# Indian Air Force X Group Mathematics Model Paper PDF 2022

** Indian Air Force X Group Math Model Papers PDF files **

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**AirForce X&Y Group Syllabus**

**Mathematics:-**

- Sets, Relations, functions
- त्रिकोणमिति[Trigonometry(Inverse Trigonometric functions)]
- सम्मिश्र संख्या(Complex Numbers)
- द्विघातीय समीकरण(Quadratic Equations)
- अनुक्रम और श्रृंखला[Sequence & Series (AP & GP)]
- क्रमचय और संचय)Permutation&Combination
- द्विपद प्रमेय(Binomial Theorem)
- Coordinate geometry – straight line, Circles, Parabola, Ellipse, Hyperbola
- Exponential&Logarithmic Series
- आव्यूह और सारणिक(Matrices & Determinants)
- Limit & Continuity
- Differentiation
- Application of Differentiation
- Indefinite&Definite Integrals
- Application of Integration
- Differential Equations
- प्रायिकता(Probability)
- सांख्यिकी(Statistics)

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**Maths Question and Answer**

**Q.1. A student has 6 pants and 7 shuts. The number of ways in which he can wear the dress in different**

**combinations are**

I. 6! x 7!

2. 13

3. 42

4. None of these

**Answer
**Correct Option – 3

**Q. 2.** **From a group of 10 men and 8 women, a team of 3 men or 2 women is to be made. How many teams can be**

**made?**

I. 148

2. 130

3. 150

4. 145

**Answer**

Correct Option – 1

**Q. 3. Find the value of tan (19n / 3)?**

I. ✓3

2. 1 / ✓3

3. 1

4. 0

**Answer**

Correct Option – 1

**Q.4. If sin 0 – cos 0 = 0, and O < 0 < 90°**

, then 0 =

1. 30°

2. 45°

3. 60°

4. 90°

**Answer**

Correct Option – 2

**Q. 5.** **A kite is flying with the string inclined 30° with the horizontal. If the string is 60 m long, then what is the height of the kite above the ground?**

1. 45

2. 30

3. 40

4. 20

**Answer**

Correct Option – 2

** Q. No.1: Determine order and degree (if defined) of differential equation (y′′′)**^{2} + (y″)^{3} + (y′)^{4} + y^{5} = 0

^{2}+ (y″)

^{3}+ (y′)

^{4}+ y

^{5}= 0

**Solution:**

Given differential equation is (y′′′)^{2} + (y″)^{3} + (y′)^{4} + y^{5} = 0

The highest order derivative present in the differential equation is y′′′.

Therefore, its order is 3.

The given differential equation is a polynomial equation in y′′′, y′′, and y′.

The highest power raised to y′′′ is 2.

Hence, its degree is 2.

**Q. No. 2: Verify that the function y = a cos x + b sin x, where, a, b ∈ R is a solution of the differential equation d ^{2}y/dx^{2 }+ y=0.**

**Solution:**

The given function is y = a cos x + b sin x … (1)

Differentiating both sides of equation (1) with respect to x,

dy/dx = – a sinx + b cos x

d^{2}y/dx^{2} = – a cos x – b sinx

LHS = d^{2}y/dx^{2 }+ y

= – a cos x – b sinx + a cos x + b sin x

= 0

= RHS

Hence, the given function is a solution to the given differential equation.

**Q. No. 3: The number of arbitrary constants in the general solution of a differential equation of fourth order is:**

**(A) 0 (B) 2 (C) 3 (D) 4**

**Solution:**

We know that the number of constants in the general solution of a differential equation of order n is equal to its order.

Therefore, the number of constants in the general equation of the fourth-order differential equation is four.

Hence, the correct answer is D.

Note: The number of constants in the general solution of a differential equation of order n is equal to zero.

**Q. No. 4: Form the differential equation representing the family of curves y = a sin (x + b), where a, b are arbitrary constants.**

**Solution:**

Given,

y = a sin (x + b) … (1)

Differentiating both sides of equation (1) with respect to x,

dy/dx = a cos (x + b) … (2)

Differentiating again on both sides with respect to x,

d^{2}y/dx^{2} = – a sin (x + b) … (3)

Eliminating a and b from equations (1), (2) and (3),

d^{2}y/ dx^{2 }+ y = 0 … (4)

The above equation is free from the arbitrary constants a and b.

This the required differential equation.

**Q. No. 5: Find the differential equation of the family of lines through the origin.**

**Solution:**

Let y = mx be the family of lines through the origin.

Therefore, dy/dx = m

Eliminating m, (substituting m = y/x)

y = (dy/dx) . x

or

x. dy/dx – y = 0

**Q. No. 6: Form the differential equation of the family of circles having a centre on y-axis and radius 3 units.**

**Solution:**

The general equation of the family of circles having a centre on the y-axis is x^{2} + (y – b)^{2} = r^{2}

Given the radius of the circle is 3 units.

The differential; equation of the family of circles having a centre on the y-axis and radius 3 units is as below:

x^{2} + (y – b)^{2} = 3^{2}

x^{2} + (y – b)^{2} = 9 ……(i)

Differentiating (i) with respect to x,

2x + 2(y – b).y′ = 0

⇒ (y – b). y′ = -x

⇒ (y – b) = -x/y′ …….(ii)

Substituting (ii) in (i),

x^{2} + (-x/y′)^{2} = 9

⇒ x^{2}[1 + 1/(y′)^{2}] = 9

⇒ x^{2 }[(y′)^{2} + 1) = 9 (y′)^{2}

⇒ (x^{2} – 9) (y′)^{2} + x^{2} = 0

Hence, this is the required differential equation.

**Q. No. 7: Find the general solution of the differential equation dy/dx =1+y ^{2}/1+x^{2}.**

**Solution:**

Given differential equation is dy/dx =1+y^{2}/1+x^{2}

Since 1 + y^{2} ≠ 0, therefore by separating the variables, the given differential equation can be written as:

dy/1+y^{2} = dx/1+x^{2} …….(i)

Integrating equation (i) on both sides,

tan^{-1}y = tan^{-1}x + C

This is the general solution of the given differential equation.

**Q. No. 8: For each of the given differential equation, find a particular solution satisfying the given condition:**

**dy/dx = y tan x ; y = 1 when x = 0**

**Solution:**

dy/dx = y tan x

dy/y = tan x dx

Integrating on both sides,

log y = log (sec x) +C

log y = log (C sec x)

⇒ y = C sec x ……..(i)

Now consider y = 1 when x= 0.

1 = C sec 0

1 = C (1)

C = 1

Substituting C = 1 in (i)

y = sec x

Hence, this is the required particular solution of the given differential equation.

**Q. No. 9: Find the equation of a curve passing through (1, π/4) if the slope of the tangent to the curve at any point P (x, y) is y/x – cos**^{2}(y/x).

^{2}(y/x).

**Solution:**

According to the given condition,

dy/dx = y/x – cos^{2}(y/x) ………….(i)

This is a homogeneous differential equation.

Substituting y = vx in (i),

v + (x) dv/dx = v – cos^{2}v

⇒ (x)dv/dx = – cos^{2}v

⇒ sec^{2}v dv = – dx/x

By integrating on both the sides,

⇒ ∫sec^{2}v dv = – ∫dx/x

⇒ tan v = – log x + c

⇒ tan (y/x) + log x = c ……….(ii)

Substituting x = 1 and y = π/4,

⇒ tan (π/4) + log 1 = c

⇒ 1 + 0 = c

⇒ c = 1

Substituting c = 1 in (ii),

tan (y/x) + log x = 1

**Q. No. 10: Integrating factor of the differential equation (1 – x ^{2})dy/dx – xy = 1 is**

**(A) -x**

**(B) x/ (1 + x ^{2})**

**(C) √(1- x ^{2})**

**(D) ½ log (1 – x ^{2})**

**Solution:**

Given differential equation is (1 – x^{2})dy/dx – xy = 1

(1 – x^{2})dy/dx = 1 + xy

dy/dx = (1/1 – x^{2}) + (x/1 – x^{2})y

dy/dx – (x/1 – x^{2})y = 1/1-x^{2}

This is of the form dy/dx + Py = Q

We can get the integrating factor as below:

Let 1 – x^{2} = t

Differentiating with respect to x

-2x dx = dt

-x dx = dt/2

Now,

I.F = **√**t = √(1- x^{2})

Hence, option C is the correct answer.

**Download notes and pdf**

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