Mathematics Test-2

Given,

\(\angle XOY=60^{\circ}\), XY = ZW

In \(\triangle WOZ\) and \(\triangle XOY\) OZ, OW and OX, OY are radius of circle.

\(\therefore~ \)\(\triangle WOZ\) and \(\triangle XOY\) are Congruent triangles

Since, two Congruent triangles have all three sides and angles equal in measurement, we conclude that \(\angle ZOW=60^{\circ}\)

\(\tan \theta=\frac{1}{\sqrt{7}}=\frac{\text { Altitude }}{\text { Base }}\)

By Pythagoras Theorem,

\((\text { Hypotenuse })^2=(\text { Base })^2+(\text { Altitude })^2\)

\(\quad \quad \quad \quad \quad\quad\quad =(\sqrt{7})^2\quad+\quad(1)^{2}\)

\(\quad \quad \quad \quad \quad \quad \quad=7+1=8\)

\(\therefore\) Hypotenuse \(=\sqrt{8}=2 \sqrt{2}\)

Now, \(\operatorname{cosec} \theta=\frac{\text { Hypotenuse }}{\text { Altitude }}=\frac{2 \sqrt{2}}{1}\)

\(\sec \theta=\frac{\text { Hypotenuse }}{\text { Base }}=\frac{2 \sqrt{2}}{\sqrt{7}}\)

Now,

\(\frac{\operatorname{cosec}^2 \theta-\sec ^2 \theta}{\operatorname{cosec}^2 \theta+\sec ^2 \theta}=\frac{\left(\frac{2 \sqrt{2}}{1}\right)^2-\left(\frac{2 \sqrt{2}}{\sqrt{7}}\right)^2}{\left(\frac{2 \sqrt{2}}{1}\right)^2+\left(\frac{2 \sqrt{2}}{\sqrt{7}}\right)^2}\)

\(=\frac{8-\frac{8}{7}}{8+\frac{8}{7}}\)

\(=\frac{\frac{56-8}{7}}{\frac{56+8}{7}}\)\(=\frac{\frac{48}{7}}{\frac{64}{7}}\)

\(=\frac{48}{7} \times \frac{7}{64}=\frac{3}{4}\)

Given,

\(3+\frac{9}{2}+6+\frac{15}{2}+\ldots .\)

\(d=\frac3 2\)

\(n=25\)

\(a=3\)

The sum of n terms of an A.P. with first term a and common difference d is given by:

\(\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2} \times[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]\)

\(S_{25}=\frac{25}{2}[2 a+(25-1) \times d]\)

\(=\frac{25}{2}\left[2 \times 3+(24) \times \frac{3}{2}\right]\)

\(=\frac{25}{2} \times 42\)

\(=25 \times 21\)

\(=525\)

As we know,

In an ellipse \(\frac{ x ^{2}}{ a ^{2}}+\frac{ y ^{2}}{ b ^{2}}=1, a >b\)

The length of the latus rectum \(=\frac{2 b ^{2}}{ a }\)

Its eccentricity is given by,

\(e=\sqrt{1-\frac{b^{2}}{a^{2}}}\)

Let the equation of the ellipse be \(\frac{ x ^{2}}{ a ^{2}}+\frac{ y ^{2}}{ b ^{2}}=1\) where \(a >b\).

According to the question,

Length of the latus rectum \(=\) Half of its minor axis

\(\frac{2 b ^{2}}{ a }=b\)

\(\Rightarrow \frac{2 b ^{2}}{b}=a\)

\(\therefore a =2 b\)

Now, the eccentricity of the ellipse \((e)=\sqrt{1-\frac{b^{2}}{a^{2}}}\)

\(=\sqrt{1-\frac{b^{2}}{ (2b)^{2}}}\)

\(=\sqrt{1-\frac{b^{2}}{ 4b^{2}}}\)

\(=\sqrt{1-\frac{1}{4}}\)

\(=\sqrt{\frac{3}{4}}\)

\(=\frac{\sqrt{3}}{2}\)

As we know,

The standard form of the equation of a circle is \(\left( x - h \right)^{2}+\left( y - k \right)^{2}= R ^{2}\).

Where \(\left(h, k\right)\) are the coordinates and the \(R\) is the radius of center of the circle.

Area of the circle \(=\pi R ^{2}\)

Given,

Area of circle \(=154\) sq.units

\(\Rightarrow \pi R ^{2}=154 \)

\(\Rightarrow R ^{2}=154 \times \frac{7}{22} \)

\(\Rightarrow R =7\)

Equation of the diameters:

\(2 x-3 y+12=0 \)...(i)

\(x+4 y-5=0\)...(ii)

Intersection of the diameters (i) - \(2 \times\) (ii), we get

\(-11 y+22=0 \)

\(\Rightarrow y = 2\)

Putting the value of \(y\) in equation (i), we get

\( 2 x -3(2)+12=0 \)

\(\Rightarrow 2 x =6\)

\( \Rightarrow x =- 3\)

The center will be \(\left(-3,2\right)\).

By the standard equation of circle

\(\left( x - h \right)^{2}+\left( y - k \right)^{2}= R ^{2}\)

\(\Rightarrow( x -(-3))^{2}+( y -2)^{2}=7^{2}\)

\(\Rightarrow( x +3)^{2}+( y -2)^{2}=49\)

\(\Rightarrow x ^{2}+9+6 x + y ^{2}+4-4 y -49=0\)

\(\Rightarrow x ^{2}+6 x + y ^{2}-4 y - 3 6 = 0\)

Total number of cases \(=6(1,2,3,4,5,6)\)

There are three even numbers \(2,4,6\)

Probability \(= \frac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ outcomes}\)

Therefore probability of getting an even number is:

\(P(\text { even })=\frac{3}{6}\)

\(\Rightarrow P(\text { even })=\frac{1}{2}\)

\(A=\left[\begin{array}{cc}4 & x+2 \\ 2 x-3 & x+1\end{array}\right]\)

Given \(A^T=A\)

\(\Rightarrow\left[\begin{array}{cc}4 & 2 x-3 \\ x+2 & x+1\end{array}\right]=\left[\begin{array}{cc}4 & x+2 \\ 2 x-3 & x+1\end{array}\right]\)

Both are equal if corresponding elements are equal,

\(\Rightarrow 2 x-3=x+2\)

\(\Rightarrow 2 x-x=3+2\)

\(\Rightarrow x=5\)

Let the position vector of \(A\) and \(C\) be \(\vec{a}\) and \(\vec{c}\) respectively. Therefore, Position vector of \(B=\vec{b}=\vec{a}+\vec{c} ~~\dots\)(i)

Also Position vector of \(E=\frac{\vec{b}+2 \vec{c}}{3}=\frac{\vec{a}+3 \vec{c}}{3}~~\dots\)(ii)

Now point \(P\) lies on angle bisector of \(\angle AOC\). Thus,

Position vector of point \(P=\lambda\left(\frac{\vec{a}}{|\vec{a}|}+\frac{\vec{b}}{|\vec{b}|}\right)~~\dots\)(iii)

Also let \(P\) divides \(EA\) in ration \(\mu: 1\). Therefore,

Position vector of \(P\)

\(=\frac{\mu \vec{a}+\frac{\vec{a}+3 \vec{c}}{3}}{\mu+1}=\frac{(3 \mu+1) \vec{a}+3 \vec{c}}{3(\mu+1)}~~\dots\)(iv)

Comparing (iii) and (iv), we get

\(\lambda\left(\frac{\vec{a}}{|\vec{a}|}+\frac{\vec{c}}{|\vec{c}|}\right)=\frac{(3 \mu+1) \vec{a}+3 \vec{c}}{3(\mu+1)}\)

\(\Rightarrow \frac{\lambda}{|\vec{a}|}=\frac{3 \mu+1}{3(\mu+1)}\) and \(\frac{\lambda}{|\vec{c}|}=\frac{1}{\mu+1}\)

\(\Rightarrow \frac{3|\vec{c}|-|\vec{a}|}{3|\vec{a}|}=\mu\)

\(\Rightarrow \frac{\lambda}{|\vec{c}|}=\frac{1}{\frac{3|\vec{c}|-\vec{a}}{3|\vec{a}|}+1}\)

\(\Rightarrow \lambda=\frac{3|\vec{a}||\vec{c}|}{3|\vec{c}|+2|\vec{a}|}\)

Let \(a_{1}, a_{2}, a_{3}\) and \(a_{4}\) be the coefficients of 4 consecutive terms viz. the rth, \((r+1)\) th, \((r+2)\) th and \((r+3)\) th terms.

Then, \(a_{1}={ }^{n} C_{r-1}, a_{2}=" C_{r}, a_{3}={ }^{n} C_{r+1}\) and \(a_{4}={ }^{n} C_{t-2}\). Now,

\(\frac{a_{1}}{a_{1}+a_{2}}+\frac{a_{3}}{a_{3}+a_{4}}=\frac{1}{1+\frac{a_{2}}{a_{1}}}+\frac{1}{1+\frac{a_{4}}{a_{3}}}\)

\(=\frac{1}{1+\frac{{ }^{n} C_{+}}{{ }^{n} C_{r-1}}}+\frac{1}{1+\frac{{ }^{n} C_{r+2}}{{ }^{n} C_{r+1}}}\)

\(=\frac{1}{1+\frac{n-r+1}{r}}+\frac{1}{1+\frac{n-r-1}{r+2}}\)

\(=\frac{r}{n+1}+\frac{r+2}{n+1}\)

\(=\frac{2 \cdot(r+1)}{n+1}\)

Given:

\(\int_{0}^{a}[f(x)+f(-x)] d x=\int_{-a}^{a} g(x) d x\),

\(g(x)=?\)

For an even function,

\(f(-x)=f(x)\) and \(\int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x\)

For the given condition \(\int_{0}^{{a}}[{f}({x})+{f}(-{x})] {dx}=\int_{-{a}}^{{a}} {g}({x}) {dx}\) to be true, \({f}({x})\) and \({g}({x})\) both must be even functions, i.e.,

\(f(x)=f(-x)\) and \(g(x)=f(x)+f(-x)\)