NDA Mock Test: Mathematics - 4 - NDA MCQ

Correct Answer : a

Explanation :  A = {(x,y) : x² + y² = 5}

=> (1,2) {(1)² + (4)² = 5}

=> {1 + 4 = 5}

=> {5 = 5}

Since T is a set of real number for it's given ordered pairs we have a condition provided that is 1 + ab is less than zero. so in the given option c if we put order of a and b then it satisfy our given condition

_xR_y => x - y + √2 is an irrational number.
Let R is a binary relation on real numbers x and y.
Clearly, R is reflexive relation
As _xR_x iff  x – x +√2 = √2 ,which is an irrational number.
Here R is not symmteric if we take x =√2  and y =1 then x – y + √2 is an irrational
number but y – x + √2 = 1, which is not irrational number
Now, R is transitive iff for all (x, y) ∈ R and (y, z) ∈ R implies (x, z) ∈ R
But here R is not transitive as we take x = 1, y = 2√2, z=√2
Given, _xR_y => x - y + √2 is irrational    ............1
and _yR_z => y - z + √2 is irrational       ............2
Add equation 1 and 2, we get
(x - y + √2) + (y - z + √2)
= x - z + √2  = 1, which is not an irrational

If A = [a_ij]_2x2 where a_ij = i + j, then, 

Let x' and y' be the reflection of x and y, therefore :



Hence, reflection is on the line - x-y = 0⇒ x + y = 0

This system of equations has a unique solution, if

A = {(6,7) (8,9)}
|A| = (6 * 9) - (8 * 7)
= 54 - 56 
|A| = -2
A^-1 = -½{(9,-7) (-8,6)}
A^-1 = {(-9/2, 7/2) (4,-3)}

2x² + 3ix + 2 = 0
Using quadratic equation;
we know, x = (-b ± √b² - 4ac)/2a
x =  [-3i ± √(3i)² - 4x2x2]/2x2
= -3i ± √-25/4
= i(-3±5)/4
x = i/2, -2i

ⁿC_{r -1} = 36, ⁿC_r = 84,ⁿC_r+1= 126
KEY CONCEPT : We know that

⇒    ....(1)

⇒ 

⇒ 2n – 5r – 3 = 0 ....(2)

Solving (1) and (2), we get n = 9 and r = 3.

Total number of words that can be formed using 5 letters out of 10 given different letters
= 10 × 10 × 10 × 10 × 10 (as letters can repeat)
= 1, 00, 000
Number of words that can be formed using 5 different letters out of 10 different letters
= ¹⁰P₅ (none can repeat)

∴ Number of words in which at least one letter is repeated
= total words–words with none of the letters repeated
= 1,00,000 – 30,240  
= 69760

 Largest coefficient in the expansion of (1+x)²⁴ 
= ²⁴C_24/2 
= ²⁴C₁₂

we can write (6ⁿ ) = (1 + 5)ⁿ
we know, according to binomial theorem,
(1 + x)ⁿ = 1 + nx + n(n-1)x²/2! + n(n-1)(n-2)x³/3! +.............∞ use this here,
(6)ⁿ = (1 + 5)ⁿ = 1 + 5n + n(n-1)5²/2! + n(n-1)(n-2)5³/3! +...........∞
= 1 + 5n + 5²{ n(n-1)/2! + n(n-1)(n-2)5/3! +.......∞}
Let P = n(n-1)/2! + n(n-1)(n-2)5/3! +.........∞
6ⁿ = 1 + 5n + 25P
6ⁿ - 5n = 1 + 25P -------(1)
but we know, according to Euclid algorithm ,
dividend = divisor × quotient + remainder ---(2)
compare eqn (1) to (2)
we observed that 6ⁿ -5 n always leaves the remainder 1 when divided by 25

 (1+a)^m+n
Coefficient of am = m+nCm = m+n
Coefficient of aⁿ = m+nCn
nCx = nC(n−x) Property
Hence, coefficients of a^m and aⁿ are equal.

 Since a,b,c are in H.P, so b = 2ac/(a+c).
Also, b,c,d are in H.P, so c = 2bd/(b+d).
Therefore, (a+c)(b+d) = 2ac/b × 2bd/c
⇒ab+cb+ad+cd = 4ad
⇒ab+bc+cd = 3ad

The 2-digit number which when divided by 3 gives remainder 1 are: 10, 13, 16, ...97
Here a = 10, d = 13 - 10 = 3
t_n = 97
nth term of an AP is t_n = a + (n – 1)d
97 = 10 + (n – 1)³
⇒ 97 = 10 + 3n – 3
⇒ 97 = 7 + 3n
⇒ 3n = 97 – 7 = 90
∴ n = 90/3 = 30
Recall sum of n terms of AP, 
= 15[20 + 87] = 15 × 107 = 1605

Let the slope of the line AB and AC are m₁ and m₂  respectively.

Then, 

Let θ be the angle between AB and BC. Then,

Slope = Change in y coordinates÷change in x coodinates
= (5-1)÷(2-x) =2
5-1 = 4-2x
0 = 2x
x=0

Let, ax + by + c = 0 (b ≠ 0) be the equation of the given straight line.
Now, convert the equation ax + by + c = 0 to its slope-intercept form.
ax + by+ c = 0
⇒ by = - ax - c
Dividing both sides by b, [b ≠ 0] we get,    y =  -a/b x - cb, which is the slope-intercept form.
Now comparing the above equation to slope-intercept form (y = mx + b) we get,
The slope of the line ax + by + c = 0 is (- a/b).
Since the required line is parallel to the given line, the slope of the required line is also (- ab).
Let k (an arbitrary constant) be the intercept of the required straight line. Then the equation of the straight line is
y = - a/b x + k
⇒ by = - ax + bk        
⇒ ax +  by = -K, Where K = bk = another arbitrary constant.

The midpoint of the line segment is (1+1/2, 3-7/2)
= (1,-2)
the equation of the line parallel to the line 2x-3y = 1 is of the form 2x-3y = k
since it passes through (1,-2)
2(1) - 3(-2) = k
k = 8
hence the required equation is 2x-3y=8

xcos 30° + y sin 30° = 3
⇒ √3x + y - 6 = 0

3x + 5y = 10, at origin
But now, it’s (2,2),
3(x-2) + 5(y-2) = 10
Hence, 3x - 6 + 5y - 10 = 10
3x + 5y = 26
So, 3x + 5y = p
=> p = 26.

T = 30 + 25 (x – 3), 3 < x < 15
It is given 200 < T < 300
200 < 30 + 25 (x – 3) < 300
170 < 25(x - 3) < 270
170/25  (x - 3) < 270/25
6.8 + 3 < x < 10.8 + 3
9.8 < x < 13.8

Let x litres of 4% boric acid solution is required to be added
Then, total mixture = (x + 750)litres
This resulting mixture is to be  more than 5% but less than 8% boric acid
Total amount of acid = 750 of 10% + x of 4%
ATQ 
5/100(750 + x) < (750*10 + 4x)/100 < 8/100(750 + x)
3750 + 5x < 7500 + 4x ; 7500 + 4x < 6000 + 8x
x < 3750  ;   x > 375 
375 and 3750

Given equation of circles are
x²+y²+6x+6y=0....(i)
and x²+y²−12x−12y=0....(ii)
Here, g1 = 3,f2 = 3, g2 = −6 and f₂ = −6
∴ Centres of circles are C₁(−3,−3) and C₂(6,6) respectively and radii are r₁ = 3√2 and r₂ = 6√2 respectively.
Now, C₁C₂ = √[(6+3)² + (6+3)²]
= 9√2
and r₁ + r₂
​= 3(2)^1/2 + 6(2)^1/2
= 9(2)^1/2
​⇒ C₁C₂ = r₁ + r₂
∴ Both circles touch each other externally.

 Parabola is x² – 8x + 2y + 7 = 0 
∴   (x – 4)² = – 2y – 7 + 16 
∴   (x – 4)² = – 2[y – (9/2)]
 ∴   x² = – 4ay  
 ⇒ x = x – 4, y = y – (9/2) and 2 = 4a
 i.e. a = (1/2) 
 Its focus is given by x = 0 and y = 0 i.e. x – 4 = 0   and     y – (9/3) = 0 
∴    x = 4    and y = (9/2) 
∴ focus [4, (9/2)].