Mathematics Test - 5

Let B, H, F denote the sets of members who are in the basket hall team, hockey team and football team respectively. 

Given n(B) =21, n(H)= 2 6, n(F) = 29. 

n(H∩B) = 14, n (H∩F) = 15, n(F∩B) = 12 and n(B ∩ H ∩ F) = 8.

We have to find n(B U H U F) i.e., The total number of members in the three athletic teams.

n(B U H U F) = n(B) + n(H).n(F) − n(B ∩ H) − n(H ∩ F) − n(F ∩ B) + n(B ∩ H ∩ F).

n(B U H U F) = (21 + 26 + 29) − (14 + 15 + 12) + 8 = 43 

Here, \(\sin ^{-1} \frac{3}{x}+\sin ^{-1} \frac{9}{x}=\frac{\pi}{2}\)

Let,

\(\sin ^{-1} \frac{9}{x}=y \cdots(1)\)

\(\Rightarrow \sin y=\frac{9}{x}\)

\(\therefore \cos ^2 y=1-\sin ^2 y\)

\(=1-\left(\frac{9}{x}\right)^2\)

\(=1-\frac{81}{x^2}\)

\(\cos y=\frac{\sqrt{x^2-81}}{x}\)

\(\Rightarrow y=\cos ^{-1} \frac{\sqrt{x^2-81}}{x}\)

So, \(\sin ^{-1} \frac{9}{x}=\cos ^{-1} \frac{\sqrt{x^2-81}}{x}\)

\(\sin ^{-1} \frac{3}{x}+\cos ^{-1} \frac{\sqrt{x^2-81}}{x}=\frac{\pi}{2}\)

Now we know \(\sin ^{-1} \mathrm{y}+\cos ^{-1} \mathrm{y}=\frac{\pi}{2}\)

So,\(\frac{3}{x}=\frac{\sqrt{x^2-81}}{x}\)

\(9=x^2-81\)

\(\Rightarrow x^2=90\)

\(x=3 \sqrt{10}\)

Given:

\(10^{\mathrm{n}}+3^{4 \mathrm{n}+2}+8\)

Put \(n=1\) in \(10^{n}+3^{4 n+2}+8\).

\(\therefore 10^{1}+3^{4+2}+8=10+729+8\)

\(=747\)

\(747=3 \times 3 \times 83\)

Prime factors of 747 are \(3,3,83\).

From the given options we can say that \(10^{n}+3^{4 n+2}+8\) is divisible by 9 ( \(\left.3 \times 3\right)\) for all positive values of \(n\).

Let us check for each option:

(A) Given:

\(f(x)=|x|, \forall x \in R\)

As we know that,

\(f(x)=|x|\)

\(\Rightarrow f(x)=\left\{\begin{array}{cc}-x, & x<0 \\ x, & x \geq 0\end{array}\right.\)

So, \(f(-1)=-(-1)=1\) and \(f(1)=1\) 

\(\Rightarrow f(-1)=f(1)\), but \(-1 \neq 1\)

\(\therefore\) The property, \(f\left(x_{1}\right)=f\left(x_{2}\right) \Rightarrow x_{1}=x_{2}\), does not hold true \(\forall x_{1}, x_{2} \in R\). 

Therefore, the function \(\mathrm{f}(x)=|x|, \forall x \in R\) is not an injective function.

(B) Given: 

\(f(x)=x^{2}, \forall x \in R\)

Let \(x_{1}=1\) and \(x_{2}=-1\)

\(\Rightarrow \mathrm{f}\left(\mathrm{x}_{1}\right)=\mathrm{x}_{1}^{2}=1\) 

and, \(\mathrm{f}\left(\mathrm{x}_{2}\right)=\mathrm{x}_{2}^{2}=1\) 

\(\Rightarrow \mathrm{f}\left(\mathrm{x}_{1}\right)=\mathrm{f}\left(\mathrm{x}_{2}\right)\), but \(-1 \neq 1\)

\(\therefore\) The property, \(f\left(x_{1}\right)=f\left(x_{2}\right)\) \( \Rightarrow x_{1}=x_{2}\), does not hold true \(\forall x_{1}, x_{2} \in R\). 

Therefore, the function \(\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}, \forall \mathrm{x} \in \mathrm{R}\) is not an injective function.

(C) Given: 

\(f(x)=-x, \forall x \in R\)

Let \(x_{1}\) and \(x_{2}\) be any two real numbers.

\(\Rightarrow f\left(x_{1}\right)=-x_{1}\) and \(f\left(x_{2}\right)=-x_{2}\)

If \(\mathrm{f}\left(\mathrm{x}_{1}\right)=\mathrm{f}\left(\mathrm{x}_{2}\right)\)

\(\Rightarrow-\mathrm{x}_{1}=-\mathrm{x}_{2}\)

\(\Rightarrow \mathrm{x}_{1}=\mathrm{x}_{2}\)

\(\therefore\) The property, \(f\left(x_{1}\right)=f\left(x_{2}\right)\)

\(\Rightarrow x_{1}=x_{2}\), holds true \(\forall x_{1}, x_{2} \in R\)

Therefore, the function \(f(x)=-x, \forall x \in R\) is an injective function.

The given point is \(P (1,0,3)\) and equation of line passing through \((4,7,1)\) and \((3,5,3)\) is given by,

\(\frac{ x -4}{1}=\frac{ y -7}{2}=\frac{ z -1}{-2}= k\) (let)...(1)

So, any point on this line is Q\(( k +4,2 k +7,-2 k +1)\).

Now direction ratios of \(PQ\) are \(k +3,2 k +7,-2 k -2\).

Also \(PQ\perp\) (1)

\(\therefore 1( k +3)+2(2 k +7)-2(-2 k -2)=0 \)

\(\Rightarrow k +3+4 k +14+4k+2=0 \)

\(\Rightarrow 9 k +21=0\)

\(\Rightarrow 9 k=-21\)

\(\Rightarrow k =\frac{-21}{7}\)

\(\Rightarrow k =\frac{-7}{3}\)

Coordinates of Q are \(x=\frac{-7}{3}+4,y=2\times \frac{-7}{3}+7,z= -2 \times\frac{-7}{3}+1\)

\(\Rightarrow x=\frac{-7+12}{3},y=\frac{-14+21}{3}, z=\frac{14+3}{3}\)

\(\Rightarrow x=\frac{5}{3},y=\frac{7}{3}, z=\frac{17}{3}\)

So, the coordinates of the foot of the perpendicular drawn from the point \(A(1,0,3)\) to the join of the points \(B(4,7,1)\) and \(C(3,5,3)\) are \(\left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right)\).

Given: \(\mathrm{P}, \mathrm{Q}, \mathrm{R}\) are three sets

\(P \cup(Q \cap R)\)

\((P \cup Q) \cap(P \cup R)\)

Venn Diagram

\(\rightarrow\) It's used to illustrate the logical relation.

\(=\) Ships between two or more sets or items.

\(\Rightarrow\) They serve to graphically organize things, highlighting how the items are similar and different.

\(=\) widely used in mathematics, statistics, logic, teaching, linguistics, computer science and business.

Conclusion:

\(P \cup(Q \cap R)=(P \cup Q) \cap(P \cup R)\)

$\begin{array}{l}z=\frac{3+2i\sin \theta }{1-2i\sin \theta }=\frac{3+2i\sin \theta }{1-2i\sin \theta }\cdot \frac{1+2i\sin \theta }{1+2i\sin \theta }\\ z=\frac{(3+2i\sin \theta )(1+2i\sin \theta )}{1+4{\sin }^2\theta }\\ =\frac{3-4{\sin }^2\theta +8i\sin \theta }{1+4{\sin }^2\theta }\end{array}$

Now, z is purely real if sin θ=0 or θ =nπ, n $\in Z$

Also, z is purely imaginary if 

$3-4{\sin }^2 \theta =0$

or $\sin \theta =\pm \frac{\sqrt{3}}{2}=\pm \sin \frac{\pi }{3}$

⇒ $g=n\pi \pm \frac{\pi }{3},n\in Z$

$\overline{z_1{z}_2}={\overline{z}}_1{\overline{z}}_2$

From above conclusion we can say that $z^n={\overline{z}}^n$ , where $n\in N$

$\left(\frac{z_1}{z_2}\right)=\frac{{\overline{z}}_1}{{\overline{z}}_2}\cdot z_2\ne 0$

Given condition is, \(3 x+8 y \leq 24\)

\({y} \leq 2, {x} \geq 0, {y} \geq 0\)

The vertices of the feasible region are,

\((0,0),(8,0),\left(\frac{8}{3}, 0\right)\) and \((0,2)\)

Find the value of \(Z\) at all points,

At O (0, 0), \(Z=4 \times 0 + 7 \times 0=0\)

At A (8, 0), \(Z=4 \times 8 + 7 \times 0=32\)

At B (\(\frac 8 3\), 0), \(Z=4 \times \frac 8 3 + 7 \times 2=\frac{74}{3}\)

At C (0, 2), \(Z=4 \times 0 + 7 \times 2=14\)

The maximum value of the objective function attains at \((8,0)\).

\(Z=4 x+7 y\)

The maximum value \(=4 \times 8+7 \times 0=32\).

Given,

\(\left|\begin{array}{ccc}1 & l & l^{2} \\ 1 & m & m^{2} \\ 1 & n & n^{2}\end{array}\right|\)

Applying \(R _2 \rightarrow R _2-R _1\),

\(\left|\begin{array}{ccc}1 & l & l^{2} \\ 0 & m-l & m^{2}-l^{2} \\ 1 & n & n^{2}\end{array}\right|\)

Applying \(R _3 \rightarrow R _3-R _1\),

\(\left|\begin{array}{ccc}1 & l & l^{2} \\ 0 & m-l & m^{2}-l^{2} \\ 0 & n-l & n^{2}-l^{2}\end{array}\right|\)

\(\left|\begin{array}{ccc}1 & l & l^{2} \\ 0 & m-l & (m-l)(m+l) \\ 0 & n-l & (n-l)(n+l)\end{array}\right|\)

\((m-l)(n-l)\left|\begin{array}{ccc}1 & l & l^{2} \\ 0 & 1 & (m+l) \\ 0 & 1 & (n+l)\end{array}\right|\)

Now, expanding from \(a_{11}\),

\((m-l)(n-l) \cdot 1 .\left[\begin{array}{cc}1 & m+l \\ 1 & n+l\end{array}\right]\)

\(=(m-l)(n-l)(n+l-m-l)\)

\(=(m-l)(n-l)(n-m)\)

Given,

Lines are:

\(x-2 y=5 \ldots \ldots \ldots .(i)\)

and \(y-2 x=5 \ldots \ldots \ldots .(ii)\)

Let \(m_1\) and \(m_2\) are the slope of the given lines

From equation (i),

\(x-5=2 y\)

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

on comparing general equation of the line (\(y = mx +c\)) we get,

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

From equation (ii) ,

\(y=2 x+5\)

\(m_{2}=2\)

Now, 

Angle between the two lines is given by:

\(\tan \theta=|\frac{(m_{1}+m_{2})}{{1+m_{1} \times m_{2}}}|\)

\(\Rightarrow \tan \theta\) \(=|\frac{(\frac{1}{2}+2)}{\{1+(\frac{1}{2}) \times 2\}}|\)

\(\Rightarrow \tan \theta\) \(=|\frac{(\frac{5}{2})}{(1+1)}|\)

\(\Rightarrow \tan \theta\) \(=|\frac{(\frac{5}{2})}{2}|\)

\(\Rightarrow \tan \theta\) \(=\frac{5}{4}\)

\(\Rightarrow \theta=\tan ^{-1}(\frac{5}{4})\)