Mathematics Test-4

Given, \(\alpha\) and \(\beta\) are the roots of the equation \(x^{2}-2 x+4=0\).

\(\Rightarrow \alpha+\beta=2\) and \(\alpha \beta=4\)

Consider, \(\alpha^{3}+\beta^{3}=(\alpha+\beta)\left(\alpha^{2}+\beta^{2}-\alpha \beta\right)\)

\(\Rightarrow \alpha^{3}+\beta^{3}=(\alpha+\beta)\left(\alpha^{2}+\beta^{2}+2 \alpha \beta-3 a \beta\right)\)

\(\left.\Rightarrow \alpha^{3}+\beta^{3}=(\alpha+\beta)\left[(\alpha+\beta)^{2}-3 a \beta\right)\right]\)

\(\Rightarrow \alpha^{3}+\beta^{3}=(2)\left[(2)^{2}-3(4)\right]\)

\(\Rightarrow \alpha^{3}+\beta^{3}=(2)(-8)\)

\(\Rightarrow \alpha^{3}+\beta^{3}=-16\)

So, if \(\alpha\) and \(\beta\) are the roots of the equation \(x^{2}-2 x+4=0\), then the value of \(\alpha^{3}+\beta^{3}=-16\).

Here, \(A=\left\{x \mid x^{2}-5 x-6=0\right\}\)

\(x^{2}-5 x-6=0\)

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

\(\Rightarrow x(x-6)+1(x-6)=0\)

\(\Rightarrow(x+1)(x-6)=0\)

\(\therefore x=-1,6\)

Which can be represented in roaster form as: \(A=\{-1,6\}\).

Similarly, \(B=\left\{y \mid y^{2}-7 y-8=0\right\}\)

\(y^{2}-7 y-8=0\)

\(\Rightarrow y^{2}-8 y+y-8=0\)

\(\Rightarrow y(y-8)+1(y-8)=0\)

\(\Rightarrow(y+1)(y-8)=0\)

\(\therefore y=-1,8\)

Which can be represented in roaster from as: \(B=\{-1,8\}\).

Let, \(C=(A \cap B)=\{-1\}\)

Now,\(A \cup(A \cap B)\)

\(=A \cup C\)

\(=\{-1,6\}\)

Given:

Correlation coefficient between \(x\) and \(y, r(x, y)=0.6\)

Covariance of x and y, Cov(x, y) \(=27\)

Variance of \(y, \operatorname{Var}(y)=\sigma^{2}(y)=25\)

Variance of \(x, \operatorname{Var}(x)=\sigma^{2}(x)\)

We know that,

\(r(x, y)=\frac{\operatorname{Cov}(x, y)}{\sigma(x) \sigma(y)}\)

\(\Rightarrow \sigma(x)=\frac{\operatorname{Cov}(x, y)}{\sigma(y) \times r(x, y)}\)

\(\Rightarrow \sigma(x)=\frac{27}{5 \times 0.6}\)

\(\Rightarrow \sigma(x)=9\)

Variance of \(x, \sigma^{2}(x)=81\)

Given:

\(P(n)=3^{3 n}-2 n+1\)

Using the principle of mathematical induction,

For \(n=1\)

\(P(1): 3^{3(1)}-2(1)+1=26=2 \times 13\), so \(P(1)\) is true

Let \(P(m)\) be true such that,

\(P(m): 3^{3(m)}-2(m)+1=2 k\)

We now need to prove that \(P(m+1)\) is also true, therefore,

\(P(m+1): 3^{3(m+1)}-2(m+1)+1\)

\(\Rightarrow 27 \times 3^{3 m}-2 m+1-2\)

\(\Rightarrow(26+1) 3^{3 m}-2 m+1-2\)

\(\Rightarrow 26 \times 3^{3 m}+\left(3^{3 m}-2 m+1\right)-2\)

\(\Rightarrow 26 \times 3^{3 m}+2 k-2\)

\(\Rightarrow 2\left(13 \times 3^{3 m}+k-1\right)\), which is divisible by 2.

Since the given function is differentiable, we can write:

\(\lim _{x \rightarrow-1^{-}} f^{\prime}(x)=\lim _{x \rightarrow-1^{+}} f^{\prime}(x)\)

\(\Rightarrow \lim _{x \rightarrow-1^{-}}(2 b x)=\lim _{x \rightarrow-1^{+}}(2 a x-b)\)

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

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

\(\Rightarrow 2 a=b \ldots .(i)\)

Since \(f(x)\) will also be continuous, we can write:

\(\lim _{x \rightarrow-1^{-}}\left(b x^{2}-a\right)=\lim _{x \rightarrow-1+}\left(a x^{2}-b x-2\right)\)

\(\Rightarrow b-a=a+b-2\)

\(\Rightarrow 2 a=2\)

\(\Rightarrow a=1\)

Using Equation \((i)\):

\(2 a=b\)

\(\Rightarrow 2 \times 1=b\)

\(\Rightarrow b=2\)

These are the roots of the equation:

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

Let there be '\(n\)' men participants.

Then, the number of games that the men play between themselves is \(2\).\(^{n} C_{2}\) and the number of games that the men played with the women is \(2.(2 n)\).

\(\therefore 2 .^{n} C_{2}-2. 2 n=66\) (given)

\(\Rightarrow n(n-1)-4 n-66=0\)

\(\Rightarrow n^{2}-5 n-66=0\)

\(\Rightarrow(n+5)(n-11)=0\)

\(\Rightarrow n=11\)

Number of participants \(=11\) men \(+2\) women \(=13\)

\(y-2 x+1+\lambda_{1}(2 y-x-1)=0\) always passes through intersection of \(y-2 x+1=0\) and \((2 y-x-1)=0\).And their intersection is \((1,1)\) \(\equiv \mathrm{Q}\)

Similarly \(3 y-x-6=0+\lambda_{2}(y-3 x+6)=0\) passes through \((3,3)\) \(\equiv P\)

Equation of \(\mathrm{PQ}\) is : \(y=x\)

According to given condition, third family of equation should always pass through the point which lies on \(y=x\)

Putting \(y=x\) in \(a x+y-2+\lambda_{3}(6 x+a y-a)=0\)

we get \(a x+x-2=0\) and \(6 x+\) ay \(-a=0\)

\(\Rightarrow \frac{2}{a+1}=\frac{a}{6+a}\)

\(\Rightarrow 12+2a = a^{2}+a\)

\( \Rightarrow a^{2}- a -12=0\)

\(\Rightarrow(a-4)(a+3)=0\)

\(\therefore a=-3\)

Given equation of the circle is:

\(x^{2}+y^{2}-4 x-6 y+11=0\)

\(\therefore \quad 2 g=-4\) and. \(2 f=-6\)

So, the centre of the circle is \(C(-g,-f) \equiv C(2,3)\) \(A(3,4)\) is one end of the diameter. Let the other end of the diameter be \(B\left(x_{1}, y_{1}\right)\)

Here, mid point of \(A B\) is \(C\).

\(\therefore 2=\frac{3+x_{1}}{2}\) and \(3=\frac{4+y_{1}}{2}\)

\(\Rightarrow x_{1}=1\) and \(y_{1}=2\)

So, the coordinates of other end of the diameter are (1,2).

Given,

\(\cot ^{-1}(\sqrt{\cos \alpha})-\tan ^{-1}(\sqrt{\cos \alpha})=x\)

\(\tan ^{-1}\left(\frac{1}{\sqrt{\cos \alpha}}\right)-\tan ^{-1}(\sqrt{\cos \alpha})=x\)

\(\Rightarrow \tan ^{-1} \frac{\frac{1}{\sqrt{\cos \alpha}}-\sqrt{\cos \alpha}}{1+\frac{1}{\sqrt{\cos \alpha}} \cdot \sqrt{\cos \alpha}}=x\) \((\because\tan^{-1} x – \tan^{-1} y = \tan^{-1}(\frac{(x-y)}{(1+xy)})\), if the value \(xy > -1)\)

\(\Rightarrow \tan ^{-1} \frac{1-\cos \alpha}{2 \sqrt{\cos \alpha}}=x\)

\(\Rightarrow \tan x=\frac{1-\cos \alpha}{2 \sqrt{\cos \alpha}}\) or \(\cot x=\frac{2 \sqrt{\cos \alpha}}{1-\cos \alpha}\)

\(=\tan ^{2} \frac{\alpha}{2}\)

\(\frac{d y}{d x}+y \cos x=3 \cos x \)

\( I F=e^{\int \cos x d x} \)

\( \Rightarrow I F=e^{\sin x}\)

Now, \(y \times( IF )=\int Q ( IF ) dx\)

\( \Rightarrow y \times e^{\sin x}=\int 3 \cos x \times e^{\sin x} d x \)

\( \Rightarrow y e^{\sin x}=\int 3 e^{\sin x} \cos x~ d x\)

Integrating, let \(\sin x=t \Rightarrow \cos x d x=d t\)

\( \Rightarrow y e^{\sin x}=\int 3 e^t d t \)

\( \Rightarrow y e^{\sin x}=3 e^t+c \)

\( \Rightarrow y e^{\sin x}=3 e^{\sin x}+c\)

(where \(c\) is integration constant)

\(\overrightarrow{a}=2 \hat{i}+\hat{j}-3 \hat{k}\) and \(\overrightarrow{b}=3 \hat{i}-2 \hat{j}-\hat{k}\)

\(\vec{a} \cdot \vec{b}=(2 \hat{i}+\hat{j}-3 \hat{k})(3 \hat{i}-2 \hat{j}-\hat{k}) \)

\(=6-2+3=7 \ldots .(1) \)

\(|\vec{a}|=\sqrt{2^{2}+1^{2}+(-3)^{2}}=\sqrt{14} \ldots(2) \)

\(|\vec{b}|=\sqrt{3^{2}+(-2)^{2}+(-1)^{2}}=\sqrt{14} \ldots .(3) \)

\(\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| \cdot|\vec{b}|}\)

Put the values from (1), (2) and (3) in above equation:

\(=\frac{7}{\sqrt{14}. \sqrt{14}}=\frac{1}{2} \)

\(\cos \theta=\cos 60^{\circ} , \theta=60^{\circ}\)