# Indian Air Force X Group Mathematics Model Paper PDF 2022

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

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### Mathematics:-

• Sets, Relations, functions
• त्रिकोणमिति[Trigonometry(Inverse Trigonometric functions)]
• सम्मिश्र संख्या(Complex Numbers)
• 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)

#### Other Competitive Exams PDF

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
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
I. 148
2. 130
3. 150
4. 145
Correct Option – 1

Q. 3. Find the value of tan (19n / 3)?
I. ✓3
2. 1 / ✓3
3. 1
4. 0
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
Correct Option – 2

#### Q. No.1: Determine order and degree (if defined) of differential equation (y′′′)2 + (y″)3 + (y′)4 + y5 = 0

Solution:

Given differential equation is (y′′′)2 + (y″)3 + (y′)4 + y5 = 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 d2y/dx+ 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

d2y/dx2 = – a cos x – b sinx

LHS = d2y/dx+ 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,

d2y/dx2 = – a sin (x + b) … (3)

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

d2y/ dx+ 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 x2 + (y – b)2 = r2

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:

x2 + (y – b)2 = 32

x2 + (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),

x2 + (-x/y′)2 = 9

⇒ x2[1 + 1/(y′)2] = 9

⇒ x[(y′)2 + 1) = 9 (y′)2

⇒ (x2 – 9) (y′)2 + x2 = 0

Hence, this is the required differential equation.

Q. No. 7: Find the general solution of the differential equation dy/dx =1+y2/1+x2.

Solution:

Given differential equation is dy/dx =1+y2/1+x2

Since 1 + y2 ≠ 0, therefore by separating the variables, the given differential equation can be written as:

dy/1+y2 = dx/1+x2 …….(i)

Integrating equation (i) on both sides,

tan-1y = tan-1x + 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 – cos2(y/x).

Solution:

According to the given condition,

dy/dx = y/x – cos2(y/x) ………….(i)

This is a homogeneous differential equation.

Substituting y = vx in (i),

v + (x) dv/dx = v – cos2v

⇒ (x)dv/dx = – cos2v

⇒ sec2v dv = – dx/x

By integrating on both the sides,

⇒ ∫sec2v 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 – x2)dy/dx – xy = 1 is

(A) -x

(B) x/ (1 + x2)

(C) √(1- x2)

(D) ½ log (1 – x2)

Solution:

Given differential equation is (1 – x2)dy/dx – xy = 1

(1 – x2)dy/dx = 1 + xy

dy/dx = (1/1 – x2) + (x/1 – x2)y

dy/dx – (x/1 – x2)y = 1/1-x2

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

We can get the integrating factor as below:

Let 1 – x2 = t

Differentiating with respect to x

-2x dx = dt

-x dx = dt/2

Now,

I.F = t = √(1- x2)

Hence, option C is the correct answer.

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