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Distance of B from origin

\(\sqrt{(-3-0)^2+(3\sqrt{3}-0)^2}\)

= \(\sqrt{(-3)^2+(-3\sqrt{3})^2}\) = \(\sqrt{9+27}\) = \(\sqrt{36}\) = 6

Distance of C from origin = \(\sqrt{(-3)^2+(-3\sqrt{3})^2}\) = \(\sqrt{9+27}\) = \(\sqrt{36}\) = 6

B(–3, 3√3) and C(–3, –3√3) are on the circle.

AB = \(\sqrt{(5-2)^2+(4-3)^2}\)

= \(\sqrt{3^2+1^2}\) = \(\sqrt{9+1}=\sqrt{10}\)

BC = \(\sqrt{(6-5)^2+(7-4)^2}=\sqrt{10}\)

CD = \(\sqrt{(5-6)^2+(6-7)^2}=\sqrt{10}\)

AD = \(\sqrt{(3-2)^2+(6-3)^2}\)

= \(\sqrt{1^2+3^2}=\sqrt{1+9}=\sqrt{10}\)

Answer is (4, 1), (4, 2)

CD = |5– (–3)| = 8

∴ CD = 8 unit

EF = 8 unit

Angle of a regular hexagon = 720 ÷ 6 = 120°

D is the origin coordinates is (0, 0) (1) G divides AD in the ratio 2 : 1

∴ Coordinates of G is (0,1)

Triangle ADB is a right triangle with 30°, 60°, 90

∴ BD = \(\frac{3}{\sqrt{3}}\) = \(\sqrt{3}\)

∴ Coordinates of B is (–√3, 0)

∴ Coordinates of C is (√3, 0)

A(4, 3), B(-4, 3)

Hence other points on AB = (-2, 3) (1, 3)

Answer is (1, 5), (2, 5)

Answer is (4, 1) (4, 5)

Let, A(2, 1), B(3, 4), C(-3, 6) be the vertices of the triangle ABC.

AB = \(\sqrt{(3-2)^2+(4-1)^2}\) = \(\sqrt{1^2+3^2}=\sqrt{10}\)

BC = \(\sqrt{(-3-3)^2+(6-4)^2}\) = \(\sqrt{6^2+2^2}=\sqrt{40}\)

AC = \(\sqrt{(2-3)^2+(1-6)^2}\) = \(\sqrt{5^2+(-5)^2}=\sqrt{50}\)

Using Pythogoras theorem AC² = AB² + BC² 

50 = (√10)² + (√40)² = 50 

∴ We obtain a right triangle.

a. B(0, 2), C(–2, 0), D(0, –2) Coordinate of vertex (0,0)

b. i. (5, 0), (–2, 0), (6, 0)

ii. (0, –2), (0, 8)

iii. (2, 3), (10, 9)

AB = \(\sqrt{(-4-4)^2+(3-3)^2}\) = \(\sqrt{-8^2}\) = \(\sqrt{64}\) = 8

As the angles are in the ratio 30 : 60: 90 the 

sides will be in the ratio 1: √3: 2 

that is A(√3, 1), B(–1, √3)

a. (10, 0), (0, 10); (–10, 0), (0, –10)

b. (9, 6); ie, all the point which satisfies.

x² + y² = 10² are on the circle, others are inside or outside the circle

Coordinates of the mid point of AB = \((\frac{4+(-4)}{2},3)\) = (0,3)

Coordinates of the mid point of AB = \((\frac{2+6}{2},1)\) = (4,1)

Answer is BC, parallel to y-axis.

The distance of a point P(– x, y) from the origin is √(x²+y²)

Assume that (x, y) is centre of the circle Distance between (9,3) and (x,y)

= \(\sqrt{(x-9)^2+(y-3)^2}\)

Distance between (7,-1) and \((x,y)\) = \(\sqrt{(x+7)^2+(y-1)^2}\)

They are equal

\(\therefore \sqrt{(x-7)^2+(y-1)^2}\) = \(\sqrt{(x-1)^2+(y-1)^2}\)

\((x-7)^2=(x-1)^2\)

\(x^2-14x+49=x^2-2x+1\)

\(\sqrt{(x-9)^2+(y-3)^2}\) = \(\sqrt{(x-1)^2+(y-1)^2}\)

when x = 4

(4 - 9)² + y² - 6y + 9 = (4 - 7)² + y² + 2y + 1

25 + y² - 6y + 9 = 9 + y² + 2y + 1

24 = 8y

\(y=\frac{24}{8}=3\)

The coordinate of the centre of the circle is (4, 3)

∴ Radius = Distance between (4, 3) and (1,-1)

= \(\sqrt{(4-1)^2+(3+1)^2}\) = \(\sqrt{3^2+4^2}=5\)

Answer is AB, parallel to x-axis.

i. y-coordinate of any point on the x-axis is zero.

ii. x-coordinate of any point on the y-axis is zero.

iii. Origin is (0, 0).

iv. y = 1

v. x = 1

The value of abscissa of point (–4, 0) situated at Cartesian plane -4.

7 sec θ – 7 tan θ  = 7.

If 3θ = 90°, then the value of tanθ is  1/√2 .

(b) b² = 4ac

If we are known with √5 is irrational than it can be proved as:

Let 3 - √5 be a rational number

 3 - √5 = p/q [ where p and q are integer , q ≠ 0 and q and p are co-prime number ]

=> √5 =  3 - p/q

=> √5 = (3q - p)/q

We know that number of form p/q is a rational number.

So, √5 is also a rational number.

But we know that √5 is irrational number. This contradicts our assumption.

Therefore, 3 - √5 is an irrational number.

Correct option:

(A) –3 

(B) (AB)^–1 = B^–1A^–1 

Correct option:

(D) 1

Longest chord of circle is Diameter.

Concept:

The standard form of a linear equation of the first order is given by:

 \(\frac{{dy}}{{dx}} + Py = Q\) 

where P, Q are arbitrary functions of x.

The integrating factor of the linear equation is given by:

\(I.F. = {e^{\smallint pdx}}\)

The solution of the linear equation is given by:

\(y\left( {I.F.} \right) = \smallint Q\left( {I.F.} \right)dx + c.\)

Calculation:

\(\left( {y + {x^2}} \right)\frac{{dx}}{{dy}} = x\)

\(x\frac{{dy}}{{dx}} = {x^2} + y\)

\(\frac{{dy}}{{dx}} - \frac{y}{x} = x\)

It is a form of \(\frac{{dy}}{{dx}} + Py = Q\)

\(I.F. = {e^{\smallint pdx}}\)

\(I.F. = {e^{ln\frac{1}{x}}} = \frac{1}{x}\)

The solution of the linear equation is given by

\(x\left( {I.F.} \right) = \smallint Q\left( {I.F.} \right)dx + c.\)

\(y\left( {\frac{1}{x}} \right) = \smallint x\left( {\frac{1}{x}} \right)dx + c\)

\(\frac{y}{x} = x + c\)

Concept:

The standard form of a linear equation of the first order is given by \(\frac{{dy}}{{dx}} + Py = Q\) where P, Q are arbitrary function of x.

The integrating factor of the linear equation is given by \(I.F. = {e^{\smallint pdx}}\)

The solution of the linear equation is given by \(y\left( {I.F.} \right) = \smallint Q\left( {I.F.} \right)dx + c.\)

Calculation:

\(\frac{{dx}}{{dy}} + \frac{x}{{co{s^2}y}} = {e^{ - \tan y}}\)

It is the linear equation of the form of \(\frac{{dx}}{{dy}} + Px = Q\)

\(I.F. = {e^{\smallint pdy}}\)

\(I.F. = {e^{\smallint se{c^2}ydy}} = {e^{\tan y}}\)

The solution of the linear equation is given by

\(x\left( {I.F.} \right) = \smallint Q\left( {I.F.} \right)dy + c\)

\(x{e^{\tan y}} = \smallint {e^{\tan y}}.{e^{ - \tan y}}dy + c\)

\(x{e^{tany}} = y + c\)

Put x = 0 and y = 0 in the above equation we get, c = 0

\(x = y{e^{ - \tan y}}\)

Put y = π / 4 in the above equation we get

\(x = \frac{\pi }{4}{e^{ - tan\frac{\pi }{4}}}\)

CONCEPT :

If roots are real and different then complimentary solution is given by F(x) = c1 em x + c2 en x

CALCULATION : 

 Differential equation  f"(x) = f(x)

 f"(x) - f(x) = 0

 [D2-1] f(x) = 0

 The corresponding auxiliary equation D2 - 1 = 0

 Root of auxiliary equation D2 = 1

⇒ D = \(±\)1

Here, m = 1 and n = - 1

Roots are real and different. So complimentary solution is given by

⇒ f(x) = c1 ex + c2 e-x

Put initial condition f(0) = 2 in the above equation we get,

⇒ 2 = c1 e0 + c2 e0

⇒ 2 = c1 + c2                                            -----(1)

∵ f(x) = c1 ex + c2 e-x

 ⇒ f`(x) = c1 ex - c2 e-x

 Put initial condition f`(0) = 3 in the above equation

⇒ 3 = c1 - c2                                    -----(2)

By adding equation (1) and equation (2) we get,

⇒ 2c1 =5  ⇒ c1 = 5/2   

Put value of c₁ in equation (1)

⇒ 2 = 5/2 + c2 ⇒c 2 = - 1/2

Solution of the given differential equation f(x) = \(\frac{5}{2}\)ex - \(\frac{1}{2}\)e-x

 \(\Rightarrow f(x) = \frac {(5e^{2x} - 1)}{2e^x}\)

So, the value of f(4)

 \(\Rightarrow f(4) = \frac {(5e^8 - 1)}{2e^4}\)

Hence, option B is the correct answer. 

Question change √5 is irrational.

Assume contrary that √5 is a rational number.

Then √5 = a/b, where a and b are co-prime and b ≠ 0.

∴ HCF (a, b) = 1

⇒ a = √5 b

Now, a² = 5b² (By squaring both sides)

⇒ 5 divides a²

⇒ 5 divides a (∵ If any prime number divides a² then that prime number must divide a)

⇒ a = 5m

⇒ a² = 25 m²

⇒ 25 m² = 5b² (∵ a² = 5b²)

⇒ b² = 5 m²

⇒ 5 divides b²

⇒ 5 divides b

Since, 5 divides both a and b.

Therefore, HCF (a, b) ≠ 1, it is a multiple of 5.

Which is contradicts the fact that a and b are co-prime numbers.

Therefore, our assumption is wrong.

i.e., √5 is an irrational number.

In mathematics, composite numbers also known as composites in Mathematics are numbers that have more than 2 factors, not like prime numbers that have only one factor, i.e. 1 and the number itself. Composite numbers are all natural numbers that are not prime numbers since they can be divided by more than two numbers.

Therefore

The smallest composite number is 4. The smallest composite number of 2 digits is 10.
4 is a composite number because there are 2 variables. One is 1 × 4 = 4 and the other is 2 × 2= 4.

There are 16 even numbers between 20 and 50, including the 20 and 50. (Excluding 20 and 50,there are 14 even numbers.)

• The numbers which are divisible by 2 are popularly known as the even numbers.

• Now, we just need to find the numbers between the 20 and 50,which are divisible by 2.

• The numbers will be 20,22,24,26,28,30,32,34,36,38,40,42,44,46,48,50.

• By counting the (including 20 and 50) numbers we get that there are a total of 16 even numbers and excluding 20,50 there will be 14 even numbers.

We have to find the smallest number by which 720 should be multiplied to get a perfect square.

Here 720 = 2 x 2 x 2 x 2 x 3 x 3 x 5

We have pairs of 2 and 3 but there is only one 5

Now if we multiply 720 by 5 then we get three complete pairs hence the resultant number will be perfect square.

Hence 720 x 5 = (2 x 2) x (2 x 2) x (3 x 3) x (5 x 5)

∵ \(\sqrt{720\times5}\) = 2 x 2 x 3 x 5 = 60