned}\)

Let \(\hat{x_i}+y_j+z \hat{k}\) be the required unit vector.

Since \(\hat{a}\) is perpendicular to \((2 \hat{i}-\hat{j}+2 \hat{k})\).

\(\therefore 2 \mathrm{x}-\mathrm{y}+2 \mathrm{z}=0 \ldots \text {..... (i) }\)

Since vector \(x_i+y \hat{j}+z \hat{k}\) is coplanar with the vector \(\hat{i}+\hat{j}-\hat{k}\) and

\( 2 \hat{i}+2 \hat{j}-\hat{k} \\\)

\( \therefore x_{\hat{i}}+y \hat{j}+z \hat{k}=p(\hat{i}+\hat{j}-\hat{k})+q(2 \hat{i}+2 \hat{j}-\hat{k})\)

where \(p\) and \(q\) are some scalars.

\(\Rightarrow x_i^{\hat{i}}+y \hat{j}+z \hat{k}=(p+2 q) \hat{i}+(p+2 q) \hat{j}-(p+q) \hat{k} \\\)

\(\Rightarrow x=p+2 q, y=p+2 q, z=-p-q\)

Now from equation (i),

\( 2 p+4 q-p-2 q-2 p-2 q=0 \\\)

\( \Rightarrow-p=0 \Rightarrow p=0 \\\)

\( \therefore x=2 q, y=2 q, z=-q\)

Since vector \(x_{\hat{i}}+y \hat{j}+z \hat{k}\) is a unit vector, therefore

\( |\hat{x}+\hat{y}+z \hat{k}|=1 \\\)

\( \Rightarrow \sqrt{x^2+y^2+z^2}=1 \\\)

\( \Rightarrow x^2+y^2+z^2=1 \\\)

\( \Rightarrow 4 q^2+4 q^2+q^2=1 \\\)

\( \Rightarrow 9 q^2=1 \Rightarrow q= \pm \frac{1}{3}\)

When \(\mathrm{q}=\frac{1}{3}\), then \(\mathrm{x}=\frac{2}{3}, \mathrm{y}=\frac{2}{3}, \mathrm{z}=-\frac{1}{3}\)

When \(\mathrm{q}=-\frac{1}{3}\), then \(\mathrm{x}=-\frac{2}{3}, \mathrm{y}=-\frac{2}{3}, \mathrm{z}=\frac{1}{3}\)

Here required unit vector is \(\frac{2 \hat{i}+2 \hat{j}-1 \hat{k}}{3}\)

or \(-\frac{2}{3} \hat{\mathrm{i}}-\frac{2}{3} \hat{\mathrm{j}}+\frac{1}{3} \hat{\mathrm{k}}\)

Given,

\(\frac{x}{4}<\frac{(5 x-2)}{3}-\frac{(7 x-3)}{5}\)

\(=\frac{x}{4}<\frac{5(5 x-2)-3(7 x-3)}{15}\)

On simplifying we get

\(=\frac{x}{4}<\frac{25 x-10-21 x+9}{15} \)

\(=\frac{x}{4}<\frac{4 x-1}{15} \)

\(=15 x<4(4 x-1) \)

\(=15 x<16 x-4 \)

\(=4

All the real numbers of \(x\) which are greater than \(4\) are the solutions of the given inequality

Hence, \((4, \infty)\) will be the solution for the given inequality.