NDA Mathematics 10 April 2022 Official Paper (Previous Year Paper)

Formula used:

(a - b)² = a² + b² - 2ab

Calculation:

Given:

$Δ_1 = \begin{vmatrix} 1 & p & q \ 1 & q & r \ 1 & r & p\end{vmatrix} \ \text{and} \ Δ_2 = \begin{vmatrix} 1 & 1 & 1 \ q & r & p \ r & p & q \end{vmatrix}$

Now, $Δ_1 = \begin{vmatrix} 1 & p & q \ 1 & q & r \ 1 & r & p\end{vmatrix}$

R₂ → R₂ - R₁ and R₃ → R₃ - R₁

⇒ $Δ_1 = \begin{vmatrix} 1 & p & q \ 0 & q-p & r-q \ 0 & r-p & p-q\end{vmatrix}$

⇒ Δ₁ = 1.{(q - p)(p - q) - (r - p)(r - q)} 

⇒ Δ1 = - (p - q)² - (r - p)(r - q)

⇒ Δ1 = - {p² - 2pq + q2} - {(r - p)(r - q)}

⇒ Δ1 = - p2 + 2pq - q2 - {r² - qr - pr + pq}

⇒ Δ1 = - p2 + 2pq - q2 - r2 + qr + pr - pq

⇒ Δ1 = - p2 - q2 - r2 + qr + pr + pq      ------(i)

Again, $Δ_2 = \begin{vmatrix} 1 & 1 & 1 \ q & r & p \ r & p & q \end{vmatrix}$

C₂→ C2 - C1 and C3 → C3 - C1

⇒ $Δ_2 = \begin{vmatrix} 1 & 0 & 0 \ q & r-q & p-q \ r & p-r & q-r \end{vmatrix}$

⇒ Δ₂ = 1.{(r - q)(q - r) - (p - q)(p - r)}

⇒ Δ2 = - (q - r)² - (p - q)(p - r)

⇒ Δ2 = -{q² + r² - 2qr} - {p² - pq - pr + qr}

⇒ Δ2 = - q2 - r2 - p2 + pq + pr + qr      ------(ii)

Adding equations (i) and (ii), we get,

Δ1 + Δ2 = - p2 - q2 - r2 + qr + pr + pq - q2 - r2 - p2 + pq + pr + qr 

⇒ Δ1 + Δ2 = -(p² + q² - 2pq + q² + r² - 2qr + r² + p² - 2pr)

⇒ Δ1 + Δ2 =  -[(p - q)² + (q - r)² + (r - p)²] which is always negative.

Δ1 + Δ2 is always negative.

Concept:

Three Basic Elementary Operations of Matrix:

Case 1: Interchange of any Two Rows or Two Columns

Any 2 columns (or rows) of a matrix can be exchanged. If the ith and jth rows are exchanged, it is shown by R_i ↔ R_j and if the ith and jth columns are exchanged, it is shown by C_i ↔ C_j.

Case 2: Multiplication of Row or Column by a Non-zero Number

The elements of any row (or column) of a matrix can be multiplied by a non-zero number. So if we multiply the ith row of a matrix by a non-zero number k, symbolically it can be denoted by R_i → kR_i.

Similarly, for the column, it is given by C_i → kC_i.

Case 3: Multiplication of Row or Column by a Non-zero Number and Add the Result to the Other Row or Column

The elements of any row (or column) can be added with the corresponding elements of another row (or column) which is multiplied by a non-zero number. So if we add the ith row of a matrix to the jth row which is multiplied by a non-zero number k, symbolically it can be denoted by R_i → R_i + kR_j Similarly, for column it is given by C_i → C_i + kC_j

Calculation:

Given:

(a - b) (b - c) (c - a) = 2 and abc = 6

Let Δ = $\begin{vmatrix}a & b & c \ a^2 & b^2 & c^2 \ a^3 & b^3 & c^3 \end{vmatrix}$

⇒ Δ = $abc\begin{vmatrix}1 & 1 & 1\ a & b & c \ a^2 & b^2 & c^2 \end{vmatrix}$

C₁ → C₁ - C₂  and C₂ → C2 - C₃

⇒ Δ = $abc\begin{vmatrix}0 & 0 & 1\ a-b & b-c & c \ a^2-b^2 & b^2-c^2 & c^2 \end{vmatrix}$

⇒ Δ = $abc\begin{vmatrix}0 & 0 & 1\ a-b & b-c & c \ (a-b)(a+b) & (b-c)(b+c) & c^2 \end{vmatrix}$

⇒ Δ = abc (a - b)(b - c) $\begin{vmatrix}0 & 0 & 1 \ 1 & 1 & c \ a+b & b+c & c^2 \end{vmatrix}$

Expanding along R₁,

⇒ Δ = abc (a - b)(b - c) [1{(b + c) - (a + b)}]

⇒ Δ = abc (a - b) (b - c) (c - a)

Putting the given values, we get, 

⇒ Δ = 6 × 2 = 12

∴ The value of $\begin{vmatrix}a & b & c \ a^2 & b^2 & c^2 \ a^3 & b^3 & c^3 \end{vmatrix}$ is 12.

Concept:

Expansion of determinant $\begin{vmatrix}a_{11} & a_{12} & a_{13} \ a_{21} & a_{22} & a_{23} \ a_{31} & a_{32} & a_{33} \end{vmatrix}$ is given by Δ = a11 a22 a33 - a11 a23 a32 - a12 a21 a33 + a12 a23 a31 + a13 a21 a32 - a13 a31 a22

Three Basic Elementary Operations of Matrix

Case 1: Interchange of any Two Rows or Two Columns

Any 2 columns (or rows) of a matrix can be exchanged. If the ith and jth rows are exchanged, it is shown by Ri ↔ Rj and if the ith and jth columns are exchanged, it is shown by Ci ↔ Cj.

Case 2: Multiplication of Row or Column by a Non-zero Number

The elements of any row (or column) of a matrix can be multiplied by a non-zero number. So if we multiply the ith row of a matrix by a non-zero number k, symbolically it can be denoted by Ri → kRi.

Similarly, for the column, it is given by Ci → kCi.

Case 3: Multiplication of Row or Column by a Non-zero Number and Add the Result to the Other Row or Column

The elements of any row (or column) can be added with the corresponding elements of another row (or column) which is multiplied by a non-zero number.

So if we add the ith row of a matrix to the jth row which is multiplied by a non-zero number k, symbolically it can be denoted by Ri → Ri + kRj.

Similarly, for column it is given by Ci → Ci + kCj.

 

Formula used:

a³ + b³ + c³ - 3abc = (a +  b + c)(a² + b² + c² - ab - bc - ca)

If (a + b + c) = 0 then a3 + b3 + c3 = 3abc 

Calculation:

Let A = $\begin{vmatrix}a & b & c \ b & c & a \ c & a & b\end{vmatrix}$

R₁ → R1 + R2 + R3

⇒ A = $\begin{vmatrix}a+b+c & a+b+c & a+b+c \ b & c & a \ c & a & b\end{vmatrix}$

⇒ A = $(a+b+c)\begin{vmatrix}1 & 1 & 1 \ b & c & a \ c & a & b\end{vmatrix}$

C2 → C2  - C1 and C3 → C3  - C1

⇒ A = (a + b + c) $\begin{vmatrix}1 & 0 & 0 \ b & c-b & a-b \ c & a-c & b-c\end{vmatrix}$

Expanding along R₁,

⇒ A = (a + b + c)[1{(c - b)(b - c) - (a - b)(a - c)}]

⇒ A = (a + b + c)[bc - c² - b² + bc - (a² - ac - ab + bc)]

⇒ A = (a + b + c)[- c2 - b2 - a2 + ac + ab + bc]

⇒ A = - (a + b + c)[a2 + b2 + c2 - ab - bc - ac]

The value of A can be zero if  (a + b + c) or (a2 + b2 + c2 - ab - bc - ac) is 0.

We know that,

a3 + b3 + c3 - 3abc = (a +  b + c)(a2 + b2 + c2 - ab - bc - ca)

⇒ a3 + b3 + c3 = 3abc 

∴ 1, 2 and 3 are correct.

Concept:

1). For the addition and subtraction of two matrices, the order of matrices should be equal.

2). For the multiplication of two matrices, the number of columns in the first matrix should be equal to the number of rows in the second matrix.

3).If the order of the matrix A is m × n and the order of matrix B is n × p, then the order of matrix AB is  m × p.

Calculation:

Given:

$A = [m ~~~ n], B = [-n \ -m] \text{and} \ C = \begin{bmatrix} m \ -m\end{bmatrix}$

Statement I: CA = CB

CA = $\displaystyle \begin{bmatrix} m \ -m\end{bmatrix} [m \ n ]$ = \(\begin{bmatrix} m^2 &mn \[0.3em] -m^2 & -mn\[0.3em] \end{bmatrix}\)

CB = $\displaystyle \begin{bmatrix} m \ -m\end{bmatrix} [-n \ -m ]$ = \(\begin{bmatrix} -mn &-m^2 \[0.3em] mn & m^2\[0.3em] \end{bmatrix}\)

CA ≠ CB

Statement I is incorrect.

Statement II: AC = BC

AC = $\displaystyle [m \ n ] \begin{bmatrix} m \ -m\end{bmatrix}$ = m² - mn

BC = $\displaystyle [-n \ -m ]\begin{bmatrix} m \ -m\end{bmatrix}$ = m2 - mn

Statement II is correct.

Statement III: C(A + B) = CA + CB

C(A + B) = $\displaystyle \begin{bmatrix} m \ -m\end{bmatrix} ([m\ \ \ n ]+[-n \ -m ])$ = $\displaystyle \begin{bmatrix} m \ -m\end{bmatrix} [m-n \ \ \ n-m ]$ = \(\begin{bmatrix} m^2-mn &mn-m^2 \[0.3em] -m^2+mn & -mn+m^2\[0.3em] \end{bmatrix}\)

CA + CB = \(\begin{bmatrix} m^2 &mn \[0.3em] -m^2 & -mn\[0.3em] \end{bmatrix}\) + \(\begin{bmatrix} -mn &-m^2 \[0.3em] mn & m^2\[0.3em] \end{bmatrix}\) = \(\begin{bmatrix} m^2-mn &mn-m^2 \[0.3em] -m^2+mn & -mn+m^2\[0.3em] \end{bmatrix}\)

Statement III is correct.

∴ Both Statement II and Statement III are correct.

)Concept:

Expansion of determinant

 $\begin{vmatrix}a_{11} & a_{12} & a_{13} \ a_{21} & a_{22} & a_{23} \ a_{31} & a_{32} & a_{33} \end{vmatrix}$ is given by 

Δ = a11 a22 a33 - a11 a23 a32 - a12 a21 a33 + a12 a23 a31 + a13 a21 a32 - a13 a31 a22

A(adj A) = ∣ A ∣ I, where I is the identity matrix.

Calculation:

Given:

 $A = \begin{bmatrix} 2 \sin θ & \cos θ & 0 \ -2\cos θ & \sin θ & 0 \ -1 & 1 & 1 \end{bmatrix}$  and I is the identity matrix.

⇒ $\mid A\mid = \begin{vmatrix} 2 \sin θ & \cos θ & 0 \ -2\cos θ & \sin θ & 0 \ -1 & 1 & 1 \end{vmatrix}$

Expanding about C₃, we get,

⇒ ∣ A ∣ =  1{(2 sinθ ×  sinθ) + 2 cos²θ}

⇒ ∣ A ∣ =  1(2 sin²θ + 2 cos² θ) 

⇒ ∣ A ∣ =  2(sin2θ + cos2θ)

⇒ ∣ A ∣ =  2          [∵ sin2θ + cos2θ = 1]

⇒ A(adj A) = ∣ A ∣ I

Putting the value of ∣ A ∣ =  2,

⇒ A(adj A) =  2I, where I is the identity matrix.

∴ A(adj A) =  2I

Concept:

1). The determinant of a Matrix with two identical rows or columns is equal to 0.

2). A matrix is singular if its determinant is 0. A singular square matrix that does not have a matrix inverse. 

Formula used:

cos2θ = cos²θ  - sin2θ = 1 - 2sin2θ = 2cos2θ - 1

Calculation:

Let A  = $\begin{bmatrix} 2\cos 2θ & 2\cos 2θ & 6 \ 1 -2 \sin^2θ & 2 \cos^2θ -1 & 3 \ k & 2k & 1 \end{bmatrix}$

A = $\begin{bmatrix} 2\cos 2θ & 2\cos 2θ & 6 \ cos\ 2θ & cos\ 2θ & 3 \ k & 2k & 1 \end{bmatrix}$

A = 2$\begin{bmatrix} \cos 2θ & \cos 2θ & 3 \ cos\ 2θ & cos\ 2θ & 3 \ k & 2k & 1 \end{bmatrix}$

So, R₁ and R₂ become the same. 

⇒|A| = 0.

∴ The value of the determinant is zero for any real value of k.

Concept:

1). A square matrix A = [a_ij]_{n x n}, is said to be a scalar matrix if;

     a_ij = 0, when i ≠ j

     a_ij = k, when i = j, for some constant k

2). Non Singular matrix is a square matrix whose determinant is a non-zero value. 

3). Null matrix or zero matrices is a matrix having zero as all its elements.

Formula used:

$\displaystyle A^{-1}=\frac{adj\ A}{∣\ A ∣ }$ 

Calculation:

Given that, A is a non-singular matrix and B = adj A

We know that

 $\displaystyle A^{-1}=\frac{adj\ A}{∣\ A ∣ }$      ------(i)

Statement I: AB = BA

Pre-multiplication by A,

​⇒ $\displaystyle AA^{-1}=\frac{A.\ adj\ A}{∣\ A ∣ }$​

⇒ $\displaystyle I=\frac{A.\ adj\ A}{∣\ A ∣ }$

⇒ I ∣A∣ = AB      [As B = adj A]      ------(ii)

Post-multiplication by A,

​⇒ $\displaystyle A^{-1}A=\frac{\ adj\ A.A}{∣\ A ∣ }$​

⇒ $\displaystyle I=\frac{\ adj\ A.A}{∣\ A ∣ }$

⇒ I ∣A∣ = BA      [As B = adj A]      ------(iii)

From equation (ii) and (iii), we have,

AB = BA

Statement I is true.

Statement II: AB is a scalar matrix

From equation (ii), we have, 

AB = ∣A∣ I 

⇒ AB is a scalar matrix

Statement III: AB can be a null matrix

A is a non-singular matrix.

⇒∣A∣ ≠ 0

A (adjA) = ∣A∣ I ≠  0

AB  ≠  0

∴ Only statement I and Statement II are correct.

Concept:

1). The null matrix is a matrix having zero as all its elements. The null matrix is also called a zero matrix.

2).  An identity matrix is a given square matrix of any order which contains on its main diagonal elements with value of one

Calculation:

Statement I: If AB is a null matrix, then at least one of A and B is a null matrix.

Let us consider two matrices,

\(A = \begin{bmatrix} 0 &0 \[0.3em] 0 & 1\[0.3em] \end{bmatrix}\ and\ B = \begin{bmatrix} 0 &1 \[0.3em] 0 &0\[0.3em] \end{bmatrix} \)

AB = \( \begin{bmatrix} 0 &0 \[0.3em] 0 & 1\[0.3em] \end{bmatrix}\ \begin{bmatrix} 0 &1 \[0.3em] 0 &0\[0.3em] \end{bmatrix}= \begin{bmatrix} 0 &0 \[0.3em] 0 & 0\[0.3em] \end{bmatrix}\ \)

Here, both A and B are non-zero matrices. But the product is zero matrices.

⇒ Statement I is incorrect.

Statement II: If AB is an identity matrix, then BA = AB.

AB = I

Pre-multiplying A-1 on both sides.

⇒ A^-1AB = A^-1I      [Considering A is invertible]

⇒ B = A^-1

Post-multiplying A on both sides.

⇒ BA = A^-1A

⇒ BA = I

⇒ Statement II is correct.

∴ Only statement II is correct.

Concept:

1). Transpose of a Matrix:

     The new matrix obtained by interchanging the rows and columns of the original matrix is called

     the transpose of the matrix.

     For example: \(A = \begin{bmatrix} a &b & c \[0.3em] x & y & z \[0.3em] \end{bmatrix}\)⇒ \(A^T = \begin{bmatrix} a &x \[0.3em] b & y \[0.3em] c & z \[0.3em] \end{bmatrix}\)

     It is denoted by A' or A^T.

2). For the addition and subtraction of two matrices, the order of matrices should be equal.

Calculation:

Given:

\(A = \begin{bmatrix} 1 &0 & 0 \[0.3em] 0& 1 &0\[0.3em] 0 & 0 & 1 \end{bmatrix}\)and \(B = A^T= \begin{bmatrix} 1 &0 & 0 \[0.3em] 0& 1 &0\[0.3em] 0 & 0 & 1 \end{bmatrix}\)

C = A + B

⇒ \(C = \begin{bmatrix} 1 &0 & 0 \[0.3em] 0& 1 &0\[0.3em] 0 & 0 & 1 \end{bmatrix}+ \begin{bmatrix} 1 &0 & 0 \[0.3em] 0& 1 &0\[0.3em] 0 & 0 & 1 \end{bmatrix}\)

⇒ \(C = \begin{bmatrix} 2 &0 & 0 \[0.3em] 0& 2 &0\[0.3em] 0 & 0 & 2 \end{bmatrix}\)

⇒ ∣C∣ = 2(4 - 0)

⇒ ∣C∣ = 8

∴ The determinant of matrix C is 8.

Concept:

Non Singular matrix is a square matrix whose determinant is a non-zero value. 

Calculation:

Given:

AB = A and BA = B

Statement I: A2 = A

We are given that A = AB

⇒ A² = (AB)2

⇒ A2 = A(BA)B

⇒ A2 = A(B)B       [∵ BA = B]

⇒ A2 = (AB)B

⇒ A2 = (A)B           [∵ AB= A]

⇒ A2 = A  

Statement I is true.

Statement II:  AB2 = A2B

We are given that B = BA

⇒ B2 = (BA)2

⇒ B2 = BABA

A is pre-multiplied both sides, we get,

⇒ AB2 = ABABA

⇒ AB2 = ABA(BA)

⇒ AB2 = ABA(B)

⇒ AB2 = (AB)AB

⇒ AB2 = (A)AB

⇒ AB2 = (AA)B

⇒ AB2 = A²B

Statement II is true.

∴ Statement I and II both are true.

Concept:

The total number of terms in the binomial expansion of (a + b)ⁿ is n + 1, i.e.

one more than the exponent n.

Calculation:

$\left(1 + \frac{2}{x}\right)^9\left(1-\frac{2}{x}\right)^9$ can be written as 

⇒ $\left[\left(1 + \frac{2}{x}\right)\left(1-\frac{2}{x}\right)\right]^9$

We know that, (a + b)(a - b) = a² - b²

⇒$\left[1-\left(\frac{2}{x}\right)^2\right]^9$

⇒ Total number of the terms = 9 + 1 = 10

∴ There are 10 terms in the expansion of $\left(1 + \frac{2}{x}\right)^9\left(1-\frac{2}{x}\right)^9.$

Concept:

1). In the expansion of (x + y)ⁿ, the term having the greatest coefficient is the middle term.

2). When n is even, there is only one middle term i.e. $\Big(\displaystyle \frac{n}{2}+1\Big)$th and the greatest coefficient is nCn/2

3). When n is odd, there are two middle terms i.e. $\Big(\displaystyle \frac{n+1}{2}\Big)$th or $\Big(\displaystyle \frac{n+3}{2}\Big)$th  and the greatest coefficient are $\displaystyle ^nC_{\frac{n-1}{2}}$ or $\displaystyle ^nC_{\frac{n+1}{2}}$

4). rth term for the binomial expression, (x + y)n, T_r = nCr-1 xn-r+1 yr-1

Calculation:

The given expansion is (x + y)10

Statement I: Among all the coefficients of the terms, the coefficient of the 6th term has the highest value

The value of n is 10 which is even. So, the term having the greatest coefficient is the $\displaystyle \frac{n+2}{2}$ = $\displaystyle \frac{10+2}{2}$ = 6

∴ The coefficient of the 6th term has the highest value.

Statement II: The coefficient of the 3rd term is equal to the coefficient of the 9th term.

We know that, $^nC_r=\frac{n!}{r!(n-r)!}$

The coefficient of the 3rd term, T₃ = 10C₂ = $\frac{10!}{8!.2!}$

The coefficient of the 9th term, T₉ = 10C₈ = $\frac{10!}{8!.2!}$ 

∴ Coefficient of the 3rd term = Coefficient of the 9th term.

∴ Both Statement I and II are correct.

Concept:

Combination

$\displaystyle ^n C_r=\frac{n !}{r ! (n-r) !}$,

where,

           ⁿC_r = number of combinations

        n = total number of objects in the set

        r = number of choosing objects from the set

Formulae

nCx = nCy ⇒ x = y or x + y = n

Calculation:

Given:

C(3n, 2n) = C(3n, 2n - 7)

⇒ ³ⁿC_2n = 3nC2n-7

⇒ 2n + 2n - 7 = 3n

⇒ 4n - 3n = 7

⇒ n = 7

The value of C(n, n - 5) = nCn-5 = ⁷C7-5 = 7C2

7C2 = $\displaystyle \frac{7!}{2!\times5!}$ = 21

The value of C(n, n - 5) = 21

Concept: 

Formulae 

• nCr ​= nC_n-r​

Calculation:

Given,

C(51, 21) - C(51, 22) + C(51, 23) - C(51, 24) + C(51, 25) - C(51, 26) + C(51, 27) - C(51, 28) + C(51, 29) - C(51, 30)

We know that nCr ​= nCn-r

So we can write,

C(51, 21) = C(51, 30)

C(51, 22) = C(51, 29)

C(51, 23) = C(51, 28)

C(51, 24) = C(51, 27)

C(51, 25) = C(51, 26)

So the above expression can be written as 

C(51, 30) - C(51, 29) + C(51, 28) - C(51, 27) + C(51, 26) - C(51, 26) + C(51, 27) - C(51, 28) + C(51, 29) - C(51, 30)

= C(51, 30) - C(51, 30) + C(51, 29) - C(51, 29) + C(51, 28) - C(51, 28) + C(51, 27) - C(51, 27) + C(51, 26) - C(51, 26)

= 0

Checking option C:

C(51, 51) - C(51, 0)

Using nCr ​= nC_n-r we can write C(51, 51) = C(51, 0),

C(51, 0) - C(51, 0) = 0

∴ The value of the expression is C(51, 51) - C(51, 0).

Concept:

A permutation is an arrangement of objects in a definite order.

$\displaystyle ^n P_r=\frac{n !}{(n-r) !}$,

where, 

           ⁿP_r = Permutation

        n = total number of objects

        r = number of objects selected

Calculation:

Numbers between 300 and 400 consist of three digits with the digit at hundred’s place equal to 3.

Since repetition is not allowed and there must be only odd numbers, one’s places can be filled using the digits 1,5,7 and 9 in ⁴P₁ ways.  

Now ten’s place can be filled by using any digits except the digit at one's and hundred's place in ⁸P₁ ways. 

Hence, the required number of numbers = 4P1 × 8P1 = 4 × 8 = 32.

∴ The total odd numbers between 300 and 400 in which none of the digits is repeated are 32.

Concept:

The number of combinations of n different thing taken k at a time is:

$^nC_k=\frac{n!}{k!(n-k)!}$ where 0 ≤ k ≤ n

Calculation:

The word 'TIGER' contains 2 vowels I and E and 3 consonants T, G, R.

If the vowels should not occupy the even position means even places should be occupied by the consonants only. 

To fill the even places by the consonants, select any 2 consonants out of 3 and arrange them.

Selection of any 2 consonants out of 3 = ³C₂

The arrangement of these two consonants is given by 2!

Number of ways to fill even places = 3C2 × 2! = 6 

Now, fill the 3 odd places with the rest of the consonants and vowels.

Number of ways to fill odd places = 3! = 6

The total number of ways = 3C2 × 2! × 3! = 6 × 6 = 36

∴ The required number of permutations = 36.

Concept :

If α and β are the roots of quadratic equation ax2 + bx + c = 0

Sum of root α + β = -b/a

Product of root αβ = c/a

Formula used :

1. a³ + b³ = (a + b)(a² + b² - ab)

2. a² + b²  = (a + b)² - 2ab

Calculation :

Given that, α and β be the roots of the equation x2 + px + q = 0

⇒ α + β = -p        -----(1)

αβ = q                 -----(2)

Using the formula (1)

α³ + β³ = (α + β)(α²  + β² - αβ) 

Again using the  formula (2)

α3 + β3 = (α + β)[(α + β)² - 3αβ]   

Put the value form equation (1) & (2)

α3 + β3 = (-p)[(-p)2 - 3 × q] 

⇒ α3 + β3 = -(p³ - 3pq)     ----(3)

Also, α3 and β3 are the roots of the equation x2 + mx + n = 0

⇒ α³ + β³ = - m    

⇒ m = (p3 - 3pq)       [From equation (3)]

α³β3 = n

⇒ n = q³         [From equation (2)]

⇒  m + n = p³ - 3pq + q³

∴  m +  n = p³ + q³ - 3pq

Formula used:

If α and β are the roots of quadratic equation ax2 + bx + c = 0

Sum of the roots: α + β = -b/a

Product of the roots: αβ = c/a

Quadratic equation with roots α and β can be written as

x² - (α + β)x + αβ = 0

Calculation: 

Given that α and β be the roots of the equation

x2 - ax - bx + ab - c = 0   

⇒ x² - (a + b)x + (ab - c) = 0 

Hence, sum of the roots 

α + β  = a + b        -----(1)

Product of the roots

αβ = ab - c

⇒ ab = αβ + c      -----(2)

Hence, quadratic equation with roots a & b is 

x2 - (a + b)x + ab = 0      

Put the value from the equations (1) & (2).

⇒ x² - (α + β)x + αβ + c = 0

∴ x² - αx - βx + α β + c = 0

Concept:

For a quadratic equation ax² + bx + c = 0,

Discriminant (D) = b² - 4ac

For equal roots, D = 0

Formula used:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Calculation:

Given that,

x2 - ax - bx - cx + bc + ca = 0

⇒ x² - (a + b + c)x + bc + ca = 0

For equal roots, D = 0

⇒ [-(a + b +  c)]²  - 4(bc + ca) = 0

⇒ a2 + b2 + c2 + 2ab + 2bc + 2ca - 4bc - 4ca = 0

⇒ a2 + b2 + c2 + 2ab - 2bc - 2ca = 0

Using the above formula, we can write

(a + b - c)² = 0

∴  a + b - c = 0

Concept:

The Standard Form of a Quadratic Equation is ax² + bx + c = 0

Sum of roots = −b/a

Product of roots = c/a

Formulae 

(α - β)2 + 4.α.β = (α + β)2 

Calculation:

Given:

x2 - 8x + q = 0 and α2 - β2 = 16

Sum of roots = α + β = −b/a = -(- 8) = 8      ------(i)

Product of roots  = α.β = q      ------(ii)

We have α2 - β2 = 16

⇒ (α + β)(α - β) = 16

⇒ 8 × (α - β) = 16

⇒ (α - β) = 2      ------(iii)

We know that, (α - β)² + 4.α.β = (α + β)2 

Putting the value of equation (i), (ii) and (iii), we get,

⇒ 2² + 4q = 8²

⇒ 4q = 64 - 4 = 60

⇒ q = 15

∴ The value of q is 15

Concept:

The highest power of the prime number p in n! = $\displaystyle \frac{n}{p^1}+\frac{n}{p^2}+\frac{n}{p^3}.....$

Calculation:

(30! + 35!)

⇒ (30! + 35 × 34 × 33 × 32 × 31 × 30!) 

⇒ 30! (1 + 35 × 34 × 33 × 32 × 31)

⇒ 30! (1 + 7 × 5 × 17 × 2 × 3 × 11 × 2⁵ × 31)

Now, the number of fives(5) in 30! is given by $\displaystyle \frac{30}{5^1}+\frac{30}{5^2}+\frac{30}{5^3}= 6+1+0=7$

So, 30! has 7 fives, and (1 + 7 × 5 × 17 × 2 × 3 × 11 × 25 × 31) has zero 5. 

∴ The max number of 5's is 7 i.e. n = 7

Concept:

1.1! + 2.2! + 3.3! + 4.4! + ..................... + n.n! = (n+1)! - 1

Calculation:

2(2 × 1) + 3(3 × 2 × 1) + 4(4 × 3 × 2 × 1) + 5(5 × 4 × 3 × 2 × 1) + ................. + 9(9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1) + 2

⇒ 2.2! + 3.3! + 4.4! + 5.5! + 6.6! + 7.7! + 8.8! + 9.9! + 2

⇒ (3 - 1).2! + (4 - 1).3! + (5 - 1).4! + (6 - 1).5! + (7 - 1).6! + (8 - 1).7! + (9 - 1).8! + (10 - 1).9! + 2

⇒ 3.2! - 2! + 4.3! - 3! + 5.4! - 4! + 6.5! - 5! + 7.6! - 6! + 8.7! - 7! + 9.8! - 8! + 10.9! - 9! + 2

⇒ 3! - 2! + 4! - 3! + 5! - 4! + 6! - 5! + 7! - 6! + 8! - 7! + 9! - 8! + 10! - 9! + 2

⇒ (3! + 4! + 5! + 6! + 7! + 8! + 9! + 10!) - (2! + 3! + 4! + 5! + 6! + 7! + 8! + 9!) + 2

⇒ 10! - 2! + 2

⇒ 10! - 2 + 2

⇒ 10!

∴ 2.2! + 3.3! + 4.4! + 5.5! + 6.6! + 7.7! + 8.8! + 9.9! + 2 = 10!

Concept:

Power set B, of any set A, is a set of all the subsets of A and the number of elements of B will be 2ⁿ where n is the number of elements of A.

Calculation:

Given:

A = {(1, 2, 3)} here n = 1

∴ Number of elements in the power set of A = 2¹ = 2

Concept:

1). The general form of terms of a GP is a, ar, ar², ar³, and so on. Here, a is the first term and r is the common ratio.

       The nth term of a GP is T_n = ar^n-1

       Common ratio = r = T_n/ T_n-1

2). If a, b and c are three quantities in GP, then and b is the geometric mean of a and c. This can be written as b² = ac or b =√ac

Calculation:

Given:

a, b, c are in GP where a > 0, b > 0, c > 0

Statement I: a2, b2, c2 are in GP

As b is G.M. of a and c, b² = ac

Squaring on both sides 

⇒ b⁴ = a²c²

⇒ (b²)² = (a²)(c²) 

⇒ b² is G.M. of a² and c²

a², b², and c² are in GP

Statement II:  $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in GP

Let the common ratio be r.

⇒ b = ar

⇒ c = br = ar²

Now, $\displaystyle \frac{\frac{1}{b}}{\frac{1}{a}}=\frac{a}{b}=\frac{a}{ar}=\frac{1}{r}$

⇒ $\displaystyle \frac{\frac{1}{c}}{\frac{1}{b}}=\frac{b}{c}=\frac{ar}{ar^2}=\frac{1}{r}$

1/a,1/b,1/c are in G.P.

Statement III:  $√ {a}, √{b}, √{c}$ are in GP

As b is G.M. of a and c, ​${b} = √{ac}$​

Taking square roots on both sides 

⇒ $√{b} = √{√{a} √{c}}$

⇒ √b is GM of √a and √c

 $√ {a}, √{b}, \ and\ √{c}$ are in GP

∴ Statement I, II, and III are correct.

Concept:

Arithmetic Progression:

If a, b and c are in AP then

b - a = c - b or 2b = a + c

Geometric Progression:

If a, b and c are in GP then

b/a = c/b  or b² = ac

Harmonic Progression:

If a, b and c are in HP then 

1/a, 1/b, and 1/c will be on AP.

Calculation:

Given:

$\displaystyle \frac{a+b}{2}, b, \frac{b+c}{2}$ are in HP

⇒ $\displaystyle \frac{2}{a+b}, \frac{1}{b}, \frac{2}{b+c}$ are in AP

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

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

⇒ $\displaystyle \frac{2}{b}=2\left[\frac{1}{b+c}\ +\frac{1}{a+b}\right]$

⇒ $\displaystyle \frac{2}{b}=2\left[\frac{a+b+b+c}{(a+b)(b+c)}\right]$

⇒ $\displaystyle \frac{1}{b}=\left[\frac{a+2b+c}{(a+b)(b+c)}\right]$

⇒ (a + b)(b + c) = ab + 2b² + bc

⇒ ab + ac + bc + b² = ab + 2b2 + bc

⇒ b2 = ac

Since b2 = ac, b is the geometric mean of a and c.

∴ a, b, c are in GP

Concept:

tan 15° = 2 - √3

cot 15° = 2 + √3

Formula used:

(a + b)² = a² + 2ab + b²

(a - b)2 = a2 - 2ab + b2

Calculation:

Let the required value is y.

⇒ y = cot2 15° + tan2 15°

⇒ y = (2 + √3)² + (2 - √3)²

Using the formula given above

⇒ y = (4 + 4√3 + 3) + (4 - 4√3 + 3) 

⇒ y = 8 + 6 = 14

∴ cot2 15° + tan2 15° = 14

Formula used :

sin 2θ = 2 sin θ cos θ 

cos C + cos D = 2cos($\frac{C+D}{2}$) cos($\frac{C-D}{2}$)

Calculaton :

sin A - cos B - cos C = 0

⇒ sin A = cos B + cos C

Using the formula given above

⇒ 2 sin(A/2) cos(A/2) = 2cos($\frac{B+C}{2}$) cos($\frac{B-C}{2}$)

We know that, for a ΔABC, ∠ A + ∠ B + ∠ C = π 

⇒ sin(A/2) cos(A/2) = cos($\frac{π - A}{2}$) cos($\frac{B-C}{2}$)

Since, cos(π/2 - θ) = sin θ 

⇒ sin(A/2) cos(A/2) =  sin($\frac{ A}{2}$) cos($\frac{B-C}{2}$)

⇒ cos(A/2) =  cos($\frac{B-C}{2}$)

⇒ A/2 = (B - C)/2

⇒ A = B - C

⇒ B = A + C

But ∠ A + ∠ B + ∠ C = π 

⇒ B = π - B

⇒ 2B = π 

⇒  B = π/2

∴ Angle  B is equal to $\frac{\pi}{2}$.

Formula used:

$tan(A-B) = \frac{tan\ A - tan \ B}{1 + tanA\ tanB}$

$tan\ 2β = \frac{2tanβ}{1-tan^2β}$

Calculation:

Given that, 2 tan α = 1

⇒ tan α = 1/2           ------(1)

Also, we have $α + β = \frac{π}{4}$

⇒ $β = \frac{π}{4}-α$

$⇒ tan\ β = tan(\frac{π}{4}-α)$

$⇒ tan\ β = \frac{tan(\frac{\pi}{4})-tan\alpha}{1+tan(\frac{\pi}{4})tan\alpha}$       

By using the formula given above

$⇒ tan\ β = \frac{1 -tanα}{1+tanα}$                                    [∵ tan(π/4) = 1]

Put the value of tanα from equation (1)

$⇒ tan\ β = \frac{1 -\frac{1}{2}}{1+\frac{1}{2}}=\frac{1}{3}$

⇒ tan β = 1/3        -----(2)

We know that, 

 $tan\ 2β = \frac{2tanβ}{1-tan^2β}$

$⇒ tan\ 2β = \frac{2(\frac{1}{3})}{1-(\frac{1}{3})^2}=\frac{\frac{2}{3}}{1-\frac{1}{9}}$

∴  tan 2β = $\frac{3}{4}$

Formula used:

tan(A + B) = $\displaystyle \frac{tan\ A+\ tan\ B}{1\ -\ tan\ A.\ tan\ B}$

sin 2θ = 2 sin θ cos θ   

Calculation:

tan (45° + θ) = 1 + sin 2θ

⇒ $\displaystyle \frac{tan\ 45^o+\ tan\ θ}{1\ -\ tan\ 45^o.\ tan\ θ}$ = 1 + sin 2θ

⇒ $\displaystyle \frac{1+\ tan\ θ}{1\ -\ tan\ θ}$ = 1 + sin 2θ

Since, tan θ = sin θ/cos θ 

⇒ $\displaystyle \frac{cos\ θ+\ sin\ θ}{cos\ θ-\ sin\ θ}$ = 1 + sin 2θ

⇒ cos θ + sin θ = cos θ - sin θ + sin 2θ(cos θ - sin θ)

⇒ 2 sin θ = sin 2θ(cos θ - sin θ)

We know that, sin 2θ = 2 sin θ cos θ   

⇒ 2 sin θ = 2 sin θ. cos θ (cos θ - sin θ)

⇒ 1 = cos θ (cos θ - sin θ)

⇒ 1 = cos²θ - cos θ.sin θ

⇒ sin2θ = - cos θ.sin θ

⇒ sin2θ + cos θ.sin θ = 0

⇒ sin θ(sin θ + cos θ) = 0

⇒ sin θ = 0 or sin θ = - cos θ

⇒ sin θ = 0 or tan θ = -1

⇒ θ = 0 or θ = $\displaystyle -\frac{\pi}{4}$

As θ lies in $\displaystyle -\frac{\pi}{4}< θ <\frac{\pi}{4},$ 

⇒ θ = 0^o

cos 2θ = cos (2 × 0) = 1

∴ cos 2θ = 1

Concept:

If A + B = 90° then

sin A = cos B

Calculation:

Given that, sin 2θ = cos 3θ

We know that sin A = cos B then A + B = 90° 

⇒ 2θ + 3θ = 90°  

⇒ 5θ = 90° 

⇒ θ = 18° 

Hence, required value 

1 + 4 sin θ  

⇒ 1 + 4 sin 18°   

⇒ 1 + 4((√5 - 1)/4)       [∵ $\sin 18^\circ =\frac{√ 5 - 1}{4}$]

⇒ 1  + √5 - 1 = √5

∴ 1 + 4 sin θ = √5

Concept:

Calculation:

Given that, 

$\tan θ = - \frac{5}{12}$ = perpendicular/Base

In triangle ABC, using the Pythagoras theorem

AC² = AB² + BC²

⇒ AC² = 5² + 12² = 13²

⇒ AC = 13 

Hence, 

sin θ = perpendicular/hypotenuse

⇒ sin θ = 5/13

Since tan θ is negative in 2^nd and 4^th quadrants.

sin θ in 2^nd quadrants = 5/13

sin θ in 4^th quadrants = - 5/13

∴ Both 5/13 and -5/13 are possible.

Concept:

Formula used:

1. cos 2θ = 2cos²θ - 1
2. cos(2π - θ) = cos θ
3. cos(π + θ) = - cos θ
4. (a - b)² = a² + b² - 2ab
5. (a + b)²  = a² +b² + 2ab

 

Calculation:

Let the required value is y.

⇒ y = $\cos^4\frac{7π}{8}+\cos^4\frac{5π}{8}$

⇒ y =  $(\cos^2\frac{7π}{8})^2+(\cos^2\frac{5π}{8})^2$

Using the formula discussed above

⇒ y = $\Big(\frac{\cos( \frac{2\times7π }{8})+1}{2}\Big)^2+\Big(\frac{\cos (\frac{2\times5π }{8})+1}{2}\Big)^2$

⇒ y = $(\frac{\cos( \frac{7π }{4})+1}{2})^2+(\frac{\cos (\frac{5π }{4})+1}{2})^2$

⇒ y = $(\frac{\cos(2\pi- \frac{π}{4})+1}{2})^2+(\frac{\cos(\pi+\frac{π}{4})+1}{2})^2$

Using the formula (2) & (3)

⇒ y = $(\frac{\cos (\frac{π}{4})+1}{2})^2+(\frac{1-\cos(\frac{π}{4})}{2})^2$

Since, cos(π/4) = 1/√2

⇒ y = $(\frac{\ \frac{1}{\sqrt 2}+1}{2})^2+(\frac{1-\frac{1}{\sqrt 2}}{2})^2$

⇒ y = $(\frac{\sqrt 2 +1}{2\sqrt2})^2+(\frac{\sqrt 2-1 }{2\sqrt2})^2$

Using the formula (4) & (5)

⇒ y = $\frac{ (2 + 1+2\sqrt 2)+(2 + 1 - 2\sqrt 2)}{8}$

⇒ y = 6/8 = 3/4

∴ $\cos^4\frac{7π}{8}+\cos^4\frac{5π}{8}$ = 3/4

Formula used:

cos($\frac{\pi}{2}-θ$) = sin θ 

cos²θ - sin²θ = cos 2θ 

Calculation:

Let the required value be y.

⇒ y = $\sin^2\left(\frac{\pi}{4}+θ\right)-\sin^2\left(\frac{\pi}{4}-θ\right)$

Since cos($\frac{\pi}{2}-θ$) = sin θ 

⇒ y = $\cos^2\left(\frac{\pi}{2}-(\frac{\pi}{4}+θ)\right)-\sin^2\left(\frac{\pi}{4}-θ\right)$

⇒ y = $\cos^2\left(\frac{\pi}{4}-θ\right)-\sin^2\left(\frac{\pi}{4}-θ\right)$

We know that, cos2θ - sin2θ = cos 2θ 

⇒ y = cos 2($\frac{\pi}{4}-θ$)

⇒ y = cos ($\frac{\pi}{2}-2θ$) = sin 2θ 

∴ $\sin^2\left(\frac{\pi}{4}+θ\right)-\sin^2\left(\frac{\pi}{4}-θ\right)$ = sin 2θ

Given:

A flagstaff of height h is surmounted on a verticle tower.

Angles of elevation of the bottom and top of the flagstaff are θ and 2θ respectively.

Formula used:

1. tan θ = Perpendicular/Base

2. tan 2θ = $\frac{(2tan θ)}{(1 - tan^2θ)}$

3. cos 2θ = $\frac{1-tan^2θ}{1+tan^2θ}$

 

Calculation:

Let the height of the tower be x.

In right triangle ABC

tan θ = AC/AB = x/AB

⇒ AB = x/tan θ          ------(1)

In right triangle ABD

tan 2θ = AD/AB = (x + h)/AB

⇒ $\frac{(2tan θ)}{(1 - tan^2θ)}$ = (x + h)/AB   [Using formula (2)]

⇒ $\frac{(2tan θ)}{(1 - tan^2θ)}$ = $\frac{ (x + h)tan\ θ}{ x}$     [From equation (1)]

⇒ $\frac{(2)}{(1 - tan^2θ)}$ = $1+\frac{ h}{ x}$ 

⇒ $\frac{h}{x} = \frac{2}{1-tan^2θ}-1$

⇒  $\frac{h}{x} = \frac{2-(1-tan^2θ)}{1-tan^2θ}$

⇒  $\frac{h}{x} = \frac{1+tan^2θ}{1-tan^2θ}$ 

⇒ $\frac{x}{h} = \frac{1-tan^2θ}{1+tan^2θ}$ = cos2θ       [Using formula (3)]

⇒ x = hcos 2θ 

∴ The height of the tower is h cos 2θ

Formula used:

1. tan θ = Perpendicular/Base

2. tan θ increases with the increase of θ when θ ∈ (0, 90°)

• tan 30° = 1/√3 ≈ 0.57
• tan 45° = 1
• tan 60° = √3 ≈ 1.732

 

Calculation:

In Δ ABC, 

tan 60° = AB/BC = √3x/BC

⇒ √3 = √3x/BC

⇒ BC = x        -----(1)

In Δ  ABD

tan θ = AB/BD = √3x/(BC + x)

⇒ tan θ = √3x/2x

⇒ tan θ = √3/2  ≈ 0.86  

Since, 0.57 < 0.86 < 1

∴ 30° < θ < 45°