Mathematics Test - 9

Given,

\(x^{m+1} y^{n+1}=c^{m+n+2}\)

As we know,

If \(f\) and \(g\) are both differentiable, then

\(\frac{d}{d x}[f(x)-g(x)]=\frac{d}{d x} f(x)-\frac{d}{d x} g(x)\)

Differentiation of a constant is always \(0\).

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

Differentiating with respect to \(x\), we get

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

\(\Rightarrow x^{m+1} \frac{d}{d x} y^{n+1}+y^{n+1} \frac{d}{d x} x^{m+1}=\frac{d}{d x} c^{m+n+2}\)

\(\Rightarrow x^{m+1}(n+1) y^{n} \frac{d y}{d x}+y^{n+1}(m+1) x^{m}=0\)

\(\Rightarrow x^{m+1}(n+1) y^{n} \frac{d y}{d x}=-y^{n+1}(m+1) x^{m}\)

\(\Rightarrow \frac{d y}{d x}=-\frac{y^{n+1}(m+1) x^{m}}{x^{m+1}(n+1) y^{n}}\)

\(\Rightarrow \frac{d y}{d x}=-\frac{y^{n+1} y^{-n}(m+1)}{x^{m+1} x^{-m }(n+1)}\)

\(\Rightarrow \frac{d y}{d x}=-\frac{y^{n+1-n}(m+1)}{x^{m+1-m}(n+1)}\)

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

First, let's find out if the given line is parallel to the plane or not.

As we can see that, the direction ratios of the given line are: (1, 1, 1)

Similarly, the direction ratios of the normal to the given plane are: (1, 1, - 2)

⇒ 1 × 1 + 1 × 1 + (- 2) × 1 = 0

So, the given line is parallel to the given plane.

Let's find out a point on the given line

As we can see that, point Q (2, 0, - 1) lies on the line

So, finding the distance between the point Q and the given plane is same as finding the distance between the given line and plane.

As we know that, distance between a point and a plane is given by:\(\left|\frac{A x_{1}+B y_{1}+C z_{1}-d}{\sqrt{A^{2}+B^{2}+C^{2}}}\right|\)

Here, x₁ = 2, y₁ = 0 and z₁ = - 1

\(\Rightarrow d=\left|\frac{1 \cdot 2+1 \cdot 0-2 \cdot(-1)+3}{\sqrt{1^{2}+1^{2}+(-2)^{2}}}\right|=\frac{7 \sqrt{6}}{6}\) units

Given: \({ }^{\mathrm{n}} \mathrm{P}_{\mathrm{r}}=2760,{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}=23\)

To find: \(\mathrm{r}=\) ?

We know, \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}=\frac{{ }^{\mathrm{n}} \mathrm{P}_{\mathrm{r}}}{\mathrm{r} !}\)

\(\Rightarrow 23=\frac{2760}{\mathrm{r} !}\)

\(\Rightarrow \mathrm{r} !=\frac{2760}{23}=120\)

\(\Rightarrow \mathrm{r} !=5 \times 24=5 \times 4 \times 6\)

\(=5 \times 4 \times 3 \times 2 \times 1\)

\(\therefore \mathrm{r}=5\)

The eqution of line is given as, \(\frac{x+2}{-3}=\frac{y-2}{4}=\frac{z+2}{1}=\lambda\)

The coordinate of point B on the given line are,

B (-3λ- 2, 4λ+ 2, λ -2)

and the given coordinate of point A  is  ( 2, 3, 7). 

\(\overrightarrow{\mathrm{AB}}=(-3 \lambda-4) \hat{\mathrm{i}}+(4 \lambda-1) \hat{\mathrm{j}}+(\lambda+5) \hat{\mathrm{k}}\)

\(\mathrm{AB}\) is perpendicular to the line, \(\frac{\mathrm{x}+2}{-3}=\frac{\mathrm{y}-2}{4}=\frac{\mathrm{z}+2}{1},\)

We know that, If two lines are Perpendicular, then  a₁a₂+b₁b₂+c₁c₂ = 0,

⇒ (-3λ- 4 ) × (-3) + (4λ-1 )× 4 + (λ + 5 )× 1 = 0

⇒ 26λ + 13 = 0

\(\Rightarrow \lambda=-\frac{1}{2}\)

\(\overrightarrow{\mathrm{AB}}=-\frac{5}{2} \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+\frac{9}{2} \hat{\mathrm{k}}\)

\(\therefore|\overrightarrow{\mathrm{AB}}|=\sqrt{\left(\frac{5}{2}\right)^{2}+3^{2}+\left(\frac{9}{2}\right)^{2}}\)

\(\Rightarrow|\overrightarrow{\mathrm{AB}}|=\sqrt{\frac{71}{2}}\)

Given: \(a * b=a+b+1\)

\(a * b=a+b+1\)

The identity element of \({b}\) for any element \({a}, {b} \in {z}\)

\(a * b=a\)

\(a+b+1=a\)

\(b=-1\)

Therefore identity element is \(-1\)

Therefore \({e}=-1\)

now for a inverse,

\(a * a^{-1}=a+a^{-1}+1=e\)

\(a+a^{-1}+1=-1\)

\(a^{-1}=-2-a\)

When \({a}=-2, {a}^{-1}=0\)

If \(S=\left(x_{1}, y_{1}\right)\) is the focus of parabola and \(P(x, y)\) is any point on parabola and directrix is given as \(a x+b y+c=\) 0 then equation of parabola is:

\(\mathrm{SP}=\) Distance of "p" from directrix.

For \(P(x, y)\) and line \(a x+b y+c=0\) the distance is:

\(D=\frac{a x+b y+c}{\sqrt{a^{2}+b^{2}}}\)

Consider the parabola as shown:

Given focus is \((-8,-2)\) and directrix \(y=2 x-9\).

\(\sqrt{(x+8)^{2}+(y+2)^{2}}=\frac{2 x-y-9}{\sqrt{(2)^{2}+(-1)^{2}}}\)

Squaring on both sides

\(\left[(x+8)^{2}+(y+2)^{2}\right] 5=[2 x-y-9]^{2}\)

\((2 x-y-9)^{2}=(2 x)^{2}+(-y)^{2}+(-9)^{2}+2(2 x)(-y)+2(-y)(-9)+2(-9)(2 x)\)

\(=4 x^{2}+y^{2}+81-4 x y+18 y-36 x \quad\quad\ldots .(1)\)

\(5\left[(x+8)^{2}+(y+2)^{2}\right]=5\left[x^{2}+64+16 x+y^{2}+4+4 y\right]\)

\(5 x^{2}+80 x+5 y^{2}+20 y+340=4 x^{2}+y^{2}+81-4 x y+18 y-36 x\)

\(x^{2}+4 y^{2}+4 x y+116 x+2 y+259=0\)

This is equation of parabola.

If the constraints in a linear programming problem are changed the problem is to be re-evaluated. The linear inequalities or equations or restrictions on the variables of a linear programming problem are called constraints. The conditions x ≥ 0, y ≥ 0 are called non-negative restrictions.

Given,

\(f(x)=\tan ^{-1}\left[\frac{\sin x}{1+\cos x}\right]\)

As we know,

\(\sin x=2 \sin \frac{x}{2} \cos \frac{x}{2}\)

\(\cos x = \cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}\)

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

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

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

\(=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\)

\(=\tan \frac{x}{2}\)

\(\therefore f(x)=\tan ^{-1}\left[\frac{\sin x}{1+\cos x}\right]\)

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

Forfirst term derivative, we have to put \(x=1\).

The first derivative of \(f(x) = \frac{1}{2}\)