Mathematics Test - 22

We know that the area of a circle of radius units is given by \(A =\pi r ^{2}\) sq.units.

\(\therefore \frac{ dA }{ dt }=\frac{ d }{ dt }\left(\pi r ^{2}\right)=2 \pi r \left(\frac{ dr }{ dt }\right)\)

Now, \(\left[\frac{ dA }{ dt }\right]_{ r =4}=8 \pi(0.01)\) m\(^{2}/\) sec \(=0.08 \pi\) m\(^{2}/\)sec

\( \alpha=\frac{(4 !) !}{(4 !)^{3 !}} \beta=\frac{(5 !) !}{(5 !)^{4 !}} \\\)

\( \alpha=\frac{(24) !}{(4 !)^6}, \beta=\frac{(120) !}{(5 !)^{24}}\)

Let 24 distinct objects are divided into 6 groups of 4 objects in each group.

No. of ways of formation of group \(=\frac{24 !}{(4 !)^6 \cdot 6 !} \in \mathrm{N}\)

Similarly,

Let 120 distinct objects are divided into 24 groups of 5 objects in each group.

No. of ways of formation of groups

\(=\frac{(120) !}{(5 !)^{24} \cdot 24 !} \in \mathrm{N}\)

Hungarian algorithm used to solveassignment problem.The assignment problem is a special case of transportation problem where the matrix must be a square matrix and every row and every column only one allocation is possible. The matrix must be square matrix.The Hungarian method is a combinatorial optimization algorithm that solves the assignment problem in polynomial time and which anticipated later primal–dual methods.

Given,

\(\sin ^{-1} {x}+\cos ^{-1} {y}=\frac{2 \pi}{5}\)

We know that, \(\sin ^{-1} {x}+\cos ^{-1} {x}=\frac{\pi}{2}\)

\(\sin ^{-1} x=\frac{\pi}{2}-\cos ^{-1} x\)....(i)

Similarly,

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

\(\cos ^{-1} {y}=\frac{\pi}{2}-\sin ^{-1} {y}\)...(i)

Now,

\(\sin ^{-1} {x}+\cos ^{-1} {y}=\frac{2 \pi}{5}\)

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

\(\Rightarrow \frac{\pi}{2}-\cos ^{-1} {x}+\frac{\pi}{2}-\sin ^{-1} {y}=\frac{2 \pi}{5}\)

\(\Rightarrow \cos ^{-1} {x}+\sin ^{-1} {y}=\frac{3 \pi}{5}\)

Given,

\(\sin x+\operatorname{sin}^{2} x=1\)

\(\Rightarrow \sin x=1-\sin ^{2} x\) \(\quad\quad (\because 1 - \sin ^{2} \theta =\cos ^{2} \theta)\)

\(\Rightarrow \sin x=\cos ^{2} x \quad\) ......(i)

Sqauring both in equation (i) we get,

\(\sin ^{2} x=\cos ^{4} x\)

\(\Rightarrow 1-\cos ^{2} x=\cos ^{4} x \) \(\quad\quad (\because \sin ^{2} \theta =1 - \cos ^{2} \theta)\)

\(\cos ^{4} x+\cos ^{2} x=1\)

\( a_1+a_3=10=a_1+d \Rightarrow 5 \\\)

\(a_1+a_2+a_3+a_4+a_5+a_6=57 \\\)

\(\Rightarrow \frac{6}{2}\left[a_1+a_6\right]=57 \\\)

\( \Rightarrow a_1+a_6=19 \\\)

\( \Rightarrow 2 a_1+5 d=19 \text { and } a_1+d=5 \\\)

\( \Rightarrow a_1=2, d=3 \\\)

\(\text { Numbers: } 2,5,8,11,14,17 \\\)

\( \text { Variance }=\sigma^2=\text { mean of squares-square of mean } \\\)

\( =\frac{2^2+5^2+8^2+(11)^2+(14)^2+(17)^2}{6}-\left(\frac{19}{2}\right)^2 \\\)

\( =\frac{699}{6}-\frac{361}{4}=\frac{105}{4} \\\)

\( 8 \sigma^2=210\)

Here, \(f : R \rightarrow R\) is defined as \(f(x)=3 x\)

Let \(x, y \in R\) such that \(f(x)=f(y)\), then:

\( 3 x=3 y\)

\( x=y\)

\(\therefore f\) is one-one.

Also, for any real number \((y)\) in co-domain \(R\), there exists \(\frac{y}{3}\) in \(R\) such that:

\(f\left(\frac{y}{3}\right)=3\left(\frac{y}{3}\right)=y\)

\(\therefore f\) is onto.

Therefore, function \(f\) is one-one and onto.

Differential equation of the form,

\(\frac{d y}{d x}+P y=Q\)

Where \(P =\tan x\) and \(Q =2 \sin x\)

Integrating factor (I.F) of the equation is given by

\(I. F=e^{\int P d x}\)

\(I. F=e^{\int \tan x d x}=e^{\log \sec x}=\sec x\)

Its solution is

\( y \times(I . F)=\int[Q \times(I . F)] d x+C \)

\( y \sec x=\int[2 \sin x \times \sec x] d x+C \)

\( =\int\left[2 \sin x \times \frac{1}{\cos x}\right] d x+C\)

\( =\int[2 \tan x] d x+C=2 \log \sec x+C \)

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

\( =\frac{\log \sec ^2 x+C}{\sec x}\)

The area of a triangle with vertices \((-3,0),(3,0)\) and \((0, k)\) is 9 sq units

Area of a triangle with vertices \(\left(x_1, y_1\right),\left(x_2, y_2\right)\) and \(\left(x_3, y_3\right)\) is given by,

\(\Delta=\frac{1}{2}\left|\begin{array}{lll}x_1 & y_1 & 1 \\x_2 & y_2 & 1 \\x_3 & y_3 & 1\end{array}\right|\)

\(\therefore \Delta=\frac{1}{2}\left|\begin{array}{ccc}-3 & 0 & 1 \\3 & 0 & 1 \\0 & k & 1\end{array}\right|\)

\(9=\frac{1}{2}[-3(-k)-0+1(3 k)]\)

\(\Rightarrow 18=3 k+3 k=6 k\)

\(\therefore k=\frac{18}{6}=3\)