Mathematics Test - 25

Given,

Equation of line \(\frac{x-1}{1}=\frac{y+2}{-1}=\frac{z-3}{-3}\) and equation of plane is \(4 x+13 y-3 z+1=\) 0

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

As we can see that, the direction ratios of the given line are: \((1,-1,-3)\)

Similarly, the direction ratios of the normal to the given plane are: \((4,13,-3)\)

\(\Rightarrow 1 \times 4+13 \times(-1)+(-3) \times(-3)=0\)

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 \(Q(1,-2,3)\) lies on the line.

So, finding the distance between the point \(\mathrm{Q}\) 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, \(x_1=1, y_1=-2\) and \(z_1=3\)

\(\Rightarrow d=\left|\frac{4 \times 1+13 \times(-2)-3 \times 3+1}{\sqrt{4^2+(-3)^2+13^2}}\right|=\frac{30}{\sqrt{194}}\) units

Let, us suppose the universal set U = A ∪ B.

Statement 1:

⇒ (A ∩ B) ∪ (A ∩ B̅) ∪ (A̅ ∩ B)

Let us draw a Venn diagram.

OR

⇒ (A ∩ B) ∪ (A ∩ B̅) ∪ (A̅ ∩ B) = {A ∪ (A ∩ B̅)} ∩ {B ∪ (A ∩ B̅)} ∪ (A̅ ∩ B)

⇒ (A ∩ B) ∪ (A ∩ B̅) ∪ (A̅ ∩ B) = {(A ∪ A) ∩ (A ∪ B̅)} ∩ {(A ∪ B) ∩ (B ∪ B̅)} ∪ (A̅ ∩ B)

⇒ (A ∩ B) ∪ (A ∩ B̅) ∪ (A̅ ∩ B) = {A ∩ U} ∩ {U ∩ U} ∪ (A̅ ∩ B)

⇒ (A ∩ B) ∪ (A ∩ B̅) ∪ (A̅ ∩ B) = A ∪ (A̅ ∩ B) = (A ∪ A̅) ∩ (A ∪ B) = A ∪ B

So, statement 1 is right.

Statement 2:

⇒ (A ∪ (A̅ ∩ B̅))

OR

⇒ (A ∪ (A̅ ∩ B̅)) = (A ∪ A̅) ∩ (A ∪ B̅)

⇒ (A ∪ (A̅ ∩ B̅)) = U ∩ (B ∩ A̅) = A̅ ∩ B

Therefore, statement 2 is wrong.

Given,

Total number of cards = 52

Two cards are lost.

So remaining cards = 50

Now one card is drawn.

The probability that it is a diamond card = \(\frac{13}{50}\)

Now probability that both lost cards are heart = \(\frac{13}{50}\times(\frac{{ }^{11} C _{2}}{{ }^{49} C _{2}})\)

\(=\frac{13}{50}×\frac{(\frac{11×10}{2})}{(\frac{49×48}{2})}\)

\(=\frac{13×11}{{5×49×48}}\)

\(=\frac{143}{{11760}}\)

So the probability that both lost cards are heart is \(\frac{143}{{11760}}\).

One of the vectex is intersection of \(x\)-axis and \(x+y+1=0\) 

\(\Rightarrow A(-1,0)\)

Let vertex \(\mathrm{B}\) be \((\alpha,-\alpha-1)\)

Line \(\mathrm{AC} \perp \mathrm{BH} \)

\(\Rightarrow \alpha=1 \)

\(\Rightarrow \mathrm{B}(1,-2)\)

Let vertex \(\mathrm{C}\) be \((\beta, 0)\)

Line \(\mathrm{AH} \perp \mathrm{BC}\)

\(\mathrm{m}_{\mathrm{AH}=-1} \)

\(\frac{1}{2} \cdot \frac{2}{\beta-1}=-1 \)

\(\Rightarrow \beta=0\)

Centroid of \(\triangle \mathrm{ABC}\) is \(\left(0,-\frac{2}{3}\right)\)

Now G(centroid) divides line joining circum centre \((O)\) and ortho centre \((H)\) in the ratio \(1: 2\).

\(2 \mathrm{~h}+1=0,~~  2 \mathrm{k}+1=-\mathrm{z} \) 

\(\mathrm{h}=-\frac{1}{2},~~  \mathrm{k}=-\frac{3}{2}\)

Circum centre is \(\left(-\frac{1}{2},-\frac{3}{2}\right)\).

Equation of circum circle is (passing through \(\mathrm{C}(0,0)\) is \(x^{2}+y^{2}+x+3 y=0\)

As we know that,

The Standard Form of a Quadratic Equation is \(a x^2+b x+c=0\)

Sum of roots \(=\frac{-b}{a}\)

Product of roots \(=\frac{c}{a}\)

Given,

\(x^2-8 x+q=0\) and \(\alpha^2-\beta^2=16\)

Sum of roots \(=a+\beta=\frac{-b}{a}=-(-8)=8\quad..(i)\)

Product of roots \(=\alpha \cdot \beta=q\quad..(ii)\)

We have \(\alpha^2-\beta^2=16\)

\((\alpha+\beta)(\alpha-\beta)=16\)

\(\Rightarrow 8 \times(\alpha-\beta)=16\)

\(\Rightarrow(\alpha-\beta)=2\quad..(iii)\)

We know that, \((\alpha-\beta)^2+4 \cdot \alpha \cdot \beta=(\alpha+\beta)^2\)

Putting the value of equation \( (i) \), \( (ii) \) and \( (iii) \), we get,

\( 2^2+4 q=8^2\)

\(\Rightarrow 4 q=64-4=60\)

\(\Rightarrow q=15\)

\(\therefore\) The value of \(q\) is \(15\).

Given:

25, -125, 625, -3125, …….

Here, first term a = 25 and common ratio r = -5.

As we know that, if a₁, a₂, …., a_n is a GP then the general term is given by:

a_n = a × r^{n - 1} where a is the first term and r is the common ratio.

Thea_nterm is:

a_n = 25 × (-5)^{n - 1} = (-1)^{n - 1} × 5^{n + 1}

Numbers divisible by 4 from 1 to 80 are:

\(4,8,12 \ldots . .80\)

Total no. of numbers divisible by 4 between 1 to 80

\(80=4+(n-1) 4\)

\(80=4 n \)

\(\Rightarrow n=20 \)

\(\therefore \text { Required probability }=\frac{{ }^{20} C_{2}}{{ }^{80} C_{2}}=\frac{19}{316}\)

Let \(\theta\) be the angle between the line and the normal to the plane. 

Converting the given equations into vector form, we have

\(\vec{r}=(-\hat{i}+3 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})\)

Here \( \vec{b}=2 \hat{i}+3 \hat{j}+6 \hat{k}\) and \(\vec{n}=10 \hat{i}+2 \hat{j}-11 \hat{k}\)

\(\sin \phi =\left|\frac{(2 \hat{i}+3 \hat{j}+6 \hat{k}) \cdot(10 \hat{i}+2 \hat{j}-11 \hat{k})}{\sqrt{2^2+3^2+6^2} \sqrt{10^2+2^2+11^2}}\right|\)

\( =\left|\frac{-40}{7 \times 15}\right|=\left|\frac{-8}{21}\right|=\frac{8}{21} \)

\( \phi=\sin ^{-1}\left(\frac{8}{21}\right)\)

The given expression is \(p \vee(\sim p \wedge q)\).

Thus, the negation of this expression can be written as:

\(\sim(p \vee(\sim p \wedge q))\)

\(=\sim p \wedge \sim(\sim p \wedge q)[\because \sim(a \vee b) \equiv \sim a \wedge \sim b]\)

\(=\sim p \wedge(p \vee \sim q)[\because \sim(a \wedge b) \equiv \sim a \vee \sim b]\)

\(=(\sim p \wedge p) \vee(\sim p \wedge \sim q)[\because c \wedge(a \vee b) \equiv(c \wedge a) \vee(c \wedge b)]\)

\(=F \vee(\sim p \wedge \sim q)[\because \sim a \wedge a \equiv F]\)

\(=(\sim p \wedge \sim q)[\because F \vee a \equiv a]\)