Placement Paper

Indian Air force model question paper for practice,Indian air Force sample placement papers,Indian Air Force mathematics and all section wise questions with answers

1.  The expression (1 + tan x + tan² x) (1 - cot x + cot² x) has the positive values for x, given by

(a) 0 ≤ x ≤ p

(b) 0 ≤ x ≤ p/2

(c) ∀ x Î R    (Ans)

(d) x ≥ 0

Solution : Now, (1 + tan x + tan² x) (1 - cot x + cot² x)

= (1 + tan x + tan² x) (1 + tan² x - tan x) / tan² x

= (1 + tan² x)² - tan² x / tan² x

Since,  (1 + tan² x)² > tan² x ∀ x Î R

Hence, given expression is always positive.

2.  The period of the function f(x) = sin px/2 + 2cos  px/3 - tan  px/4 is

(a) 3

(b) 4

(c) 6

(d) 12   (Ans)

Solution :  Given f(x) =  sin px/2 + 2cos  px/3 - tan  px/4

Period of sin   px/2  =  2p/p/2 =  4

Period of cos   px/3  =  2p/p/3  = 6

Period of tan   px/4 = p/p/4  =  4

 ∴ Period of  f(x) = LCM (4, 6, 4) = 12

3.  If in a ∆ ABC cos A + 2 cos B + cos C = 2, then a, b, c are in

(a) HP

(b) GP

(c) AP   (Ans)

(d) AGP

Solution :  Given,  cos A + 2 cos B + cos C = 2

⇒   cos A + cos C = 2 (1 - cos B)

⇒  2 cos (A + C / 2) cos (A - C / 2) = 4 sin²B/2

⇒  2 cos (A - C / 2) = 4 sin B/2

⇒  2 cos B/2 cos(A - C / 2) = 2 (2 sin B/2 cos B/2)

⇒  2 sin (A + C / 2) cos (A - C / 2) = 2 sin B

⇒  sin A +sin C  = 2 sin B

⇒  a + c = 2b    (by sine rule's)

4.  The set of values of a such that a² - 6 sin x - 5a ≤ 0, ∀ x is

(a) [-1, 6]

(b) [2, 3]   (Ans)

(c) (-1, 6)

(d) None of these

Solution : Given, a² - 6 sin x - 5a ≤ 0

⇒  a² - 5a / 6  ≤ sin x

⇒  a² - 5a / 6  ≤ - 1

⇒  a² - 5a + 6 ≤ 0

⇒  (a - 2) (a - 3) ≤ 0

⇒ a Î [2, 3] .

5.  cos 170⁰. cos 150⁰. cos 130⁰. cos 110⁰ is equal to

(a) 1/16

(b) 3/16  (Ans)

(c) 5/16

(d) 7/16

Solution : cos 170⁰. cos 150⁰. cos 130⁰. cos 110⁰

=  cos 10⁰. cos 30⁰. cos 50⁰. cos 70⁰

=  √3/4  cos 50⁰ (cos 80⁰ + cos 60⁰)

=  √3/8 (cos 130⁰ + cos 30⁰ + cos 50⁰) = 3/16

6.  A flag staff stands in the centre of a rectangular field whose diagonal is 1200 m and subtends angles 15⁰ and 45⁰ at the mid points of the sides of the field.  The height of the flag staff is

(a) 200 m

(b) 300 √2 + √3  m

(c) 300 √2 - √3  m   (Ans)

(d) 400 m

Solution : Let OP be the flag staff of height h standing at the centre O of the field.

image - Practice Paper - Mathematics - Ans No. 06

In  ∆ OEP,  OE =  h cot 15⁰ = h (2 + √3)

and in  ∆ OFP, OF = h cot 45⁰ = h

∴       EF   =  h √1 + (2 + √3)²

                =  2h √2 + √3

⇒      2EF  =  4h √2 + √3

⇒     1200 =   4h √2 + √3            [∵ 2EF  = BD]

⇒          h=  300/√2 + √3

               =  300 √2 - √3  m

7.  The domain of definition of the function f(x) = 3e ^√x2-1  log (x - 1) is

(a) (-∞, -1) È (1, ∞)

(b) [-1, ∞]

(c) (1, ∞)  (Ans)

(d) None of these

Solution :   Given, f(x) = 3e ^√x2-1  log (x - 1)

For f(x) to be defined

             x² - 1 ≥ 0 and x - 1 > 0

⇒     - 1≤ x  or  x ≥ 1 and x > 1

⇒   x Î (1, ∞)

8.  The function f :  (-∞, -1]  → (0, e⁵] defined by f(x) = e^{x3 - 3x + 2} is

(a) One-one into   (Ans)

(b) one-one onto

(c) many-one onto

(d) many-one into

Solution : Given, f(x) = e^{x3 - 3x + 2}

Let   g(x)  =  x³ - 3x + 2

⇒    g'(x)  = 3x² - 3 = 3 (x² - 1)

For  xÎ (-∞, -1], g'(x) ≥ 0

∴   g(x)  is an increasing function.

We know as x is increasing,  e^x is increasing. 

∴    f(x)is an increasing function.

∴    f(x) is one-one function.

For  xÎ(-∞, -1], range of f(x) is (0, e⁴)

But it is given (0, e⁵),therefore it is into function.

9.  Out of 800 boys in a school, 224 played in cricket, 240 played hockey and 336 played basketball, of the total 64 played both basketball and 40 played cricket and hockey, 24 played all the three games.  The number of boys who did not play any game is

(a) 128

(b) 160  (Ans)

(c) 216

(d) 240

Solution : Given n(C) = 224,   n(H) = 240,    n(B) = 336,   

n(B Ç H) = 64, n(H Ç C) = 40, n(C Ç H Ç B) = 24

∴    n(C^c Ç H^c Ç B^c) = n[(C È H È B)^c]

       = n (U) - n(C È H È B)

       =800 - [n(C) +  n(H)  + n(B)  -  n(H Ç C) - n(H Ç B) - n(C Ç B) + n(C Ç H Ç B)]

       = 800 - [224 + 240 + 336 - 64 - 80 - 40 + 24]

       = 800 - 640 = 160

10.  The relation R defined on the set N of natural numbers by x Ry ⇔ 2x² - 3xy + y² = 0 is

(a) symmetric but not reflexive

(b) only symmetric

(c) not symmetric but reflexive   (Ans)

(d) None of the above

Solution :  xR x ⇔ 2x² - 3x . x + x² = 0

⇒  0 = 0

∴  R is reflexive.

(ii)  At  x = 1,

2(1)² - 3(1) . y + y² = 0

⇒   y² - 3y + 2 = 0

⇒   c1, 2

At    x = 2,

        2(2)² - 3(2)y + y² = 0

⇒   y² - 6y + 8 = 0

⇒   y = 4, 2

∴  1 R 2 ≠  2 R 1

∴  R is not symmetric.

11.  The root of the equation 1 + z + z³ + z⁴ = 0 are represented by the vertices of

(a) a square

(b) an equilateral triangle    (Ans)

(c) a rhombus

(d) None of the above

Solution :  Given, 1 + z + z³ + z⁴ = 0

⇒  1(1 + z) + z³ (1 + z) = 0

⇒  (1 + z)(1 + z³) = 0

⇒  z = - 1, z³ = - 1

⇒  z = - 1,  z = - 1,  1 ± √3 i / 2

⇒  z = - 1,  - w², -w

Let  A(z) = - 1, B(z) = - w² and C(z) = - w

Now, AB = |-1 + w²| = |w²(w - 1)|

= |w²||w - 1| = |w - 1|

BC = |w² - w| = |w||w - 1| = |w - 1|

CA = |w - 1|

⇒  AB  =  BC  =  CA

12.  If f(z) = 7 - z / 1- z², where z = 1 + 2i, then |f(z)| is equal to

(a) (|z|) / 2   (Ans)

(b) |z|

(c) 2|z|

(d) None of these

Solution :  Given,  f(z) = 7 - z / 1- z², where z = 1 + 2i

Now, |z| = √1 + 4 = √5

∴  f(z) = 7 - (1 + 2i) / 1 - (1 + 2i)²

= 6 - 2i / 1 - (1 + 4 + 4i)

= 3 - i / -2 - 2i

⇒  |f(z)| = |3 - i / 2 + 2i| = √9 + 1 / √4 + 4

= √10 / 2√2  =  √5/2  = |z| / 2

13.  The number of solutions of |[x] - 2x| = 4,  where [x] is the greatest integer ≤ x is

(a) 1

(b) 2

(c) 3

(d) 4   (Ans)

Solution : Given, |[x] - 2x| = 4

If  x = nÎI

∴   |n - 2n| = 4  ⇒  |n| = 4

 ⇒ n = ± 4

If x = n + k, nÎI  and 0 < k < 1

then |n - 2 (n + k)| = 4

⇒ |- n - 2 k| = 4.

It is possible,  if  k = 1/2.  Then,

|- n - 1| = 4  ie  n + 1  = ± 4.

∴   n = 3, - 5

Hence, number of solutions are 4

14.  If both the roots of the equation x² - ax + 2 = 0 lie in the interval (0, 3),then a lies in

(a) (-1, -3)

(b) (-2, 7)

(c) (-11/3,  -2/√2)

(d) None of these    (Ans)

Solution :  Given,  x² - ax + 2 = 0

(i)    D = a²  - 4(2) > 0

⇒    a < - 2 √2  or  a > 2 √2

(ii)   - B² / 2A = a/2 Î (0, 3)

⇒    a Î (0, 6)

(iii)  f(0) = 2 > 0

       and f(3) = 9 - 3a + 2 > 0

⇒    a < 11/3

From Eqs.  (i),(ii) and (iii),we get 2√2 < a < 11 / 3

15.  If the roots of the equation  1/ x + p  +  1/ x + q  = 1/r are equal in magnitude but opposite in sign, then the product of the roots will be 

(a) p²  +  q²/ 2

(b)  - (p²  +  q²)/ 2   (Ans)

(c) p²  -  q²/ 2

(d)  - (p²  -q²)/ 2

Solution : Given equation can be rewritten as

x² + x (p + q - 2r) + (pq - pr - qr) = 0        ........(i)

Let roots are a and - a.

∴  Sum, a - a  = - p + q - 2r/ 1

⇒    r =  p + q / 2

Product a (-a) = pq - p + q / 2  (p + q)

= - 1/2 {(p + q)²- 2pq}

= - (p² + q²) / 2