Mathematics Test - 14

Given:

\(y=x \frac{d y}{d x}+\left(\frac{d y}{d x}\right)^{-1}\)

The given differential equation can be written as:

\(\Rightarrow y=x \frac{d y}{d x}+\frac{1}{\frac{d y}{d x}}\)

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

As we know that, the highest order derivative occurring in a differential equation is called the order of a differential equation

So, for the given differential equation order is 1.

As we know that, the power of the highest order derivative which occurs in it, after it is made free from radicals and fractions is called the degree of a differential equation.

So, for the given differential equation degree is 2.

Given,

\(\tan ^{2} x-3 \sec ^{2} x+3=0 \)

\(\text { Put } \sec ^{2} x=1+\tan ^{2} x \)

\(\Rightarrow \tan ^{2} x-3\left(1+\tan ^{2} x\right)+3=0 \)

\(\Rightarrow \tan ^{2} x-3-3 \tan ^{2} x+3=0 \)

\(\Rightarrow-2 \tan ^{2} x=0 \)

\(\Rightarrow \tan^ 2 x=0 \)

\(\Rightarrow \tan x=0\left(\tan x=0, \text { when } x=0^{\circ}\right) \)

Given,

\(a^{2} x^{2}-3 a b x+2 b^{2}=0\)

\(\Rightarrow a^{2} x^{2}-3 a b x+2 b^{2}=0\)

\(\Rightarrow a^{2} x^{2}-2 a b x-a b x+2 b^{2}=0\)

\(\Rightarrow a x(a x-2 b)-b(a x-2 b)=0\)

\(\Rightarrow(a x-2 b)(a x-b)=0\)

\(\Rightarrow(a x-2 b)=0 \text { or }(a x-b)=0\)

\(\Rightarrow x=\frac{2 b}{a} \text { or } x=\frac{b}{a}\)

\(\therefore\) The roots are \(\frac{2 b}{a}\) and \(\frac{b}{a}\).

The total number of caps \(=2+4+5+1=12\)

The number of ways for selecting four caps \(={ }^{12} \mathrm{C}_{4}\)

The total number of caps other than green \(=12-5=7\)

The number of ways for selecting four caps other than green \(={ }^{7} \mathrm{C}_{4}\)

The probability of selecting four caps and none of them are green P(A')\(=\frac{\text { The number of ways for selecting four caps other than green }}{\text { The number of ways for selecting four caps }}\)

\( \mathrm{P}\left(\mathrm{A}^{\prime}\right)=\frac{{ }^{7} \mathrm{C}_{4}}{{ }^{12} \mathrm{C}_{4}}\)

\(\mathrm{P}\left(\mathrm{A}^{\prime}\right)=\frac{7 !(12-4) ! 4 !}{4 !(7-4) ! 12 !}\)

\(\mathrm{P}\left(\mathrm{A}^{\prime}\right)=\frac{7}{99}\)

As we know,

The sum of the roots of a quadratic equation \(a x^{2}+b x+c=0, a \neq 0\) is given by, \(\frac{\text { coefficient of } x}{\text { coefficient of } x^{2}}=-\left(\frac{b}{a}\right)\)

Let us verify the options,

(A) \(2 x^{2}-3 x+6=0\)

Sum of the roots \(=\frac{-b}{a}=-\left(\frac{-3}{2}\right)=\frac{3}{2}\)

(B) \(-x^{2}+3 x-3=0\)

Sum of the roots \(=\frac{-b}{a}=-\left(\frac{3}{-1}\right)=3\)

(C) \(\sqrt{2} x^{2}-\frac{3}{\sqrt{2 x}}+1=0\)

\(2 x^{2}-3 x+\sqrt{2}=0\)

Sum of the roots \(=\frac{-b}{a}=-\left(\frac{-3}{2}\right)=\frac{3}{2}\)

(D) \(3 x^{2}-3 x+3=0\)

Sum of the roots \(=\frac{-b}{a}=-\left(\frac{-3}{3}\right)=1\)

So, the equation which has the sum of its roots as \(3\) is \(-x^{2}+3 x-3=0\)

Given,

\(2 y+x^{2}=3\)

Differentiating both sides, we get

\(2 d y+2 x \cdot d x=0 \)

\(\Rightarrow x \cdot d x=-d y \)

\(\Rightarrow \frac{d x}{d y}=\frac{-1}{x}\)

If \(y=f(x)\) then, slope of normal of given curve \(=\frac{d x}{d y}\).

So, slope of normal \((m)=\frac{-d x}{d y}=-\frac{-1}{x}=\frac{1}{x}\) at \((1,1)\).

Slope of normal, \(m=\frac{-d x}{d y}=\frac{1}{1}=1\)

\(\therefore m=1\)

Equation of normal is \(y-y_{1}=m\left(x-x_{1}\right)\).

Here \(\left(x_{1}, y_{1}\right)\) is point lies on the normal of curve and \(m\) is slope of normal.

\(m=1\) and \(\left(x_{1}, y_{1}\right)=(1,1)\)

So, \((y-1)=1(x-1)\)

\(x-y=0\)

Therefore, equation of normal is \(x-y=0\)

Given

Objective function

Maximize, \(Z=x_{1}+2 x_{2}\)

Constraints

\(x_{1} \leq 2 \)

\(x_{2} \leq 2 \)

\(x_{1}+x_{2} \leq 2 \)

Non neagative constarints

\(\mathrm{x}_{1}, \mathrm{x}_{2} \geq 0\)

The above equations can be written as,

\(\frac{x_{1}}{2} \leq 1 \)

\(\frac{x_{2}}{2} \leq 1 \)

\(\frac{x_{1}}{2}+\frac{x_{2}}{2} \leq 1\)

Plot the above equations on \(x_{1}-x_{2}\) graph and find out the solution space.

Now, find out the value of the objective function at every extreme point of solution space.

\(Z=x_{1}+2 x_{2}\)

\(Z_{0}=0+2 \times 0=0 \)

\(Z_{A}=0+2 \times 2=4 \)

\(Z_{B}=2+2 \times 0=2\)

Since the value of the objective function is maximum at \(A\). There \(A(0,2)\) is the optimal solution.

Given:

Arithmetic mean of \(9\) terms \(=49\)

Mean of first \(7\) terms \(=37\)

Mean \(=\frac{\text { Sum of terms }}{\text { Number of terms }}\)

Let sum of last \(2\) terms be \(x\).

Sum of all \(9\) terms \(=49 \times 9=441\)

Sum of first \(7\) terms \(=37 \times 7=259\)

Sum of all \(9\) terms \(=\) Sum of first 7 terms \(+\) Sum of last 2 terms

\(441=259+x\)

\(x=182\)

Mean of last \(2\) terms \(=\frac{182}{2}=91\)

\(\therefore\) The mean of last \(2\) terms is \(91\).

Given:

Number of house \((\mathrm{m})=3\)

Number of people applying for house(n) \(=3\)

Total number of way three people can apply for house \(=3^{3}=27\)

Number of way in which all people apply for same house \(=3\)

\(\mathrm{P}(\) all three applying for same house \()=\frac{3}{27}=\frac{1}{9}\)

Given,

\(3 x^{2}+7 x y+2 y^{2}+5 x+5 y+k=0\)

Compare it with second-degree equation

\(a x^{2}+2 h x y+b y^{2}+2 g x+2 f y+c=0\)

So, \(a=3, h=3.5, b=2, g=2.5, f=2.5, c=k\)

The given equation represents the pair of straight lines. So,

\(\Delta=0\)

As \(\Delta=a b c+2 f g h-a f^{2}-b g^{2}-c h^{2}\)

\(\therefore a b c+2 f g h-a f^{2}-b g^{2}-c h^{2}=0\)

\(\Rightarrow 3 \times 2 \times k+2 \times 2.5 \times 2.5 \times 3.5-3 \times(2.5)^{2}-2 \times(2.5)^{2}-k \times(3.5)^{2}=0\)

\(\Rightarrow 6 k+43.75-18.75-12.5-12.25 k=0\)

\(\Rightarrow 6.25 k=12.5\)

\( k=2\)