NDA Mock Test: Mathematics - 3 - NDA MCQ

We have f(x) = ax+b, g(x) = cx+d
Therefore, f{g(x)} = g{f(x)} 
⇔ f(cx+d) = g(ax+b)
⇔ a(cx+d)+b = c(ax+b)+d
⇔ ad+b = cb+d 
⇔ f(d) - g(b)

let y = (4x + 3)/(6x - 4)
⇒ y * (6x - 4) = 4x + 3
⇒ 6xy - 4y = 4x + 3
⇒ 6xy - 4y - 4x = 3
⇒ 6xy - 4x = 3 + 4y   
⇒ x(6y - 4) = 4y + 3
⇒ x = (4y + 3)/(6y - 4)
So, f^-1 (x) = (4x + 3)/(6x - 4)

n(A) = 3
n(A × A × A) = n(A) * n(A) * n(A)
n(A × A × A) = 3 * 3 * 3 = 27

A = {1, 2} and B = {5, 6, 7} 
A x B = {(1, 5), (1, 6), (1, 7), (2, 5), (2, 6), (2, 7)} 

z = a + ib
a = -1, b = -√3
(-1, -√3) lies in third quadrant.
Arg(z) = -π + tan^-1(b/a)
= - π + tan^-1(√3)
= - π + tan^-1(tan π/3)
= - π + π/3
= -2π/3

x = |z| cos θ and y = |z| sin θ satisfies infinite values of θ and for any infinite values of θ is the value of Arg z. Thus, for any unique value of θ that lies in the interval - π < θ ≤ π and satisfies the above equations x = |z| cos θ and y = |z| sin θ is known as the principal value of Arg z or Amp z and it is denoted as arg z or amp z.
 
We know that, cos (2nπ + θ) = cos θ and sin (2nπ + θ) = sin θ (where n = 0, ±1, ±2, ±3, .............), then we get,
 
Amp z = 2nπ + amp z where - π < amp z ≤ π

Let us say that the two particular friends are A and B.
If A is invited among six guests and B is not, then:  number of  combinations to select 5 more guests from the remaining 8 friends:
          C(8, 5) =  8 ! / (5! 3!)  = 56
If B is invited among the six guests and A is not , then the number of ways of selecting the remaining 5 guests =  C(8, 5) =  56
Suppose both A and B are not included in the six guests list : then the number of such combinations =  C(8, 6) = 7 * 8 /2 = 28
So the total number of sets of guests that can be selected =  140.

Mixed doubles includes 2 men and 2 women.
Since 2 men are selected of 6 men
⇒ number of ways = ⁶C₂
Also 2 women are selected of 4 women
⇒ number of ways = ⁴C₂
But they also can be interchanged in 2 ways
∴ Total no. of ways 

Total no of players = 14 out of which 5 are fixed.
So, 11-5 = 6
Remaining players = 14 - 6
= 9 players
9C6 = 9!/(3!*6!)
= 84 

No. of ways to select 4 senior and 3 U-19 players = ⁶C₄ * ⁵C3 = 150
No. of ways to select 5 senior and 2 U-19 players = ⁶C₅ * ⁵C2 = 60
No. of ways to select 6 senior and 1 U-19 players = ⁶C₆ * ⁵C1 = 5 
Total no. of ways to select the team = 150 + 60 + 5 = 215

 n = 44, p = 2y q = -4x
General term of (p+q)n is given by
T(r+1) = ⁿC_r . p^r . q^(n-r)
= ⁴⁴C_r . (2y)^r . (-4x)^(44-r)

x⁴ can be achieved in the following ways: 
x⁴ . 1^(n-4) . (x²)⁰ . (x³)⁰
Hence, coefficient will be  ⁿC₄ .
x² . 1^(n-3) . (x²)¹ . (x3)⁰
Hence, coefficient will be 3ⁿC₃.
x¹ . 1^(n-2) . (x²)⁰ . (x³)¹
Hence, coefficient will be 2ⁿC₂.
x⁰ . 1^(n-2) .(x²)² .(x³)⁰
Hence, coefficient will be ⁿC₂ .
Hence, the required coefficient will be 
ⁿC₄ + 3ⁿC₃ + 3ⁿC₂
= ⁿC₄ + 3(ⁿC₃ + ⁿC₂).
= ⁿC₄  + 3(ⁿ⁺¹C₃).
= ⁿC₄  + ⁿC₂ + ⁿC₁ . ⁿC₂

(1+x)^m (1−x)ⁿ = (1 + mx + m(m−1)x²/2!)(1 − nx + n(n−1)x²/2! ) 
= 1 + (m−n)x + [(n² − n)/2 − mn + (m2 − m)/2] x² 
Given m − n = 3 or n = m − 3
Hence (n² − n)/2 − mn + (m² − m)/2 = 6
⇒ [(m−3)(m−4)]/2 − m(m−3) + (m² − m)/2 = −6
⇒ m2 − 7m + 12 − 2m2 + 6m + m² − m + 12 = 0
⇒ − 2m + 24 = 0
⇒ m = 12.

We know that, the coefficients in a binomial expansion is obtained by replacing each variable by unit in the given expression.
Therefore, sum of the coefficients in (a+b)^n
= 4096=(1+1)ⁿ
⇒ 4096=(2)ⁿ
⇒ (2)¹²=(2)ⁿ
⇒ n=12
Here n is even, so the greatest coefficient is nCn/2  i.e., 12C6

a_n = (n²)/2ⁿ
⇒ a₅ = [(5)²]/2⁽⁵⁾
⇒ a₅ = 25/32

a1 = 2 
a2 = 2a1 + 1
=> 2(2) + 1 = 5
a3 = 2a2 + 1
=> 2(5) + 1 = 11
a4 = 2a3 + 1
=> 2(11) + 1 = 23
Hence, the required series is : 2,5,11,23………

 6 + 6 + 6.........∞
6^1/2 * 6^1/4 * 6^1/8………..∞
= 6^{1/2 + 1/(2*2)} + 1/(2*2*2)   (sum of infinte G.P.= a/(1−r))
= (1/2)/(6^(1-½))
= (1/2)/(6^1/2)
= 6

 (1 + 1/5²) + (1/2 + 1/5²) + (1/2² + 1/54)+......
= (1 + 1/2 + 1/2² +....) + (1/5² + 1/5⁴ + 1/5⁶ +......)
= r = (1/4)/(1/2) , r = (1/5⁴)/(1/5²)
⇒ r = 1/2 + 1/25
⇒ S = 1 + [(1/2)/(1-1/2)],    S = [(1/25)/(1-1/25)]
⇒ S = 1 + 1,    S = 1/24
Total sum = 2 + 1/24
⇒ 49/24 

As A(3,y) and B(2,7) is parallel to C(-1,4) and D(0,6)
∴ Their slopes are equal
so, (y-7)/(3-1) = (4-6)/(-1-0)
y-7 = 2
y = 9

The equation of line which cuts intercepts of equal lengths on the axes is:

Equation of line passing through point (a,b) and parallel to line Ax + By + C = 0
(y-y1) = -A/B(x-x1)
(y-b) = -A/B(x-a)
A (x – a) + B (y – b) = 0

The given point is P(0,0) and the given line is x + y - 2 = 0
Let d be the length of the perpendicular from P(0,0) to the line x + y - 2 = 0
Then,
d = |(1 × 0) + (3 × 0) − 2|/(√12 + 12)
= 2/√2 
= (2/√2) * (√2/√2)
= √2

If p isthe length of the normal from the origin to a line and ωis the angle made by the normal with the positive direction of thex-axis,then the equation of the line is given by xcosω +ysinω= p.
Here, p = 8 units and ω= 60°
Thus, therequired equation of the given line is
xcos 60° + y sin 60° = 8
x(1/2) + y(√3/2) = 8
x/2 + √3y/2 = 8
x + √3y = 16

Since the line is equally inclined the slope of the line should be -1, because it makes 135^o in the positive direction of the X axis
This implies the equation of the line is y= -x +c 
i.e; x+y - c =0
distance of the line from the origin is given as √2
Therefore √2 = |c|/(√1²+1²)
This implies c = 2
Hence the equation of the line is x+y=2

Given Lines:-
L1 : y=mx
L2: y+2x=0
L3 : y=2x+k 
L4 : y+mx=k
To form a rhombus, opposite sides must be parallel to each other.
∵ Slopes of parallel lines are always equal
L1∥L3
L2 ∥ L4
​
Therefore,m = 2

Given the equation is,
2x²+3y²−8x−18y+35=K
Or, 2{x²−4x+4} + 3{y²−6y+9}=K
Or, 2(x−2)² + 3(y−3)² =K.
From the above equation it is clear that if K>0 then the given equation will represent an ellipse and for K<0, no geometrical interpretation.
Also if K=0 then the given equation will be reduced to a point and the point will be (2,3).