NDA Mock Test: Mathematics - 6 - NDA MCQ

A = {(a, b) ∈ R × R : |a − 5| < 1 and |b − 5| < 1}

⇒ a ∈ (4, 6) and b ∈ (4, 6)

Therefore, A = {(a, b): a ∈ (4, 6) and b ∈ (4, 6)}

Now, B = {a, b) ∈ R × R : 4 (a − 6) ² + 9(b−5) ² ≤ 36}

From the conditions above for set A, the maximum value of B is:

4 (a − 6) ² + 9(b−5) ² = 4 (4 − 6)² + 9(6−5)² = 25 ≤ 36

We can check for other values as well.

Elements of set A satisfy the conditions in Set B.

Hence, A is a subset of B.

A = {2, 3, 4, 8, 10}
B = {3, 4, 5, 10, 12}
C = {4, 5, 6, 12, 14}
► (A ∩ B) = {3,4,10}
► (A ∩ C) = {4}
► (A ∩ B) U (A ∩ C) = {3,4,10}

y is defined  If -x ≥ 0, i.e.

By definition, The Signum function =  

The given function is defined as Signum function. i.e.

domain is R and Range is {-1 , 0 , 1}.

f(a+1)−f(a−1)
=(a+1)−(a+1)2−{(a−1)−(a−1)2} = 2−4a(a−1)2} = 2−4a

A minor, Mij, of the element aij is the determinant of the matrix obtained by deleting the ith row and jth column.

Δ = Sum of products of element of row(or column)  with their corresponding co-factors.
Δ = a11 A11 + a21 A21 + a31 A31

Concept:
If a number ω = a + ib is purely imaginary, then

• a = 0 
• and ω =

Calculation:
Given, z is a complex number such that is purely imaginary,
Let = a + ib, then a = 0
And

As per question,

⇒ n(n - 1) (n + 1 -n + 2)= 126
⇒ n (n – 1) = 42 ⇒ n (n – 1) = 7 × 6  ⇒ n = 7.

If we see the blocks in terms of lines then there are 2m vertical lines and 2n horizontal lines. To form the required rectangle we must select two horizontal lines, one even numbered (out of 2, 4, .....2n) and one odd numbered (out of 1, 3....2n–1) and similarly two vertical lines. The number of rectangles is
^mC₁ .  ^mC₁ . ⁿC₁ . ⁿC₁ = ^m2n₂

Lets first place the men (M). '*' here indicates the linker of round table
 
* M -M - M - M - M *
which is in (5-1)! ways
So we have to place the women in between the men which is on the 5 empty seats ( 4 -'s and 1 linker i.e * )
So 5 women can sit on 5 seats in (5)! ways or
1st seat in 5 ways
2nd seat 4
3rd seat 3
4th seat 2
5th seat 1
i.e 5*4*3*2*1 ways
So the answer is 5! * 4! = 2880

Total no of balls = 9
red balls = 4
yellow balls = 3
green balls = 2
Total no. of arrangements = 9!/(4!*3!*2!)
= 1260

T_r+1 = ²ⁿCr(x)^r (1)^2n-r
T_r+3 = ²ⁿC_r+2 (x)^r+2 (1)^2n-r-2
T_r+1 : T_r+3 
= ²ⁿC_r = ²ⁿC_r+2
=> 2n!/(2n-r)!r! = 2n!/(r+2)!(2n-r-2)!
=> 1/(2n-r)(2n-r-1) = 1/(r+2)(r+1)
=> (r+2)(r+1) = (2n-r)(2n-r-1)
=> r2 + 3r + 2 = 4n2 - 2nr - 2n - 2nr + r^2 + r
=> 0 = 4n2 - 4nr - 2n + r - 3r - 2
=> 0 = 4n2 - 4nr - 2n + r - 3r - 2 - 2 + 2
=> 0 = 4n2 - 4nr - 4 - [2n + 2r - 2]
=> 4n(n-r-1) -2(n-r-1)
Therefore n - r - 1 = 0
=> n = r+1

Given that, T_m = 1/n
⇒ a + (m - 1) d = 1/n ......(i)
and T_n = 1/m
⇒ a + (n - 1)d = 1/m  ......(ii)
On solving Eqs. (i) and (ii), we get
a = d = 1/mn
⇒ a - d = 0

Given that,

 .........(ii)

Now, a,b,c are in AP.

⇒ -a, -b, -c are in AP.
⇒ 1 - a, 1 - b, 1 - c are  also in AP.
 are in AP.
⇒ x, y,z are in HP.

√3x + y - 8
√3x + y = 8
Dividing by √[(√3)2 + (1)2]
= √[3+1]
= √4
= 2
√3x/2 + y/2 = 8/2
√3x/2 + y/2 = 4
x(√3/2) + y(1/2) = 4.....(1)
Normal form of any line : xcos w + ysin w = p....(2)
Comparing (1) and (2)
p = 4

• IQ = (MA*100)/CA
• and 80 ≤ IQ ≤ 140
• 80 ≤ (MA*100)/CA ≤ 140
• 80 ≤ (MA*100)/12 ≤ 140
• 960 ≤ (MA*100) ≤ 1680
• 9.6 ≤ MA ≤ 16.8

Numbers obtained in three throws are 3, 5 and 6.
Let the number obtained in fourth throw be x.
Now, Sum > 20
=> 5 + 3 + 6 + x > 20
=> 14 + x > 20
=> x > 20 – 14
=> x > 6 (no one wins)

and substitute (7, 1)

Given directrix of parabola ⇒ x+y=2
and force is origin vertex is A(0,0)
We know that perpendicular distance from vertex  of the parabola to directrix is equal to 'a'  where 4a is the Latus Rectum of the 
Parabola 

x² − 2 = −2cost
⇒ x² = 2 − 2cost
⇒x² = 2(1−cost)
⇒x² = 2(1−(1−2sin² t/2))
⇒x² = 4sin² t/2
We have y = 4cos² t/2
⇒cos² t/2= y/4
We know the identity, sin² t/2 + cos² t/2 = 1
⇒ x²/4 + y/4 = 1
⇒ x² = 4−y represents a parabolic profile.

Since the major axis is along the y-axis.
∴ Equation of tangent is x = my + [b²m² + a]^1/2
Slope of tangent = 1/m = −4/3    
⇒ m = −3/4
Hence, equation of tangent is 4x+3y=24 or  
x/6 + y/8 = 1
Its intercepts on the axes are 6 and 8.
Area (ΔAOB) = 1/2×6×8
= 24 sq. unit

Equation of normal at P (θ) is

ax+bycosecθ = (a² +b²)secθ  .......(1)
And at Q (φ) is
ax + bycoseφ = (a² +b²) secφ
⇒ ax +bysecθ = (a² +b²)cosecθ   ......(2)
From (1) and (2) 

Equation of normal at any point (x₁, y₁) is  it passes through 
P (h, k) then  thus (x₁, y₁) lie on the conic