NDA Mock Test: Mathematics - 2 - NDA MCQ

Clearly, mappings given in options (a),(b) and (c ) satisfy the given conditions and are one-one onto.

Let X, Y be sets, and let f : X → Y be a function. We say that f is injective (sometimes called one-to-one) if ∀x₁, x₂ ∈ X, f(x₁) = f(x₂) 
⇒ x₁ = x₂.

X is a natural number and x < 5, the number is 1, 2, 3, 4
y = 2x

► x(1), y = 2 × 1 = 2
► x(2), y = 2 × 2 = 4
► x(3), y = 2 × 3 = 6
► x(4), y = 2 × 4 = 8

Range = {2, 4, 6, 8}

A = {1, 2, 3, 5}
R = {(x, y): x + y >10: x, y ∈ A}
For any value of x, y ∈ A, the condition does not hold.
Therefore, R = Φ (Empty set)

By R₃ → R₃ – (xR₁ + R₂);

= (ax² + 2bx + c)(b² – ac) = (+)(–) = -ve.

For homogeneous system of equations to have non zero solution, Δ = 0

On simplification, 

∴ a,b,c are  in Harmonic Progression.

 a(-c) - a(b-c) + c(c) = 0
 - ac - ab + ac + c² = 0
c² = ab

The determinant of a square matrix is the same as the determinant of its transpose.

 A ={(1,2) (4,3)} B = {(3,2) (-1,1)}
AB= {[(1*3)+(2*(-1)) (1*2)+(2*1)] [(4*3)+(3 *(-1)) (4*2)+(3*1)]} 
= {(1,4) (9,11)}

If A is a symmetric matrix, then B'AB is a symmetric metrix. So, B'AB is a symmetric matrix.

z̅  = x - iy 
Real Part of = Real Part of = Real Part of
⇒Real Part of

⇒ x² - x - y² + y = 2x² + 2y² - 4y + 2
⇒ x² + 3y² + x - 5y + 2 = 0
The above equation does not represent a circle or Straight line.
So No Option is Correct.
Option (2) is correct only when we use Z in place of Z̅ in the numerator of 

Total flags = 4
Different colour = 3
red = 4
Total no. of signals = 7P3
= 7!/(4!)
= 210

(n + 1)! = 20 (n – 1)!
n (n + 1) = 20
(n – 4) (n + 5) = 0          
Since, (n – 1)! exists, n ≥ 1
So, n = 4 

(x+a)ⁿ = ⁿC₀ xⁿ + ⁿC₁ x^(n-1) a¹ + ⁿC₂ x^(n-2) a² + ..........+ ⁿC_(n-1) xa^(n-1) + ⁿC_n  aⁿ
Now, ⁿC₀ = ⁿC_n, ⁿC₁ = ⁿC_n-1,    ⁿC₂ = ⁿC_n-2,........
therefore, ⁿC_r = ⁿC_n-r
The binomial coefficients equidistant from the beginning and the end in the expansion of (x+a)n are equal.

The expression depicts the ratio of two numbers, if the ratio between the numbers is constant, then it will definately form a GP.

g_n = 6( 3^n-1) it is a geometric expression with coefficient of constant as 3^n-1.So it is GP with common ratio 3.

Let AP be a, a + d, a + 2d, .....
∴ T₃ = a + 2d = 7 ....(i)
& T₇ = a + 6d = 2 + 3 (7) = 23 .....(ii)
On solving (i) & (ii), we get
d = 4 and a = (–1)

x₁, x₂, x₃ and y₁, y₂, y₃ are in GP with same common ratio,
∴ (x₁,y₁) ⇒ p(x₁,y₁)
(x₂,y₂) ⇒ Q(x₁r,y₁r)
(x₃,y₃) ⇒ R(x₁r²,y₁​r²)

∴ P, Q, R are collivear

Slope of line passing through (a,b) and (−a,−b) is given by (b+b)/(a+a) = b/a
So equation of line passing is given by (using slope point form)
y−b = b/a(x−a)
⇒ ay − ab = bx − ab
⇒ ay = bx
Clearly the point (a²,ab) lie on the above line

This question can be done by picking the options.
a₁/a₂ = b₁/b₂ = c₁/c₂
Equation : 2x + y = 3
(d) 4x + 2y = 6
a₁ = 2⇒, b₁ = 1, c₁ = 3
a₂ = 4, b₂ = 2, c₂ = 6
⇒ 2/4 = 1/2 = 3/6
=> 1/2 = 1/2 = 1/2

We know that, If p is the length of the normal from origin to a line and w is the angle made by the normal with the positive direction of x-axis then the equation of the line is given by xcosw + ysinw = p.
Here, p=2a units and w=60°
Thus, the required equation of the given line is 
xcos60° + ysin60° = 2a
x(1/2) + y(√3/2) = 2a
x + √3y = 4a

(7x-5)/(8x+3) > 4
(7x-5)/(8x+3) - 4 >0
7x - 5 - 4 ( 8x + 3 ) / 8x + 3 > 0
- 25 x - 17 / 8x + 3 > 0
Now furthermore solving for general range :
x ∈ ( -17/ 25, - 3/8)

|4 − x| + 1 < 3
⇒ 4 − x + 1 < 3
Add −4 and −1 on both sides, we get
4 − x + 1 − 4 − 1 < 3 − 4 − 1
⇒ − x < −2
Multiply both sides by −1, we get
x > 2
Also,|4−x| + 1 < 3
⇒ −(4−x) + 1 < 3
⇒ − 4 + x + 1 < 3
Add 4 and −1 on both sides, we get
− 4 + x + 1 + 4 − 1 < 3 + 4 − 1
⇒ x < 6
Thus, x ∈ (2,6).

y = x is tangent to the parabola
y=ax²+c
if a= then c=?
y′ =2ax
y’ = 2(7/2)x  =1
x = 1/7
1/7 = 2(1/7)² + c
c = 1/7 * 2/49
c = 7/2

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).

The circle intercept the co-ordinate axes at a and b. it means x - intercept at ( a, 0) and y-intercept at (0, b) .
Now, we observed that circle passes through points (0, 0) , (a, 0) and (0, b) .
we also know, General equation of circle is
x² + y² + 2gx + 2fy + C = 0
when point (0,0)
(0)² + (0)² + 2g(0) + 2f(0) + C = 0
0 + 0 + 0 + 0 + C = 0
C = 0 -------(1)
when point (a,0)
(a)² + (0)² + 2g(a) + 2f(0) + C = 0
a² + 2ag + C = 0
from equation (1)
a² + 2ag = 0
a(a + 2g) = 0
g = -a/2
when point ( 0, b)
(0)² + (b)² + 2g(0) + 2f(b) + C = 0
b² + 2fb + C = 0
f = -b/2
Now, equation of circle is
x² + y² + 2x(-a/2) + 2y(-b/2) + 0 = 0 { after putting values of g, f and C }
x² + y² - ax - by = 0
As we know that, a=2, b=4
x^2 + y^2 - 2x - 4y = 0