Mathematics Test - 21

Given:

A = {x ϵ Z : x³ – 1 = 0} and B = {x ϵ Z: x² + x + 1= 0}

As we know that, the roots of the x³ – 1 = 0 are 1, ω and ω².

⇒ A = {1, ω, ω²}

As we know that, the roots of x² + x + 1= 0 are ω and ω2.

⇒ B = {ω, ω²}

⇒ A ∩ B = B

Radius of circle \(S\) touching \(x\)-axis and centre \((\alpha, \beta)\) is \(|\beta|\). According to given conditions

\(\alpha^2+(\beta-1)^2=(|\beta|+1)^2 \)

\(\alpha^2+\beta^2-2 \beta+1=\beta^2+1+2|\beta| \)

\(\alpha^2=4 \beta \text { as } \beta>0\)

\(\therefore\) Required louse is \(L: x^2=4 y\)

\(\text { The area of shaded region } =2 \int_0^4 2 \sqrt{y} d y \)

\(=4 \cdot\left[\frac{y^{\frac{3}{2}}}{\frac{3}{2}}\right]_0^4 \)

\(=\frac{64}{3}\) square units

The parabola \({y}^{2}=4 {x}\) has axis along \({x}\) axis and opens in positive direction of \({x}\) axis. The line \({x}+{y}+4=0\) cut the \({x}\) and \({y}\) axis at points (-4,0) and (0,4) respectively. 

The graph is shown below:

The given equation of a parabola is \(y^{2}=4 x\).

Compare it with general form \({y}^{2}=4 {ax}\) to get \({a}=1\) 

So, its vertex is (0,0) and focus is (1,0) 

It is given that the line \({x}+{y}+4=0\) intersect the image of parabola \({y}^{2}=4 {x}\) at points \(\mathrm{A}\) and \(\mathrm{B}\). 

Note that \(\mathrm{AB}\) is the latus rectum of the new parabola.

So, we only need to find the image of vertex (0,0) and focus (1,0) 

First find the image of vertex (0,0) about the line \(x+y+4=0\).

Let the image of vertex (0,0) be \(\left({x}_{1}, {y}_{1}\right),\) then

\(\frac{{x}_{1}-0}{1}=\frac{{y}_{1}-0}{1}=-2 \frac{(0+0+4)}{1^{2}+1^{2}}\)

\(\frac{{x}_{1}-0}{1}=\frac{{y}_{1}-0}{1}=-2 \times \frac{4}{2}\)

\(\frac{{x}_{1}-0}{1}=\frac{{y}_{1}-0}{1}=-4\)

\({x}_{1}=-4, {y}_{1}=-4\)

So, image of vertex is (-4,-4).

Now find the image of focus (1,0) about the line \({x}+{y}+4=0\).

Let it be \(\left({x}_{2}, {y}_{2}\right),\) then

\(\frac{{x}_{2}-1}{1}=\frac{{y}_{2}-0}{1}=-2 \frac{(1+0+4)}{1^{2}+1^{2}} \)

\(\frac{{x}_{2}-1}{1}=\frac{{y}_{2}-0}{1}=-2 \times \frac{5}{2} \)

\({x}_{2}-1={y}_{2}-0=-5 \)

\({x}_{2}=-5+1=-4, {y}_{2}=-5\)

So, image of focus is (-4,-5).

So, the focus of the new parabola is at a distance of 1 unit from the vertex on the axis. 

So, we have \(\mathrm{a}=1\) 

Therefore, the distance between \(\mathrm{A}\) and \(\mathrm{B}\) is, \(\mathrm{A B}=\) latus rectum of new parabola

\(=4 \mathrm{a} \)

\(=4 \times 1 \)

\(=\mathrm{4}\)

Given: The point of intersection of diagonals of a square \(A B C D\) is at the origin and one of its vertices is at \(A(4,2)\).

So, the diagonal \(AC\) passes through the origin

As we know that, the slope of the line joining the points \(\left( x _{1}, y _{1}\right)\) and \(\left( x _{2}, y _{2}\right)\) is: \(m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\) The slope of line \(AC\) is given by

\(\frac{2-0}{4-0}=\frac{1}{2}\)

In square \(A B C D\), the diagonals \(A C\) and \(B D\) are perpendicular to each other.

\(\Rightarrow\) Slope of \(A C \times\) slope of \(B D=-1\).

So, the slope of \(BD\) is \(- 2\).

As we know that, the equation of a line passing through the point \(\left(x_{1}, y_{1}\right)\) and having the slope ' \(m\) ' is given as:

\(y-y_{1}=m \cdot\left(x-x_{1}\right)\)

The equation of \(BD\) whose slope is \(-2\) and passes through origin is given by:

\(y-0=(-2)(x-0) \)

\(\Rightarrow 2 x+y=0\)

Concept:

The distance of a point \((p, q, r)\) from the plane \(a x+b y+c z+d=0\)

\(D=\left|\frac{a p+b q+c r+d}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}\right|\)

Calculation:

Given the equation of the plane is

\(2 x-3 y+6 z-42=0\)

The distance of the plane from the origin \((0,0,0)\)

\(D=\left|\frac{2 \times 0-3 \times 0+6 \times 0-42}{\sqrt{2^2+(-3)^2+6^2}}\right|\)

\(\Rightarrow D=\left|\frac{-42}{\sqrt{4+9+36}}\right|=\left|\frac{-42}{7}\right|\)

\(\Rightarrow D=6\)

Given here:

\(\lim _{x \rightarrow 0} \frac{\sqrt{4+x}-2}{3 x}\)

On substituting the limit value in the given function, we get:

\(\lim _{x \rightarrow 0} \frac{\sqrt{4+x}-2}{3 x}\)

\(=\frac{\sqrt{(4+0)}-2}{3(0)}\)

\(=\frac{0}{0}\)

Conjugate of \(\sqrt{4+x}-2\) is \(\sqrt{4+x}+2\)

Multiply and divide by \(\sqrt{4+x}+2\) in the given expression, we get:

\(\therefore \lim _{x \rightarrow 0} \frac{\sqrt{4+x}-2}{3 x} \times \frac{\sqrt{4+x}+2}{\sqrt{4+x}+2}\)

\(=\lim _{x \rightarrow 0} \frac{4+x-4}{3 x \sqrt{4+x}+2} \quad\left(\because(a+b)(a-b)=\left(a^{2}-b^{2}\right)\right)\)

\(= \lim _{x \rightarrow 0} \frac{x}{3 x(\sqrt{4+x}+2)}\)

\(=\lim _{x \rightarrow 0} \frac{1}{3(\sqrt{4+x}+2)}\)

Now, putting the value of x in the limit, we get:

\(=\frac{1}{3(2+2)}\)

\(=\frac{1}{12}\)

Given: \(\sin ^{-1} \frac{2 p}{1+p 2}-\cos ^{-1} \frac{1-q^{2}}{1+q^{2}}=\tan ^{-1} \frac{2 x}{1-x^{2}}\)

Let, p = tan A, q = tan B and x = tan C

\(\Rightarrow \sin ^{-1} \frac{2 \tan A}{1+\tan ^{2} A}-\cos ^{-1} \frac{1-\tan ^{2} B}{1+\tan ^{2} B}=\tan ^{-1} \frac{2 \tan C}{1-\tan ^{2} C}\)

\(\Rightarrow 2 A-2 B=2 C\)

⇒ A – B = C

⇒ tan^-1 p – tan^-1 q = tan^-1 x

As we know that,\(\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\)

\(\Rightarrow \tan ^{-1} p-\tan ^{-1} q=\tan ^{-1}\left(\frac{p-q}{1+p q}\right)=\tan ^{-1} x\)

\(\Rightarrow x=\frac{p-q}{1+p q}\)

The equation to the plane passing through \(P \left( x _1, y _1, z _1\right)\) and having direction ratios \(( a , b , c )\) for its normal is

\(a\left(x-x_1\right)+b\left(y-y_1\right)+c\left(z-z_1\right)=0\)

Since the plane \(\frac{2 x}{k}+\frac{2 y}{3}+\frac{z}{3}=2\) pass through the point \((2,3,-6)\); it satisfies the given equation.

\(\Rightarrow \frac{2 \times 2}{k}+\frac{2 \times 3}{3}+\frac{(-6)}{3}=2 \)

\(\Rightarrow k =2\)

The given equation of the plane is \(\frac{2 x}{k}+\frac{2 y}{3}+\frac{z}{3}=2\) Putting the value of \(k =2\), we get,

\(\frac{2 x}{2}+\frac{2 y}{3}+\frac{z}{3}=2 \)

\(\Rightarrow 3 x +2 y + z =6\)

The direction ratios of normal to the plane are\(<3, 2, 1>\).

\(\therefore\) The direction ratios of normal to the plane are\(<3, 2, 1>\).

The equation of the plane whose intercepts are \(a, b, c\) on the \(x, y, z\) axes, respectively, is

\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1 \quad( a , b , c \neq 0)\)

Since the plane \(\frac{2 x}{k}+\frac{2 y}{3}+\frac{z}{3}=2\) pass through the point \((2,3,-6)\);

So, it satisfies the given equation.

So, \(\frac{2 \times 2}{k}+\frac{2 \times 3}{3}+\frac{(-6)}{3}=2\) \(\Rightarrow k =2\)

The given equation of the plane is \(\frac{2 x}{k}+\frac{2 y}{3}+\frac{z}{3}=2\)

Putting the value of \(k=2\), we get,

\(\frac{2 x}{2}+\frac{2 y}{3}+\frac{z}{3}=2 \)

\(\Rightarrow 3 x +2 y + z =6\)

The equation of the plane in intercepts form,

\(\frac{3 x}{6}+\frac{2 y}{6}+\frac{z}{6}=1 \)

\(\Rightarrow \frac{x}{2}+\frac{y}{3}+\frac{z}{6}=1\)

2,3 , and 6 are the intercepts made by the plane on the coordinate axes, respectively.

So, Sum of the intercept \(=2+3+6=11\).

\(\therefore\) The value of \((p+q+r)\) is 11.