Indian-Airforce-AFCAT Aptitude-Analytical Contributed by AFCAT-Aptitude updated on Aug 2019
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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 sin2 B/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) = ex3 - 3x + 2 is

(a) One-one into   (Ans)

(b) one-one onto

(c) many-one onto

(d) many-one into

Solution : Given, f(x) = ex3 - 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,  ex 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(Cc Ç Hc Ç Bc) = 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 :  x R 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