NDA Mathematics 19 April 2015 Official Paper (Previous Year Paper)

Concept:

Let R be a binary relation on a set A.

1. Reflexive: Each element is related to itself.

• R is reflexive if for all x ∈ A, xRx.

2. Symmetric: If any one element is related to any other element, then the second element is related to the first.

• R is symmetric if for all x, y ∈ A, if xRy, then yRx.

3. Transitive: If any one element is related to a second and that second element is related to a third, then the first element is related to the third.

• R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

4. R is an equivalence relation if A is nonempty and R is reflexive, symmetric and transitive.

Calculation:

Given: x < y if y at least 5 years older than x

⇒ y ≥ x + 5

For Reflexive: (x, x) should ∈R for all x ∈ X

Now, x cannot be 5 years older than himself. So the relation is not reflexive.

For Symmetric: If (x, y) ∈ R ⇒(y, x) ∈ R

(x, y) ∈ R ⇒ y is at least 5 years older than x.

(y, x) ∈ R ⇒ x is at least 5 years older than y. This contradicts the above statement. Hence the relation is not symmetric

For Transitive: If (x, y) ∈ R and (y, z) ∈ R ⇒ (x, z) ∈ R.

(x, y) ∈ R ⇒ y is at least 5 years older than x.

(y, z) ∈ R ⇒ z is at least 5 years older than y.

Then, (x, z) ∈ R ⇒ z is at least 5 years older than x.

Since, z is at least 10 years older than x. The relation is transitive.

Concept:

• Elementary matrix: An elementary matrix is a matrix which differs from the identity matrix by one single elementary row operation
• An elementary matrix has each diagonal element 1.

Calculation:

Let's check option b,

Let E = \left[ {\begin{array}{*{20}{c}} 1&5&0\ 0&1&0\ 0&0&1 \end{array}} \right]

Apply R₁ → R₁ – 5R₂

\Rightarrow {\rm{E;}} = \left[ {\begin{array}{*{20}{c}} 1&0&0\ 0&1&0\ 0&0&1 \end{array}} \right] = ;{{\rm{I}}_{3{\rm{;}} \times 3}}

We can see that option B converted in to an identity matrix by one elementary operation.

Hence Option B is correct.

Shortcut Method:

We know that an elementary matrix has each diagonal element 1.

Only B has diagonal element is 1. (By Definition)

So, Option B is correct.

Concept:

Let us consider the standard form of a quadratic equation, ax² + bx + c =0

Where a, b and c are constants with a ≠ 0

• ${\bf{x}} = ;\frac{{ - {\bf{b}}; \pm ;\sqrt {{{\bf{b}}^2} - 4{\bf{ac}}} }}{{2{\bf{a}}}}$

Calculation:

Given:

${\left( {{x^2} + 2} \right)^2} + 8{x^2} = 6x\left( {{x^2} + 2} \right)$

Let x² + 2 = t

⇒ t² + 8x² = 6xt

⇒ t² - 6xt + 8x² = 0

⇒ t² - 4xt - 2xt + 8x² = 0

⇒ t (t – 4x) – 2x (t – 4x) = 0

⇒ (t – 2x) (t – 4x) = 0

Now, putting the value of t in above equation, we get

⇒ (x² + 2 – 2x) (x² + 2 – 4x) = 0

Now, (x² – 2x + 2) = 0 and (x² – 4x + 2) = 0

⇒ x² – 2x + 2 = 0

${\rm{x}} = {\rm{;}}\frac{{ - \left( { - 2} \right){\rm{;}} \pm {\rm{;}}\sqrt {{{\left( { - 2} \right)}^2} - 4{\rm{;}} \times 1 \times 2} }}{{2{\rm{;}} \times 1}} = {\rm{;}}\frac{{2{\rm{;}} \pm {\rm{;}}\sqrt { - 4} }}{2} = {\rm{;}}\frac{{2{\rm{;}} \pm {\rm{;}}2{\rm{i}}}}{2} = 1{\rm{;}} \pm {\rm{i}}$

Roots are imaginary.

Now,

⇒ (x² – 4x + 2) = 0

${\rm{x}} = {\rm{;}}\frac{{ - \left( { - 4} \right){\rm{;}} \pm {\rm{;}}\sqrt {{{\left( { - 4} \right)}^2} - 4{\rm{;}} ;\times ;1 ;\times; 2} }}{{2{\rm{;}}; \times ;1}} = {\rm{;}}\frac{{4{\rm{;}} \pm {\rm{;}}\sqrt 8 }}{2} = {\rm{;}}\frac{{4{\rm{;}} \pm {\rm{;}}2\sqrt 2 }}{2} = 2{\rm{;}} \pm \sqrt 2$

Roots are real.

We can say that not all roots of the equation are complex.

So statement 1^st is wrong.

Now,

Sum of all the roots = (1 + i) + (1 – i) + (2 + √2) + (2 - √2) = 6

So statement 2^nd is true.

Concept:

Standard form of a quadratic equation can be written as:

• ax² + bx + c = x² – (sum of roots) x + (product of roots) = 0

Calculation:

Let f(x) = x² – (sum of roots) x + (product of roots) = 0

Given: one student makes a mistake in the constant term and obtains 8 and 2 for roots.

That means 1^st student obtain correct sum of roots

∴ Sum of roots = 8 + 2 = 10

Given: Another student makes a mistake only in the coefficient of first-degree term and finds -9 and -1 for roots.

That means 2^nd student obtain correct product of roots

∴ Product of roots = -9 × -1 = 9

Now,

f(x) = x² – 10x + 9 = 0

Concept:

• If \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} {{{\rm{a}}{11}}}&{{{\rm{a}}{12}}}\ {{{\rm{a}}{21}}}&{{{\rm{a}}{22}}} \end{array}} \right]\) then determinant of A is given by: |A| = (a­₁₁ × a₂₂) – (a₁₂ – a₂₁).
• ${{\rm{A}}^{ - 1}} = {\rm{;}}\frac{{{\rm{Adj;}}\left( {\rm{A}} \right)}}{{\left| {\rm{A}} \right|}}$

Note:

For 2 × 2 matrix,

Let A = \(\left[ {\begin{array}{{20}{c}} {\rm{a}}&b\ {\rm{c}}&{\rm{d}} \end{array}} \right]\) than Adj A = \(\left[ {\begin{array}{{20}{c}} {\rm{d}}&{ - b}\ { - {\rm{c}}}&{\rm{a}} \end{array}} \right]\) 

Note: Interchange the diagonal elements and change the sign of remaining elements**.**

Calculation:

Given: A = ;\left[ {\begin{array}{*{20}{c}} 2&7\ 1&5 \end{array}} \right]

⇒ |A| = 10 – 7 = 3

We know that for 2 × 2 matrix, interchanging the diagonal elements and changing the sign of remaining elements gives an adjoint matrix.

\therefore {\rm{Adj;}}\left( {\rm{A}} \right){\rm{;}} = \left[ {\begin{array}{*{20}{c}} 5&{ - 7}\ { - 1}&2 \end{array}} \right]

Now,

{{\rm{A}}^{ - 1}} = {\rm{;}}\frac{{{\rm{Adj;}}\left( {\rm{A}} \right)}}{{\left| {\rm{A}} \right|}} = ;\frac{{\left[ {\begin{array}{*{20}{c}} 5&{ - 7}\ { - 1}&2 \end{array}} \right]}}{3}

\Rightarrow 3{{\rm{A}}^{ - 1}} = {\rm{;}}\left[ {\begin{array}{*{20}{c}} 5&{ - 7}\ { - 1}&2 \end{array}} \right]

We have to find the value of A + 3A^-1

⇒ A + 3A^-1 = \(\left[ {\begin{array}{{20}{c}} 2&7\ 1&5 \end{array}} \right] + ;\left[ {\begin{array}{{20}{c}} 5&{ - 7}\ { - 1}&2 \end{array}} \right] = ;\left[ {\begin{array}{{20}{c}} 7&0\ 0&7 \end{array}} \right] = 7\left[ {\begin{array}{{20}{c}} 1&0\ 0&1 \end{array}} \right]\) 

∴ A + 3A^-1 = 7I

Calculation:

Total number of students in class = 60

Students who like music = A + B = 45     …. (1)

Students who like dancing = B + C = 50     …. (2)

Students who like nothing =5

Students who like both music and dancing = B

We know that A + B + C + 5 = 60

From equation 1^st,

⇒ 45 + C + 5 = 60

⇒ C = 10

Put the value of C in equation 2^nd, we get

⇒ B = 40

Hence, 40 students like both music and dancing.

Concept:

Logarithm properties

1. Product rule: The log of a product equals the sum of two logs.

${\log _{\rm{a}}}\left( {{\rm{mn}}} \right) = {\rm{;}}{\log _{\rm{a}}}{\rm{m}} + {\rm{;}}{\log _{\rm{a}}}{\rm{n}}$

1. Quotient rule: The log of a quotient equals the difference of two logs.

${\log _{\rm{a}}}\frac{{\rm{m}}}{{\rm{n}}} = {\rm{;}}{\log _{\rm{a}}}{\rm{m}} - {\rm{;}}{\log _{\rm{a}}}{\rm{n}}$

1. Power rule: In the log of a power the exponent becomes a coefficient.

${\log _{\rm{a}}}{{\rm{m}}^{\rm{n}}} = {\rm{n}}{\log _{\rm{a}}}{\rm{m}}$

• If a, b, c are in A.P ⇔ b = $\frac{{{\rm{a;}} + {\rm{;c}}}}{2}$ or 2b = a + c

Calculation:

Given: log₁₀ 2, log₁₀ (2^x - 1) and log₁₀ (2^x + 3) are three consecutive terms of an AP

We know that if a, b, c are in A.P than 2b = a + c

⇒ 2 × log₁₀ (2^x- 1) = log₁₀ 2 + log₁₀ (2^x + 3)

⇒ log₁₀ (2^x- 1)² = log₁₀ (2 × (2^x + 3))

⇒ (2^x- 1)² = 2 × (2^x+ 3)

Let 2^x = t

⇒ (t - 1)² = 2 × (t + 3)

⇒ t² – 2t + 1 = 2t + 6

⇒ t² – 4t - 5 = 0

⇒ t² – 5t + t - 5 = 0

⇒ t (t – 5) + 1 (t – 5) = 0

⇒ (t – 5) (t + 1) = 0

∴ t = 5, -1

⇒ 2^x ≠ -1         (∵2^x > 0)

Now,

⇒ 2^x = 5

∴ x = log₂ 5

Concept:

1. Symmetric Matrix: Square matrix A is said to be symmetric if the transpose of matrix A is equal to matrix A itself ⇔ A^T = A or A’ = A

2. Skew-Symmetric Matrix or Anti-symmetric: Square matrix A is said to be skew-symmetric if transpose of matrix A is equal to negative of matrix A ⇔ A^T = −A

3. Hermitian Matrix:  A Hermitian matrix is a complex square matrix that is equal to its own conjugate transpose ⇔ ${{\bf{A}}^{\bf{T}}} = ;{\bf{\bar A}}$

4. Skew-Hermitian: ${{\bf{A}}^{\bf{T}}} = - {\bf{\bar A}}$

5. Let z = x + iy be a complex number.

• Conjugate of z =

 = x – iy

Calculation:

Let A = \left[ {\begin{array}{*{20}{c}} 0&{ - 4 + i}\ {4 + i}&0 \end{array}} \right]

Now, transpose of matrix A

 A^T = \left[ {\begin{array}{*{20}{c}} 0&{4 + i}\ { - 4 + i}&0 \end{array}} \right]

⇒ A^T ≠ A

∴ given matrix is not symmetric

A^T = \left[ {\begin{array}{*{20}{c}} 0&{4 + i}\ { - 4 + i}&0 \end{array}} \right] \ne - {\rm{A}}

∴ given matrix is not skew-symmetric.

Now, Conjugate of matrix A

\({\rm{\bar A}} = ;\left[ {\begin{array}{{20}{c}} 0&{\overline { - 4 + i} }\ {\overline {4 + i} }&0 \end{array}} \right] = ;\left[ {\begin{array}{{20}{c}} 0&{ - 4 - i}\ {4 - i}&0 \end{array}} \right] = ; - \left[ {\begin{array}{*{20}{c}} 0&{4 + i}\ { - ;4 + i}&0 \end{array}} \right]\)

 

$\Rightarrow {\rm{\bar A}} = - ;{{\rm{A}}^{\rm{T}}}$

∴ given matrix is skew-hermitian matrix.

Concept:

Let R be a binary relation on a set A.

1. Reflexive: Each element is related to itself.

• R is reflexive if for all x ∈ A, xRx.

2. Symmetric: If any one element is related to any other element, then the second element is related to the first.

• R is symmetric if for all x, y ∈ A, if xRy, then yRx.

3. Transitive: If any one element is related to a second and that second element is related to a third, then the first element is related to the third.

• R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

4. R is an equivalence relation if A is nonempty and R is reflexive, symmetric and transitive.

Calculation:

A relation is defined on Z such that aRb ⇒ (a − b) is divisible by 5,

For Reflexive: (a, a) ∈ R.

Since, (a − a) = 0 is divisible by 5. 

Therefore, the relation is reflexive.

For symmetric: If (a, b) ∈ R ⇒ (b, a) ∈ R.

(a, b)∈ R ⇒ (a − b) is divisible by 5.

Now, (b − a) = − (a − b) is also divisible by 5.

Therefore, (b, a) ∈ R

Hence, the relation is symmetric.

For Transitive: If (a, b) ∈ R and (b, a) ∈ R ⇒ (a, c) ∈ R.

(a, b) ∈ R ⇒ (a − b) is divisible by 5.

(b, c) ∈ R ⇒ (b − c) is divisible by 5.

Then,

(a − c) = (a – b + b −c)

(a − c)  = (a − b) + (b − c)

We know that (a − b) is divisible by 5 and (b − c) is divisible by 5 then (a − c) is also divisible by 5.  Therefore, (a, c) ∈ R.

Hence, the relation is transitive.

∴ the relation is equivalent.

Now, depending upon the remainder obtained when dividing (a−b) by 5 we can divide the set Z into 5 equivalent classes and they are disjoint i.e., there are no common elements between any two classes.

Concept:

Let z = x + iy be a complex number.

• Modulus of z = $\left| {\rm{z}} \right| = {\rm{;}}\sqrt {{{\rm{x}}^2} + {{\rm{y}}^2}} = {\rm{;}}\sqrt {{\rm{Re}}{{\left( {\rm{z}} \right)}^2} + {\rm{;Im;}}{{\left( {\rm{z}} \right)}^2}}$
• arg (z) = arg (x + iy) = ${\tan ^{ - 1}}\left( {\frac{{\rm{y}}}{{\rm{x}}}} \right)$

Calculation:

Given:

${\rm{z}} = {\rm{;}} - \frac{{2\left( {1 + 2{\rm{i}}} \right)}}{{3 + {\rm{i}}}}$

$\Rightarrow {\rm{z}} = {\rm{;}} - \frac{{2\left( {1 + 2{\rm{i}}} \right)}}{{3 + {\rm{i}}}} \times \frac{{3 - {\rm{i}}}}{{3 - {\rm{i}}}}$

$\Rightarrow {\rm{z}} = {\rm{}} - \frac{{2\left( {3 + 6{\rm{i}} - {\rm{i}} - 2{{\rm{i}}^2}} \right)}}{{{3^2} - {{\rm{i}}^2}}}$                 (∵ (a + b) (a – b) = (a² – b²))

$\Rightarrow {\rm{z}} = {\rm{}} - \frac{{2\left( {3 + 5{\rm{i}} + 2} \right)}}{{9 + 1}} = ; - \frac{{2\left( {5 + 5{\rm{i}}} \right)}}{{10}} = ; - 1 - i$

Here z lies in 3^rd quadrant,

∴ Argument θ = -π + tan^-1 (1/1) = -π + π/4 = -3π/4

Concept:

Let us consider the standard form of a quadratic equation, ax² + bx + c =0

Let α and β be the two roots of the above quadratic equation. 

• The sum of the roots of a quadratic equation are:

${\rm{\alpha }} + {\rm{\beta }} = - \frac{{\rm{b}}}{{\rm{a}}} = - \frac{{{\rm{coefficient;of;x}}}}{{{\rm{coefficient;of;}}{{\rm{x}}^2}}}$

• The product of the roots is given by:

${\rm{\alpha \beta }} = \frac{{\rm{c}}}{{\rm{a}}} = \frac{{{\rm{constant;term}}}}{{{\rm{coefficient;of;}}{{\rm{x}}^2}}}$

Calculation:

(x + p) (x + q) – k = 0

⇒ x² + (p + q) x + pq – k = 0

Given m and n are the roots of this equation

So,

Sum of roots = m + n = − (p + q)    …. (1)

Product of the roots = mn = pq − k

⇒ pq = mn + k      …. (2)

Consider, (x−m) (x−n) + k = 0

⇒ x²− (m + n) x + mn + k = 0

From equation 1 and 2,

⇒ x²+ (p + q) x + pq = 0

⇒ x²+ px + qx + pq = 0

⇒ x (x + p) + q(x + p) = 0

⇒ (x + p) (x + q) = 0

∴ x = -p, -q

Hence, the roots of the new equation are −p, −q

Concept:

Let us consider sequence a₁, a₂, a₃ …. a_n is an G.P.

• Common ratio = r = \(\frac{{{{\rm{a}}_2}}}{{{{\rm{a}}_1}}} = \frac{{{{\rm{a}}3}}}{{{{\rm{a}}2}}} = \ldots = \frac{{{{\rm{a}}{\rm{n}}}}}{{{{\rm{a}}{{\rm{n}} - 1}}}}\)
• n^th  term of the G.P. is an = arⁿ⁻¹
• Sum of n terms = s = $\frac{{{\rm{a;}}\left( {{{\rm{r}}^{\rm{n}}} - 1} \right)}}{{{\rm{r}} - {\rm{;}}1}}$; where r >1
• Sum of n terms = s = $\frac{{{\rm{a;}}\left( {1 - {\rm{;}}{{\rm{r}}^{\rm{n}}}} \right)}}{{1 - {\rm{;r}}}}$; where r <1

Calculation:

Let S = 0.5 + 0.55 + 0.555 + … to n terms

⇒ S = 5 [0.1 + 0.11 + 0.111 + … to n terms]

⇒ S = (5/9) × [0.9 + 0.99 + 0.999 + … to n terms]

⇒ S = (5/9) × [(1 - 0.1) + (1 - 0.01+ (1 - 0.001) + … to n terms]

⇒ S = (5/9) × [(1 + 1 + 1 + … to n terms) – (0.1 + 0.01 + 0.001 + … to n terms)]

⇒ S = (5/9) × [n – (0.1 + 0.01 + 0.001 + … to n terms)]      …. (1)

Let S₁ = 0.1 + 0.01 + 0.001 + … to n terms

Sequence is in GP,

Here a = 0.1 = 1/10 and r = 0.1 = 1/10 < 1

Sum = $\frac{{{\rm{a;}}\left( {1 - {\rm{;}}{{\rm{r}}^{\rm{n}}}} \right)}}{{1 - {\rm{;r}}}}$

$\therefore {{\rm{S}}_1} = ;\frac{{\frac{1}{{10}}\left( {1 - ;\frac{1}{{{{10}^n}}}} \right)}}{{\left( {1; - ;\frac{1}{{10}};} \right)}} = \frac{1}{9}\left( {1 - ;\frac{1}{{{{10}^n}}}} \right)$          

Put the value of S₁ in equation 1^st,

$\therefore {\rm{;S;}} = \frac{5}{9}\left[ {{\rm{n}} - \frac{1}{9}\left( {1 - \frac{1}{{{{10}^{\rm{n}}}}}} \right)} \right]$

Concept:

If 1, ω, ω² are the cube roots of unity

i). 1 + ω + ω² = 0

ii). ω³ = 1

Calculation:

We have to find the value of (1 + ω) (1 + ω²) (1 + ω⁴) (1 + ω⁸)

⇒ (1 + ω) (1 + ω²) (1 + ω⁴) (1 + ω⁸)

= [(1 + ω) (1 + ω²)] × [(1 + ω³ × ω) (1 + (ω³)² ω²)]

= [(1 + ω) (1 + ω²)] × [(1 + ω) (1 + ω²)]      (∵ω³ = 1)

= (1 + ω + ω² + ω³) × (1 + ω + ω² + ω³)

= (0 + 1) × (0 + 1)

= 1 × 1 = 1

Concept:

Combination: Selecting r objects from given n objects.

• The number of selections of r objects from the given n objects is denoted by ${{\rm{;}}^{\rm{n}}}{{\rm{C}}_{\rm{r}}}{\rm{;}}$
• ${{\rm{;}}^{\rm{n}}}{{\rm{C}}_{\rm{r}}}{\rm{}} = {\rm{}}\frac{{{\rm{n}}!}}{{{\rm{r}}!\left( {{\rm{n;}} - {\rm{;r}}} \right)!}}$

Note: Use combinations if a problem calls for the number of ways of selecting objects.

Calculation:

Number of elements in A = 10

Number of subsets of A containing exactly two elements = Number of ways we can select 2 elements from 10 elements

⇒ Number of ways we can select 2 elements from 10 elements = ¹⁰C₂ = 45

∴ Number of subsets of A containing exactly two elements = 45

Concept:

Equality of complex numbers.

• Two complex numbers z₁ = x₁+ iy₁ and z₂ = x₂ + iy₂ are equal if and only if x₁ = x₂ and y₁ = y₂

Or Re (z₁) = Re (z₂) and Im (z₁) = Im (z₂).

Calculation:

Let √i = x + iy

Squaring both sides, we get

⇒ i  = (x + iy)²

⇒ i  = x² + i2xy + (iy)²

⇒ x² – y² + i2xy = i

Compare real and imaginary parts,

∴ x² – y² = 0 and 2xy = 1

2xy = 1 ⇒ y = 1/2x       …. (1)

Now, x² – y² = 0

⇒ x² – (1/4x² ) = 0

⇒ 4x⁴ – 1 = 0

⇒ x⁴ = 1/4

∴ x = ± (1/√2)

Put the value of x in equation 1^st,

∴ y = ± (1/√2)

So, square root of i = √i = x + iy =  $\frac{{1 + {\rm{i}}}}{{\sqrt 2 }}$

Alternate method:

$\sqrt {\rm{i}} = {\rm{;}}\sqrt {2{\rm{;}} \times {\rm{;i;}} \times {\rm{;}}\frac{1}{2}} = {\rm{;}}\sqrt {2{\rm{;}} \times {\rm{;}}\frac{{\rm{i}}}{{\sqrt 2 }}{\rm{;}} \times {\rm{;}}\frac{1}{{\sqrt 2 }}}$

$\Rightarrow \sqrt {\rm{i}} = \sqrt {{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2} - {{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2} + {\rm{;}}\left( {2{\rm{;}} \times {\rm{;}}\frac{{\rm{i}}}{{\sqrt 2 }}{\rm{;}} \times {\rm{;}}\frac{1}{{\sqrt 2 }}} \right)}$

$\Rightarrow \sqrt {\rm{i}} = \sqrt {{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2} + {{\left( {\frac{{\rm{i}}}{{\sqrt 2 }}} \right)}^2} + {\rm{;}}\left( {2{\rm{;}} \times {\rm{;}}\frac{{\rm{i}}}{{\sqrt 2 }}{\rm{;}} \times {\rm{;}}\frac{1}{{\sqrt 2 }}} \right)}$        (∵ i² = 1)

$\Rightarrow \sqrt {\rm{i}} = \sqrt {{{\left( {\frac{1}{{\sqrt 2 }} + {\rm{;}}\frac{{\rm{i}}}{{\sqrt 2 }}} \right)}^2}} = {\rm{;}} \pm \left( {\frac{1}{{\sqrt 2 }} + {\rm{;}}\frac{{\rm{i}}}{{\sqrt 2 }}} \right)$                 (∵ (a + b) ² = a² + b² + 2ab)

Convert decimal to binary

Conversion steps:

• Divide the number by 2.
• Get the integer quotient for the next iteration.
• Get the remainder for the binary digit.
• Repeat the steps until the quotient is equal to 0.

Calculation:

 

∴ 127₁₀ = (1111111)₂

Now,

0.25 × 2 = 0.5 → 0

0.5 × 2 = 1 → 1

Hence (127.25)₁₀ = (1111111.01)₂

Concept:

A square matrix is nonsingular if its determinant is nonzero.

If \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} {{{\rm{a}}{11}}}&{{{\rm{a}}{12}}}\ {{{\rm{a}}{21}}}&{{{\rm{a}}{22}}} \end{array}} \right]\) then determinant of A is given by:

|A| = (a­₁₁ × a₂₂) – (a₁₂ × a₂₁).

${{\rm{A}}^{ - 1}} = {\rm{;}}\frac{{{\rm{Adj;}}\left( {\rm{A}} \right)}}{{\left| {\rm{A}} \right|}}$

Note:

For 2 × 2 matrix,

Let A = \left[ {\begin{array}{*{20}{c}} {\rm{a}}&{\rm{b}}\ {\rm{c}}&{\rm{d}} \end{array}} \right] 

than, Adj A = \left[ {\begin{array}{*{20}{c}} {\rm{d}}&{ - {\rm{b}}}\ { - {\rm{c}}}&{\rm{a}} \end{array}} \right] 

Note: Interchange the diagonal elements and change the sign of the remaining elements**.**

Calculation:

Let {\rm{A;}} = \left[ {\begin{array}{*{20}{c}} 2&1\ 1&1 \end{array}} \right] and,

{\rm{B;}} = {\rm{;}}\left[ {\begin{array}{*{20}{c}} 2&1\ 1&2 \end{array}} \right]

Now,

 \({\rm{A}} + {\rm{B}} = \left[ {\begin{array}{{20}{c}} 2&1\ 1&1 \end{array}} \right] + \left[ {\begin{array}{{20}{c}} 2&1\ 1&2 \end{array}} \right] = {\rm{;}}\left[ {\begin{array}{*{20}{c}} 4&2\ 2&3 \end{array}} \right]{\rm{;}}\)

{\rm{A;}} = \left[ {\begin{array}{*{20}{c}} 2&1\ 1&1 \end{array}} \right]

⇒ det A = (2 – 1) = 1

{\rm{B;}} = {\rm{;}}\left[ {\begin{array}{*{20}{c}} 2&1\ 1&2 \end{array}} \right]

⇒ det B = (4 – 1) = 3

det A + det B = 1 + 3 = 4

{\rm{A}} + {\rm{B}} = \left[ {\begin{array}{*{20}{c}} 4&2\ 2&3 \end{array}} \right]

⇒ det (A + B) = (12 – 4) = 8

∴ det (A+ B) ≠  det A + det B

Statement 1^st is wrong.

We know that for 2 × 2 matrix, interchange the diagonal elements and change the sign of remaining elements gives adjoint matrix.

{\rm{adj;}}\left( {\rm{A}} \right){\rm{;}} = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}\ { - 1}&2 \end{array}} \right]

{\rm{adj;}}\left( {\rm{B}} \right){\rm{;}} = {\rm{;}}\left[ {\begin{array}{*{20}{c}} 2&{ - 1}\ { - 1}&2 \end{array}} \right]

{\rm{adj;}}\left( {{\rm{A}} + {\rm{B}}} \right) = \left[ {\begin{array}{*{20}{c}} 3&-2\ -2&4\end{array}} \right]

We know that, ${{\rm{A}}^{ - 1}} = {\rm{;}}\frac{{{\rm{Adj;}}\left( {\rm{A}} \right)}}{{\left| {\rm{A}} \right|}}$

\( \Rightarrow {{\rm{A}}^{ - 1}} = {\rm{;}}\frac{{\left[ {\begin{array}{{20}{c}} 1&{ - 1}\ { - 1}&2 \end{array}} \right]}}{1} = {\rm{;}}\left[ {\begin{array}{{20}{c}} 1&{ - 1}\ { - 1}&2 \end{array}} \right]\)

\( \Rightarrow {{\rm{B}}^{ - 1}} = {\rm{;}}\frac{{\left[ {\begin{array}{{20}{c}} 2&{ - 1}\ { - 1}&2 \end{array}} \right]}}{3} = {\rm{;}}\left[ {\begin{array}{{20}{c}} {\frac{2}{3}}&{ - \frac{1}{3}}\ { - \frac{1}{3}}&{\frac{2}{3}} \end{array}} \right]\)

Now,

\({{\rm{A}}^{ - 1}} + {\rm{;}}{{\rm{B}}^{ - 1}} = {\rm{;}}\left[ {\begin{array}{{20}{c}} 1&{ - 1}\ { - 1}&2 \end{array}} \right] + {\rm{;}}\left[ {\begin{array}{{20}{c}} {\frac{2}{3}}&{ - \frac{1}{3}}\ { - \frac{1}{3}}&{\frac{2}{3}} \end{array}} \right] = {\rm{;}}\left[ {\begin{array}{*{20}{c}} {\frac{5}{3}}&{ - \frac{4}{3}}\ { - \frac{4}{3}}&{\frac{8}{3}} \end{array}} \right]\)

\({\left( {{\rm{A}} + {\rm{B}}} \right)^{ - 1}} = {\rm{;}}\frac{{{\rm{adj;}}\left( {{\rm{A}} + {\rm{B}}} \right)}}{{\det \left( {{\rm{A}} + {\rm{B}}} \right)}} = {\rm{;}}\frac{{\left[ {\begin{array}{{20}{c}} 3&-2\ -2&4 \end{array}} \right]}}{8} = {\rm{;}}\left[ {\begin{array}{{20}{c}} {\frac{3}{8}}&{\frac{-2}{8}}\ {\frac{-2}{8}}&{\frac{4}{8}} \end{array}} \right] = {\rm{;}}\left[ {\begin{array}{*{20}{c}} {\frac{3}{8}}&{\frac{-1}{4}}\ {\frac{-1}{4}}&{\frac{1}{2}} \end{array}} \right]\)

∴ (A + B)^-1 ≠ A^-1 + B^-1

Statement 2^nd is wrong.

Concept:

• If \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} {{{\rm{a}}{11}}}&{{{\rm{a}}{12}}}\ {{{\rm{a}}{21}}}&{{{\rm{a}}{22}}} \end{array}} \right]\) then determinant of A is given by: |A| = (a­₁₁ × a₂₂) – (a₁₂ – a₂₁).
• ${{\rm{A}}^{ - 1}} = {\rm{;}}\frac{{{\rm{Adj;}}\left( {\rm{A}} \right)}}{{\left| {\rm{A}} \right|}}$
• A A^-1 = I

Note:

For 2 × 2 matrix,

Let A = \(\left[ {\begin{array}{{20}{c}} {\rm{a}}&{\rm{b}}\ {\rm{c}}&{\rm{d}} \end{array}} \right]\) than Adj A = \(\left[ {\begin{array}{{20}{c}} {\rm{d}}&{ - {\rm{b}}}\ { - {\rm{c}}}&{\rm{a}} \end{array}} \right]\) 

Note: Interchange the diagonal elements and change the sign of remaining elements**.**

Calculation:

AX = B

Post multiple by X^-1, we get

⇒ AX X^-1 = B X^-1

⇒ AI = B X^-1                        (∵A A^-1 = I)

⇒ A = B X^-1      …. (1)

{\rm{X}} = \left[ {\begin{array}{*{20}{c}} 3&{ - 4}\ 1&{ - 1} \end{array}} \right]

det (X) = (-3 + 4) = 1

We know that for 2 × 2 matrix, interchange the diagonal elements and change the sign of remaining elements gives adjoint matrix.

{\rm{adj;}}\left( {\rm{x}} \right){\rm{;}} = \left[ {\begin{array}{*{20}{c}} { - 1}&4\ { - 1}&3 \end{array}} \right]

We know that, ${{\rm{x}}^{ - 1}} = {\rm{;}}\frac{{{\rm{Adj;}}\left( {\rm{x}} \right)}}{{\left| {\rm{x}} \right|}}$

\( \Rightarrow {{\rm{x}}^{ - 1}} = {\rm{;}}\frac{{\left[ {\begin{array}{{20}{c}} { - 1}&4\ { - 1}&3 \end{array}} \right]}}{1} = {\rm{;}}\left[ {\begin{array}{{20}{c}} { - 1}&4\ { - 1}&3 \end{array}} \right]\)

Put in equation 1^st,

A = B X^-1     

⇒ A = \(\left[ {\begin{array}{{20}{c}} 5&2\ { - 2}&1 \end{array}} \right] \times {\rm{;}}\left[ {\begin{array}{{20}{c}} { - 1}&4\ { - 1}&3 \end{array}} \right] = {\rm{;;}}\left[ {\begin{array}{{20}{c}} { - 5 - 2}&{20 + 6}\ {2 - 1}&{ - 8 + 3} \end{array}} \right] = {\rm{;;}}\left[ {\begin{array}{{20}{c}} { - 7}&{26}\ 1&{ - 5} \end{array}} \right]\)

Concept:

• ⁿC_n = 1
• ⁿC₁ = n

Calculation:

We have to find the value of \mathop \sum \limits_{{\rm{r;}} = {\rm{;}}0}^1 {{\rm{;}}^{{\rm{n}} + {\rm{r}}}}{{\rm{C}}_{\rm{n}}}

⇒ \mathop \sum \limits_{{\rm{r;}} = {\rm{;}}0}^1 {{\rm{;}}^{{\rm{n}} + {\rm{r}}}}{{\rm{C}}_{\rm{n}}}

⇒ ⁿ⁺⁰C_n + ⁿ⁺¹C_n

⇒ nCn + n+1Cn 

⇒ 1 + (n + 1)

⇒ (n + 2)

∴  The value of \mathop \sum \limits_{{\rm{r;}} = {\rm{;}}0}^1 {{\rm{;}}^{{\rm{n}} + {\rm{r}}}}{{\rm{C}}_{\rm{ n}}} is ^{n + 2}C_1.

Concept:

• Suppose a set of n objects has n₁ of one kind of object, n₂ of a second kind, n₃ of a third kind, and so on with n = n₁ + n₂ + n₃  +...+ n_k then the number of distinguishable  permutations of the n objects is  = \(\frac{{{\rm{n}}!}}{{{{\rm{n}}_1}! \times {{\rm{n}}_2}! \times {{\rm{n}}3}! \ldots \ldots \ldots {{\rm{n}}{\rm{k}}}!}}\)

Calculation:

The given word is 'NATION'

Number of letters repeating in the word =2N

Vowels in the word are {A, I, O}

Total number of words that can be formed from given word 'NATION' = $\frac{{6!}}{{2!}} = 360$

Consider A, I, O as one group

Then the no. of words formed by this group and remaining letters is 4!

The three vowels can be arranged among themselves in 3! Ways

∴ Number of words where all the vowels come together = $\rm \frac {(4!) (3!)}{2}$ =72

Number of words formed so that all the three vowels are never together = Total number of words formed using all the letters in the word NATION − Number of words where all the vowels come together

Thus, the number of words formed so that all the three vowels are never together = 360 – 72 = 288

Concept:

1. Let us consider the standard form of a quadratic equation, ax² + bx + c =0

Where a, b and c are constants with a ≠ 0

• ${\rm{x}} = {\rm{;}}\frac{{ - {\rm{b;}} \pm {\rm{;}}\sqrt {{{\rm{b}}^2} - 4{\rm{ac}}} }}{{2{\rm{a}}}}$

2. x³ – y³ = (x – y) (x² + y² + xy)
3. Cube root of unity:

• ${\rm{\omega }} = {\rm{;}}\frac{{ - 1 + {\rm{i}}\sqrt 3 }}{2}{\rm{;}}$
• ${{\rm{\omega }}^2} = \frac{{ - 1 - {\rm{i}}\sqrt 3 }}{2}$
• ω³ = 1

 

Calculation:

We know that x³ – y³ = (x – y) (x² + xy + y²)

∴ (x³ – 1) = (x – 1) (x² + x + 1)       …. (1)

Now, Factor of (x² + x + 1)

${\rm{x}} = {\rm{;}}\frac{{ - 1{\rm{;}} \pm {\rm{;}}\sqrt {{1^2} - 4{\rm{;}} \times 1{\rm{;}} \times 1} }}{{2{\rm{;}} \times 1}} = {\rm{;}}\frac{{ - 1{\rm{;}} \pm {\rm{i}}\sqrt 3 }}{2}$

$\Rightarrow {\rm{x}} = {\rm{;}}\frac{{ - 1 + {\rm{i}}\sqrt 3 }}{2}{\rm{;or;;}}\frac{{ - 1 - {\rm{i}}\sqrt 3 }}{2}$

⇒ x = ω or ω²

∴ (x² + x + 1) = (x – ω) (x – ω²)

Putting the value in equation 1,

(x³ – 1) = (x – 1) (x² + x + 1)

⇒ (x³ – 1) = (x – 1) (x – ω) (x – ω²)

Concept:

1. Euler's Formula on Complex Numbers:

• e^ix = cos x + i sin x
• e^-ix = cos x - i sin x

2. Trigonometry formulas:

• 1 – cos θ = 2 sin² (θ/2)
• sin θ = 2 sin (θ/2) cos (θ/2

Calculation:

We have to find the value of ${\left[ {\frac{{\sin \frac{{\rm{\pi }}}{6} + {\rm{i}}\left( {1 - \cos \frac{{\rm{\pi }}}{6}} \right)}}{{\sin \frac{{\rm{\pi }}}{6} - {\rm{i}}\left( {1 - \cos \frac{{\rm{\pi }}}{6}} \right)}}} \right]^3}$

$\Rightarrow {\left[ {\frac{{\sin \frac{{\rm{\pi }}}{6} + {\rm{i}}\left( {1 - \cos \frac{{\rm{\pi }}}{6}} \right)}}{{\sin \frac{{\rm{\pi }}}{6} - {\rm{i}}\left( {1 - \cos \frac{{\rm{\pi }}}{6}} \right)}}} \right]^3} = {\rm{;}}{\left[ {\frac{{2 \times \sin \frac{{\rm{\pi }}}{{12}} \times \cos \frac{{\rm{\pi }}}{{12}} + {\rm{i}}\left( {{\rm{;}}2{{\sin }^2}\frac{{\rm{\pi }}}{{12}}} \right)}}{{2 \times \sin \frac{{\rm{\pi }}}{{12}} \times \cos \frac{{\rm{\pi }}}{{12}} - {\rm{i}}\left( {{\rm{;}}2{{\sin }^2}\frac{{\rm{\pi }}}{{12}}} \right)}}} \right]^3}{\rm{;}}$

$= {\rm{;}}{\left[ {\frac{{2 \times \sin \frac{{\rm{\pi }}}{{12}} \times \left( {\cos \frac{{\rm{\pi }}}{{12}} + {\rm{i}}\sin \frac{{\rm{\pi }}}{{12}}} \right)}}{{2 \times \sin \frac{{\rm{\pi }}}{{12}} \times \left( {\cos \frac{{\rm{\pi }}}{{12}} - {\rm{i}}\sin \frac{{\rm{\pi }}}{{12}}} \right)}}} \right]^3}$

$= {\rm{;}}{\left[ {\frac{{\left( {\cos \frac{{\rm{\pi }}}{{12}} + {\rm{i}}\sin \frac{{\rm{\pi }}}{{12}}} \right)}}{{\left( {\cos \frac{{\rm{\pi }}}{{12}} - {\rm{i}}\sin \frac{{\rm{\pi }}}{{12}}} \right)}}} \right]^3}$                          (∵e^ix = cos x + i sin x and e^-ix = cos x - i sin x)

$= {\rm{;}}{\left[ {\frac{{{{\rm{e}}^{\frac{{{\rm{i\pi }}}}{{12}}}}}}{{{{\rm{e}}^{\frac{{ - {\rm{i\pi }}}}{{12}}}}}}} \right]^3} = {\left[ {{{\rm{e}}^{\frac{{{\rm{i\pi }}}}{{12}}}}{\rm{;}} \times {\rm{;}}{{\rm{e}}^{\frac{{{\rm{i\pi }}}}{{12}}}}} \right]^3} = {\left[ {{{\rm{e}}^{\frac{{{\rm{i\pi }}}}{6}}}{\rm{;}}} \right]^3} = {\rm{;}}{{\rm{e}}^{\frac{{{\rm{i\pi }}}}{2}}}$             

= cos (π/2) + i sin (π/2) = 0 + i  = i

Concept:

Step to find matrix multiplication:

First make sure the number of columns of first matrix is equal to number of rows of second matrix.

Multiply elements of each row of first matrix to elements of each column of second matrix.

Let A = [a_ij]_{m × p} & B = [b_ij]_{p × n} then AB = [c_ij]_{m × n} where \({{\rm{c}}{{\rm{ij}}}} = \mathop \sum \limits{{\rm{k}} = 1}^{\rm{p}} \left( {{{\rm{a}}{{\rm{ik}}}}{{\rm{b}}{{\rm{kj}}}}} \right)\).

 

Calculation:

Given: \({\rm{A}} = \left[ {\begin{array}{{20}{c}} {{\rm{x}} + {\rm{y}}}&{\rm{y}}\ {2{\rm{x}}}&{{\rm{x}} - {\rm{y;}}} \end{array}} \right]\), \({\rm{B}} = \left[ {\begin{array}{{20}{c}} 2\ { - 1} \end{array}} \right]\) and {\rm{C}} = \left[ {\begin{array}{*{20}{c}} 3\ 2 \end{array}} \right] and AB = C

So, \({\rm{AB}} = \left[ {\begin{array}{{20}{c}} {{\rm{x}} + {\rm{y}}}&{\rm{y}}\ {2{\rm{x}}}&{{\rm{x}} - {\rm{y;}}} \end{array}} \right] \times \left[ {\begin{array}{{20}{c}} 2\ { - 1} \end{array}} \right]\) 

\Rightarrow {\rm{AB}} = \left[ {\begin{array}{*{20}{c}} {2\left( {{\rm{x}} + {\rm{y}}} \right) - {\rm{y}}}\ {2\left( {2{\rm{x}}} \right) - \left( {{\rm{x}} - {\rm{y}}} \right)} \end{array}} \right]

\Rightarrow {\rm{AB}} = \left[ {\begin{array}{*{20}{c}} {2{\rm{x}} + {\rm{y}}}\ {3{\rm{x}} + {\rm{y}}} \end{array}} \right]{\rm{;}}

On comparing, the elements with matrix C we get,

2x + y = 3     ...1)

3x + y = 2     ...2)

Subtracting (1) from (2)

x = -1

Putting x = -1  in equation (1)

So, y = 5

So, {\rm{A}} = \left[ {\begin{array}{*{20}{c}} 4&5\ { - 2}&{ - 6} \end{array}} \right]

\({{\rm{A}}^2} = \left[ {\begin{array}{{20}{c}} 4&5\ { - 2}&{ - 6} \end{array}} \right] \times \left[ {\begin{array}{{20}{c}} 4&5\ { - 2}&{ - 6} \end{array}} \right]\)

\Rightarrow {{\rm{A}}^2} = \left[ {\begin{array}{*{20}{c}} {4 \times 4 + 5 \times \left( { - 2} \right)}&{4 \times 5 + 5 \times \left( { - 6} \right)}\ {\left( { - 2} \right) \times 4 + \left( { - 6} \right) \times \left( { - 2} \right)}&{\left( { - 2} \right) \times 5 + \left( { - 6} \right) \times \left( { - 6} \right)} \end{array}} \right]

\Rightarrow {{\rm{A}}^2} = \left[ {\begin{array}{*{20}{c}} 6&{ - 10}\ 4&{26} \end{array}} \right]

Concept:

Elementary row or column transformations do not change the value of the determinant of a matrix.

Calculation:

\left| {\begin{array}{*{20}{c}} 1&1&1\ 1&{1 + {\rm{x}}}&1\ 1&1&{1 + {\rm{y}}} \end{array}} \right|

Applying R₂ → R₂ – R₁, R₃ → R₃ – R₁, we get

= \left| {\begin{array}{*{20}{c}} 1&1&1\ 0&{\rm{x}}&0\ 0&0&{\rm{y}} \end{array}} \right|

Now, Expanding along C₁

= 1 (xy – 0) – 0 + 0 = xy

Concept:

• Null set is the set that does not contain anything.
• In mathematical sets, the null set, also called the empty set,
• It is symbolized or { }.

Calculation:

Given:

A = {x : x is a multiple of 3}

∴ A = {3, 6, 9, 12, 15, 18, 24…}

B = {x : x is a multiple of 4}

∴ B = {4, 8, 12, 16, 20, 24, 28, 32 ...}

C = {x : x is a multiple of 12}

∴ C = {12, 24, 36, 48, 60, 72, 84, 96 ...}

Now,

A∩B = {12, 24 ...} = C

∴ (A∩B) \ C = { } = Null set

Concept:

Convert Binary to Decimal:

For binary numbers with n digits;

D_n-1 …. D₃D₂D₁D₀

The decimal number is equal to the sum of binary digits (D_n) times their power of 2 (2ⁿ)

Decimal = D₀ × 2⁰ + D₁ × 2¹ + D₂ × 2² + ….

Calculation:

We have to find the value of (11101011)₂

(11101011)₂ = 1 × 2⁷ + 1 × 26 +1 × 25 + 0 + 1 × 23 + 1 × 21 + 1 × 20

⇒ 128 + 64 + 32 + 8 + 2 + 1 = 235

(11101011)₂ = 235₁₀

Concept:

Euler's Formula on Complex Numbers:

• e^ix = cos x + i sin x
• e^-ix = cos x - i sin x

Calculation:

(sin x + icos x)³

Take i common, we get

(sin x + icos x)³

= ${{\rm{i}}^3}{\left( {\frac{{\sin {\rm{x}}}}{{\rm{i}}} + {\rm{;}}\cos {\rm{x}}} \right)^3}$

= -i  × (-i sin x + cos x)³                

(∵ i³ = -i and 1/i = -i)

= -i × (cos x - i sin x) ³

$= {\rm{}} - {\rm{i;}} \times {\rm{}}{\left( {{{\rm{e}}^{ - {\rm{ix}}}}} \right)^3}$      

$= {\rm{}} - {\rm{i;}} \times {\rm{;}}{{\rm{e}}^{ - {\rm{i}}3{\rm{x}}}}{\rm{;}}$

(∵e^-ix = cos x - i sin x)

= -i (cos 3x – i sin 3x)

= (-i cos 3x + i² sin 3x)

= -sin3x – i cos 3x

∴ Real part = -sin 3x

Concept:

Step to find matrix multiplication:

First make sure the number of columns of first matrix is equal to number of rows of second matrix.

Multiply elements of each row of first matrix to elements of each column of second matrix.

Let A = [a_ij]_{m × p} & B = [b_ij]_{p × n} then AB = [c_ij]_{m × n} where \({{\rm{c}}{{\rm{ij}}}} = \mathop \sum \limits{{\rm{k}} = 1}^{\rm{p}} \left( {{{\rm{a}}{{\rm{ik}}}}{{\rm{b}}{{\rm{kj}}}}} \right)\).

Calculation:

Given: {\rm{E}}\left( {\rm{\theta }} \right) = \left[ {\begin{array}{*{20}{c}} {\cos {\rm{\theta }}}&{\sin {\rm{\theta }}}\ { - \sin {\rm{\theta ;}}}&{\cos {\rm{\theta }}} \end{array}} \right]

To find: E(α) E(β)

{\rm{E}}\left( {\rm{\alpha }} \right) = \left[ {\begin{array}{*{20}{c}} {\cos {\rm{\alpha }}}&{\sin {\rm{\alpha }}}\ { - \sin {\rm{\alpha ;}}}&{\cos {\rm{\alpha }}} \end{array}} \right]

{\rm{E}}\left( {\rm{\beta }} \right) = \left[ {\begin{array}{*{20}{c}} {\cos {\rm{\beta }}}&{\sin {\rm{\beta }}}\ { - \sin {\rm{\beta ;}}}&{\cos {\rm{\beta }}} \end{array}} \right]

Now,

E(α)E(β) = 

\(\left[ {\begin{array}{{20}{c}} {\cos {\rm{\alpha }}}&{\sin {\rm{\alpha }}}\ { - \sin {\rm{\alpha ;}}}&{\cos {\rm{\alpha }}} \end{array}} \right] \times \left[ {\begin{array}{{20}{c}} {\cos {\rm{\beta }}}&{\sin {\rm{\beta }}}\ { - \sin {\rm{\beta ;}}}&{\cos {\rm{\beta }}} \end{array}} \right]\)

\left[ {\begin{array}{*{20}{c}} {\cos {\rm{\alpha }} \times \cos {\rm{\beta }} + \sin {\rm{\alpha }} \times \left( { - \sin {\rm{\beta }}} \right)}&{\cos {\rm{\alpha }} \times \sin {\rm{\beta }} + \sin {\rm{\alpha }} \times \cos {\rm{\beta }}}\ {\left( { - \sin {\rm{\alpha }}} \right) \times \cos {\rm{\beta }} + \cos {\rm{\alpha }} \times \left( { - \sin {\rm{\beta }}} \right)}&{\left( { - \sin {\rm{\alpha }}} \right) \times \sin {\rm{\beta }} + \cos {\rm{\alpha }} \times \cos {\rm{\beta }}} \end{array}} \right]

 

\left[ {\begin{array}{*{20}{c}} {\cos \left( {{\rm{\alpha }} + {\rm{\beta }}} \right)}&{\sin \left( {{\rm{\alpha }} + {\rm{\beta }}} \right)}\ { - \sin \left( {{\rm{\alpha }} + {\rm{\beta }}} \right)}&{\cos \left( {{\rm{\alpha }} + {\rm{\beta }}} \right)} \end{array}} \right]

E(α) E(β) = E(α +β)

Concept:

Let set A has n (A) elements and set B has n (B) elements.

Number of distinct relations from A to B = 2^{Number of elements in set A × Number of elements in set B}

Calculation:

Let A = {x, y, z} and B = {p, q, r, s}.

Number of elements in set A = n (A) = 3

Number of elements in set B = n (B) = 4

∴ Number of distinct relations from B to A = 2^{n (B) × n (A)} = 2^{4 × 3} = 2¹² = 4096

Calculation:

Given: 2p + 3q = 18 and 4p² + 4pq – 3q² – 36 = 0

To find: (2p + q)

2p + 3q = 18

⇒ 2p = 18 – 3q

4p² + 4pq – 3q² – 36 = 0     ...1)

Put the value of 2p in equation (1) we get,

⇒ (18 – 3q)² + 2(18 – 3q)(q) – 3q² – 36 = 0

⇒ (324 + 9q² – 108q) + 36q – 6q² – 3q² – 36 = 0

⇒ 288 – 72q = 0

⇒ q = 4

Now, 2p = 18 – 3(4)

⇒ 2p = 6

So, 2p + q = 6 +4

= 10

Calculation:

Let number of degree in θ be x and let the number of radians in θ be y.

So, we have given,

$\frac{{\rm{x}}}{{\rm{y}}} = \frac{{180^\circ }}{{\rm{\pi }}}$     ...1)

${\rm{xy}} = \frac{{125^\circ {\rm{\pi }}}}{9}$     ...2)

${\rm{x}} = \frac{{180^\circ }}{{\rm{\pi }}} \times {\rm{y}}$

Putting the value of x in equation (2) we get,

$\frac{{180^\circ }}{{\rm{\pi }}} \times {\rm{y}} \times {\rm{y}} = \frac{{125^\circ {\rm{\pi }}}}{9}$

$\Rightarrow {{\rm{y}}^2} = \frac{{125{{\rm{\pi }}^2}}}{{180 \times 9}}$

$\Rightarrow {{\rm{y}}^2} = \frac{{25{{\rm{\pi }}^2}}}{{9 \times 4 \times 9}}$

⇒ y = 5π / 18

Now, ${\rm{x}} = \frac{{180^\circ }}{{\rm{\pi }}} \times {\rm{y}}$ 

$\Rightarrow {\rm{x}} = \frac{{180^\circ }}{{\rm{\pi }}} \times \frac{{5{\rm{\pi }}}}{{18}}$

⇒ x = 50°

Concept:

Consider a triangle ABC,

Cosine rule:

$\cos {\rm{C}} = \frac{{{{\rm{a}}^2} + {{\rm{b}}^2} - {{\rm{c}}^2}}}{{2{\rm{ab}}}}$

Sine rule:

$\frac{{\rm{a}}}{{\sin {\rm{A}}}} = \frac{{\rm{b}}}{{\sin {\rm{B}}}} = \frac{{\rm{c}}}{{\sin {\rm{C}}}}$

Calculation:

Given: a = (1 + √ 3) cm, b = 2 cm and ∠C = 60°

We know that, $\cos {\rm{C}} = \frac{{{{\rm{a}}^2} + {{\rm{b}}^2} - {{\rm{c}}^2}}}{{2{\rm{ab}}}}$

$\Rightarrow \cos 60^\circ = \frac{{{{\left( {1 + \sqrt 3 } \right)}^2} + {2^2} - {{\rm{c}}^2}}}{{2\left( {1 + \sqrt 3 } \right)\left( 2 \right)}}$

⇒ 2(1 + √3) = 1 + 3 + 2√3 + 4 – c²

⇒ c² = 6

⇒ c = √6

Using sin rule we get,

$\frac{{\rm{a}}}{{\sin {\rm{A}}}} = \frac{{\rm{b}}}{{\sin {\rm{B}}}} = \frac{{\rm{c}}}{{\sin {\rm{C}}}}$

$\Rightarrow \frac{{1 + \sqrt 3 }}{{\sin {\rm{A}}}} = \frac{2}{{\sin {\rm{B}}}} = \frac{{\sqrt 6 }}{{\sin 60^\circ }}$

⇒ sin B = (2 / √6) × sin 60°

$\Rightarrow \sin {\rm{B}} = \frac{1}{{\sqrt 2 }}$

⇒ B = 45°

Now, sum of all angles of triangles = 180°

⇒ A + B + C = 180°

⇒ A = 180° – 45° – 60°

⇒ A = 75°

Hence, other angles are 45° and 75°.

Concept:

Let ax² + bx + c = 0 be a quadratic equation, then the roots of the equation is given as:

${\rm{x}} = \frac{{ - {\rm{b}} \pm \sqrt {{{\rm{b}}^2} - 4{\rm{ac}}} }}{{2{\rm{a}}}}$

Calculation:

Given: 25 cos² θ + 5 cos θ – 12 = 0, where $\frac{{\rm{\pi }}}{2} < {\rm{\alpha }} < {\rm{\pi }}$

To find: tan α

25 cos² θ + 5 cos θ – 12 = 0

$\Rightarrow \cos {\rm{\theta }} = \frac{{ - 5 \pm \sqrt {{5^2} - 4\left( {25} \right)\left( { - 12} \right)} }}{{2\left( {25} \right)}}$

$\Rightarrow \cos {\rm{\theta }} = \frac{{ - 5 \pm 35}}{{50}}$

⇒ cos θ = -4 / 5 or cos θ = 3 / 5

Now, $\frac{{\rm{\pi }}}{2} < {\rm{\alpha }} < {\rm{\pi }}$. So, cos α is negative.

∴ cos α = -4 / 5

So, cos α = -4 / 5, where α is in second quadrant.

∴ sin α should be positive.

$\sin {\rm{\alpha }} = \sqrt {1 - {{\cos }^2}{\rm{\alpha }}}$

$\Rightarrow \sin {\rm{\alpha }} = \sqrt {1 - \frac{{16}}{{25}}}$

⇒ sin α = 3 / 5

Now, $\tan {\rm{\alpha }} = \frac{{\sin {\rm{\alpha }}}}{{\cos {\rm{\alpha }}}}$ 

⇒ tan α = -3 / 4

Concept:

Formulas used:

• sin 2α = 2(sin α)(cos α)

Calculation:

To find: sin 2α

We know that sin 2α = 2(sin α)(cos α)

$\Rightarrow sin 2 \alpha=2 \times \frac35 \times \frac{-4}{5}$

$\Rightarrow sin 2 \alpha=\frac{-24}{25}$

Concept:

Angle of elevation: The term angle of elevation denotes the angle from the horizontal upward to an object. An observer's line of sight would be above the horizontal.

The term angle of depression denotes the angle from the horizontal downward to an object. An observer's line of sight would be below the horizontal.

Note that the angle of elevation and the angle of depression are congruent.

Calculation:

Given: Angle of elevation = θ = 45° and point is 20 m away from base.

Let the height of tower be h.

Rough diagram:

Now, $\tan {\rm{\theta }} = \frac{{\rm{h}}}{{20}}$

⇒ 20 tan (45°) = h

⇒ h = 20 m

So, the height of tower is 20 m.

Concept:

Formulas used:

• ${\tan ^{ - 1}}{\rm{A}} + {\tan ^{ - 1}}{\rm{B}} = {\tan ^{ - 1}}\left( {\frac{{{\rm{A}} + {\rm{B}})}}{{1 - {\rm{AB}}}}} \right)$ where A > 0, B > 0 & AB < 1

 

Calculation:

Given: ${\tan ^{ - 1}}\left( {1 + {\rm{x}}} \right) + {\tan ^{ - 1}}\left( {1 - {\rm{x}}} \right) = \frac{{\rm{\pi }}}{2}$

We now that, ${\tan ^{ - 1}}{\rm{A}} + {\tan ^{ - 1}}{\rm{B}} = {\tan ^{ - 1}}\left( {\frac{{{\rm{A}} + {\rm{B}})}}{{1 - {\rm{AB}}}}} \right)$ where A > 0, B > 0 & AB < 1

${\tan ^{ - 1}}\left( {1 + {\rm{x}}} \right) + {\tan ^{ - 1}}\left( {1 - {\rm{x}}} \right) = \frac{{\rm{\pi }}}{2}$

$\Rightarrow {\tan ^{ - 1}}\left( {\frac{{1 + {\rm{x}} + 1 - {\rm{x}}}}{{1 - \left( {1 + {\rm{x}}} \right)\left( {1 - {\rm{x}}} \right)}}} \right) = {\tan ^{ - 1}}\left( {\frac{1}{0}} \right)$

$\Rightarrow \frac{2}{{1 - 1 + {{\rm{x}}^2}}} = \frac{1}{0}$

⇒ x² = 0

⇒ x = 0

Concept:

Angle of elevation: The term angle of elevation denotes the angle from the horizontal upward to an object. An observer's line of sight would be above the horizontal.

The term angle of depression denotes the angle from the horizontal downward to an object. An observer's line of sight would be below the horizontal.

Note that the angle of elevation and the angle of depression are congruent.

Calculation:

Let the height of tower be h.

Rough diagram:

Now, $\tan 47^\circ = \frac{{\rm{h}}}{{36}}$     ...1)

And also, $\tan 43^\circ = \frac{{\rm{h}}}{{49}}$

$\Rightarrow \tan \left( {90^\circ - 47^\circ } \right) = \frac{{\rm{h}}}{{49}}$

$\Rightarrow \cot 47^\circ = \frac{{\rm{h}}}{{49}}$     ...2)

Multiplying (1) and (2) we get,

$\tan 47^\circ \times \cot 47^\circ = \frac{{\rm{h}}}{{36}} \times \frac{{\rm{h}}}{{49}}$

⇒ 49 × 36 = h²

⇒ h = 7 × 6

⇒ h = 42 m

Hence, the height of the tower is 42 m.

Concept:

sin²A + cos²A = 1

(a + b)² = a² + b² + 2ab

Calculation:

(1 – sin A + cos A)² = [(1 – sin A) + cos A]²

= (1 – sin A) ² + cos²A + 2(1 – sin A) cos A

= 1 + sin²A − 2sin A + cos²A + 2(1 – sin A) cos A

= 1 + (sin²A + cos²A) − 2sin A + 2(1-sin A) cos A

= 1 + 1 − 2sin A + 2(1 – sin A) cos A

= 2 − 2sin A + 2(1 – sin A) cos A

= 2(1 – sin A) + 2(1 – sin A) cos A

= 2(1 – sin A) (1 + cos A)

Alternate Method We have to find value of,

(1 – sin A + cos A) 2

Using Formula:

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

⇒ 1 + sin² A + cos² A - 2sin A - 2sin A cos A + 2cos A

⇒ 1 + 1 - 2sin A - 2sin A cos A + 2cos A

⇒ 2 - 2sin A - 2sin A cos A + 2cos A

⇒ 2[(1 - sin A) + cos A(1 - sin A )]

Hence,

(1 – sin A + cos A) 2 = 2(1 – sin A)(1 + cos A)

Formula Used:

$tan θ=\frac{1}{cot θ}=\frac{sinθ}{cosθ}$

a2 - b2 = (a - b) (a + b)

Calculation:

We have to find the value of,

$\frac{{\cos {\rm{θ }}}}{{1 - \tan {\rm{θ }}}} + \frac{{\sin {\rm{θ }}}}{{1 - \cot {\rm{θ }}}}$

tan θ & cot θ can be written as,​

$\frac{{\cos {\rm{θ }}}}{{1 - (\frac{sin \theta}{cos\theta}})} + \frac{{\sin {\rm{θ }}}}{{1 - (\frac{cos \theta}{sin\theta}})}$

⇒ $\frac{cos \ \theta }{\frac{cos\ \theta-sin\ \theta}{cos\ \theta}}+\frac{sin \ \theta }{\frac{sin\ \theta-cos\ \theta}{sin\ \theta}}$

⇒ $\frac{{{{\cos }^2}{\rm{θ }}}}{{\cos {\rm{θ }} - {\rm{;}}\sin {\rm{θ }}}} + {\rm{;}}\frac{{{{\sin }^2}{\rm{θ }}}}{{\sin {\rm{θ }} - {\rm{;}}\cos {\rm{θ }}}}$

⇒ $\frac{{{{\cos }^2}{\rm{θ }}}}{{\cos {\rm{θ }} - {\rm{;}}\sin {\rm{θ }}}} -( {\rm{;}}\frac{{{{\sin }^2}{\rm{θ }}}}{{\cos {\rm{θ }} - {\rm{;}}\sin {\rm{θ }}}})$

⇒ $\frac{cos^2\theta-sin^2\theta}{cos\theta -sin\theta }=\frac{(cos\theta +sin\theta)(cos\theta -sin\theta)}{(cos\theta -sin\theta)}$

Hence,

$\frac{{\cos {\rm{θ }}}}{{1 - \tan {\rm{θ }}}} + \frac{{\sin {\rm{θ }}}}{{1 - \cot {\rm{θ }}}}$ = (sin θ + cos θ)

Concept:

$2 {\tan ^{ - 1}}x = {\tan ^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} \right);,; - 1 < x < 1$

Calculation:

Given:

 $x = 4{\tan ^{ - 1}}\left( {\frac{1}{5}} \right)$

$\Rightarrow x = 4{\tan ^{ - 1}}\left( {\frac{1}{5}} \right) = 2 \times 2 \times ;{\tan ^{ - 1}}\left( {\frac{1}{5}} \right)$

As we know, $2 \times {\tan ^{ - 1}}x = {\tan ^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} \right);,; - 1 < x < 1$

$\Rightarrow x = 4{\tan ^{ - 1}}\left( {\frac{1}{5}} \right) = 2 \times \left( {{{\tan }^{ - 1}}\left( {\frac{{\frac{2}{5}}}{{1 - {{\left( {\frac{1}{5}} \right)}^2}}}} \right)} \right) = 2 \times {\tan ^{ - 1}}\left( {\frac{5}{{12}}} \right)$

As we know, $2 \times {\tan ^{ - 1}}x = {\tan ^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} \right);,; - 1 < x < 1$

$\Rightarrow x = 4{\tan ^{ - 1}}\left( {\frac{1}{5}} \right) = 2 \times {\tan ^{ - 1}}\left( {\frac{5}{{12}}} \right) = {\tan ^{ - 1}}\left( {\frac{{\frac{5}{6}}}{{1 - {{\left( {\frac{5}{{12}}} \right)}^2}}}} \right) = {\tan ^{ - 1}}\left( {\frac{{120}}{{119}}} \right)$

Hence option 2 is the correct answer.