NDA Mock Test: Mathematics - 5 - NDA MCQ

f(x) = 3x + 1,  g(x) = x² + 2 
f-g = (3x+1) - (x² + 2)
= 3x + 1 - x² - 2
= 3x - x² -1

f(g(2)) compare f(gx) x=2
on comparing g(x)=x, g(2)=2
f(g(x)) = f(2) = x² = 2²= 4

• We have, A^c in X = The set of elements in X which are not in A = 0,1
• {0} ∈ A^c w.r.t. X is false, because {0} is not an element of A^c in X.
• ϕ ∈ A^c in X is false because ϕ is not an element of A^c in X.
• {0} ⊂ A^c in X is correct because the only element of {0} namely 0 also belong to A^c in X.
• 0 ⊂ A^c in X is false because 0 is not a set.

• P contains {0, 1, 2, 3, 4} and Q contain natural numbers which start with 1 and P = {0, 1, 2, 3, 4} Q = {1, 2, 3}
• Here P isn't a subset of Q because all the elements of P are not in Q.

A = {1, 2, 3, 4} 
B = {1, 3, 5}
(a, b) ∈ element of R ⇔ a < b for all a ∈ A, b ∈ B
(a, b) pairs satisfying the condition of R are:
(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)
So, 
R = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)}

 R be the relation in the set {1, 2,3, 4] given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}
it is seen that (a, a) ∈ R for every a ∈ {1, 2, 3, 4}
so,R is reflexive.
it is seen that (a, b) = (b, a) ∈ R 
because, (1, 2)∈ R but (2, 1) ∉ R
so, R is not symmetric.
it is seen that (a, b), (b, c) ∈ R ⇒ (a, c) ∈ R for all a, b, c ∈ {1, 2, 3, 4}.
so, R is transitive.
Hence, R is reflexive and transitive but not symmetric.

∵ a₁, a₂ ,a₃ , ....... are in G.P..
∴ Using a_n = ar ^{n -1} ,we get the given determinant, as

Operating C₃ - C₂ and C₂-C₁ and using

= 0 (two columns being identical)

Hence, AB = BA only when a = b
∴ There can be infinitely many B's for which AB = BA

Hence, D is divisible by both x and y

R₁ ------> R₁ + R₂, taking (a+b+c) common from the first row, we get the resultant matrix.

(126)1^1/3
=  (125 + 1)^1/3
=  [125 (1 + (1/125))]^1/3
=  125^1/3(1 + (1/125))^1/3         1/125 < 1
= 5 [1 + (1/3)(1/125) + ..........]
= 5 [1 + (1/3)(0.008)]
= 5 [1 + 0.002666]
= 5.013

The letter of word COCHIN in alphabetic order are C, C, H, I, N, O.
Fixing first letter C and keeping C at second place, rest 4 can be arranged in 4! ways.
Similarly the words starting with CH, CI, CN are 4! in each case.
Then fixing first two letters as CO next four places when filled in alphabetic order give  the word COCHIN.
∴ Numbers of words coming before COCHIN are
4 × 4! = 4 × 24 = 96

We h ave to for m 7 digit n umber s, using th e digits 1, 2 and 3 only, such that the sum of the digits in a number = 10.
This can be done by taking 2, 2, 2, 1, 1, 1, 1, or by taking 2, 3, 1, 1, 1, 1, 1.
∴ Number of ways

∵ Each person gets at least one ball.
∴ 3 Per son s can have 5 balls in th e followin g systems

or

The number of ways to distribute the balls in first system =  
⁵C₁ x ⁴C₁ x ³C₃

Also 3, persons having 1, 1 and 3 balls can be arranged in   ways.

∴ No. of ways to distribute 1, 1, 3 balls to the three persons

⁵C₁ x ⁴C₁ x ³C₃ x

Similarly the total no. of ways to distribute 1, 2, 2 balls to the three persons = ⁵C₁ x ⁴C₂ x ²C₂ x

∴ The required number of ways = 60 + 90 = 150

⇒ a,b,c are in AP
⇒ sin A, sin B, sin C are in AP

Where d be a common difference of an AP.

Let d be the common difference of AP.


On adding all of these, we get

On putting the value of d in Eq. (i), we get

5x – 4y + 3 = 0
 5x – 4y = -3 
(5/4)x + (-4/5)y = -3
Line is parallel m = 5/4
Points (2,5)
= 5 = (5/4)(2) + b
5 = 5/2 + b
b = 5/2
y = 5x/4 - 5/2
4y = 5x - 10
5x - 4y + 10 = 0

When we form the equation it would be like this:
√3x + y + 2 = 0
There is another way to write this equation: √3x − y = 2
When we divide both sides by 2 we will get this
√3x/2 − 1/2y = 1
Cosθ = √3/2 and sinθ = 1/2, and p=1
We know that both cosθ and sinθ are negatives, so we will get this:
θ = π + π/6
= 7π/6

 x + 2y – 3 = 0,(1)
 2x + 3y – 4 = 0,(2)
 3x + 4y – 5 = 0,(3) 
4x + 5y – 6 = 0,(4)
 Solving equation (1) and (2), we get
 3x + 6y − 9 = 0 [Multiplying (1) by 3] 
3x + 4y − 7 = 0 
 ⇒ 2y − 2 = 0  
 y = 1 
Putting value of y in (1), we get 
x + 2 − 3 = 0 
x = 1 
The point (1, 1) lies on 3x + 4y − 7 = 0 but not on 2x + 3y − 4 = 0 and 4x + 5y − 6 = 0.
Hence, they are neither concurrent, nor they can form a quadrilateral nor parallel.

4(640 + x) ≤ 8 × 640 + 2x ≤ 6(640 + x)

2560 + 4x ≤ 5120 + 2x ≤ 3840 + 6x

2560 + 4x ≤ 5120 + 2x

⇒ 2x ≤ 2560

⇒ x ≤ 1280 ...(i)

5120 + 2x ≤ 3840 + 6x

⇒ – 4x ≤ –1280

⇒ 320 ≤ x ...(ii)

From (i) and (ii),

320 ≤ x ≤ 1280

F = 68°F

C = 5 / 9 (F – 32)

= 5 / 9 (68 – 32) = 20°C

ax²+by²+2hxy+2gx+2fy+c = 0
In the circle equation,
Coefficients of x² & y² are the same and non-zero.
Therefore, a = b ≠ 0
There is no term for xy. Therefore, h = 0.
Radius of the circle: r² = g² + f² – c,
So, g² + f² – c > 0