Mathematics Test - 29

First, graph the "equals" line, then shade in the correct area.

There are three steps:

Rearrange the equation so "y" is on the left and everything else on the right.

Plot the "y = " line (make it a solid line for y ≤ or y ≥, and a dashed line for y < or y >)

Shade above the line for a "greater than" (y > or y ≥) or below the line for a "less than" (y < or y ≤).

Given: 

The constraints –x + y ≤ 1, −x + 3y ≤ 9 and x, y ≥ 0

–x + y ≤ 1

⇒ y ≤ 1 + x

So shade below the line.

−x + 3y ≤ 9

3y ≤ 9 + x

⇒ y ≤ 3 + \(\frac{x}{3}\)

So shade below the line.

We can see through above graph region is unbounded feasible space

It is given that,

\(\vec{a}=\hat{i}-2 \hat{j}+\hat{k}\) and \(\vec{b}=4 \hat{i}-4 \hat{j}+7 \hat{k}\)

\(\Rightarrow \vec{a} \cdot \vec{b}=(\hat{i}-2 \hat{j}+\hat{k}) \cdot(4 \hat{i}-4 \hat{j}+7 \hat{k})=4+8+7=19\)

\(\Rightarrow|\vec{b}|=\sqrt{4^{2}+(-4)^{2}+7^{2}}=9\)

\(\therefore\) The scalar projection of vector \(\vec{a}\) on \(\vec{b}=\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}=\frac{19}{9}\)

Equation of the given lines \(\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}\) and \(\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}\)
It is known that the shortest distance between the two lines,
\(\frac{x+x_{1}}{a_{1}}=\frac{y+y_{1}}{b_{1}}=\frac{z+z_{1}}{c_{1}}\) and \(\frac{x-x_{2}}{a_{2}}=\frac{y-y_{2}}{b_{2}}=\frac{z-z_{2}}{c_{2}}\) is given by,
\(d=\frac{\left|\begin{array}{ccc}x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2}\end{array}\right|}{\sqrt{\left(b_{1} c_{2}-b_{2} c_{1}\right)^{2}+\left(c_{1} a_{2}-c_{2} a_{1}\right)^{2}+\left(a_{1} b_{2}-a_{2} b_{1}\right)^{2}}}\quad\quad\)....(i)
Comparing the given equations, we have
\(x_{1}=-1 ~~~~~~~~ y_{1}=-1 ~~~~~~~~z_{1}=-1 ~~~~~~ a_{1}=7 ~~~~~~~ b_{1}=-6~ ~~~ c_{1}=1 \)
\( x_{2}=3 ~~~~~~~~~~~ y_{2}=5 ~~~~~~~~~~~ z_{2}=7 ~~~~~~~~~ a_{2}=1 ~ ~~~~~~~b_{2}=-2~~~~ c_{2}=1\)
Then \(\left|\begin{array}{ccc}x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2}\end{array}\right|=\left|\begin{array}{ccc}4 & 6 & 8 \\ 7 & -6 & 1 \\ 1 & -2 & 1\end{array}\right|\)
\(=4(-6+2)-6(7-1)+8(-14+6)=-16-36-64=-116\)
and \(\sqrt{\left(b_{1} c_{2}-b_{2} c_{1}\right)^{2}+\left(c_{1} a_{2}-c_{2} a_{1}\right)^{2}+\left(a_{1} b_{2}-a_{2} b_{1}\right)^{2}}=\sqrt{(-6+2)^{2}+(1+7)^{2}+(-14+6)^{2}}\) \(=\sqrt{16+36+64}=\sqrt{116}=2 \sqrt{29}\)
Substituting all the values in equation (1), we have \(d=\frac{-116}{2 \sqrt{29}}=\frac{-58}{\sqrt{29}}=-2 \sqrt{29}\)

The given trigonometric equation is,

$\sin \theta (1+\tan \theta )+\cos \theta (1+\cot \theta )$

$=\sin \theta (1+\frac{\sin \theta }{\cos \theta })+\cos \theta (1+\frac{\cos \theta }{\sin \theta })$

$=\frac{\sin \theta (\cos \theta +\sin \theta )}{\cos \theta }+\frac{\cos \theta (\sin \theta +\cos \theta )}{\sin \theta }$

$=(\sin \theta +\cos \theta )[\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta }]$

$=(\sin \theta +\cos \theta )\left[\frac{{\sin }^2\theta +{\cos }^2\theta }{\cos \theta \sin \theta }\right]$

$=\frac{\sin \theta +\cos \theta }{\cos \theta \sin \theta }=\frac{\sin \theta }{\cos \theta \sin \theta }+\frac{\cos \theta }{\cos \theta \sin \theta }$

$=\frac{1}{\cos \theta }+\frac{1}{\sin \theta }=\sec \theta +\mathrm{cosec}\theta$

\((\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=|\vec{a}|^{2}-|\vec{a}| \times|\vec{b}| \times \cos \theta+|\vec{a}| \times|\vec{b}| \times \cos \theta-|\vec{b}|^{2}=|\vec{a}|^{2}-|\vec{b}|^{2}\)

Therefore, statement 1 is true.

\((|\vec{a}+\vec{b}|)(|\vec{a}-\vec{b}|)=\sqrt{|\vec{a}|^{2}+|\vec{b}|^{2}+2 \times|\vec{a}| \times|\vec{b}| \times \cos \theta} \times \sqrt{|\vec{a}|^{2}+|\vec{b}|^{2}-2 \times|\vec{a}| \times|\vec{b}| \times \cos \theta}\)

If \(\cos \theta=1\)

\(\Rightarrow(|\vec{a}+\vec{b}|)(|\vec{a}-\vec{b}|)=|\vec{a}|^{2}-|\vec{b}|^{2}\)

But if \(\cos \theta \neq 1\)

\(\Rightarrow(|\vec{a}+\vec{b}|)(|\vec{a}-\vec{b}|) \neq|\vec{a}|^{2}-|\vec{b}|^{2}\)

Therefore, statement 2 is not true.

\(|\vec{a} \cdot \vec{b}|^{2}+|\vec{a} \times \vec{b}|^{2}=|\vec{a}|^{2} \times|\vec{b}|^{2} \times \cos ^{2} \theta+|\vec{a}|^{2} \times|\vec{b}|^{2} \times \sin ^{2} \theta=|\vec{a}|^{2}|\vec{b}|^{2}\)

Therefore, statement 3 is true.

Given,

\(\frac{(x-3)}{1}=\frac{(y-2)}{-1}=\frac{(z+\lambda)}{-2}=k\)

Let \(P\) be any point on the plane,

\(P=(k+3,-k-2,-2 k-\lambda)\)

\(P\) lie on the plane, \(2 x-4 y+3 z=2,\) Which means \(P\) satisfy the equationn of plane.

So, \(2(k+3)-4(-k-2)+3(-2 k-\lambda)=2\)

Solving above equation, we have \(\lambda=4\)

Now, the shortest distance between the below lines:

\(\frac{(x-3)}{1}=\frac{(y-2)}{-1}=\frac{(z+4)}{-2}\) and \(\frac{(x-1)}{12}=\frac{y}{9}=\frac{z}{4}\) is:

\(\begin{aligned} \mathrm{d}^{2} &=\left|\begin{array}{ccc}x_{1}-x_{2} & y_{1}-y_{2} & z_{1}-z_{2} \\ l_{1} & m_{1} & n_{1} \\ l_{2} & m_{2} & n_{2}\end{array}\right| \\ &=\left|\begin{array}{ccc}3-1 & -2-0 & -4-0 \\ 1 & -1 & -2 \\ 12 & 9 & 4\end{array}\right| \end{aligned}\)

\(=\mid2(14)+2(28)-4(21) \mid\)

\(=0\)

\(d^{2}=0\) or \({d}=0\)

Given term \(\sec ^{-1} \tan x\)

Let us consider \(f(x)=\sec ^{-1} \tan x\)

As we know that the differentiation of \(\sec ^{-1} x\) is given as

\(\frac{d\left(\sec ^{-1} x\right)}{d x}=\frac{1}{x \sqrt{x^{2}-1}}\)

In order to solve the problem we will use the chain rule.

According to the chain rule, we have:

\(\frac{d p}{d q}=\frac{d p}{d u} \cdot \frac{d u}{d q}\)

Using the chain rule let us proceed

So, by using the formula, we get

\(\frac{d f(x)}{d x}=\frac{d\left(\sec ^{-1} \tan x\right)}{d x}\)

\(=\frac{1}{\tan x \sqrt{\tan ^{2} x-1}} \times \frac{d \tan x}{d x}\)

\(\frac{1}{\tan x \sqrt{\tan ^{2} x-1}} \times \sec ^{2} x \quad \quad\left[\because \frac{d \tan x}{d x}=\sec ^{2} x\right]\)

\(=\frac{1}{\frac{\sin x}{\cos x} \sqrt{\frac{\sin ^{2} x}{\cos ^{2} x}-1}} \times \frac{1}{\cos ^{2} x}\)

\(=\frac{1}{\frac{\sin x}{\cos x} \sqrt{\frac{\sin ^{2} x-\cos ^{2} x}{\cos ^{2} x}}} \times \frac{1}{\cos ^{2} x}\)

By simplifying above equation, we get

\(=\frac{1}{\frac{\sin x}{\cos x} \frac{\sqrt{\sin ^{2} x-\cos ^{2} x}}{\cos x}} \times \frac{1}{\cos ^{2} x}\)

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

Thus, the differentiation of \(\sec ^{-1} \tan x\) is \(\frac{1}{\sin x \sqrt{\sin ^{2} x-\cos ^{2} x}}\)
 

Given

\(f(x)=\frac{x\left(3^{x}-1\right)}{1-\cos x}\)

\(\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{x\left(3^{x}-1\right)}{1-\cos x}\)

\(\Rightarrow \lim _{x \rightarrow 0} \frac{3^{x}-1}{x} \cdot \frac{x^{2}}{1-\cos x}\)

\(\Rightarrow \log 3 \cdot\left(\frac{1}{\frac{1}{2}}\right)\)

\(\Rightarrow2 \log 3\)

Given:

Mean of 20 observations \(=15\)

Sum of 20 observations \(=15 \times 20=300\)

Correct sum \(=300+8+4-3-6=303\)

\(\therefore\) Correct mean \(=\frac{303}{20}=15.15\)

Given, 

\(m\left[\begin{array}{ll}-3 & 4\end{array}\right]+n\left[\begin{array}{ll}4 & -3\end{array}\right]=\left[\begin{array}{ll}10 & -11\end{array}\right]\)

\(\Rightarrow\left[\begin{array}{lll}-3 & m+4 n & 4 m-3 n\end{array}\right]=\left[\begin{array}{ll}10 & -11\end{array}\right]\)

By equating above matrices, we get

\(-3 m+4 n=10 \Rightarrow 12 m-16 n=-40 \ldots \ldots\)...(i)

\(4 m-3 n=-11 \Rightarrow 12 m-9 n=-33 \ldots \ldots\) (ii)

Solving equation (i) and (ii) , we get,

\(-7 n=-7\)

\(\therefore n=1\) and \(m=-2\)