Mathematics Test - 16

Let the perpendicular OM is drawn from the origin to AB

Condition for Parallel & Perpendicular Lines

Equation of parallel and perpendicular lines to a given line

(1) Equation of a line which is parallel to ax+by+c=0 is ax+by+λ=0

(2) Equation of a line which is perpendicular to ax+by+c=0 is bx-ay+λ=0. The value of λ in both cases is obtained with the help of additional information given in the problem.

Point Slope form of Line

Equation of a line throught the point (x₁, y₁) and having slope m is y - y₁ = m(x - x₁).

Given:

\(x^{2}-x-1=0\)

We know that,

If the given equation is \(a x^{2}+b x+c=0\)

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

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

Now,

As \(\alpha\) and \(\beta\) are roots of \(x^{2}-x-1=0\), then,

\(\alpha+\beta=-(-1)\)

\(\Rightarrow \alpha+\beta=1\)

\(\Rightarrow \alpha \beta=-1\)

Now, if \((\frac{\alpha}{\beta})\) and \((\frac{\beta}{\alpha})\) are roots then,

Sum of roots \(=(\frac{\alpha}{\beta})+(\frac{\beta}a)\)

\(=\frac{\left(\alpha^{2}+\beta^{2}\right)}{a \beta}\)

\(=\frac{\left[(\alpha+\beta)^{2}-2 a \beta\right]}{ a \beta}\)

\(=\frac{\left[(1)^{2}-2(-1)\right]}{(-1)}\)

\(=-3\)

Product of roots \(=(\frac{\alpha}{\beta}) \times(\frac{\beta}{\alpha})=1\)

Now, then the equation is,

\(x^{2}-(\)Sum of roots\() x+\) Product of roots \(=0\)

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

\(\Rightarrow x^{2}+3 x+1=0\)

On equating the divisor with 0, we get x = 1.

On substituting the value of x in the given quadratic equation, we get f(x) = 4. (With the help of remainder theorem)

Given, \(f(x)=\left(9-x^{2}\right)^{2}\)
\(\Rightarrow f^{\prime}(x)=2\left(9-x^{2}\right)(-2 x)\)
\(\Rightarrow f^{\prime}(x)=4 x(x+3)(x-3)\)
\(f^{\prime}(x) \frac{-1}{-3}+\frac{1}{0} \frac{1}{3}\)
\(\therefore \quad f(x)\) is increasing in (-3,0)\(\cup(3, \infty)\)

We know that,

Standard equation of a circle:

\(({x}-{h})^{2}+({y}-{k})^{2}={r}^{2}\)

Where centre is \(({h}, {k})\) and radius is \({r}\).

The distance between a point on a circle and the center is the radius of the circle.

Distance between 2 points \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\) is:

\(D=\sqrt{\left({y}_{2}-{y}_{1}\right)^{2}+\left({x}_{2}-{x}_{1}\right)^{2}}\)

Let coordinates of the centre are \(({h}, {k})\).

Given the circle passes through \((3,3)\) and \((-1,-1)\).

Now the radius,

\(r=\sqrt{\left(y_{2}-y_{1}\right)^{2}+\left(x_{2}-x_{1}\right)^{2}}\)

\(\Rightarrow r=\sqrt{(h-3)^{2}+(k-3)^{2}}\)

Also, \({r}=\sqrt{({h}+1)^{2}+({k}+1)^{2}}\)

Comparing the 2 values of \(r\) we get:

\(h+k-2=0\quad\dots(1)\)

Also, the center is on the given line so follow the equation,

\(4 k-3 h-1=0\quad\dots(2)\)

Substracting (2) from \(4 \times({1})\)

\(7 {~h}-7=0\)

\(\Rightarrow {h}=1\)

Putting \(h=1\) in equation (1),

\(k=1\)

\(\therefore\) Coordinates of the centre are \((1,1)\).

Also,

\(r=\sqrt{(h+1)^{2}+(k+1)^{2}}\)

\(\Rightarrow {r}=\sqrt{(1+1)^{2}+(1+1)^{2}}\)

\(\Rightarrow {r}=\sqrt{8}\)

\(\Rightarrow {r}=2 \sqrt{2}\)

\(\therefore\) Diameter \(=2 r=4 \sqrt{2}\)

The equation could be written as,

sin(dy / dx)=3−e^x

dy / dx=sin⁻¹(3−ex)

The degree of the equation is 1.

Degree of Differential Equation

The degree of a differential equation is the power of the highest order derivative when differential coefficients are made free from radicals and fractions.

We can solve this problem using fundamental principle of counting.

Fundamental Principle of Multiplication:

If there are two jobs such that one of them can be completed in m ways, and when it has been completed in any one of these m ways, second job can be completed in n ways; then two jobs in succession can be completed in m × n ways.

By using this definition we can find the answer.

Let the 7 people be A, B, C, D, E, F and G. There are 7 positions to be filled D, E, F and G can occupy any of the 7 positions.

Thus :

Number of options for D = 7

(Any of the 7 position)

Number of options for E = 6

(Any of the 6 remaining positions)

Number of options for F = 5

(Any of the 5 remaining positions)

Number of options for G = 4

(Any of the 4 remaining positions)

Since, A must speak before B and B must speak before C, the 3 remaining positions must be occupied as follows : A - B - C

Thus,

number of options for A = 1

(A must occupy the leftmost remaining position)

Number of options for C = 1

(C must occupy the right most remaining positions)

Number of options for B = 1

(Only 1 position left)

To combine all of these options, we multiply :

7 × 6 × 5 × 4 × 1 × 1 × 1 = 840

Hence, the number of ways in which 7 persons can address a meeting so that out of the three persons A, B and C, A will speak before B and B before C, is 840.

Another method :

The same problem can be done by permutation also. If you do it by permutation we get

We have

Let the 7 people be A, B, C, E, F and G. Out of the three persons A, B and C. A will speak before B and B before C. Then arrange the remaining 4 peoples in 7 places. This is done in ⁷P₄ ways. Then arrange ABC is

⁷P₄ = 7 × 6 × 5 × 4 × 1 × 1 × 1 = 840