NDA Mathematics 16 April 2023 Official Paper (Previous Year Paper)

Concept:

If ω is a non-real cube root of 1, that is, ω³ = 1, then

1 + ω + ω² = 0  ------(1)

and $\omega = {-1 + i √{3} \over 2}$

If Z = x + iy then |Z| = √(x² + y²)

Calculation:

Given, ω is a non-real cube root of 1, then

⇒ $\left|\frac{1-\omega}{\omega+\omega^2}\right| = \left|\frac{1-\omega}{-1}\right|$ ....{From equation (1)}

⇒ $\left|\frac{1-\omega}{\omega+\omega^2}\right| = \left|{\omega - 1}\right|$

⇒ $\left|\frac{1-\omega}{\omega+\omega^2}\right| = \left|{{-1 + i √{3} \over 2} - 1}\right|$

⇒ $\left|\frac{1-\omega}{\omega+\omega^2}\right| = \left|{{-3 + i √{3} \over 2} }\right|$

⇒ $\left|\frac{1-\omega}{\omega+\omega^2}\right| = \sqrt{[{9 \over 4} + { 3 \over 4}]} = \sqrt 3$

∴ The correct option is (1).

Explanation:

If a 6-digit number is formed only by using 0, 1, 2, 3, 4 and 5 (each once), then every digit is used once as six digits are given.

Since the number is 6-digit so 0 should not be at starting place.

and the number is divisible by 6, so it is divisible by 2 and 3.

So only even digit {0,2,4} occur at last place for divisible by 2.

And the number always contains the digits 0,1,2,3,4,5 once, so sum of digits = 15, so such a number will always be divisible by 3.

Case 1: The number ends with 0.

Number of choices for first place = 5

And for second, third, fourth and fifth number of choices will be 4, 3, 2, and 1 respectively.

So, the number of 6 digit numbers which are ending with 0 with given conditions = 5 × 4 × 3 × 2 × 1 = 120

Case 2: The number ends with 2.

Number of choices for first place = 4 {excluding 2 and 0}

And for second, third, fourth and fifth number of choices will be 4, 3, 2, and 1 respectively.

So, the number of 6 digit numbers which are ending with 0 with given conditions = 4 × 4 × 3 × 2 × 1 = 96

Case 3: The number ends with 4.

Number of choices for first place = 4 {excluding 4 and 0}

And for second, third, fourth and fifth number of choices will be 4, 3, 2, and 1 respectively.

So, the number of 6 digit numbers which are ending with 0 with given conditions = 4 × 4 × 3 × 2 × 1 = 96

So total number of ways = 120 + 96 + 96 = 312

∴ The correct option is (4).

Concept:

Decimal number is with base 10 and binary number is with base 2.

Calculation:

Divide the number repeatedly by 2 until the quotient becomes 0.

Write the remainder from bottom to top.

(1011)₁₀ = (1111110011)₂

∴ The correct option is (4).

Concept:

If A is a matrix of order n × n, then

• |k A| = kⁿ |A|
• |adj(A)| = |A|^n-1

 

Calculation:

Given, A is the matrix of order 3 × 3

then, |2adj(3A)| = 2³ |adj(3A)|   {∵ order of adj(3A) is 3 × 3}

⇒ |2adj(3A)| = 23 |3A|^3-1 =  23 |3A|2 

⇒ |2adj(3A)| = 23 (3³|A|)²  = 23 36 4²

⇒ |2adj(3A)| = 2⁷3⁶ = 2α 3β, 

then α = 7 and β = 6

⇒ α + β = 7 + 6 = 13

∴ The correct option is (2).

Concept:

1) The roots of unity (ω):

ω = ${-1 + i√{3} \over 2}$, ω² = ${-1 - i√{3} \over 2}$

2)  1 + ω + ω² = 0

3)  |x + iy| = √(x² + y²)

Calculation:

Given equation x2 - x + 1 = 0,

⇒ $x = {1 \pm √{1-4} \over 2}$

⇒ $x = {1 \pm i√{3} \over 2}$

⇒ $x =- ω, -ω^2$

where $ω$ is the cube root of the unity, that is $ω$³ = 1

Let α = -$ω$ and β = -$ω$²

⇒  α¹⁰⁰ = $ω^{100} = ω^{99}.ω = (ω^3)^{33}.ω = ω$ 

and β¹⁰⁰ = $ω^{200} = ω^{198}.ω^2 = (ω^3)^{66}.ω^2 = ω^2$

So, $\left|\frac{\alpha^{100}+\beta^{100}}{\alpha^{100}-\beta^{100}}\right|$ = $\left|\frac{ω+ω^{2}}{ω-ω^{2}}\right|$

So, $\left|\frac{\alpha^{100}+\beta^{100}}{\alpha^{100}-\beta^{100}}\right|$ = $\left|\frac{ω(1+ω)}{ω(1-ω)}\right|$

⇒ $\left|\frac{\alpha^{100}+\beta^{100}}{\alpha^{100}-\beta^{100}}\right|$ = $\left|\frac{1+ω}{1-ω}\right|$

⇒ $\left|\frac{\alpha^{100}+\beta^{100}}{\alpha^{100}-\beta^{100}}\right|$ = $\left|\frac{1+{-1 + i√{3} \over 2}}{1-{-1 + i√{3} \over 2}}\right|$

⇒ $\left|\frac{\alpha^{100}+\beta^{100}}{\alpha^{100}-\beta^{100}}\right|$ = $\left|\frac{1 + i√{3} }{3 - i√{3} }\right|$

Rationalizing the denominator,

⇒ $\left|\frac{\alpha^{100}+\beta^{100}}{\alpha^{100}-\beta^{100}}\right|$ = $\left|\frac{1 + i√{3} }{3 - i√{3} }\right|\left|\frac{3+ i√{3} }{3+ i√{3} }\right|$

⇒ $\left|\frac{\alpha^{100}+\beta^{100}}{\alpha^{100}-\beta^{100}}\right|$ = $\left|\frac{3 + i√{3}+3i\sqrt {3}-3 }{9+3 }\right|$

⇒ $\left|\frac{\alpha^{100}+\beta^{100}}{\alpha^{100}-\beta^{100}}\right|$ = $\left|\frac{4i\sqrt 3}{12}\times \frac{i}{i}\right|=\left|\frac{i}{\sqrt 3}\right|$

⇒ $\left|\frac{\alpha^{100}+\beta^{100}}{\alpha^{100}-\beta^{100}}\right|$ = $\sqrt{[0 + (\frac{1}{\sqrt 3})^2]} = \frac{1}{√ 3}$

∴ The correct option is (4).

Concept:

• If A is a symmetric matrix then, A = A', where A' is the transpose of A
• If A is a skew-symmetric matrix then, A = -A', where A' is the transpose of A
• (AB)' = B'A'
• The diagonal of a skew-symmetric matrix consists of zero elements and therefore the sum of elements in the main diagonals is equal to zero.
• The sum of non-diagonal entries of the skew-symmetric matrix is zero.

 

Calculation:

Given, A and B be symmetric matrices of the same order,

⇒ A = A' and B = B' __(1)

Now (AB - BA)' = (AB)' - (BA)' 

⇒ (AB - BA)' = B'A' - A'B'

⇒ (AB - BA)' = BA - AB  (From (1))

⇒ (AB - BA)' = - (AB - BA) 

So (AB - BA) is a skew-symmetric matrix

By using the properties discussed above, 

Diagonal entries of (AB - BA) are zero.

The sum of non-diagonal entries of the skew-symmetric matrix is zero.

∴ The correct statement is (2).

Explanation:

Statement 1. AB = AC ⇒ B = C if A is non-singular

If AB = AC, and A is non singular,

Multiply both sides by A^-1 on left,

⇒ A^-1AB = A^-1AC

⇒ B = C

So statement 1 is true.

Statement 2. If BX = CX for every column matrix X having n rows then B = C

If BX = CX for every column matrix X having n rows 

let X be from the Euclidean basis B = {e₁,e₂,....e_n}

then Ae_i = Be_i 

⇒ i^th column of A = i^th column of B for 1 ≤ i ≤ n

Hence A = B

So statement 2 is true.

∴ The correct option is (3).

Given  - 

x + 2y + z = 4, 2x + 4y + 2z = 8 and 3x + 6y + 3z = 10

Concept - 

ρ (A) =  number of no zero row of the matrix A.

if $ρ(A) \neq ρ(A:B)$ then no solution exist.

Explanation - 

matrix representation of the given equations -

$\left[ \begin{array}{ccc|c} 1 & 2 & 1 & 4 \ 2 & 4 & 2 & 8 \ 3 & 6 & 3 & 10 \end{array} \right]$

Now use $R_2 \to R_2-2R_1 \ \ and \ \ R_3 \to R_3-3R_1$

$\left[ \begin{array}{ccc|c} 1 & 2 & 1 & 4 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & -2 \end{array} \right]$

So ρ (A) = 1 because in the first part of the matrix only one non zero row exist.

and ρ(A:B) = 3 because in the  matrix three non zero row exist.

Hence $ρ(A) \neq ρ(A:B)$

So no solution exist.

Important Points Crammer's rule is applied only when determinant of the matrix is not Zero.

Concept:

In a system of linear equations, if we have two distinct solutions, say X₁ and

X₂, then any linear combination of these solutions, such as aX₁ + bX₂, will

also, be a solution to the system if certain conditions are met.

Calculation:

Given that X₁ and X₂ are distinct solutions of the system AX = B, we can

write the following equations:

AX1 = B        ---(1)

AX2 = B        ---(2)

Now, let's consider the combination aX₁ + bX₂.

We want to check if this combination satisfies the equation AX = B.

Substituting X = aX₁ + bX₂ into AX = B, we have:

A × (aX₁ + bX₂) = B

Using the linearity property of matrix multiplication, we can distribute the

matrix A over the linear combination:

a × (AX₁) + b(AX₂) = B

Substituting equations (1) and (2) into the above equation, we get:

aB + bB = B

a + b = 1

Therefore, the relation between a and b is that their sum is equal to 1.

Calculation:

Given, $\left|\begin{array}{ccc} 0 & x-a & x-b \ 0 & 0 & x-c \ x+b & x+c & 1 \end{array}\right|=0$

Expanding along column 1,

0 + 0 + (x + b)[(x - a)(x - c) - 0(x - b)] = 0

⇒ (x + b)(x - a)(x - c) = 0

So, the roots of the equation are -b, a, c.

⇒ The sum of the roots = a - b + c

∴ The correct option is (2).

Concept:

The roots of a quadratic equation with real coefficients are either real or occur in conjugate pairs(if complex).

For the quadratic equation ax² + bx + c = 0, the sum of roots = -b/a and product of roots = c/a

Calculation:

Given, 2 - i√3 where i = √-1 is a root of the equation x2 + ax + b = 0,

Since complex roots occur in conjugate pairs, 

⇒ 2 + i√3 is also the root of the equation.

So the roots of the equation are 2 + i√3 and 2 - i√3.

⇒ The sum of roots = 4 = -a/1

⇒ a = -4

And the product of roots = (2 + i√3)(2 - i√3) = (4 + 3) = b/1

⇒ b = 7

⇒ a + b = -4 + 7 = 3

∴ The correct option is (4).

Concept:

If z = a + ib then arg(z) = tan^-1(b/a)

Calculation:

Given, $z=\frac{1+i √{3}}{1-i √{3}}$

Multiplying numerator and denominator by (1 + i√3)

⇒ $z=\frac{1+i √{3}}{1-i √{3}} \times \frac{1+i √{3}}{1+i √{3}}$

⇒ $z=\frac{(1+i √{3})^2}{1-(i √{3})^2}$

⇒ $z=\frac{1+(\sqrt 3 i)^2+ 2 i √{3}}{1+ 3}$

⇒ $z=\frac{1 - 3+ 2 i √{3}}{4}$

⇒ $z={-1 \over 2} + i {\sqrt 3 \over 2}$ 

So, arg(z) = $\tan^{-1} {{\sqrt 3 \over 2} \over { 1\over 2} } = \tan^{-1} \sqrt 3$ in second quadrant

(where x is negative and y is positive, that is second quadrant.)

⇒ arg(z) = π - $\frac{\pi}{3}$ = $\frac{2\pi}{3}$

∴ The correct option is (2).

Concept:

If three terms a, b, c are in AP, then 2b = a + c

Calculation:

Given, a, b, c are in AP, then 2b = a + c

Given determinant,

$\left|\begin{array}{lll} x+1 & x+2 & x+3 \ x+2 & x+3 & x+4 \ x+a & x+b & x+c \end{array}\right|$

Apply C₁ → C₁ + C₃ - 2C₂,

⇒ $\left|\begin{array}{lll} 0 & x+2 & x+3 \ 0 & x+3 & x+4 \ a+ c- 2b & x+b & x+c \end{array}\right|$

Since a + c = 2b

⇒ \(\left|\begin{array}{lll} 0 & x+2 & x+3 \ 0 & x+3 & x+4 \0 & x+b & x+c \end{array}\right|\)

If any row or column of matrix is 0 then, detreminant equals to 0.

∴ The correct option is (2).

Concept:

If three terms p, q, r are in GP, then pr = q2.

 Property of log- $\log_q p = \frac{\log p}{\log q}$ for usual log

Calculation:

Given, logxa, ax and logbx are in GP, 

⇒ $(a^x)^2=\log_x a . \log_b x$

⇒ $a^{2x}={\log a \over \log x}.{ \log x \over \log b}$ 

⇒ $a^{2x}={\log a \over \log b}$

⇒ $a^{2x}={\log_b a}$

⇒ ${2x}=\log_a({\log_b a})$

⇒ x = $\frac{\log _a\left(\log _b a\right)}{2}$

∴ The correct option is (3).

Concept:

If three terms p, q, r are in GP, then pr = q².

If three terms a, b, c are in AP, then 2b = a + c

Calculation:

Given, $2^{\frac{1}{c}}, 2^{\frac{b}{a c}}, 2^{\frac{1}{a}}$ are in GP, 

⇒ $(2^{\frac{b}{ac}})^2= 2^{\frac{1}{ c}}. 2^{\frac{1}{a}}$

⇒ $2^{\frac{2b}{ac}}= 2^{\frac{1}{ c}+{\frac{1}{a}}}$

⇒ ${\frac{2b}{ac}}= {\frac{1}{ c}+{\frac{1}{a}}}$

⇒ 2b = a + c

⇒ a, b, c are in AP

∴ The correct option is (1).

Concept:

If a is the first term of AP and d is the common difference between terms then, 

n^th term of the AP is given by, a_n = a + (n - 1)d

Calculation:

Given,

The first and the second terms of an AP are $\frac{5}{2}$ and $\frac{23}{12}$ respectively

⇒ a = $\frac{5}{2}$ and d =  $\frac{23}{12}$ - $\frac{5}{2}$ = $\frac{-7}{12}$

The largest negative term will be the first negative term.

T_n < 0

⇒ a + (n - 1) < 0

⇒ $\frac{5}{2}$ + (n - 1)$\frac{-7}{12}$ < 0

⇒ $\frac{5}{2}$ + $\frac{7}{12}$ - $\frac{7n}{12}$ < 0

⇒ (30 + 7)/12 - 7n/12 < 0

⇒ 37/12 - 7n/12 < 0

⇒ 37 - 7n < 0

⇒ 7n - 37 > 0

⇒ n > 37/7 or n > 5.2

But n will be the integer (as it is the term of AP)

⇒ n = 6 

∴ The correct option is (2).

Concept:

For ax² + bx + c = 0 to have real roots, 

Δ = b² - 4ac ≥ 0

Calculation:

Given, x2 - 4x + k = 0

For roots to exist, (- 4)² - 4k ≥ 0

⇒ 16 - 4k ≥ 0

⇒ k ≤ 4 ...(i)

Equation given is, x2 - 4x + k = 0

⇒ x = ${4 ± \sqrt{16-4k} \over 2}$

According to the question, the roots lie between (0, 5)

⇒ 0 < ${4 ± \sqrt{16-4k} \over 2}$ < 5

⇒ 0 < $4 ± \sqrt{16-4k}$ < 10

⇒ - 4 < $± \sqrt{16-4k}$ < 6

⇒ 0 < 16 - 4k < 36

⇒ - 16 < - 4k < 20

⇒ - 4 < - k < 5

⇒ - 5 < k < 4 ...(ii)

Using (i) and (ii), the possible values of k are  1, 2, 3.

∴ Number of possible values of k is 3

Concept:

For an AP with first term a and common difference d,

n^th term is a_n = a + (n - 1)d

And sum of first n terms = S_n = ${n \over 2} [2a + (n-1)d]$

Calculation:

Given, first term = x and S_n = 0

⇒ ${n \over 2} [2x + (n-1)d]$ = 0

⇒ 2x + (n - 1)d = 0 __(1)

Sum of next m terms = S_m+n - S_n = S_m

⇒ Sum of next m terms = Sm+n = ${m+n \over 2} [2x + (m+n-1)d]$

⇒ Sum of next m terms = ${m+n \over 2} [2x + (n-1)d + md]$

⇒ Sum of next m terms = ${m+n \over 2} [md]$   { From equation (1)}

Putting the value of d from (1),

⇒ Sum of next m terms = ${m+n \over 2}m {(-2x) \over n-1}$ 

⇒ Sum of next m terms = $\frac{m x(m+n)}{1-n}$

∴ The correct option is (2).

Calculation:

Statement:1 

25! = 25 × 24 × 23 × .....× 13 .....× 2 × 1

Since 13 & 2 are the factor of 25!, it is divisible by 26.

Now,

Remainder of (25! + 1)/26 = Rem (25!/26) + Rem(1/26)

Remainder of (25! + 1)/26 = 0 + 1 = 1

∴ 25! + 1 is not divisible by 26.

Statement:2

6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

6! + 1 = 720 + 1 = 721 

Which is divisible by 7.

∴ Only statement 2 is correct.

Concept:

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

• a = 0
• and ω = -$\overline {\omega}$

Calculation:

Given, z is a complex number such that $\frac{z-1}{z+1}$ is purely imaginary,

Let $\frac{z-1}{z+1}$ = a + ib, then a = 0

And $\frac{z-1}{z+1} = - \overline{(\frac{z-1}{z+1})}$ ⇒ $\frac{z-1}{z+1} = -(\frac{ \overline{z}-1}{ \overline{z}+1})$

⇒ $({z-1})({ \overline{z}+1}) = -({ \overline{z}-1})({z+1})$ ⇒ $({z \overline{z}+z }- { \overline{z}-1}) = -({z \overline{z}-z }+ { \overline{z}-1})$ 

⇒ $|z|^2 -1 = -|z|^2 + 1$ ⇒ |z|2 = 1 ⇒ |z| = 1

∴ The correct option is (3).

Explanation: 

Case 1: x < 4

In this case, both |x - 4| and |x - 7| are negative. So,

⇒ -(x - 4) - (x - 7) = 15

⇒ -2x + 11 = 15

⇒  x = -2  which belongs in the range x < 4

Case:2  4 ≤ x ≤  7

In this case, |x - 4| is nonnegative, and |x - 7| is negative.

⇒ (x - 4) - (x - 7) = 15

⇒ 7 = 15, which is not possible.

Case:3 x > 7

In this case, both |x - 4| and |x - 7| are nonnegative.

 (x - 4) + (x - 7) = 15

2x = 26, or x = 13

Therefore, there is only two real number solution.

Concept:

Codomain - A codomain is the group of possible values that the dependent variable can take.

Range - The range is all the elements from set B that have the corresponding pre-image in set A.

Calculation:

Given, $f(x)=\frac{2 x+3}{3 x+5}, x \in A$

For f(x) to be defined, 3x + 5 ≠ 0

⇒ x = $-\frac{5}{3}$

∴ A = \(R \backslash\left{-\frac{5}{3}\right}\)

Now, y = f(x) is onto

⇒ y = $\frac{2 x+3}{3 x+5}$

⇒ 3xy + 5y = 2x + 3

⇒ 3xy - 2x = 3 - 5y

⇒ x(3y - 2) = 3 - 5y

⇒ x = $\frac{3-5y}{3y-2}$

For x to be defined for \(R \backslash\left{\frac{2}{3}\right}\)

Since, for onto functions co-domain = range

∴ B = \(R \backslash\left{\frac{2}{3}\right}\)

∴ \(A=R \backslash\left{-\frac{5}{3}\right} \text { and } B=R \backslash\left{\frac{2}{3}\right}\)

Concept:

In the quadratic equation ax² + bx + c = 0 

Sum of root (α +  β) = -b/a

Product of root (αβ) = c/a

Formula used:

(x + y)² = x² + y² + 2xy

(x - y)2 = x2 + y2 - 2xy

Explanation:

We have  x2 + ax + b = 0

By using the above concept

α + β = - a            -----(1)

α.β =  b              -----(2)

To find α, we need to consider the given statements.

Statement 1: 

α + β = 0 or β = - α 

Also, α2 + β2 = 2

⇒ α2 +  (-α)2 = 2

⇒ 2α² = 2

⇒ α = ± 1

Statement 2: αβ² = -1, a = 0

From equation (1)

α + β = - a

α + β = 0               [∵ a = 0]

⇒ β = - α 

Now

⇒ αβ² = -1

⇒ α (-α)² = -1

⇒ α³ = -1

⇒ α = -1  

∴  The correct answer is option (3).

Formula used:

\({{\rm{T}}{\left( {{\rm{r;}} + {\rm{;}}1} \right)}} = {\rm{;}}{{\rm{;}}^{\rm{n}}}{{\rm{C}}{\rm{r}}} × {{\rm{a}}^{{\rm{n}} - {\rm{r}}}} × {{\rm{b}}^{\rm{r}}}\)

ⁿC_r = $\frac{n!}{r!(n-r)!}$ 

Calculation:

Given expression is $\left(x^{-\frac{8}{3}}+x^2 \log _{10} x\right)^8$

Here, r = 5, n = 8

According to the question

$T_6= ~^ 8C_5 × {(x^{-8/3})^{8-5}} × {(x^2.log_{10}~x)}^5$

Simplifying, we get:

T₆ = ⁸C₅ x^-8 × x¹⁰ (log₁₀x)⁵

5600 = 8C5 x²(log10x)5       [∵ T₆ = 5600]

Since, nCr = $\frac{n!}{r!(n-r)!}$ 

5600 = 56 × x2(log10x)5 

x2(log10x)5 = 100

From the above option, let x = 10

100 × (log1010)5 = 100 

(log1010)5 = 1

1 = 1      [log₁₀10 = 1]

Therefore, the value of x that satisfies the equation is x = 10.

Formula used:

Number of term in the (x + y)ⁿ = ^n+2-1C_2-1

Here, 2 is used as there are two terms x & y.

Calculation:

We have (3x - y)4(x + 3y)4

This can be written as 

[(3x - y)(x + 3y)]⁴

[3x² + 9xy - xy - 3y²]⁴

[3x² + 8xy - 3y²]⁴

Here, n = 4 & r = 3

Number of terms ^4+3-1C_3-1 = ⁶C2

⇒ (6 × 5)/(2 × 1) = 15

∴ Number of term = 15

Concept used:

Four terms of AP are: a -3d, a -d, a + d, a + 3d

Calculation:

Let, p = a - 3d, q = a - d, r = a + d, s = a + 3d

Difference between largest and smallest numbers

|(a - 3d) - (a + 3d)| = 6d

Given that  p + s = 8, 

⇒ 2a = 8

⇒ a = 4

Also, qr = 15

⇒ (a - d)(a + d) = 15

⇒ a² - d² = 15

⇒ 16 - d² = 15

⇒ d² = 16 - 15

⇒ d = 1   (absolute value)

⇒ 6d = 6 × 1 = 6

∴ The difference between the largest and smallest numbers is 6.

Concept:

(x + y)n = nC0 xn yo + nC1 xn-1 y1 +  nC2 xn-2 y2 +...... + nCn x0 yn

Where, n ≥ 0 and 

nCr = $\frac{n!}{r!(n-r)!}$ = Binomial coefficient  

Case: I n is even, greatest binomial coefficient is given by ⁿC_n/2 

It means, n = 2r

Case: II n be odd, the greatest binomial coefficient is given by

ⁿC_(n+1)/2 or ⁿC_(n-1)/2

It means,  n = 2r - 1 & n = 2r + 1 

Explanation: 

For n = 2r:

n will be the even number of every positive integer r.

Hence statement I is correct.

For n =  n = 2r - 1 & 2r + 1:

n will be odd for every positive value of r. 

Hence statement II is correct. 

∴ Correct answer is option (3)

Formula used:

If m parallel lines in a plane are intersected by n parallel lines, what is the number of parallelograms that are formed

Number of parallelograms = mn(m-1)(n-1)/4

Calculation:

By using the above concept

mn(m-1)(n-1)/4 = 60

mn(m-1)(n-1) = 240

m.(m - 1).n.(n - 1) = 240

Here, we have one equation & two variables.

So, to find the value of m & n by by hit & trial method.

m.(m - 1).n.(n - 1) = 5 × 3 × 2 × 2 × 2 × 2

This can be re-written as  

m.(m - 1).n.(n - 1) = 5 × 4 × 4 × 3 

On comparing both sides,

m = 5 & n = 4

∴ m + n = 5 + 4 = 9  

Thus, the value of (m + n) is 9.

Given that

x is the number of permutations of the word "PERMUTATIONS "

y is the number of permutations of the word "COMBINATIONS"

The number of permutations in any word is given by

= factorial of total number of letters in word/ number of repetitive letters!

Calculation:

 

In the word PERMUTATIONS, the letter "T" is repetitive i.e., it repeats 2 times.

So, the number of permutations for the word PERMUTATIONS  would be 

= factorial of the total number of letters in PERMUTATIONS/ 2!

= 12!/(2!)

x = 12!/2 ....(1)

Similarly, in the word COMBINATIONS,

I repeat 2 times,

O repeats 2 times

and N repeats 2 times.

the total number of permutations will be =  12!/(2! 2! 2!)

y = 12!/8 ....(2)

From 1 and 2

x = 4y

The correct answer is an option (3)

Calculation:  

Firstly, we need to determine the total number of 5-digit numbers that can be created with the digits 0, 1, 2, 4, 5 without repetition.

As the first position cannot be 0 (as that would result in a 4-digit number), there are 4 choices (1, 2, 4, or 5) for the first digit.

For each of the remaining positions, we have one fewer choice because we can't repeat digits.

Therefore,

Total possible 5-digit numbers = 4 × 4 × 3 × 2 × 1 = 96

Next, we need to determine how many of these 5-digit numbers are greater than 50,000.

The first digit of such a number must be either 5 (1 choice) because 0, 1, 2, and 4 would result in a number less than 50,000.

After the first digit has been decided, we can choose any of the remaining 4 digits for the second position, then 3 for the third, 2 for the fourth, and 1 for the fifth

Total possible 5-digit numbers are greater than 50,000

 1 × 4 × 3 × 2 × 1 = 24.

∴ Required % = $\frac{24}{96} \times 100=25 \%$.

Concept:

GM of a & b is √(ab)

Formula used:

sin²x + cos² = 1

cos 2x = 1 - 2sin²x

Calculation:

Given that, sin β is the GM of sin α & cos α then

sin β = √( sin α × cos α)

sin²β = (sin α × cos α)

2sin²β = 2sin α × cos α

1 - 2sin2β = 1 - 2sin α × cos α

By using the above formula

cos 2β = sin²α + cos²α - 2sin α × cos α

Since, (a - b)² = a² + b² - 2ab

∴ cos 2β = (cosα - sin α)²

Concept:

AM of two number a & b is (a + b)/2

Formula used:

1. 1 + tan² θ =  sec² θ
2. sin ²θ + cos²θ = 1
3. sec θ = 1/cos θ
4. 2 cos² θ - 1 = cos 2 θ

Calculation:

Since tan γ is the arithmetic mean (AM) of sin α and cos α,

$tan γ =\frac{sin α + cos α }{2}$

⇒ tan²γ = (sin α + cos α)²/4

⇒ tan2γ = (sin α + cos α)2/4

tan2γ = (sin²α + cos²α + 2sin α.cos α)2/4

By using the formula (1)

⇒ tan2γ = (1 + sin 2α)/4

Adding 1 on both sides

⇒ 1 + tan2γ = 1 + [(1 + sin 2α)/4]

⇒ 1 + tan2γ = (4 + 1 + sin 2α)/4

⇒ 1 + tan2γ = (5 + sin 2α)/4

But 1 + tan2 θ = sec2 θ

⇒ sec2γ = (5 + sin 2α)/4

But, sec θ = 1/cos θ

​⇒ cos2γ = 4/(5 + sin 2α)

​⇒ 2cos2γ = 2 × 4/(5 + sin 2α)

​⇒ 2cos2γ = 8/(5 + sin 2α)

​⇒ 2cos2γ - 1 = 8/(5 + sin 2α) - 1

​⇒ cos²γ = [8/(5 + sin 2α)] - 1

​⇒ cos²γ = [8 - (5 + sin 2α)] / (5 + sin 2α)

​⇒ cos²γ = [3 - sin 2α)] / (5 + sin 2α)

⇒ 1/cos²γ = (5 + sin 2α)/(3 - sin 2α)

But, sec θ = 1/cos θ

**∴ sec²γ = (5 + sin 2α)/(3 - sin 2α)**​​.

Formula used:

tan θ = Perpendicualr / Base

tan (x + y) = (tan x + tan y)/(1 - tan x.tan y)

Calculation:

Given that 

α = tan-10.5

⇒ tan α = 1/2      ------(1)

In an ΔPAB

tan θ = 10/x       ------(2)

In an ΔPAC

tan (α + θ) = (20 + 10)/x

⇒ tan (α + θ) = 30/x

By using the above formula 

(tan α + tan θ)/(1 - tan α.tan θ) = 30/x

Using equations (1) & (2)

⇒ [1/2 + 30/x]/[1 - (1/x2)(30/x)]

⇒ $\frac{1/2 + 10/x}{1 - (1/2)(10/x)}=30/x$

⇒ (x -20)(2x - 10) = 30/x

⇒ x² - 20x = 60x - 300

⇒ x² - 40x + 300 = 0

⇒ x² - 30x - 10x + 300 = 0

⇒ x(x - 30) - 10(x - 30) = 0

⇒ (x - 10)(x - 30) = 0

⇒ x = 10 & 30

Statement:1 x can take two values which are in the ratio 1 : 3

Clearly, the ratio of the value of x = 1/3

Statement:2 x can be equal to the height of the flagstaff.

x = 10 & 30. So x ≠ 20

∴ Only statement 1 is correct.

Formula used:

tan θ = Perpendicualr / Base

tan (x + y) = (tan x + tan y)/(1 - tan x.tan y)

Calculation:

Given that 

α = tan-10.5

⇒ tan α = 1/2      ------(1)

In an ΔPAB

tan θ = 10/x       ------(2)

In an  ΔPAC

tan (α + θ) = (20 + 10)/x

⇒ tan (α + θ) = 30/x

By using the above formula 

(tan α + tan θ)/(1 - tan α.tan θ) = 30/x

Using equations (1) & (2)

⇒ [1/2 + 30/x]/[1 - (1/x2)(30/x)]

⇒ $\frac{1/2 + 10/x}{1 - (1/2)(10/x)}=30/x$

⇒ (x -20)(2x - 10) = 30/x

⇒ x2 - 20x = 60x - 300

⇒ x2 - 40x + 300 = 0

⇒ x2 - 30x - 10x + 300 = 0

⇒ x(x - 30) - 10(x - 30) = 0

⇒ (x - 10)(x - 30) = 0

⇒ x = 10 & 30

From equation (2)

tan θ = 10/10 & 10/30

⇒ tan θ =1 & 1/3

∴ Option 3 is correct.

Concept used:

Sine Rule: In the triangle ABC, where sides BC, CA & AB are a, b & c

respectively, we can write 

sin A/a = sin B/b = sin C/c

Where A, B & C are the angles opposite to the sides a, b & c

Formula used:

Arithmetic Mean of three number p, q & r is (P + q + r)/3

The perimeter of the triangle is the sum of three sides.

Calculation:

Let the side of the triangle be BC, CA & AB are a, b & c which are opposite to

its angle A, B & C respectively.

According to the question

a + b + c = 6[(sin A + sin B + sin C)/3]

a + b + c = 2(sin A + sin B + sin C)     ----(1)

But from sine rule

sin A/a = sin B/b = sin C/c = k (let)

sin A = ak, sin B = bk, sin C = ck

From equation (1)

a + b + c = 2(ak + bk + ck) 

a + b + c = 2k(a + b + c)

k = 1/2 

But side BC (i.e. a) = √3 & CA (i.e. b) = 1

Since, sin A = ak

sin A = (√3) × (1/2) = √3/2 

A = 60° 

Similarly,  sin B = bk

sin B = 1/2 

B = 30° 

As, A + B + C = 180°

60° + 30° + C = 180° 

C = 90° 

From equation (1)

a + b + c = 2(sin 60° + sin 30° + sin 90°) 

a + b + c = 2(√3/2 + 1/2 + 1) 

∴ a + b + c = 3 + √3

Concept used:

Sine Rule: In the triangle ABC, where sides BC, CA & AB are a, b & c

respectively, we can write 

sin A/a = sin B/b = sin C/c

Where A, B & C are the angles opposite to the sides a, b & c

Formula used:

Arithmetic Mean of three number p, q & r is (P + q + r)/3

The perimeter of the triangle is the sum of three sides.

Calculation:

Let the side of the triangle be BC, CA & AB are a, b & c which are opposite to

its angle A, B & C respectively.

According to the question

a + b + c = 6[(sin A + sin B + sin C)/3]

a + b + c = 2(sin A + sin B + sin C)     ----(1)

But from sine rule

sin A/a = sin B/b = sin C/c = k (let)

sin A = ak, sin B = bk, sin C = ck

From equation (1)

a + b + c = 2(ak + bk + ck) 

a + b + c = 2k(a + b + c)

k = 1/2 

But side BC (i.e. a) = √3 & CA (i.e. b) = 1

Since, sin A = ak

sin A = (√3) × (1/2) = √3/2 

A = 60° 

Similarly,  sin B = bk

sin B = 1/2 

B = 30° 

As, A + B + C = 180°

60° + 30° + C = 180° 

C = 90° 

Now, 

Statement: 1 ABC is a right-angled triangle.

As, ∠C = 90°. therefore, ΔABC is a right-angled triangle.

Statement: 1 The angles of the triangle are in the AP

The angles of the triangle are 30°, 60° & 90° which are in AP.

∴ Both statements are correct.

Formula used:

Minimum value of x + 1/x is 2 & it comes at x = 1

Calculation:

Given that 

$x=\frac{\sin ^2 A+\sin A+1}{\sin A}$ where 0 < A ≤ $\frac{π}{2}$

This can be written as 

x = sin A + 1/sin A + 1

For a minimum value of x,

⇒ sin A = 1

⇒ x_min  = 1 + 1/1 + 1 = 3

∴ The required answer is 3.

Formula used:

Minimum value of x + 1/x is 2 & it comes at x = 1

Calculation:

Given that 

$x=\frac{\sin ^2 A+\sin A+1}{\sin A}$ where 0 < A ≤ $\frac{π}{2}$

This can be written as 

x = sin A + 1/sin A + 1

For a minimum value of x,

⇒ sin A = 1

⇒ A = π/2

∴ The required answer is π/2.

Concept:

Cosine Rule:

If a, b, c are three sides of a triangle opposite to angles A, B, C then:

cos A = $\frac{b^2+c^2-a^2}{2bc}$

cos B = $\frac{a^2+c^2-b^2}{2ac}$

cos C = $\frac{a^2+b^2-c^2}{2ab}$

Calculation:

Using cosine rule, b² + c² - a² = 2bc cos A ...(i)

a² + c2 - b2 = 2ac cos B ...(ii)

a2 + b2 - c2 = 2ab cos C ...(iii)

Adding (i), (ii) and (iii), we get:

a² + b² + c² = 2ab cos C + 2bc cos A + 2ca cos B

According to the question, a2 + b2 + c2 = ac + √3bc

Comparing the coefficients of both equations:

2cos C = 0 ⇒ cos C = 0 ⇒ C = 90^∘

2cos B = 1 ⇒ cos B = 1/2 ⇒ B = 60^∘ 

2cos A = √3 ⇒ cos A = √3 /2 ⇒ A = 30^∘

⇒ The angles of triangles are A = 30^∘, B = 60^∘, C = 90^∘

∴ The triangle is a Right angled triangle.

Concept:

Cosine Rule:

If a, b, c are three sides of a triangle opposite to angles A, B, C then:

cos A = $\frac{b^2+c^2-a^2}{2bc}$

cos B = $\frac{a^2+c^2-b^2}{2ac}$

cos C = $\frac{a^2+b^2-c^2}{2ab}$

Calculation:

Using cosine rule,

b2 + c2 - a2 = 2bc cos A ...(i)

a2 + c2 - b2 = 2ac cos B ...(ii)

a2 + b2 - c2 = 2ab cos C ...(iii)

Adding (i), (ii) and (iii), we get:

a2 + b2 + c2 = 2ab cos C + 2bc cos A + 2ca cos B

According to the question, a2 + b2 + c2 = ac + √3bc

Comparing the coefficients of both equations:

2cos C = 0 ⇒ cos C = 0 ⇒ C = 90∘

2cos B = 1 ⇒ cos B = 1/2 ⇒ B = 60∘ 

2cos A = √3 ⇒ cos A = √3 /2 ⇒ A = 30∘

⇒ The angles of triangles are A = 30∘, B = 60∘, C = 90∘

Given, c = 8

sin A = sin 30^∘ = $\frac{a}{c}$

⇒ $\frac{1}{2}$ = $\frac{a}{c}$

⇒ a = 4

sin B = sin 60^∘ = $\frac{b}{c}$

⇒ $\frac{√{3}}{2}$ = $\frac{b}{c}$

⇒ b = 4√3

∴ Area of triangle = $\frac{1}{2}$ × a × b

= $\frac{1}{2}$ × 4 × 4√3

= 8√3

∴ Area of the triangle is 8√3.

Concept:

Modulus function:

It is a function that gives the absolute value of a variable (x). It is defined as such that |x| = x, if x ≥  0 else |x| = - x

Calculation:

f(x) = |x - 2| + |3 - x| + |4 - x|, where x ∈ R.

For x < 2, f(x) = 2 - x + 3 - x + 4 - x = 9 - 3x

For 2 ≤ x < 3, f(x) = x - 2 + 3 - x + 4 - x = 5 - x

For 3 ≤ x < 4, f(x) = x - 2 + x - 3 + 4 - x = x - 1

For x ≥ 4, f(x) = x - 2 + 3 - x + x - 4 = 3x - 9

∴ Minimum value of f(x) is given by:

fmin =  \(\left{\begin{matrix} 3, x < 2 \3, 2 \leq x< 3 \2, 3 \leq x < 4 \3, 4 \leq x \end{matrix}\right.\)

∴ Minimum value of f(x) is at x = 3.

Concept:

Modulus function:

It is a function that gives the absolute value of a variable (x). It is defined as such that |x| = x, if x ≥  0 else |x| = - x

Calculation:

f(x) = |x - 2| + |3 - x| + |4 - x|, where x ∈ R.

For x < 2, f(x) = 2 - x + 3 - x + 4 - x = 9 - 3x

For 2 ≤ x < 3, f(x) = x - 2 + 3 - x + 4 - x = 5 - x

For 3 ≤ x < 4, f(x) = x - 2 + x - 3 + 4 - x = x - 1

For x ≥ 4, f(x) = x - 2 + 3 - x + x - 4 = 3x - 9

∴ Minimum value of f(x) is given by:

f_min =  \(\left{\begin{matrix} 3, x < 2 \3, 2 \leq x< 3 \2, 3 \leq x < 4 \3, 4 \leq x \end{matrix}\right.\)

∴ Minimum value of f(x) is 2.

Concept:

To find the remainder when the sum S is divided by 8, we need to calculate the value of S and then take the modulus of S with 8.

First, let's calculate the value of S:

S = 0! + 1! + 2! + 3! + 4! + ... + 100!

We can see that the factorial of any number greater than or equal to 5 will have at least one factor of 5, and therefore, it will have at least one factor of 2. This means that the factorial of any number greater than or equal to 5 will be divisible by 8.

4! is also divisible by 8

So, when we add up the factorials from 4! to 100!, they will all contribute a multiple of 8 to the sum. The remaining part of the sum, from 0! to 3!, can be calculated as follows:

0! + 1! + 2! + 3! = 1 + 1 + 2 + 6 = 10

Now, let's take the modulus of S with 8:

S = 10 (mod 8)

To find the remainder, we can divide 10 by 8:

10 = (1 × 8) + 2

So, the remainder when S is divided by 8 is 2.

∴ The correct answer is Option 03.

Calculation:

To find the remainder when S is divided by 60, we need to calculate the value of S modulo 60.

First, let's calculate the factorials from 0! to 9! to understand the pattern:

0! = 1

1! = 1

2! = 2

3! = 6

4! = 24

5! = 120

6! = 720

7! = 5040

8! = 40320

9! = 362880

Notice that for n ≥ 5, n! is divisible by 60 because it contains at least two factors of 2 and one factor of 5 (and 60 = 2 × 2 × 3 × 5). Therefore, for n ≥ 5, n! ≡ 0 (mod 60).

Now, let's calculate the sum S by considering the factorials from 0! to 4!:

S = 0! + 1! + 2! + 3! + 4!

= 1 + 1 + 2 + 6 + 24

= 34

Since S = 34, the remainder when S is divided by 60 is also 34.

Therefore, the correct answer is option 4) 34.

Concept:

Cosine Rule:

If a, b, c are three sides of a triangle opposite to angles A, B, C then:

cos A = $\frac{b^2+c^2-a^2}{2bc}$

cos B = $\frac{a^2+c^2-b^2}{2ac}$

cos C = $\frac{a^2+b^2-c^2}{2ab}$

Calculation:

Given, cos P = 1/3

Using, cosine rule, cos P = $\frac{PQ^2+PR^2-QR^2}{2\times PQ \times PR}$

⇒ $\frac{1}{3}$ = $\frac{(2n+2)^2+(2n+4)^2-(2n+6)^2}{2\times (2n+2) \times (2n+4)}$

⇒ $\frac{1}{3}$ = $\frac{4n^2-16}{2(4n^2+12n+8)}$

⇒ 8n² + 24n + 16 =12n² - 48

⇒ 4n² - 24n - 64 = 0

⇒ n² - 6n - 16 = 0

⇒ n2 - 8n + 2n - 16 = 0

⇒ n(n - 8) + 2(n - 8) = 0

⇒ (n - 8)(n - 2) = 0

⇒ n = 8

∴ The value of n is 8.