Mathematics Test - 8

\(\overrightarrow{a}=2 \hat{i}+\hat{j}-3 \hat{k}\) and \(\overrightarrow{b}=3 \hat{i}-2 \hat{j}-\hat{k}\)

\(\vec{a} \cdot \vec{b}=(2 \hat{i}+\hat{j}-3 \hat{k})(3 \hat{i}-2 \hat{j}-\hat{k}) \)

\(=6-2+3=7 \ldots .(1) \)

\(|\vec{a}|=\sqrt{2^{2}+1^{2}+(-3)^{2}}=\sqrt{14} \ldots(2) \)

\(|\vec{b}|=\sqrt{3^{2}+(-2)^{2}+(-1)^{2}}=\sqrt{14} \ldots .(3) \)

\(\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| \cdot|\vec{b}|}\)

Put the values from (1), (2) and (3) in above equation:

\(=\frac{7}{\sqrt{14}. \sqrt{14}}=\frac{1}{2} \)

\(\cos \theta=\cos 60^{\circ} , \theta=60^{\circ}\)

Let \(3^{x}=t ; t>0\) \(\Rightarrow t(t-1)+2=|t-1|+|t-2|\) \(\Rightarrow t^{2}-t+2=|t-1|+|t-2|\)
Case-I : \({t}<1\), \(t^{2}-t+2=1-t+2-t\)
\(t^{2}+2=3-t\) \(\Rightarrow t^{2}+t-1=0\)
\(t=\frac{-1 \pm \sqrt{5}}{2}\) \(=\frac{\sqrt{5}-1}{2}\) is only acceptable.
Case-II : \(1 \leq {t}<2\), \(t^{2}-t+2=t-1+2-t\)
\(t^{2}-t+1=0\), \({D}<0\) no real solution
Case-III : \(t \geq 2\), \({t}^{2}-{t}+2={t}-1+{t}-2\)
\(t^{2}-3 t \quad 5=0 \Rightarrow D<0\), no real solution
\(3^{x}\left(3^{x}-1\right)+2=\left|3^{x}-1\right|+\left|3^{x}-2\right|\)
put \(3^{x}=t\)
\({t}({t}-1)+2=|{t}-1|+|{t}-2|\)
\({t}^{2}-{t}+2=|{t}-1|+|{t}-2|\)
from graph

let \(\alpha\) is real solutions, \(\alpha=3^{x}\) \(\Rightarrow x=\log _{3} \alpha\)         (\(\therefore\) only one solution)
\(\therefore\) singleton set

Given: \(f(x)=\log x\)

To Find: \(A M\) of \(f(x y)\) and \(f(\frac{x}{y})\)

\(f(x y)=\log (x y)=\log x+\log y \quad(\because \log m n=\log m+\log n) \)

\( f\left(\frac{x}{y}\right)=\log \left(\frac{x}{y}\right)=\log x-\log y\)

Now, AM of \(f(x y)\) and \(f(\frac{x}{y})=\frac{f(x y)+f\left(\frac{x}{y}\right)}{2}\)

\( =\frac{\log x+\log y+\log x-\log y}{2} \)

\( =\frac{2 \log x}{2} \)

\( =\log x\)

Here, 5 letters are there in AGAIN but A should always be at beginning.

So number of ways to arrange 4 letters = 4! = 4 × 3 × 2 = 24

Let \(\overrightarrow{ a }= a _{1} \overrightarrow{ i }+ b _{1} \overrightarrow{ j }+ c _{1} \overrightarrow{ k }, \overrightarrow{ b }= a _{2} \overrightarrow{ i }+ b _{2} \overrightarrow{ j }+ c _{2} \overrightarrow{ k }\) and \(\overrightarrow{ c }= a _{3} \overrightarrow{ i }+ b _{3} \overrightarrow{ j }+ c _{3} \overrightarrow{ k }\) be the three vectors.

Condition for coplanarity:

\(\overrightarrow{ a } \cdot(\vec{b} \times \vec{c})=\left|\begin{array}{lll}a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3}\end{array}\right|=0\)

Here, \(2 \hat{ i }-\hat{ j }+\hat{ k }, \hat{ i }+2 \hat{ j }-3 \hat{ k }\) and \(3 \hat{ i }+ m \hat{ j }+5 \hat{ k }\) are coplanar.

\(\left|\begin{array}{rrr}2 & -1 & 1 \\ 1 & 2 & -3 \\ 3 & m & 5\end{array}\right|=0\)

\(\Rightarrow 2(10+3 m)+1(5+9)+1(m-6)=0\)

\(\Rightarrow 20+6 m+14+m-6=0\)

\(\Rightarrow 7 m+28=0\)

\(\Rightarrow m=-4\)

\(\therefore\) The value of \(m\) is \(-4\).

Given: \(|\overrightarrow{\mathrm{a}}|=\frac{1}{\sqrt{3}},|\overrightarrow{\mathrm{b}}|=2\) and \(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}\) is a unit vector

Let the angle between the vectors \(\vec{a}\) and \(\vec{b}\) is \(\theta\)

We know that, \(|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \sin \theta\)

\(\Rightarrow 1=\frac{1}{\sqrt{3}} \times 2 \times \sin \theta\)

\(\Rightarrow \sqrt{3}=2 \times \sin \theta\)

\(\Rightarrow \frac{\sqrt{3}}{2}=\sin \theta\)

⇒ θ​ = 60°

L'Hospital's Rule:

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

We have,

\(\lim _{x \rightarrow 0} \frac{(1+\sin x-\cos x+\ln (1-x))}{x \tan ^{2} x}\)

Apply L'Hospital's Rule, we get

\(\lim _{x \rightarrow 0} \frac{\cos x-(-\sin x)-\frac{1}{1-x}}{\tan ^{2} x+2 x \sec ^{2} x \tan x}\quad\quad\)\(\left[\because \ln x = \frac{1}{x}, \frac{d(\sin x)}{d x}=\cos x,\frac{d(\cos x)}{d x}=-\sin x, \frac{d(\tan x)}{d x}=\sec ^{2} x \text { and } d(xy)= x\ dy + y\ dx\right]\)

\(\lim _{x \rightarrow 0} \frac{\sin x+\cos x-\frac{1}{1-x}}{\tan ^{2} x+2 x \sec ^{2} x \tan x}\)

Apply L'Hopital's rule, we get

\(\lim _{x \rightarrow 0} \frac{\cos x-\sin x-\frac{1}{(1-x)^{2}}}{\operatorname{tanx}\left(4 {x \sec}^{2} x \tan x+3 \sec ^{2} x\right)+\sec ^{2} x\left(\tan x+2 x \sec ^{2} x\right)}\)

Apply L'Hopital's rule, we get

\(\lim _{x \rightarrow 0} \frac{-\sin x-\cos x-\frac{2}{(1-x)^{3}}}{4 \sec ^{2} x {\tan x}\left(2 x \tan ^{2} x+2 \tan x+{x \sec}^{2} x\right)+2 \sec ^{2} x\left(2 \tan ^{2} x+6 x \sec ^{2} x \tan x+3 \sec ^{2} x\right)}\)

\(=\frac{-\sin (0)-\cos (0)-\frac{2}{(1-0)^{3}}}{4 \sec ^{2}(0) \tan (0)[0+2 \tan (0)+0]+2 \sec ^{2}(0)\left(2 \tan ^{2}(0)+0+3 \sec ^{2}(0)\right)}\)

\(=\frac{-0-1-2}{0+2(0+3)}\)

\(=\frac{-3}{6}\)

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

Equation of line with Z-axis is,

\(\overrightarrow{\mathrm{b}}=\hat{\mathrm{k}}\)

Equation of plane is x -3y + 5z -8 = 0

or in vector form \(, \overrightarrow{\mathrm{n}}=\hat{\mathrm{i}}-3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\)

\(\therefore \sin \alpha=\left|\frac{\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{n}}}{|\overrightarrow{\mathrm{b}}||\overrightarrow{\mathrm{n}}|}\right|\)

\(\Rightarrow \sin \alpha=\frac{(\hat{\mathrm{i}}-3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}) \cdot(0 \hat{\mathrm{i}}+0 \hat{\mathrm{j}}+\hat{\mathrm{k}})}{\sqrt{1^{2}+(-3)^{2}+5^{2}} \cdot \sqrt{1^{2}}}=\frac{5}{\sqrt{35}}\)

\(\Rightarrow \sin \alpha=\frac{5}{\sqrt{35}}\)

\(\Rightarrow \alpha=\sin ^{-1}\left(\frac{5}{\sqrt{35}}\right)\)

\(\Rightarrow \alpha=\frac{5}{\sqrt{35}}\)

We have \(\log _3 72^5=\frac{\log 72^5}{\log 3}=\frac{5 \log 72}{\log 3}=\frac{5\{\log (8 \times 9)\}}{\log 3}=\frac{5(\log 8+\log 9)}{\log 3}\)

\( =\frac{5\left(\log 2^3+\log 3^2\right)}{\log 3}=\frac{5(3 \log 2+2 \log 3)}{\log 3}=\frac{5(3 \times 0.301+2 \times 0.4771)}{0.4771} \)

\( =\frac{5(0.903+0.9542)}{0.4771}=\frac{5 \times 1.8572}{0.4771}=\frac{9.286}{0.4771}=19.46\)