Mathematics Test - 3

Sum of the roots: \(a+\beta=\lambda-2\)

Product of the roots: \(\alpha \cdot \beta=10-\lambda\)

Therefore,

\(a^{3}+\beta^{3}=(a+\beta)^{3}-3 a \beta(a+\beta)\)

\(=(\lambda-2)^{3}+3(\lambda-10)(\lambda-2)\)

\(=(\lambda-2)\left(\lambda^{2}-\lambda-26\right)\)

For minimum, differentiating w.r.t. \(\lambda\)

\(\frac{d\left(a^{3}+\beta^{3}\right)}{d \lambda}=\left(\lambda^{2}-\lambda-26\right)+(\lambda-2)(2 \lambda-1)\)

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

\(\Rightarrow(\lambda-4)(\lambda+2)=0\)

So, it will have minima at \(\lambda=4\)

Now,

\((a-\beta)^{2}=(a+\beta)^{2}-4 a \beta\)

\(\Rightarrow(\alpha-\beta)^{2}=(\lambda-2)^{2}-4(10-\lambda)\)

\(\Rightarrow(\alpha-\beta)^{2}=2^{2}-4 \times 6=-20\)

\(\Rightarrow a-\beta=\pm i 2 \sqrt{5}\)

So, the magnitude will be \(2 \sqrt{5}\).

Let A be the set of people who like coffee and B be the set of people who like tea.

Given that: 

n(A) = 37, n(B) = 52 and n(A ⋃ B) = 70.

Since, every person likes at least one drink (0 elements outside A and B),

We know that:

n(A ⋃ B) = n(A) + n(B) - n(A ⋂ B)

⇒ 70 = 37 + 52 - n(A ⋂ B)

⇒ n(A ⋂ B) = 89 - 70 = 19

People who like coffee and NOT tea is given by: 

n(A - B) = n(A) - n(A ⋂ B) 

= 37 - 19 

= 18 

Given:

The function \(f: R \rightarrow R\) defined by \(\sqrt{x^{2}-x-110}\)

We know that the domain of a function is the complete set of possible values of the independent variable.

To find the domain

\(=x^{2}-x-110 \geq 0 \)

\(=x^{2}-11 x+10 x-110 \geq 0 \)

\(=x(x-11)+10(x-11) \geq 0 \)

\(=(x+10)(x-11) \geq 0 \)

\(=x \leq-10 \text { or } x \geq 11 \)

\(=x \in(-\infty,-10] \cup[11, \infty)\)

Hence, the domain of the function \(f: R \rightarrow R\) defined by \(f(x)=\sqrt{x^{2}-x-110}\) is \((-\infty,-10] \cup[11, \infty)\)

Since the distance from the focus to thevertex is \(5~cm\). We have, \(a =~5\). If the origin is taken atthe vertex and the axis of the mirror lies along thepositive \(x-axis\), the equation of the parabolic section is \(y^{2}=4(5)x=20\)

Note that \(x=45\).

Thus, \(y^{2}=900\)

\(y= \pm 30\)

Therefore \(AB=2y=2 \times 30=60~cm\).

Given:

\(Z=x_{1}+x_{2}\)

\(x_{1}+x_{2} \leq 10,-2 x_{1}+3 x_{2} \leq 15, x_{1} \leq 6, x_{1}, x_{2} \geq 0\)

The point of intersection of lines \(x_{1}+x_{2}=10\)

and \(-2 x_{1}+3 x_{2}=15\) is \(B(3,7)\) and point of intersection of lines \(x_{1}=6\) and \(x_{1}+x_{2}=10\) is \(C(6,4)\).

The feasible region is \(O A B C D\). The corner points of the feasible region are:

\(O(0,0), A(0,6), B(3,7), C(6,4)\) and \(D(6,0)\)

\(\begin{array}{|l|l|} \hline At O(0,0) & Z=0+0=0 \\ \hline A(0,6) & Z=0+6=6 \\ \hline B(3,7) & Z=3+7=10 \\ \hline C(6,4) & Z=6+4=10 \\ \hline D(6,0) & Z  = 6+0= 6 \\ \hline \end{array}\)

So, \(Z\) is maximum at every point of the segment combined two points \(B(3,7)\) and \(C(6,4)\).

\(\begin{aligned} & A=\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right], B=\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right] \\ & \Rightarrow A B=\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right] \\ & \Rightarrow A B=\left[\begin{array}{cc}1-1 & 1-1 \\ -1+1 & -1+1\end{array}\right] \\ & \Rightarrow A B=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]\end{aligned}\)

The coefficient matrix for the given system of equations is:

\(\operatorname{det}(A)=1(10-4)+1(16-20)+1(2-4)=6-4-2=0\)

Therefore, either there is no solution (inconsistent) or there are infinitely many solutions (consistent and dependent).

Let, us convert the augmented matrix into the row echelon form to find its solutions. The augmented matrix is:

\(A \mid B =\left[\begin{array}{ccccc}1 & 1 & 1 & \mid & 1 \\ 2 & 1 & 4 & \mid & k \\ 4 & 1 & 10 & \mid & k ^{2}\end{array}\right]\)

\(R _{3} \rightarrow R _{3}-4 R _{1}\)

\(R _{2} \rightarrow R _{2}-2 R _{1}\)

\(A \mid B =\left[\begin{array}{ccccc}1 & 1 & 1 & \mid & 1 \\ 0 & -1 & 2 & \mid & k -2 \\ 0 & -3 & 6 & \mid & k ^{2}-4\end{array}\right]\)

\(R _{3} \rightarrow R _{3}-3 R _{2}\)

\(A \mid B =\left[\begin{array}{ccccc}1 & 1 & 1 & \mid & 1 \\ 0 & -1 & 2 & \mid & k -2 \\ 0 & 0 & 0 & \mid & k ^{2}-3 k +2\end{array}\right]\)

The rank of both the coefficient matrix and the augmented matrix must be equal for the system to be consistent.

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

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

\(\Rightarrow k ( k -2)-( k -2)=0\)

\(\Rightarrow( k -2)( k -1)=0\)

\(\Rightarrow k -2=0\) OR \(k -1=0\)

\(\Rightarrow k =2\) or \(k =1\)

Let \(A\) and \(B\) be the positions of the two flag posts and \(P(x, y)\) be the position of the man.

Accordingly, \(PA+PB=10\)

We know that if a point moves in a plane in such a way that the sum of its distances from two fixed points is constant, then the path is an ellipse and this constant value is equal to the length of the major axis of the ellipse.

Therefore, the path described by the man is an ellipse where the length of the major axis is \(10~m\), while points \(A\) and \(B\) are the foci.

Taking the origin of the coordinate plane as the center of the ellipse, while taking the major axis along the \(x\)-axis, the ellipse can be diagrammatically represented as

The equation of the ellipse will be of the form \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where \(a\) is the semi-major axis

Accordingly, \(2 a=10\)

\(\Rightarrow a=5\)

Distance between the foci \((2 c)=8\)

\(\Rightarrow c=4\)

On using the relation \(c=\sqrt{a^{2}-b^{2}}\), we obtain

\(4=\sqrt{25-b^{2}}\)

\(\Rightarrow 16=25-b^{2}\)

\(\Rightarrow b^{2}=25-16=9\)

Thus, the equation of the path traced by the man is \(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\)

Statement 1: The cross product of two unit vectors is always a unit vector.

Let \(\vec{a}\) and \(\vec{b}\) are two unit vectors.

i.e., \(|\vec{a}|=1\) and \(|\vec{b}|=1\)

As we know that, the cross product of two vectors \(\vec{a}\) and \(\vec{b}\) is given by \(\vec{a} \times \vec{b}=|\vec{a}| \cdot|\vec{b}| \sin \theta \hat{n}\) and \(|\vec{a} \times \vec{b}|=|\vec{a}| \cdot|\vec{b}| \sin \theta\)

\(\Rightarrow|\vec{a} \times \vec{b}|=|\vec{a}| \cdot|\vec{b}| \sin \theta=\sin \theta\)

The range of \(\sin \theta\) is \([-1,1]\)

So, it is not necessarily true that the cross product of two unit vectors is always a unit vector.

Therefore, statement 1 is false.

Statement 2. The dot product of two unit vectors is always unity.

Let \(\vec{a}\) and \(\vec{b}\) are two unit vectors.

i.e., \(|\vec{a}|=1\) and \(|\vec{b}|=1\)

As we know that, the scalar product of two vectors \(\vec{a}\) and \(\vec{b}\) is given by \(\vec{a} \cdot \vec{b}=|\vec{a}| \times|\vec{b}| \cos \theta\) \(\Rightarrow|\vec{a} \cdot \vec{b}|=\cos \theta\)

The range of \(\cos \theta\) is \([-1,1]\).

So, it is not necessarily true that the dot product of two unit vectors is always a unit vector.

Therefore, statement 2 is false.

Statement 3: The magnitude of sum of two unit vectors is always greater than the magnitude of their difference.

Let \(\vec{a}=\hat{i}\) and \(\vec{b}=\hat{j}\)

As we can see that, the vectors \(\vec{a}\) and \(\vec{b}\) are two unit vectors

\(\Rightarrow|\hat{i}+\hat{j}|=\sqrt{2}\) and \(|\hat{i}-\hat{j}|=\sqrt{2}\)

\(\Rightarrow|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|\)

So, statement 3 is also false.

We have to find angle between the lines.

Now, let the angle between the given lines 7x - 4y = 0 and 3x - 11y + 5 = 0 is θ

First we need to find the slope of both the lines.

7x - 4y = 0

\(\Rightarrow \mathrm{y}=(\frac{7 }{ 4}) \mathrm{x}\)

Therefore, the slope of the line \(7 x-4 y=0\) is \(m_{1}=\frac{7 }{ 4}\)

Again, \(3 x-11 y+5=0\)

\(\Rightarrow y=(\frac{3 }{ 11}) x+(\frac{5 }{ 11})\)

Therefore, the slope of the line \(3 x-11 y+5=0\) is \(m_{2}=\frac{3 }{ 11}\)

We know that \(\tan \theta=\left|\frac{m_{2}-m_{1}}{1+m_{1} m_{2}}\right|\)

\(\Rightarrow \tan \theta=\left|\frac{\frac{3}{11}-\frac{7}{4}}{1+\frac{3}{11} \times \frac{7}{4}}\right|=\left|\frac{12-77}{44+21}\right|=\left|\frac{-65}{65}\right|=1\)

⇒ tan θ = 1 = tan 45°

∴ θ = 45°

Therefore, the required acute angle between the given lines is 45°.