Mathematics Test - 17

\( \left(x^2+y^2\right) d x=2 x y d y \)

\(\Rightarrow \frac{d y }{d x}=\frac{\left(x^2+y^2\right) }{ 2 x y} \) is homogeneous.

Put \(y=v x\)

\(\Rightarrow \frac{d y }{d x} = v + x\frac {dv }{dx} \)

\( \Rightarrow v+x \frac{d v}{d x}=\frac{x^2+v^2 x^2}{2 v x^2}=\frac{x^2\left(1+v^2\right)}{2 v x^2} \)

\( \Rightarrow x \frac{d v}{d x}=\frac{1+v^2}{2 v}-v=\frac{1-v^2}{2 v}\)

\( \Rightarrow \frac{2 v}{1-v^2} d v=\frac{1}{x} d x\)

\( \Rightarrow-\log \left(1-v^2\right)=\log x+\log c \)

\( \Rightarrow \log \left[\frac {1 }{\left(1-v^2\right)}\right]=\log x+\log c \)

\( \Rightarrow\left[\frac {1 }{\left(1-v^2\right)}\right]=c x \)

\( \Rightarrow\left[\frac {1 }{\left(1-\frac{y^2 } {x^2}\right)}\right]=c x \)

\( \Rightarrow\left[\frac{x^2 }{\left(x^2-y^2\right)}\right]=c x\)

\(\Rightarrow x^2=c x\left(x^2-y^2\right)\) is the general solution of the differential equation.

Let, \(v=\cos ^{-1}\left(\frac{1}{2 x^{2}-1}\right)\) and \(u=\sqrt{1-x^{2}}\)

Substituting \(x=\cos \theta\), 

\(\Rightarrow \theta=\cos ^{-1} x\),  we get:

\(v=\cos^{-1}\left(\frac{1}{2 \cos ^{2} \theta-1}\right)\)

\(=\cos ^{-1}\left(\frac{1}{\cos ^{2} \theta-\sin ^{2} \theta}\right)\)

\(=\cos ^{-1}\left(\frac{1}{\cos 2 \theta}\right)\)

\(=\cos ^{-1}(\sec 2 \theta)\)

\(=\cos^{-1}\left(\cos \left(90^{\circ}-2 \theta\right)\right)\)

\(=90^{\circ}-\) \(2 \theta\).

And, \(u=\sqrt{1-x^{2}}\)

\(=\sqrt{1-\cos ^{2} \theta}\)

\(=\sqrt{\sin ^{2} \theta}=\sin \theta\)

Now, \(\frac{d v}{d \theta}=-2\) 

And, \(\frac{d u}{d \theta}=\cos \theta\)

\(\therefore \frac{\mathrm{dv}}{\mathrm{du}} =\frac{\mathrm{dv}}{\mathrm{d} \theta} \times \frac{\mathrm{d} \theta}{\mathrm{d} u}\)

\(=\frac{-2}{\cos \theta}=\frac{-2}{x}\)

\(\therefore\) The required area is given by:

\(A = \)

\(\int_{-4}^4\left(\sqrt{16-x^2}\right) d x\left[\because \int \sqrt{a^2-x^2} d x=\frac{x \sqrt{a^2-x^2}}{2}+\frac{a^2}{2} \sin ^{-1}\left(\frac{x}{a}\right)\right] \)

\(=2 \int_0^4\left(\sqrt{4^2-x^2}\right) d x \)

\(=2\left[\frac{x \sqrt{4^2-x^2}}{2}+\frac{4^2}{2} \sin ^{-1}\left(\frac{x}{4}\right)\right]_0^4 \)

\(=2\left(0+\frac{8 \pi}{2}-0-0\right) \)

\(=8 \pi \) sq units. 

\(\therefore\) The area of the region bounded by the curve \(y =\sqrt{16- x ^2}\) and the \(x\)-axis is \(8 \pi\) sq units.

Given,

\(-2<2 x-1<2\)

\(\Rightarrow-2+1<2 x<2+1 \)

\(\Rightarrow-1<2 x<3 \)

\(\Rightarrow \frac{-1}{2}

\(\Rightarrow x \in\left(\frac{-1}{2}, \frac{3}{2}\right)\)

Given function:

\(f:\mathrm{R} \rightarrow\{0,1\}\) such that \(f(x)=\left\{\begin{array}{c}1 \text { if } x \text { is rational } \\ 0 \text { if } x \text { is irrational }\end{array}\right.\)

Let f(x) be any function.

f (x) is onto if range of f (x) = Codomain

Codomain = {0, 1}

Since, on taking a straight line parallel to the x-axis, the group of given function intersect it at many points.

\(\Rightarrow f(x)\) is many-one.

Range of function is \(\{0,1\}\)

As range of \(f(x)=\) Codomain

\(\Rightarrow f(x)\) is onto.

Therefore, \(f(x)\) is many-one onto.

\(2 x+y \leq 20, x+2 y \leq\) \(20, x \geq 0, y \geq 0\)

The points in the feasible region are,

\(\mathrm{A}(0,0), \mathrm{B}(0,10), \mathrm{C}\left(\frac{20}{3}, \frac{10}{3}\right)\) and \(\mathrm{D}(10,0)\)

Objective function is \((P=x+3 y)\)

At point \(A, P=0+3(0)=0\)

At point \(\mathrm{B}, \mathrm{P}=0+3(10)=30\)

At point \(\mathrm{C}, \mathrm{P}=\frac{20}{3}+\frac{3(10)}{3}=\frac{50}{3}\)

At point \(D, P=10+3(0)=10\)

\(\therefore\) Maximum value of \(P=30\) at \(B(0,10)\).

Given:

Mean = 5 and Standard deviation = 2

If 5 is added to each value,

We know that adding a constant to each value Mean will be changed by adding a constant value and standard deviation remain the same.

Therefore, Mean = 5 + 5 = 10

Standard deviation will not change.

Now,

Coefficient of variation \(=\frac{\text { standard deviaiton }}{\text { Mean }} \times 100=\frac{2}{10} \times 100\)

∴ Coefficient of variation = 20

\(\sum\left(\mathrm{x}^2\right)=\sum \mathrm{x}^2 \mathrm{P}(\mathrm{x})=2\left[\frac{25}{36}+\frac{32}{36}+\frac{27}{36}+\frac{16}{36}+\frac{5}{36}\right] \\\)

\( =\frac{105}{18}=\frac{35}{6} \\\)

\( \mu=\sum(\mathrm{x})=0 \text { as data is symmetric } \\\)

\( \sigma^2=\sum\left(\mathrm{x}^2\right)=\sum \mathrm{x}^2 \mathrm{P}(\mathrm{x})=\frac{35}{6} \mathrm{P}=35=5 \times 7 \\\)

Sum of divisors = \(\left(5^0+5^1\right)\left(7^0+7^1\right)=6 \times 8=48\)

The passenger can go from Delhi to Mumbai in ⁴C₁ = 4 ways and from Mumbai to Mangalore in³C₁ = 3 ways.

Thus, the number of ways to travel from Delhi to Mangalore = 4 × 3 = 12