Mathematics Test - 20

\(Z_{\max }=3 x_{1}+2 x_{2}\)

Subjected to:

\(x_{1} \leq 4 \ldots . .(i) \)

\(x_{2} \leq 6 \ldots(i i)\)

\(3 x_{1}+2 x_{2} \leq 18 \ldots(i i i) \)

\(\text { and } x_{1}, x_{2} \geq 0\)

Now the intersections of the lines are denoted by \(E, F\)

For \(E, 3 x_{1}+2 x_{2}=18\)

\(x_{2}=6\)

So,

\(3 x_{1}+(2 \times 6)=18\)

\(\boldsymbol{x}_{1}=2\)

\(E(2,6)\)

\(x_{1}=4\)

\((3 \times 4)+2 x_{2}=18\)

\(x_{2}=3\)

\(F(4,3)\)

So the feasible points are \((0,0),(4,0),(4,3),(2,6),(0,6)\) \(Z(0,0)=0\)

\(Z(4,0)=12\)

\(Z(4,3)=(3 \times 4)+(2 \times 3)=18\)

\(Z(2,6)=(3 \times 2)+(2 \times 6)=18\)

\(Z(0,6)=12\)

So the objective function has multiple solution.

We know that:

To arrange \(n\) things in an order of a number of objects taken \(r\) things is:

\( { }^{n} P_{r}=\frac{n!}{(n-r)!}\)

or, \( { }^{n} P_{r}=\frac{n \times (n-1)\times(n-2)\times \cdots \times (n-r)\times \cdots \times3\times2\times1}{(n-r)!}\)

Given:

Number of signals that can be sent using 1 flag is:

\({ }^{4} \mathrm{P}_{1}=\frac{4!}{(4-1)!}=4\)

Number of signals that can be sent using 2 flags is:

\({ }^{4} \mathrm{P}_{2}=\frac{4!}{(4-2)!}=12\)

Number of signals that can be sent using 3 flags is:

\({ }^{4} P_{3}=\frac{4!}{(4-3)!}=24\)

Number of signals that can be sent using 4 flags is:

\({ }^{4} P_{4}=\frac{4!}{(4-4)!}=24\)

\(\therefore\) Total number of signals that can be sent at a time \(=4+12+24+24=64\)

We know that:

Sum of roots \(=\frac{-b}{a}\)

Product of roots \(= \frac{c}{a}\)

Given: \(\cos ^{2} x+a \cos x+b=0 \quad \ldots \ldots(1)\)

\(\sin ^{2} x+p \sin x+q=0\quad \ldots \ldots(2)\)

\(x=\alpha\) and \(x=\beta\) satisfy the equations.

Since both the equations are in the quadratic form, (in 'sin' and 'cos'), we have its roots, let us say the roots of the first equation be (sinα, sin β), and that of the second equation be (cosα, cos β).

So, for equation (1).

Sum of roots \(=\sin \alpha+\sin \beta=-a\quad\quad.....(3)\)

Product of roots \(=\sin \alpha \cdot \sin \beta={b}\quad\quad.....(4)\)

For equation (2),

Sum of roots \(=\cos \alpha+\cos \beta=-p\quad\quad.....(5)\)

Product of roots \(=\cos \alpha . \cos \beta={q}\quad\quad.....(6)\)

Squaring and adding (3) and (5), we get,

\((\sin \alpha+\sin \beta)^{2}+(\cos \alpha+\cos \beta)^{2}=(-a)^{2}+(-p)^{2}\)

\(\sin ^{2} \alpha+\sin ^{2} \beta+2 \sin \alpha \cdot \sin \beta+\cos ^{2} \alpha+\cos ^{2} \beta+2 \cos \alpha \cdot \cos \beta=a^{2}+p^{2}\)

\(1+1+2(\sin \alpha \cdot \sin \beta+\cos \alpha \cdot \cos \beta)=a^{2}+p^{2}\)

\(2(b+q)=a^{2}+p^{2}-2\)

Thus, we conclude that the relation is \(2(b+q)=a^{2}+p^{2}-2\)

Given,A box contains \(4\) tennis balls, \(6\) season balls and \(8\) dues balls

We know that,Probability \(=\frac{\text { Favourable outcomes }}{\text { Total outcomes }}\)

Let us assume that all balls are unique.

There are a total of \(18\) balls.

Number of all combinations of \(n\) things, taken \(r\) at a time, is given by \({ }^{n} C_{r}=\frac{n !}{(r) !(n-r) !}\)

Total ways \(=3\) balls can be chosen in \({ }^{18} \mathrm{C}_{3}\) ways

\(=\frac{18 !}{3 ! \times 15 !}\)

\(=\frac{18 \times 17 \times 16}{3 \times 2 \times 1}\)\(=816\)

There are \(4\) tennis balls, \(6\) season balls and \(8\) dues balls, \(1\) tennis ball, \(1\) season ball and \(1\) dues Ball drawn.

Therefore, favorable ways \(=4 \times 6 \times 8\)\(=192\)

Probability \(=\frac{192}{816}\)\(=\frac{4}{17}\)

The lines \(\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}\) and \(\frac{x-1}{3 k}=\frac{y-5}{1}=\frac{z-6}{-5}\) are mutually perpendicular.

As we know, if \(a, b, c\) are the direction ration ratios of a line passing through the point \(\left(x_{1}, y_{1}, z_{1}\right)\), then the equation of line is given by,

\(\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\)

Let, \(L_{1}\) be \(\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}\) and \(L_{2}\) be \(=\frac{x-1}{3 k}=\frac{y-5}{1}=\frac{z-6}{-5}\) respectively.

Now by comparing \(L_{1}\) and \(L_{2}\) with \(\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\), we get

\( a _{1}=-3, b _{1}=2 k , c _{1}=2, a _{2}=3 k , b _{2}=1\) and \(c _{2}=-5\)

As we know that if two lines are perpendicular, then

\(a _{1} \cdot a _{2}+ b _{1} \cdot b _{2}+ c _{1} \cdot c _{2}=0\)

\(\therefore (-3) \cdot 3k+2k \cdot 1+2 \cdot (-5)=0\)

\(\Rightarrow -9 k+2 k-10=0 \)

\(\Rightarrow-7 k=10\Rightarrow k=-\frac{10 }{ 7}\)

Let,

Set A of students who like mathematics, so:

n(A)=20

Set \(B\) of students who like science, so:

\(n(B)=15\)

Given:

\(n(A \cap B)=5\)

Set of students who like at least one of the 2 given subjects is \(A \cup B\), then:

\(n(A \cup B)=n(A)+n(B)-n(A \cap B)\)

\(n(A \cup B)=20+15-5=30\)

Students who do not like either of the subjects = Total students -\(n(A \cup B)\)

\(=50-30\)

\(=20\)

Given,

\(S=7^{2 n}+16 n-1\)

For \(n=1\),

\(S=49+16-1=64\)

For \(n=2\),

\(S=2401+32-1\)

\(=2432\)

\(=64 \times 38\)

As we can say that\(2432\)is divisible by \(64\).

Given:

Total consumers \(=1000\)

Consumers who like product \(A =720\)

Consumers who like product \(B =450\)

We know that:

\(n(A \cup B)=n(A)+n(B)-n(A \cap B)\)

Least number that like both the products are \(=n(A \cap B)\)

\(\Rightarrow 1000=720+450- n ( A \cap B )\)

\(\Rightarrow 1000=1170- n ( A \cap B )\)

\(\Rightarrow n ( A \cap B )=170\)

\(\therefore 170\) consumers like both the products \(A\) and \(B\).

\(\lim _{x \rightarrow \frac{1}{\sqrt{2}}} \frac{\sin \left(\cos ^{-1} x\right)-x}{1-\tan \left(\cos ^{-1} x\right)} \\\)

\( \text { Let } \cos ^{-1} \mathrm{x}=\mathrm{t} \\\)

\( \Rightarrow \mathrm{x}=\cos \mathrm{t} \\\)

\( \text { When } \mathrm{x} \rightarrow \frac{1}{\sqrt{2}} \text {, then } \mathrm{t} \rightarrow \cos ^{-1}\left(\frac{1}{\sqrt{2}}\right) \rightarrow \frac{\pi}{4} \\\)

\(\therefore \lim _{t \rightarrow \frac{\pi}{4}} \frac{\sin t-\cos t}{1-\tan (t)} \\\)

\( =\lim _{t \rightarrow \frac{\pi}{4}} \frac{\sin t-\cos t}{1-\frac{\sin t}{\cos t}} \\\)

\( =\lim _{t \rightarrow \frac{\pi}{4}} \frac{(\sin t-\cos t)(\cos t)}{(\cos t-\sin t)} \\\)

\( =\lim _{t \rightarrow \frac{\pi}{4}}-\cos t \\\)

\( =-\lim _{t \rightarrow \frac{\pi}{4}} \cos t \\\)

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

Unit vector \(\hat{\mathrm{u}}=\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}\)

\( \vec{p}_1=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{k}_2 \vec{p}_2=\frac{1}{\sqrt{2}} \hat{j}+\frac{1}{\sqrt{2}} \hat{k} \\\)

\( \vec{p}_3=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}\)

Now angle between \(\hat{\mathrm{u}}\) and \(\overrightarrow{\mathrm{p}}_1=\frac{\pi}{2}\)

\(\hat{\mathrm{u}} \cdot \overrightarrow{\mathrm{p}}_1=0 \Rightarrow \frac{\mathrm{x}}{\sqrt{2}}+\frac{\mathrm{z}}{\sqrt{2}}=0 \\\)

\( \Rightarrow \mathrm{x}+\mathrm{z}=0\)

Angle between \(\hat{\mathrm{u}}\) and \(\overrightarrow{\mathrm{p}}_2=\frac{\pi}{3}\)

\(\hat{\mathrm{u}} \cdot \overrightarrow{\mathrm{p}}_2=|\hat{\mathrm{u}}| \cdot\left|\overrightarrow{\mathrm{p}}_2\right| \cos \frac{\pi}{3} \\\)

\(\Rightarrow \frac{\mathrm{y}}{\sqrt{2}}+\frac{\mathrm{z}}{\sqrt{2}}=\frac{1}{2} \Rightarrow \mathrm{y}+\mathrm{z}=\frac{1}{\sqrt{2}}\)

Angle between \(\hat{\mathrm{u}}\) and \(\overrightarrow{\mathrm{p}}_3=\frac{2 \pi}{3}\)

\(\hat{\mathrm{u}} \cdot \overrightarrow{\mathrm{p}}_3=|\hat{\mathrm{u}}| \cdot\left|\overrightarrow{\mathrm{p}}_3\right| \cos \frac{2 \pi}{3} \\\)

\(\Rightarrow \frac{\mathrm{x}}{\sqrt{2}}+\frac{4}{\sqrt{2}}=\frac{-1}{2} \Rightarrow \mathrm{x}+\mathrm{y}=\frac{-1}{\sqrt{2}}\)

from equation (i), (ii) and (iii) we get

\(x=\frac{-1}{\sqrt{2}} \quad y=0 \quad z=\frac{1}{\sqrt{2}}\)

Thus \(\hat{\mathrm{u}}-\overrightarrow{\mathrm{v}}=\frac{-1}{\sqrt{2}} \hat{\mathrm{i}}+\frac{1}{\sqrt{2}} \hat{\mathrm{k}}-\frac{1}{\sqrt{2}} \hat{\mathrm{i}}-\frac{1}{\sqrt{2}} \hat{\mathrm{j}}-\frac{1}{\sqrt{2}} \hat{\mathrm{k}}\)

\(\hat{\mathrm{u}}-\overrightarrow{\mathrm{v}}=\frac{-2}{\sqrt{2}} \hat{\mathrm{i}}-\frac{1}{\sqrt{2}} \hat{\mathrm{j}}\)

\(\therefore|\hat{u}-\vec{v}|^2=\left(\sqrt{\frac{4}{2}+\frac{1}{2}}\right)^2=\frac{5}{2}\)