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NDA Mock Test: Mathematics - 9 - NDA MCQ


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30 Questions MCQ Test NDA (National Defence Academy) Mock Test Series 2024 - NDA Mock Test: Mathematics - 9

NDA Mock Test: Mathematics - 9 for NDA 2024 is part of NDA (National Defence Academy) Mock Test Series 2024 preparation. The NDA Mock Test: Mathematics - 9 questions and answers have been prepared according to the NDA exam syllabus.The NDA Mock Test: Mathematics - 9 MCQs are made for NDA 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for NDA Mock Test: Mathematics - 9 below.
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NDA Mock Test: Mathematics - 9 - Question 1

Directions : Read the following information carefully and answer the questions given below.


=

Q. 

(α+ β) is equal to

NDA Mock Test: Mathematics - 9 - Question 2

Suppose ω and ω2 are the complex cube root of unity which are given as

Q. 

(1 – ω) (1 – ω2) (1 + ω4) (1 + ω8) is equal to

Detailed Solution for NDA Mock Test: Mathematics - 9 - Question 2

Correct Answer :- d

Explanation : (1-w)(1-w2)(1+w4)(1+w8)......(1)

w4 = w3 * w

w4 = 1*w = w.......(2)

w8 = w3 * w3 * w2

= w2.........(3)

Put the values of (2) and (3) in eq (1)

(1-w)(1-w2)(1+w)(1+w2)

(1-w)(1+w)(1-w2)(1+w2)

=> (1-w2) (1-w)

From eq(2) w4 = w

1 - w - w2 + w3 

As we know that 1 + w + w^2 = 0

w2 + w = -1……..(4)

Putting in eq(4)

= 1 + 1 + 1 

= 3

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NDA Mock Test: Mathematics - 9 - Question 3

Suppose ω and ω2 are the complex cube root of unity which are given as

Q. 

The value of expression

NDA Mock Test: Mathematics - 9 - Question 4

What is the value of   where n

NDA Mock Test: Mathematics - 9 - Question 5

If 2x = 3 + 5i, then what is the 2x3 + 2x2 – 7x + 72?

Detailed Solution for NDA Mock Test: Mathematics - 9 - Question 5

NDA Mock Test: Mathematics - 9 - Question 6

285 is summation of 3 numbers. Ratio between 2nd and 3rd numbers is 6:5. Ratio between 1st and 2nd numbers is 3:7. The 3rdnumber is?

Detailed Solution for NDA Mock Test: Mathematics - 9 - Question 6
Given:
- The sum of 3 numbers is 285.
- The ratio between the 2nd and 3rd numbers is 6:5.
- The ratio between the 1st and 2nd numbers is 3:7.
To find:
The value of the 3rd number.

Let's assume the 1st number is 3x and the 2nd number is 7x (according to the given ratio).
The ratio between the 2nd and 3rd numbers is 6:5, so we can write it as:
(7x)/(5y) = 6/5
Cross-multiplying, we get:
35x = 6y
Similarly, the sum of the 3 numbers is given as 285, so we can write it as:
3x + 7x + 5y = 285
Combining the equations, we have:
3x + 7x + (35x/6) = 285
Simplifying the equation, we get:
(18x + 42x + 35x)/6 = 285
(95x)/6 = 285
95x = 1710
x = 1710/95
x = 18
Now, we can find the value of the 3rd number (5y) using the ratio between the 2nd and 3rd numbers:
(7x)/(5y) = 6/5
(7*18)/(5y) = 6/5
126/5y = 6/5
Cross-multiplying, we get:
30y = 630
y = 630/30
y = 21
Therefore, the value of the 3rd number is 5y = 5*21 = 105.
Answer:
The 3rd number is 105.
NDA Mock Test: Mathematics - 9 - Question 7

If α and β are the roots of the equation x2 + x + 1 = 0, then which of the following are the roots of the equation x2 – x + 1 = 0? 

Detailed Solution for NDA Mock Test: Mathematics - 9 - Question 7

To find the roots of the equation x2 - x + 1 = 0, we need to find the values of x that satisfy this equation.
Given that α and β are the roots of the equation x2 + x + 1 = 0, we can use the fact that the sum of roots of a quadratic equation is equal to the negation of the coefficient of x divided by the coefficient of x2.
The sum of the roots of x2 + x + 1 = 0 is -1/1 = -1.
Therefore, the sum of the roots of x2 - x + 1 = 0 is also -1.
Since the sum of the roots remains the same, it means that the roots of x2 - x + 1 = 0 are not α and β.
Hence, the answer is D: None of these.
NDA Mock Test: Mathematics - 9 - Question 8

If the equation x2 – bx + 1 = 0 does not possess real roots, then which one of the following is correct? 

Detailed Solution for NDA Mock Test: Mathematics - 9 - Question 8

NDA Mock Test: Mathematics - 9 - Question 9

Consider the equation (x – p) (x – 6) + 1 = 0 having integral coefficients. If the equation has integral roots, then what values can p have?

Detailed Solution for NDA Mock Test: Mathematics - 9 - Question 9

To find the values of p, we need to consider the equation (x – p)(x – 6) + 1 = 0.
Expanding the equation, we get x^2 - (6+p)x + 6p + 1 = 0.
Since the equation has integral coefficients, the discriminant of the quadratic equation must be a perfect square.
The discriminant is given by D = (6+p)^2 - 4(6p + 1).
Simplifying, we have D = 36 + 12p + p^2 - 24p - 4.
Combining like terms, we get D = p^2 - 12p + 32.
To find the values of p, we need to find the perfect squares that can be represented by p^2 - 12p + 32.
The perfect squares between 32 and 64 are 36 and 49.
Therefore, the possible values of p are:
- p = 6: (6)^2 - 12(6) + 32 = 0, which is not a perfect square.
- p = 8: (8)^2 - 12(8) + 32 = 64 - 96 + 32 = 0, which is a perfect square.
So the answer is p = 8.
Therefore, the correct option is A: 4 or 8.
NDA Mock Test: Mathematics - 9 - Question 10

The roots of the equation (x – p) (x – q) = r2, where p, q and r are real, are 

Detailed Solution for NDA Mock Test: Mathematics - 9 - Question 10

To find the roots of the equation (x - p)(x - q) = r^2, we can expand the equation and set it equal to zero:
(x - p)(x - q) = r^2
x^2 - qx - px + pq = r^2
x^2 - (p + q)x + pq - r^2 = 0
We can now use the quadratic formula to find the roots of the equation:
x = (-(-p - q) ± √((-p - q)^2 - 4(pq - r^2))) / 2
x = (p + q ± √((p + q)^2 - 4(pq - r^2))) / 2
x = (p + q ± √(p^2 + 2pq + q^2 - 4pq + 4r^2)) / 2
x = (p + q ± √(p^2 - 2pq + q^2 + 4r^2)) / 2
x = (p + q ± √((p - q)^2 + 4r^2)) / 2
From the quadratic formula, we can see that the roots of the equation will be real if the discriminant ((p - q)^2 + 4r^2) is greater than or equal to zero. Let's analyze the different cases:
1. If r^2 > 0:
- The discriminant ((p - q)^2 + 4r^2) will always be greater than zero, as the sum of two positive numbers is always positive.
- Therefore, the roots will always be real.
2. If r^2 = 0:
- The discriminant ((p - q)^2 + 4r^2) simplifies to (p - q)^2.
- The roots will be real if (p - q)^2 is greater than or equal to zero, which is always true.
- Therefore, the roots will always be real.
3. If r^2 < 0:
- The discriminant ((p - q)^2 + 4r^2) will always be negative, as the sum of a positive number and a negative number is always negative.
- Therefore, the roots will always be complex.
Based on the analysis of the different cases, we can conclude that the roots of the equation (x - p)(x - q) = r^2 will always be real (option B).
NDA Mock Test: Mathematics - 9 - Question 11

is equal to

NDA Mock Test: Mathematics - 9 - Question 12

If A and B are two matrices such that AB = B and BA = A, then A2 + B2 is equal to

Detailed Solution for NDA Mock Test: Mathematics - 9 - Question 12

2 + B2 = AA + BB
=A(BA) + B(AB)     (Given AB = B and BA = A)
=(AB)A + (BA)B
=BA + AB
=A + B

NDA Mock Test: Mathematics - 9 - Question 13

 then F(α) . F(β) is equal to

NDA Mock Test: Mathematics - 9 - Question 14

If A and B are two matrices of same order, then

Q. 

(AB)n = AnBn is / (AB)n = AnBn

Detailed Solution for NDA Mock Test: Mathematics - 9 - Question 14

To determine whether the statement (AB)n = AnBn is true, we need to consider different cases:
Case 1: AB = BA (i.e., A and B commute)
In this case, the statement (AB)n = AnBn is true. This can be proved using mathematical induction.
Case 2: AB ≠ BA (i.e., A and B do not commute)
In this case, the statement (AB)n = AnBn is not true. This can be shown by providing a counterexample.
Counterexample:
Let A = [[1, 0], [0, 2]] and B = [[2, 0], [0, 3]]. Here, AB ≠ BA.
If we calculate (AB)2 and A2B2, we get:
(AB)2 = [[4, 0], [0, 18]]
A2B2 = [[4, 0], [0, 36]]
Since (AB)2 ≠ A2B2, the statement (AB)n = AnBn is false when AB ≠ BA.
Therefore, the correct answer is C: true only when AB = BA.
NDA Mock Test: Mathematics - 9 - Question 15

If A and B are two matrices of same order, then

Q. 

If A and B are symmetric matrices, then (ABA)T is

Detailed Solution for NDA Mock Test: Mathematics - 9 - Question 15

Given:
- A and B are symmetric matrices of the same order.
To find:
- The nature of the matrix (ABA)T.
Explanation:
- A symmetric matrix is a matrix that is equal to its transpose. In other words, the elements of a symmetric matrix remain the same when they are reflected along the main diagonal.
- The transpose of a matrix is obtained by interchanging its rows with columns.
- A matrix multiplied by its transpose is always a symmetric matrix.
Now, let's analyze the given expression (ABA)T:
1. First, perform the multiplication ABA:
- A * B = C, where C is a matrix of the same order as A and B.
- C * A = D, where D is also a matrix of the same order.
2. Take the transpose of D: DT.
3. As mentioned earlier, the product of a matrix and its transpose is always a symmetric matrix.
- Therefore, (ABA)T is a symmetric matrix.
Hence, the correct answer is Option A: symmetric matrix.
NDA Mock Test: Mathematics - 9 - Question 16

The marked price of a watch is Rs. 800. A shopkeeper gives two successive discounts and sells the watch at Rs. 612. If the first discount is 10%, the second discount is

Detailed Solution for NDA Mock Test: Mathematics - 9 - Question 16

NDA Mock Test: Mathematics - 9 - Question 17

If each element in a row of a determinant is multiplied by the same factor r, then the value of the determinant

Detailed Solution for NDA Mock Test: Mathematics - 9 - Question 17
If each element in a row of a determinant is multiplied by the same factor r, then the value of the determinant will be multiplied by r.
Explanation:
To understand why the answer is option D, let's first review the concept of determinants:
- A determinant is a scalar value that can be calculated for a square matrix.
- The value of a determinant is used to determine various properties of the matrix, such as whether it is invertible or singular.
- The determinant of a matrix can be calculated by expanding along any row or column using cofactor expansion.
Now, let's consider the scenario where each element in a row of a determinant is multiplied by the same factor r. The key point to note here is that we are multiplying the elements of a single row, not the entire determinant.
When we multiply each element in a row by r, the determinant can be expanded along that row. The cofactors of the expanded determinant will have the same row, except for the elements that were multiplied by r. These elements will now be multiplied by r as well.
Here's a step-by-step explanation:
1. Let's assume the determinant is of size n x n, and we are multiplying the elements of the ith row by r.
2. The cofactor expansion of the determinant along the ith row will give us a sum of terms, where each term is a product of an element from the ith row, its cofactor, and the determinant of the remaining (n-1) x (n-1) matrix.
3. Since each element in the ith row has been multiplied by r, all the terms in the cofactor expansion will include a factor of r.
4. Therefore, when we simplify the expanded determinant, we can factor out r from each term.
5. The remaining cofactors and the determinant of the remaining (n-1) x (n-1) matrix will be unchanged.
6. Thus, the value of the determinant will be multiplied by r.
In conclusion, when each element in a row of a determinant is multiplied by the same factor r, the value of the determinant is multiplied by r. Therefore, the answer is option D.
NDA Mock Test: Mathematics - 9 - Question 18

What is the value of the determinant 

Detailed Solution for NDA Mock Test: Mathematics - 9 - Question 18



NDA Mock Test: Mathematics - 9 - Question 19

What is the value of  where ω is the cube root of unity?

NDA Mock Test: Mathematics - 9 - Question 20

NDA Mock Test: Mathematics - 9 - Question 21

Detailed Solution for NDA Mock Test: Mathematics - 9 - Question 21

(cotx+cosecx)-1/(cotx-cosecx+1)
=(cosecx+cotx)-(cosec²x-cot²x)/(cotx-cosecx+1)
=(cosecx+cotx)[1-(cosecx-cotx)]/(cotx-cosecx+1)
=(cosecx+cotx)(cotx-cosecx+1)/(cotx-cosecx+1)
=(cosecx+cotx)
=(1/sinx+cosx/sinx)
=1+cosx/sinx

NDA Mock Test: Mathematics - 9 - Question 22

What is the value of sin A cos A tan A + cos A sin A cot A? 

Detailed Solution for NDA Mock Test: Mathematics - 9 - Question 22

To find the value of sin A cos A tan A cos A sin A cot A, we can simplify each term step by step.
1. sin A cos A:
- The product of sin A and cos A can be simplified using the identity sin 2A = 2sin A cos A.
- Therefore, sin A cos A = 1/2 * sin 2A.
2. tan A cos A:
- The product of tan A and cos A can be simplified using the identity tan 2A = 2tan A / (1 - tan^2 A).
- Therefore, tan A cos A = 1/2 * tan 2A.
3. cos A sin A:
- The product of cos A and sin A is the same as sin A cos A.
- Therefore, cos A sin A = 1/2 * sin 2A.
4. cos A sin A cot A:
- The product of cos A sin A and cot A can be simplified using the identity cot 2A = (cot^2 A - 1)/2cot A.
- Therefore, cos A sin A cot A = 1/2 * (cot^2 A - 1).
Now, let's combine all the simplified terms:
sin A cos A tan A cos A sin A cot A
= (1/2 * sin 2A) * (1/2 * tan 2A) * (1/2 * sin 2A) * (1/2 * (cot^2 A - 1))
= 1/16 * sin 2A * tan 2A * sin 2A * (cot^2 A - 1)
= 1/16 * sin^2 2A * (cot^2 A - 1)
Finally, we can simplify the expression further:
sin^2 2A = 1 - cos^2 2A (using the Pythagorean identity)
= 1 - (1 - sin^2 2A)
= sin^2 2A
Therefore, the expression becomes:
1/16 * sin^2 2A * (cot^2 A - 1)
= 1/16 * sin^2 2A * (1/cos^2 A - 1)
= 1/16 * sin^2 2A * (1 - cos^2 A) / cos^2 A
= 1/16 * sin^2 2A * sin^2 A / cos^2 A
= 1/16 * sin^4 A / cos^2 A
= (1/16) * (sin^4 A / cos^2 A)
Using the identity sin^2 A = 1 - cos^2 A, we can simplify further:
(1/16) * (sin^4 A / cos^2 A)
= (1/16) * ((1 - cos^2 A)^2 / cos^2 A)
= (1/16) * (1 - 2cos^2 A + cos^4 A) / cos^2 A
= (1/16) * (1/cos^2 A - 2 + cos^2 A)
= (1/16) * (sec^2 A - 2
NDA Mock Test: Mathematics - 9 - Question 23

If cos A + cos B = m and sin A + sin B = n, where m, n ≠ 0, then what is sin (A + B) equal to?

Detailed Solution for NDA Mock Test: Mathematics - 9 - Question 23

Given:
- cos A cos B = m
- sin A sin B = n
- m, n ≠ 0
To find: sin (A B)
Step 1: Rewrite the given equations using the trigonometric identities:
- cos A cos B = m ----- (1)
=> cos A cos B = (1/2)(2m)
=> cos(A + B) + cos(A - B) = 2m ----- (2)
- sin A sin B = n ----- (3)
=> -cos(A + B) + cos(A - B) = 2n ----- (4)
Step 2: Add equations (2) and (4):
- 2cos(A - B) = 2m + 2n
=> cos(A - B) = m + n ----- (5)
Step 3: Rewrite equation (5) using the trigonometric identity:
- sin^2(A - B) = 1 - cos^2(A - B)
=> sin^2(A - B) = 1 - (m + n)^2
=> sin^2(A - B) = 1 - (m^2 + 2mn + n^2)
=> sin^2(A - B) = 1 - m^2 - 2mn - n^2 ----- (6)
Step 4: Rewrite equation (6) using the trigonometric identity:
- sin^2(A - B) = sin^2(A) - 2sin(A)sin(B) + sin^2(B)
=> sin^2(A - B) = (1 - cos^2(A)) - 2sin(A)sin(B) + (1 - cos^2(B))
=> sin^2(A - B) = 2 - cos^2(A) - cos^2(B) - 2sin(A)sin(B) ----- (7)
Step 5: Compare equations (6) and (7):
- Equating the coefficients of sin(A)sin(B) from equations (6) and (7), we get:
=> 2mn = 2sin(A)sin(B)
=> sin(A)sin(B) = mn ----- (8)
Step 6: Rewrite equation (8) using the trigonometric identity:
- sin(A + B) = sin(A)cos(B) + cos(A)sin(B)
=> sin(A)sin(B) + cos(A)cos(B) = sin(A + B)
=> mn + cos(A)cos(B) = sin(A +
NDA Mock Test: Mathematics - 9 - Question 24

If cos x ≠ – 1, then what is  equal to?

Detailed Solution for NDA Mock Test: Mathematics - 9 - Question 24

NDA Mock Test: Mathematics - 9 - Question 25

If angles A, B and C are in AP, then what is sin A + 2 sin B + sin C equal to? 

Detailed Solution for NDA Mock Test: Mathematics - 9 - Question 25

Since, A, B, C are in AP.

NDA Mock Test: Mathematics - 9 - Question 26

What is the value of + cos 

NDA Mock Test: Mathematics - 9 - Question 27

 If p = sin (989o) cos (991o), then which one of the following is correct? 

Detailed Solution for NDA Mock Test: Mathematics - 9 - Question 27

To solve this problem, we need to find the value of p = sin (989°) cos (991°).
First, let's simplify the expression:
sin (989°) = sin (360° + 629°) = sin (629°)
cos (991°) = cos (360° + 631°) = cos (631°)
Now, let's look at the values of sin and cos for angles greater than 360°:
sin (629°) = sin (629° - 360°) = sin (269°) = -sin (91°) (since sin (269°) = -sin (91°))
cos (631°) = cos (631° - 360°) = cos (271°) = -cos (89°) (since cos (271°) = -cos (89°))
Now, let's substitute these values back into the expression:
p = sin (989°) cos (991°) = -sin (91°) * (-cos (89°))
Since sin (91°) and cos (89°) are both positive values, the product of two positive values is positive.
Therefore, p is finite and positive.
So, the correct answer is:
A: p is finite and positive
NDA Mock Test: Mathematics - 9 - Question 28

Q. 

NDA Mock Test: Mathematics - 9 - Question 29

Q. 

 is equal to

NDA Mock Test: Mathematics - 9 - Question 30

If  

Q. cos (α + β + λ) is equal to 

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