Mathematics Test - 43

Given, |x| = -5

Since |x| is always positive or zero

So, it can not be negative

Therefore, given inequality has no solution.

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

\(\vec{u}=4 \hat{i}-4 \hat{j}+7 \hat{k}\)

\(\vec{v} \cdot \vec{u}=(\hat{i}-2 \hat{j}+\hat{k}) \cdot(4 \hat{i}-4 \hat{j}+7 \hat{k})\)

\(\Rightarrow \vec{v} \cdot \vec{u}=4+8+7\)

\(\overrightarrow{|u|}=\sqrt{4^{2}+(-4)^{2}+7^{2}}=\sqrt{81}=9\)

\(\therefore\) The projection of the vector \(\hat{i}-2 \hat{j}+\hat{k}\) on the vector \(4 \hat{i}-4 \hat{j}+7 \hat{k}=\frac{19}{9}=2 \frac{1}{9}\)

Given,

\(\cos \left(-1230^{\circ}\right) \times \cot \left(-315^{\circ}\right)\)

\(\cos \left(-1230^{\circ}\right)=\cos \left(1230^{\circ}\right)\)\(\quad\quad(\because \cos (-\theta)=\cos \theta)\)

\(=\cos \left(3 \times 360^{\circ}+150\right)\)\(\quad\quad(\because\cot (2 n \pi-\theta)=-\cot (\theta))\)

\(=\cos \left(150^{\circ}\right)\)

\(=\cos \left(180^{\circ}-30^{\circ}\right)\)\(=-\cos 30^{\circ} \quad(\because \cos (\pi-\theta)=-\cos (\theta))\)

\(=\frac{-\sqrt{3}}{2}\)

Now,

\(\cot \left(-315^{\circ}\right)=-\cot \left(315^{\circ}\right) \quad(\because \cot (-\theta)=-\cot (\theta))\)

\(=-\cot \left(360^{\circ}-45^{\circ}\right) \)

\(=\cot 45^{\circ} \quad(\because \cot (2 n \pi-\theta)=-\cot (\theta))\)

\(=1\)

\(\cos \left(-1230^{\circ}\right) \times \cot \left(-315^{\circ}\right)=\frac{-\sqrt{3}}{2}\)

\(=1 × \frac{-\sqrt{3}}{2}\)

\( g(\alpha)=\int_{\pi / 6}^{\pi / 3} \frac{\sin ^\alpha x}{\cos ^\alpha x+\sin ^\alpha x} d x \ldots \text {.. (i) } \\\)

\( \text { Applying } \int_a^b f(x) d x=\int_a^b f(a+b-x) d x \\\)

\( g(\alpha)=\int_{\pi / 6}^{\pi / 3} \frac{\sin ^\alpha(\pi / 2-x)}{\cos ^\alpha\left(\frac{\pi}{2}-x\right)+\sin ^\alpha\left(\frac{\pi}{2}-x\right)} d x \\\)

\( g(\alpha)=\int_{\pi / 6}^{\pi / 3} \frac{\cos ^\alpha x}{\cos ^\alpha x+\sin ^\alpha x} d x \text {.... (ii) }\)

Adding Eqs. (i) and (ii),

\(2 g(\alpha)=\int_{\pi / 6}^{\pi / 3} \frac{\sin ^\alpha x+\cos ^\alpha x}{\sin ^\alpha x+\cos ^\alpha x} d x \\\)

\(\begin{aligned} & 2 g(\alpha)=\int_{\pi / 6}^{\pi / 3} 1 \cdot d x=[x]_{\pi / 6}^{\pi / 3}=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6} \\ & \therefore g(\alpha)=\frac{\pi}{12}\end{aligned}\)

\(g(\alpha)\) is constant function.

\(\therefore\) It is even function.

Given that \(\vec{a} \cdot \vec{d} \neq 0, \vec{a} \cdot \vec{d}=0\)

Now, \(\vec{b} \times \vec{c}=\vec{b} \times \vec{d}\)

\(\Rightarrow \vec{a} \times(\vec{b} \times \vec{c})=\vec{a} \times(\vec{b} \times \vec{d}) \\\)

\( \Rightarrow(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}=(a \cdot \vec{d}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{d} \\\)

\( \Rightarrow(\vec{a} \cdot \vec{b}) \vec{d}=-(\vec{a} \cdot \vec{c}) \vec{b}+(\vec{a} \vec{b}) \vec{c} \\\)

\( \vec{d}=\vec{c}-\left(\frac{\vec{a} \cdot \vec{c}}{\vec{a} \cdot \vec{b}}\right) \vec{b}\)

We know mirror image of point \(\left(\mathrm{x}_1: \mathrm{y}_1, \mathrm{z}_1\right)\) in the plane \(\mathrm{ax}+\mathrm{by}+\mathrm{cz}=\mathrm{d}\)

\(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=\frac{-2\left(\mathrm{ax}_1+b y_1+c z_1-d\right)}{a^2+b^2+c^2}\)

Here given point \((2,4,7)\) and plane \(3 x-y+4 z=2\) then mirror image is

\( \frac{x-2}{3}=\frac{y-4}{-1}=\frac{z-7}{4}=\frac{-2(6-4+28-2)}{9+1+16} \\\)

\( \Rightarrow \frac{x-2}{3}=\frac{y-4}{-1}=\frac{z-7}{4}=-\frac{28}{13} \\\)

\( \therefore x=-\frac{58}{13}=a \\\)

\( y=\frac{80}{13}=b \\\)

\( z=-\frac{21}{13}=c \\\)

\( \therefore 2 a+b+2 c \\\)

\( =2\left(-\frac{58}{13}\right)+\frac{80}{13}+2\left(-\frac{21}{13}\right) \\\)

\( =\frac{-116+80-42}{13}=\frac{-78}{13}=-6\)

If we fix a girl, then the rest 4 girls can be arranged in the ways \(=4 !\)

There is 1 position between each pair of girl, which means only 5 positions are there.

Ways to arrange 5 boys in those 5 position will be \(=5!\)

The total number of ways, \(N=5 ! \times 4!\)

\(N=120 \times 24\)

\(N=2880\) ways

Given here:

\(f(2)=2\) and \(f(x)=2\).

Let us check the form by putting \(x = 2\), we get:

\(\lim _{x \rightarrow 2} \frac{x f(2)-2 f(x)}{x-2}\)

\(=\frac{2(2)-2(2)}{2-2}\)

\(=\frac{0}{0}\)

Apply L’ hospital’s rule,

\(\lim _{x \rightarrow a} \frac{\mathrm{f}(\mathrm{x})}{\mathrm{g}(x)}=\lim _{\mathrm{x} \rightarrow \mathrm{a}} \frac{\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{f}(\mathrm{x}))}{\frac{d}{d x}(\mathrm{~g}(\mathrm{x}))}=\lim _{\mathrm{x} \rightarrow \mathrm{a}} \frac{\mathrm{f}^{\prime}(x)}{\mathrm{g}^{\prime}(x)}\)

\(\lim _{x \rightarrow 2} \frac{x f(2)-2 f(x)}{x-2}\)

Differentiating the numerator and the denominator in above, we get:

\(=\lim _{x \rightarrow 2} \frac{\frac{d}{d x}(x f(2)-2 f(x))}{\frac{d}{d x}(x-2)}\)

\(=\lim _{x \rightarrow 2} \frac{f(2)-2 f^{\prime}(x)}{1}\)

Now, take limit at x = 2

\(\lim _{x \rightarrow 2}=f(2)-2 f^{\prime}(2)\)

\(=2-2(2)\)

\(=-2\)

Given,

\(4^{\text {th }}\) term of a G. P is 8 and \(10^{\text {th }}\) term is \(27\)

\(n^{\text {th }}\) term of the G.P. is \(T_{n}=a r^{n-1}\)

\(\therefore \mathrm{T}_{4}=\mathrm{a} \cdot \mathrm{r}^{3}=8 \quad \ldots(1)\)

\(T_{10}=a r^{9}=27 \quad \ldots(2)\)

Equation (2) \(\div(1)\), we get

\(r^{6}=\frac{27}{8}\)

\(\left(\mathrm{r}^{2}\right)^{3}=\left(\frac{3}{2}\right)^{3}\)

\(\therefore r^{2}=\frac{3}{2}\)

\(\mathrm{T}_{6}=a \mathrm{r}^{5}\)

\(=a r^{3} \cdot r^{2}\)

\(=8 \times \frac{3}{2}\)

\(=12\)