NDA Mathematics 9 Sept 2018 Official Paper (Previous Year Paper)

Concept:

Logarithm properties

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

${\log _a}\left( {mn} \right) = ;{\log _a}m + ;{\log _a}n$

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

${\log _a}\frac{m}{n} = ;{\log _a}m - ;{\log _a}n$

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

${\log _a}{m^n} = n{\log _a}m$

1. Change of base rule

${\log _m}n = \frac{{{{\log }_a}n}}{{{{\log }_a}m}}$

If m = n;

⇒ ${\log _m}m = \frac{{{{\log }_a}m}}{{{{\log }_a}m}} = 1$

1. ${\log _m}n = \frac{1}{{{{\log }_n}m}}$

 

Calculation:

Here, we have to find the value of ${\log _7}{\rm{;}}{\log _7}\sqrt {7\sqrt {7\sqrt 7 } }$

${\log _7}{\rm{;}}{\log _7}\sqrt {7\sqrt {7\sqrt 7 } }$

= log₇ log₇ (7^1/2 × 7^1/4 × 7^1/8)

= log₇ log₇ (7(^{1/2 + 1/4 + 1/8)})

= log₇ log₇ (7^{(4 + 2 + 1)/8})

= log₇ log₇ (7^7/8)

From power rule;

= log₇ (7/8) log₇7

= log₇ (7/8) × 1 = log₇ (7/8) = log₇ 7 – log₇ 8

= 1 – log₇ 8 = 1 – log₇ 2³

= 1 – 3 log₇ 2

Concept:

If a, ar, ar², …. is an infinite GP, then the sum of infinite geometric series is given by.

Sum of infinite GP = ${{\rm{s}}_\infty } = {\rm{;}}\frac{{\rm{a}}}{{1{\rm{;}} - {\rm{;r}}}}{\rm{;}}$; |r| < 1

Calculation:

Given:

First term of GP = x

Sum of infinite GP = 5

We know that;

S_∞ = a/ (1 - r)

⇒ 5 = x/ (1 - r)

⇒ 5 × (1 - r) = x

⇒ 5 – 5r = x

⇒ 5 – x = 5r

∴ r = (5 – x)/5

For infinite GP |r| < 1

|r| < 1

⇒ -1 < r < 1

⇒ -1 < (5 – x)/5 < 1

⇒ -5 < 5 - x < 5

⇒ -10 < -x < 0

⇒ 0 < x < 10

Concept:

Rational expressions: A rational expression is nothing more than a fraction in which the numerator and/or the denominator are polynomials.

A rational expression is considered simplified if the numerator and denominator have no factors in common.

Steps for obtaining a rational expression.

1. Factor the numerator and the denominator.
2. List restricted values.
3. Cancel common factors.
4. Simplify and note any restricted values not implied by the expression.

 

Calculation:

1. $x + {x^2} - \frac{1}{x} = {\rm{;}}\frac{{{x^2} + {\rm{;}}{x^3}{\rm{;}} - 1{\rm{;}}}}{x}$
2. By definition of rational function, it is a rational expression.
3. $\sqrt {a{x^2} + bx + x - c + \frac{d}{x} - \frac{e}{{{x^2}}}}$
4. It is not a rational expressions because it is not a polynomials.
5. 3x² - 5x + ab
6. It is in form of a fraction of denominator 1 so it is a rational expression.
7. $\frac{2}{{{x^{2{\rm{;}}}} - {\rm{;}}ax{\rm{;}} + {\rm{;}}{b^2}}}$
8. By definition of rational function it is a rational expressions.
9. 1/x - 1/x + 5 = (x + 5 - x) / x(x + 5) = 5 / x(x + 5) = 5 / x² + 5x
10. By definition of rational function it is a rational expressions.

 

∴ Option 1^st is correct.

Concept:

Orthogonal matrix: When the product of a matrix to its transpose gives identity matrix.

Suppose A is a square matrix with real elements and of n x n order and A^T or A’ is the transpose of A.

AA^T = I

Calculation:

Suppose A is a square matrix with real elements and of n x n order and A^T or A’ is the transpose of A.

Then according to the definition;

AA^T = I

Pre multiplication by A^-1

A^-1 AA^T = A^-1 I

IA^T = A^-1

A^T = A^-1 or A’ = A^-1

then A is orthogonal matrix.

∴ Option 2 is correct

Concept:

If A, B and C are subsets of a set X. Then

Commutative property: A ⋃ B = B ⋃ A

Associative property: (A ⋃ B) ∪ C = A ⋃ (B ⋃ C)

Distributive property: A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) or A ⋂ (B ⋃ C) = (A ⋂ B) ⋃ (A ⋂ C)

Some others property

I. A ∪ A = A, and A ∩ A = A

II. A ∩ (A ∪ B) = A, A ∪ (A ∩ B) = A

III. (A ∩ B) ∪ C = (A∩C) ∪ (B∩C)

IV. (A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C)

V. (A ⋃ B)’ = A’ ⋂ B’ and (A ⋂ B)’ = A’ ⋃ B’

VI. (A’)’ = A

Calculation:

1. A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

According to Distributive property option 1 is true.

2. L.H.S = A’ ∪ (A ∪ B) = A ∪ B

 R.H.S = (B’ ∩ B)’ ∪ A’ = ((B’)’ ⋃ B’) ⋃ A’ = (B ⋃    B’) ⋃ A’ = (A ∪ B) ⋃ A’ = A ∪ B

 So, option 2 is correct.

3. L.H.S = A’ ∪ (B ∪ C)

 R.H.S = (C’ ∩ B)’ ∩ A’ = ((C’)’ ⋃ B) ⋂ A’ = (C ⋂ B) ⋂ A’

 L.H.S ≠ R.H.S

 So, option 3 is incorrect.

Calculation:

Total number of integers lying between 2999 and 8001 = 8001 - 2999 - 1 = 5001

These are four digits numbers.

For distinct digits, starting digit can be filled by 5 ways (3, 4, 5, 6 or 7)

Total number of all distinct digits = 5 × 9 × 8 × 7 = 2520

Now, Total number which have at least two digits equal = 5001 - 2520 = 2481

∴ x = 2481

Concept:

If a, ar, ar², …. is an infinite GP, then the sum of infinite geometric series is given by.

Sum of infinite GP = ${s_\infty } = ;\frac{a}{{1; - ;r}}$; |r| < 1

Calculation:

Here, we have to find the sum of the series $3 - 1 + \frac{1}{3} - \frac{1}{9} + \ldots$

As, we can see that sequence is an GP with a = 3 and r = - 1/3

Sum of the series = S_∞ = a/ (1 - r)

Sum of the series = $\frac{3}{{1; - ;\frac{{ - 1}}{3}}} = ;\frac{3}{{1 + ;\frac{1}{3}}} = ;\frac{3}{{\left( {\frac{4}{3}} \right)}} = ;\frac{9}{4}$

Concept:

Let A, B and C denote the three set of elements.

n(A), n(B) and n(C) are the number of elements present in sets A, B and C.

n(A ⋃ B) is the number of elements present in either of the sets A or B.

n(A ⋂ B), n(B ⋂ C) and n(A ⋂ C) are the number of elements present in both the sets (A and B) ,

(B and C) and (A and C).

1. No. of elements present in at least one set

n(A ⋃ B ⋃ C) = n(A) + n(B) + n(C) - n(A ⋂ B) - n(B ⋂ C) - n(A ⋂ C) + n(A ⋂ B ⋂ C)

2. Number of elements present in exactly only one set

n(A) + n(B) + n(C) - 2 × [n(A ⋂ B) + n(B ⋂ C) + n(A ⋂ C)] + 3 × n(A ⋂ B ⋂ C)]

3. Number of elements present in exactly two of the sets

n(A ⋂ B) + n(B ⋂ C) + n(A ⋂ C)] - 3 × n(A ⋂ B ⋂ C)

Calculation:

Let A, B, and C be the set of persons who can like to play cricket, football and tennis

Given: n(A ⋃ B ⋃ C) = 300, n(A) = 125, n(B) = 145, n(C) = 90

n(A ⋃ B ⋃ C) = n(A) + n(B) + n(C) - n(A ⋂ B) - n(B ⋂ C) - n(A ⋂ C) + n(A ⋂ B ⋂ C)

⇒ 300 = 125 + 145 + 90 - n(A ⋂ B) - n(B ⋂ C) - n(A ⋂ C) + n(A ⋂ B ⋂ C)

⇒ n(A ⋂ B) + n(B ⋂ C) + n(A ⋂ C) = 360 - 300 + n(A ⋂ B ⋂ C)

⇒ n(A ⋂ B) + n(B ⋂ C) + n(A ⋂ C) = 60 + n(A ⋂ B ⋂ C)     ….(1)

Again;

32 students like to play exactly two games out of the three games.

n(A ⋂ B) + n(B ⋂ C) + n(A ⋂ C) - 3 × n(A ⋂ B ⋂ C) = 32

⇒ n(A ⋂ B) + n(B ⋂ C) + n(A ⋂ C) = 32 + 3 × n(A ⋂ B ⋂ C)     ….(2)

Put this value in equation 1^st

n(A ⋂ B) + n(B ⋂ C) + n(A ⋂ C) = 60 + n(A ⋂ B ⋂ C)

⇒ 32 + 3 × n(A ⋂ B ⋂ C) = 60 + n(A ⋂ B ⋂ C)

⇒ 2 × n(A ⋂ B ⋂ C) = 60 - 32

⇒ n(A ⋂ B ⋂ C) = 28/2

∴ n(A ⋂ B ⋂ C) = 14

Calculation:

Number of students like to play exactly only one game = n(A) + n(B) + n(C) - 2 × [n(A ⋂ B) + n(B ⋂ C) + n(A ⋂ C)] + 3 × n(A ⋂ B ⋂ C)

From equation 2

= 125 + 145 + 90 - 2 × [32 + 3 × n(A ⋂ B ⋂ C)] + 3 × 14

= 360 - 2 × [32 + 3 × 14] + 42

= 360 - 2 × 74 + 42

= 402 - 148

= 254

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

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:

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

Greatest value of a quadratic equation = $\frac{{ - D}}{{4a}} = \frac{{ - ;\left( {{b^2} - 4ac} \right)}}{{4a}}$

Calculation:

Given: x² + αx - β = 0

The sum of the roots = α + β = - α     …. (1)

The product of the roots = α × β = - β

⇒ α β + β = 0

⇒ (α + 1) β = 0

∴ (α + 1) = 0 and β ≠ 0

∴ α = - 1 and β ≠ 0

Now put the value of α in equation 1^st

α + β = - α

⇒ - 1 + β = - ( - 1)

∴ β = 2

Again; - x² + α x + β = - x² - x + 2

Greatest value of - x² + α x + β = - D/4a

Greatest value = - (1 + 8) / - 4 = 9/4

Concept:

General term: General term in the expansion of (x + y)ⁿ is given by

${T_{\left( {r; + ;1} \right)}} = ;{;^n}{C_r} \times {x^{n - r}} \times {y^r}$

Middle terms: The middle terms is the expansion of (x + y) ⁿ depends upon the value of n.

• If n is even, then the total number of terms in the expansion of (x + y) ⁿ is n +1. So there is only one middle term i.e. $\left( {\frac{n}{2} + 1} \right){{\rm{;}}^{th}}$ term is the middle term.

${T_{\left( {\frac{n}{2}; + ;1} \right)}} = ;{;^n}{C_{\frac{n}{2}}} \times {x^{\frac{n}{2}}} \times {y^{\frac{n}{2}}}$

• If n is odd, then the total number of terms in the expansion of (x + y) ⁿ is n + 1. So there are two middle terms i.e. ${\left( {\frac{{n; + ;1}}{2}} \right)^{th}};$and ${\left( {\frac{{n; + ;3}}{2}} \right)^{th}}$ are two middle terms.

 

Calculation:

Here, we have to find the coefficient of the middle term in the binomial expansion of (2 + 3x) ⁴

Here n = 4 (n is even number)

∴ Middle term = $\left( {\frac{n}{2} + 1} \right) = \left( {\frac{4}{2} + 1} \right) = 3^rd;term$

T₃ = T _{(2 + 1)} = ⁴C₂ × (2) ^{(4 - 2)} × (3x) ²

T₃ = 6 × 4 × 9x² = 216 x²

∴ Coefficient of the middle term = 216

Concept:

If A is a square matrix;

1. (AB) ^{- 1} = B ^{- 1} A ^{- 1}
2. (A ^{- 1}) ^{- 1} = A
3. (A^T) ^{- 1} = (A ^{- 1})^T
4. (KA ^{- 1}) = $\frac{1}{k};{A^{ - 1}}$ for any K ≠ 0
5. (Aⁿ) ^{- 1} = (A ^{- 1})ⁿ
6. AA ^{- 1} = A ^{- 1}A = I

Calculation:

1. If matrix A is square matrix and invertible matrix. Such that AA ^{- 1} = A ^{- 1}A = I also (AB) ^{- 1} = B ^{- 1} A ^{- 1}

AA ^{- 1} = I

(AA ^{- 1}) ^{- 1} = I ^{- 1}

We know that (AB) ^{- 1} = B ^{- 1} A ^{- 1}

∴ (A ^{- 1}) ^{- 1} A ^{- 1} = I ^{- 1} = I

Post multiple by A

⇒ (A ^{- 1}) ^{- 1} A ^{- 1} A = IA

⇒ (A ^{- 1}) ^{- 1} I = A

∴ (A ^{- 1}) ^{- 1} = A

∴ Option 1 is correct.

2. We know that det (AB) = det (A) det (B)

If matrix A is square matrix and invertible matrix. Such that AA ^{- 1} = A ^{- 1}A = I

det (AA ^{- 1}) = det (I)

det (A) det (A ^{- 1}) = 1

∴ $\det \left( {{A^{ - 1}}} \right) = \frac{1}{{detA}}$

∴ Option 2 is correct.

3. Given: (λA) ^{- 1} = λA ^{- 1} where λ is a scalar

We know that (KA ^{- 1}) = $\frac{1}{k};{A^{ - 1}}$ for any K ≠ 0

∴ Option 3 is wrong.

Concept:

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

Calculation:

\left| {\begin{array}{*{20}{c}} x&y&3\ {{x^2}}&{5{y^3}}&9\ {{x^3}}&{10{y^3}}&{27} \end{array}} \right|

Apply C₁ → C₁ - C₃

\Rightarrow \left| {\begin{array}{*{20}{c}} {x - 3}&y&3\ {{x^2} - 9}&{5{y^3}}&9\ {{x^3} - 27}&{10{y^3}}&{27} \end{array}} \right|

\(\Rightarrow ;\left| {\begin{array}{{20}{c}} {x - 3}&y&3\ {\left( {{\rm{x}} - 3} \right)\left( {x + 3} \right)}&{5{y^3}}&9\ {\left( {{\rm{x;}} - 3} \right)\left( {{x^2} + 9 + 3x} \right)}&{10{y^3}}&{27} \end{array}} \right| = \left( {x - 3} \right)\left| {\begin{array}{{20}{c}} 1&y&3\ {\left( {x + 3} \right)}&{5{y^3}}&9\ {\left( {{x^2} + 9 + 3x} \right)}&{10{y^3}}&{27} \end{array}} \right|\)

∴ (x - 3) is a factors of the expansion of the determinant.

Concept:

Let $\rm A; = ;\left[ {{a_{ij}}} \right]$ be a square matrix.

Steps for obtaining adjoint of the matrix

1. We need to calculate cofactors of all elements

⇒ $\rm {C_{ij}};or;{A_{ij}} = {\left( { - 1} \right)^{i + j}};{M_{ij}}$

Here, $\rm {M_{ij}}$ = Minor of matrix A

2. Transpose of cofactor is called adjoint of the matrix

⇒ adj (A) = C^T

Calculation:

⇒ cos ( - θ) = cos θ

⇒ sin ( - θ) = sin θ

A = \(\left( {\begin{array}{{20}{c}} {\cos \left( { - \theta } \right)}&{ - \sin \left( { - \theta } \right)}\ { - \sin \left( { - \theta } \right)}&{\cos \left( { - \theta } \right)} \end{array}} \right)\) = \(\left( {\begin{array}{{20}{c}} {\cos \theta }&{\sin \theta }\ {\sin \theta }&{\cos \theta } \end{array}} \right)\)

Cofactors of matrix A,

A₁₁ = ( - 1)^{1 + 1} cos θ = cos θ

A12 = ( - 1)^{1 + 2} sin θ = - sin θ

A21 = ( - 1)^{2 + 1} sin θ = - sin θ

A22 = ( - 1)^{2 + 2} cos θ = cos θ

Cofactors of matrix A = = {A_{ij}} = ;\left( {\begin{array}{*{20}{c}} {\cos \theta }&{ - \sin \theta }\ { - \sin \theta }&{\cos \theta } \end{array}} \right)

adj (A) = {\left[ {{A_{ij}}} \right]^T} = \left( {\begin{array}{*{20}{c}} {\cos \theta }&{ - \sin \theta }\ { - \sin \theta }&{\cos \theta } \end{array}} \right)

∴ Option 1 is correct.

Shortcut Method

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

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**.**

Concept:

Cube Roots of unity are 1, ω and ω²

Here, ω = $\frac{{ - ;1; + ;i\sqrt 3 }}{2}$ and ω² = $\frac{{ - ;1; - ;i\sqrt 3 }}{2}$

Property of cube roots of unity

1. ω³ = 1
2. 1 + ω + ω² = 0
3. ω = 1 / ω ² and ω² = 1 / ω
4. ω³ⁿ = 1

Calculation:

We know that ω = $\frac{{ - ;1; + ;i\sqrt 3 }}{2}$ and ω² = $\frac{{ - ;1; - ;i\sqrt 3 }}{2}$

${\left( {\frac{{ - 1 + {\rm{i}}\sqrt 3 }}{2}} \right)^{3{\rm{n}}}} + {\left( {\frac{{ - 1 - {\rm{i}}\sqrt 3 }}{2}} \right)^{3{\rm{n}}}} = {\rm{;}}{{\rm{\omega }}^{3{\rm{n}}}} + {({{\rm{\omega }}^2})^{3{\rm{n}}}} = ;{\left( {{\omega ^3}} \right)^n} + ;{\left( {{\omega ^3}} \right)^{2n}}$ = 1 + 1 = 2

Concept:

Combination: Selecting r objects from given n objects.

The number of selections of r objects from the given n objects is denoted by

${^n}{C_r}; = ;\frac{{n!}}{{r!\left( {n; - ;r} \right)!}}$

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

Calculation:

There are 17 cricket players, out of which 5 players can bowl.

We have to make a team of 11 players so to include 3 bowlers

So, we select 3 bowlers out of 5 players and remaining 8 players must be selected from 12 players

Total ways = C (5, 3) × C (12, 8)

∴ Option 4 is correct.

Concept:

Logarithm properties

Product rule:

The log of a product equals the sum of two logs.

${\log _a}\left( {mn} \right) = ;{\log _a}m + ;{\log _a}n$

Quotient rule:

The log of a quotient equals the difference of two logs.

${\log _a}\frac{m}{n} = ;{\log _a}m - ;{\log _a}n$

Power rule:

In the log of power, the exponent becomes a coefficient.

${\log _a}{m^n} = n{\log _a}m$

Change of base rule

${\log _m}n = ;\frac{{{{\log }_a}n}}{{{{\log }_a}m}}$

If m = n;

⇒ ${\log _m}m = ;\frac{{{{\log }_a}m}}{{{{\log }_a}m}} = 1$

 

Calculation:

log₉ 27 + log₈ 32

= $\frac{{\log 27}}{{\log 9}} + ;\frac{{\log 32}}{{\log 8}}$

= $\frac{{\log {3^3}}}{{\log {3^2}}} + ;\frac{{\log {2^5}}}{{\log {2^3}}}$

$= ;\frac{{3;\log 3}}{{2;\log 3}} + ;\frac{{5;\log 2}}{{3;\log 2}}$

$= ;\frac{3}{2}; + ;\frac{5}{3}$

$= ;\frac{{9 + 10}}{6} = ;\frac{{19}}{6}$

Concept:

If A is a square matrix;

(AB) ^{- 1} = B ^{- 1} A ^{- 1}

Calculation:

We know that AA ^{- 1} = I

So, (AB) (AB) ^{- 1} = I

Pre multiply by A ^{- 1},

⇒ (A ^{- 1} AB) (AB) ^{- 1} = A ^{- 1} I

⇒ (I B) (AB) ^{- 1} = A ^{- 1}

⇒ B (AB) ^{- 1} = A ^{- 1}

Pre multiply by B ^{- 1},

⇒ B ^{- 1} B (AB) ^{- 1} = B ^{- 1} A ^{- 1}

⇒ I (AB) ^{- 1} = B ^{- 1} A ^{- 1}

∴ (AB) ^{- 1} = B ^{- 1} A ^{- 1}

Note: (AB) ^{- 1} ≠ A ^{- 1} B ^{- 1}

Concept:

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

Calculation:

\left| {\begin{array}{*{20}{c}} {a - x}&c&b\ c&{b - x}&a\ b&a&{c - x} \end{array}} \right| = 0

Apply R₁ → R₁ + R₂ + R₃

\left| {\begin{array}{*{20}{c}} {a + b + c - x}&{a + b + c - x}&{a + b + c - x}\ c&{b - x}&a\ b&a&{c - x} \end{array}} \right| = 0

Take common (a + b + c - x) from row R₁

⇒ (a + b + c - x) \left| {\begin{array}{*{20}{c}} 1&1&1\ c&{b - x}&a\ b&a&{c - x} \end{array}} \right| = 0

Now; (a + b + c - x) = 0 and \left| {\begin{array}{*{20}{c}} 1&1&1\ c&{b - x}&a\ b&a&{c - x} \end{array}} \right| = 0

(a + b + c - x) = 0

⇒ x = a + b + c

Given: a + b + c = 0

∴ x = 0

Concept:

Consider a matrix A and let its inverse be A ^{- 1}

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

Here; adj (A) is adjoint of matrix A and det (A) is determinant of matrix A.

⇒ If det (A) ≠ 0, so inverse of matrix is exist.

⇒ If det (A) = 0, so inverse of matrix is not exist.

If A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}\ {{a_{21}}}&{{a_{22}}} \end{array}} \right] then determinant of A is given by: |A| = (a­₁₁ × a₂₂) - (a₁₂ - a₂₁).

Calculation:

Let A = \left( {\begin{array}{*{20}{c}} 2&4\ { - 8}&{\rm{x}} \end{array}} \right)

Given Matrix A does not have an inverse.

So, det (A) = 0

|A| = (2x) - (4 × - 8) = 0

⇒ 2x - ( - 32) = 0

⇒ 2x = - 32

∴ x = - 16

Concept

Let the system of equations be,

a₁x + b₁y + c₁z = d₁

a₂x + b₂y + c₂z = d₂

a₃x + b₃y + c₃z = d₃

\(\Rightarrow \left[ {\begin{array}{{20}{c}} {{a_1}}&{{b_1}}&{{c_1}}\ {{a_2}}&{{b_2}}&{{c_2}}\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right];\left[ {\begin{array}{{20}{c}} x\ y\ z \end{array}} \right] = ;\left[ {\begin{array}{*{20}{c}} {{d_1}}\ {{d_2}}\ {{d_3}} \end{array}} \right]\)⇒ AX = B

⇒ X = A ^{- 1} B = $\frac{{{\rm{adj;}}\left( {\rm{A}} \right)}}{{\det {\rm{;}}({\rm{A}})}};B$

⇒ If det (A) ≠ 0, system is consistent having unique solution.

⇒ If det (A) = 0 and (adj A). B = 0, system is consistent, with infinitely many solutions.

⇒ If det (A) = 0 and (adj A). B ≠ 0, system is inconsistent (no solution)

If A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}}\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}}\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}} \end{array}} \right] then determinant of A is given by:

|A| = a₁₁ × {(a₂₂ × a₃₃) - (a₂₃ × a₃₂)} - a₁₂ × {(a₂₁ × a₃₃) - (a₂₃ × a₃₁)} + a₁₃ × {(a₂₁ × a₃₂) - (a₂₂ × a₃₁)}

Calculation:

Given: The system of equations

2x + y - 3z = 5

3x - 2y + 2z = 5 and

5x - 3y - z = 16

So, A = \left[ {\begin{array}{*{20}{c}} 2&1&{ - 3}\ 3&{ - 2}&2\ 5&{ - 3}&{ - 1} \end{array}} \right]

det (A) = |A| = 2 × {( - 2 × - 1) - ( - 3 × 2)} - 1 × {(3 × - 1) - (2 × 5)} + ( - 3) × {(3 × - 3) - (5 × - 2)}

⇒ |A| = 2 × (2 + 6) - 1 × ( - 3 - 10) - 3 × ( - 9 + 10)

⇒ |A| = 16 + 13 - 3 = 26

∴ |A| ≠ 0

So, system is consistent having unique solution.

Concept

1. Cube Roots of unity are 1, ω and ω²

Here, ω = $\frac{{ - ;1; + ;i\sqrt 3 }}{2}$ and ω² = $\frac{{ - ;1; - ;i\sqrt 3 }}{2}$

2. Let A(x₁, y₁) and B(x₂, y₂)

⇒ Distance between A and B = $\sqrt {{{\left( {{y_2} - ;{y_1}} \right)}^2} + ;{{\left( {{x_2} - ;{x_1}} \right)}^2}}$

Calculation:

Let A = 1, B = $\frac{{ - ;1; + ;i\sqrt 3 }}{2}$ and C = $\frac{{ - ;1; - ;i\sqrt 3 }}{2}$

Now coordinate of points A, B and C in argand plane are A (1, 0), B ( - 1/2, √3/2) and C ( - 1/2, - √3/2)

AB = $\sqrt {{{\left( {\frac{{\sqrt 3 }}{2} - ;0} \right)}^2} + ;{{\left( {\frac{{ - ;1}}{2} - ;1} \right)}^2}}$

= $;\sqrt {{{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}; + {{\left( {\frac{{ - ;3}}{2}} \right)}^2}}$

= $;\sqrt {\frac{3}{4} + ;\frac{9}{4}} = ;\sqrt {\frac{{12}}{4}} = ;\sqrt 3$

BC = $\sqrt {{{\left( {\frac{{ - ;\sqrt 3 }}{2} - ;\frac{{\sqrt 3 }}{2}} \right)}^2} + ;{{\left( {\frac{{ - 1}}{2} - ;(\frac{{ - 1}}{2}}) \right)}^2}}$

= $;\sqrt {{{\left( {\frac{{ - ;2\sqrt 3 }}{2}} \right)}^2}; + {{\left( 0 \right)}^2}} ; = ;\sqrt 3$

AC = $\sqrt {{{\left( {\frac{{ - ;\sqrt 3 }}{2} - 0} \right)}^2} + ;{{\left( {\frac{{ - 1}}{2} - ;1} \right)}^2}}$

= $;\sqrt {{{\left( {\frac{{ - ;\sqrt 3 }}{2}} \right)}^2}; + {{\left( {\frac{{ - ;3}}{2}} \right)}^2}}$

= $;\sqrt {\frac{3}{4} + ;\frac{9}{4}} = ;\sqrt {\frac{{12}}{4}} = ;\sqrt 3$

⇒ AB = BC = AC

∴ They form an equilateral triangle.

Concept

1. Let a be the first term of the GP and r is the common ratio.

So, n^th term of a GP = a r^{n - 1}

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

${\log _a}\left( {mn} \right) = ;{\log _a}m + ;{\log _a}n$

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

${\log _a}\frac{m}{n} = ;{\log _a}m - ;{\log _a}n$

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

${\log _a}{m^n} = n{\log _a}m$

Calculation:

Let a is the first term of the GP and R is the common ratio, then we are given,

Given: u, v and w (all positive) are the p^th, q^th and r^th terms of a GP

So, u = aR^{p - 1}, v = aR^{q - 1} and w = aR^{r - 1}

\(\left[ {\begin{array}{{20}{c}} {\ln {\rm{u}}}&{\rm{p}}&1\ {\ln {\rm{v}}}&{\rm{q}}&1\ {\ln {\rm{w}}}&{\rm{r}}&1 \end{array}} \right] = ;\left[ {\begin{array}{{20}{c}} {\ln ;(a{R^{p; - ;1}})}&{\rm{p}}&1\ {\ln ;(a{R^{q; - ;1}})}&{\rm{q}}&1\ {\ln (a{R^{r; - ;1}})}&{\rm{r}}&1 \end{array}} \right]\)

= \left[ {\begin{array}{*{20}{c}} {\ln a + \left( {p - 1} \right)\ln R}&{\rm{p}}&1\ {\ln a + \left( {{\rm{q}} - 1} \right)\ln R}&{\rm{q}}&1\ {\ln a + \left( {{\rm{r}} - 1} \right)\ln R}&{\rm{r}}&1 \end{array}} \right]

Now, Applying R₁ → R₁ - R₃ and R₂ → R₂ - R₃

= \left[ {\begin{array}{*{20}{c}} {\left( {{\rm{p}} - {\rm{r}}} \right){\rm{ln;R}}}&{{\rm{p}} - {\rm{r}}}&0\ {\left( {{\rm{q}} - {\rm{r;}}} \right)\ln {\rm{R}}}&{{\rm{q}} - {\rm{r;}}}&0\ {\ln a{\rm{;}} + {\rm{;}}\left( {{\rm{r}} - 1} \right)\ln R}&{\rm{r}}&1 \end{array}} \right]

Now, Expanding along C₃

= 1 × {(p - r) ln R × (q - r) - (p - r) × (q - r) ln R}

= (p - r) (q - r) (ln R - ln R)

= 0

Concept

1. General term: General term in the expansion of (x + y) ⁿ is given by

${T_{\left( {r; + ;1} \right)}} = ;{;^n}{C_r} \times {x^{n - r}} \times {y^r}$

2. Middle terms: The middle terms is the expansion of (x + y) ⁿ depends upon the value of n.

• If n is even, then total number of terms in the expansion of (x + y) ⁿ is n+1. So there is only one middle term i.e. $\left( {\frac{n}{2} + 1} \right){{\rm{;}}^{th}}$ term is the middle term.

 

${T_{\left( {\frac{n}{2}; + ;1} \right)}} = ;{;^n}{C_{\frac{n}{2}}} \times {x^{\frac{n}{2}}} \times {y^{\frac{n}{2}}}$

• If n is odd, then total number of terms in the expansion of (x + y) ⁿ is n+1. So there are two middle terms i.e. ${\left( {\frac{{n; + ;1}}{2}} \right)^{th}};$ and ${\left( {\frac{{n; + ;3}}{2}} \right)^{th}}$ are two middle terms.

3. ${;^n}{C_r} + ;{;^n}{C_{r; + ;1}} = ;{;^{n; + ;1}}{C_{r; + ;1}}$

 

Calculation:

Middle term of (1 + x) ²ⁿ

Since 2n is even, Middle term = ${\left( {\frac{{2n}}{2} + 1} \right)^{th}}$ = (n + 1)^th = T_{n + 1}

T_{n + 1} = ${;^{2n}}{C_n};{x^n}$

Hence, coefficient of the middle term of the binomial expansion of (1 + x) ²ⁿ = α = ${;^{2n}}{C_n}$

Again, Middle term of (1 + x) ^{2n - 1}

Since 2n - 1 is odd, there will be two middle terms

Middle term = $\frac{{2{\rm{n;}} - {\rm{;}}1{\rm{;}} + {\rm{;}}1}}{2} = n;and;\frac{{2{\rm{n;}} - {\rm{;}}1{\rm{;}} + {\rm{;}}3}}{2} = n + 1;$

T_n = T_{n - 1 + 1} = ${;^{2n; - ;1}}{C_{n; - ;1}};{x^{n; + ;1}};$

Hence, coefficient = ${;^{2n; - ;1}}{C_{n; - ;1}}$ = β

T_{n + 1} = ${;^{2n; - ;1}}{C_{n;}};{x^{n;}}$

Hence, coefficient = ${;^{2n; - ;1}}{C_{n;}}$ = γ

⇒ β + γ = ${;^{2n; - ;1}}{C_{n; - ;1}} + ;{;^{2n; - ;1}}{C_{n;}} = ;{;^{2n}}{C_n} = \alpha ;$

∴ Option 3 is correct.

Concept:

Function is a relation between sets that associates to every element of a first set exactly one element of the second set.

Or, A function can only have one output, for each unique input.

Vertical line test: vertical line test is a visual way to determine if a curve is a graph of a function or not.

Calculation:

Given: S = {(x, y): x2 + y2 = 1,  - 1 ≤ x ∈ R ≤ 1 and - 1 ≤ y ∈ R ≤ 1}

x² + y² = 1, It represents the equation of circle.

As we can see through vertical line test it gives two output for unique input.

∴ It is not a function.

Concept:

Let a is the first term of the AP and d is the common difference,

So, n^th term of a AP = T_n = a + (n - 1) d

Calculation:

Given: T_m = 1/n

⇒ a + (m - 1) d =1/n …. (1)

Again, T_n = 1/m

⇒ a + (n - 1) d =1/m …. (2)

Now, subtracting equation (1) from equation (2)

{a + (n - 1) d} - {a + (m - 1) d} = 1/m - 1/n

⇒ a + nd - d - a - md + d = (n - m)/mn

⇒ d (n - m) = (n - m)/mn

∴ d = 1/mn

Put the value of d in equation 1^st

a + (m - 1) d =1/n

⇒ a + (m - 1) × 1/mn = 1/n

⇒ a + 1/n - 1/mn = 1/n

∴ a = 1/mn =d

Now, T_mn = a + (mn - 1) d

⇒ T_mn = a + mnd - d

⇒ T_mn = mn × 1/mn = 1

∴ Option 3 is correct.

Concept:

1. Consider the quadratic equation p(x) = ax² + bx + c = 0 where a, b and c ∈ R & a ≠ 0

Discriminant D = b² - 4ac

If D < 0 ⇔ Roots are imaginary and p(x) > 0

2. f’(x) is derivative of a function f(x) and f’’(x) is second derivative of a function f(x).

Calculation:

Given: f(x) is a quadratic expression and it is positive for all real x

So, f(x) = ax² + bx + c > 0

If f(x) > 0 then D < 0

So, b² - 4ac < 0 …. (1)

f(x) = ax² + bx + c

f’(x) = 2ax + b

f’’(x) = 2a

g(x) = f(x) + f'(x) + f”(x)

⇒ g(x) = ax² + bx + c + 2ax + b + 2a

⇒ g(x) = ax² + x(b + 2a) + (c + b +2a)

Discriminant = D = (b + 2a)² - 4 × a × (c + b +2a)

D = b² + 4a² + 4ab - 4ac - 4ab - 8a²

D = b² - 4a² - 4ac

D = (b² - 4ac) - 4a²

From equation 1^st

b² - 4ac < 0 and we know that square is always positive so 4a² > 0

∴ D < 0

Thus, g(X) > 0 for all real x.

Concept:

1. Transpose of a sum: The transpose of the sum of two matrices is equivalent to the sum of their transposes:

(A + B)’ = A’ + B’

∴ (A + B + C)’ = A’ + B’ + C’

So, statement 1 is correct.

2. Transpose of a product: The transpose of the product of two matrices is equivalent to the product of their transposes in reversed order:

(AB)’ = B’A’

Or, (ABC)’ = C’B’A’

So, Statement 3 is correct.

∴ Option 3 is correct.

Concept:

1. Adding two binary numbers: There are four rules that need to be followed when adding two binary numbers.

• 0 + 0 = 0.
• 1 + 0 = 1.
• 1 + 1 = 10 (binary for 2)
• 1 + 1 + 1 = 11 (binary for 3)

 

2. Binary Subtraction:

• 0 - 0 = 0
• 1 - 0 = 1
• 1 - 1 = 0
• 0 - 1 = 1 with a borrow of 1

 

Calculation:

(11011)₂ + (10110110)₂ + (10011x0y)₂ = (101101101)₂

⇒ (10011x0y)₂ = (101101101)₂ - (10110110)₂ - (11011)₂

101101101 - 10110110 = 10110111

10110111 - 11011 = 10011100 = 10011x0y

∴ x = 1 and y = 0

Concept:

Consider a matrix A and A is a square matrix.

A.adj (A) = adj (A).A = |A|I

Calculation:

Given:

B = adj (A)

I = Identity matrix

|A| = k

Now, AB = A.adj (A) = |A|I =KI

∴ Option 2 is correct.

Concept:

Logarithm properties

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

${\log _a}\left( {mn} \right) = ;{\log _a}m + ;{\log _a}n$

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

${\log _a}\frac{m}{n} = ;{\log _a}m - ;{\log _a}n$

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

${\log _a}{m^n} = n{\log _a}m$

1. Change of base rule

${\log _m}n = ;\frac{{{{\log }_a}n}}{{{{\log }_a}m}}$

If m = n;

⇒ ${\log _m}m = ;\frac{{{{\log }_a}m}}{{{{\log }_a}m}} = 1$ (Example, log₁₀ 10 = 1)

1. ${\log _m}n = ;\frac{1}{{{{\log }_n}m}}$

 

Calculation:

Given:

(0.2)^x = 2

Taking log both sides,

log₁₀ (0.2)^x = log₁₀ 2

⇒ x log₁₀ (2/10) = 0.3010

⇒ x (log₁₀ 2 - log₁₀ 10) =0.3010

⇒ x (0.3010 - 1) = 0.3010

⇒ x ( - 0.6990) = 0.3010

⇒ x = 0.3010/ ( - 0.6990)

∴ x = - 0.43

So, option 3 is correct.

Calculation:

Total number of possible digit = 10

(They are 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9)

Number of choices for the 1^st digit = 9

(Excluding the digit 0, since a number cannot start with zero)

Number of choices for the 2^nd digit = 9

(Any digit except the one that is already used as the 1^st digit)

Number of choices for the 3^rd digit = 8

(Any digit except the ones that are already used as the 1^st and 2^nd digit)

Number of choices for the 4^th digit = 7

(Any digit except the ones that are already used as the 1^st, 2^nd and 3^rd digit)

Number of choices for the 5^th digit = 6

(Any digit except the ones that are already used as the 1^st, 2^nd, 3^rd and 4^th digit)

Total possible combinations = 9 x 9 x 8 x 7 x 6 = 27216

Concept:

1. Elementary row or column transformations do not change the value of the determinant of a matrix.
2. If A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}}\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}}\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}} \end{array}} \right] then determinant of A is given by:

|A| = a₁₁ × {(a₂₂ × a₃₃) - (a₂₃ × a₃₂)} - a₁₂ × {(a₂₁ × a₃₃) - (a₂₃ × a₃₁)} + a₁₃ × {(a₂₁ × a₃₂) - (a₂₂ × a₃₁)}

Calculation:

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

Now, Applying R₁ → R₁ + R₂ + R₃

= ;\left| {\begin{array}{*{20}{c}} {x + ;y; + z}&{x + ;y; + z}&{2\left( {x + ;y; + z} \right)}\ z&x&{z + x}\ y&z&{x + y} \end{array}} \right|

= \left( {x + ;y; + z} \right);\left| {\begin{array}{*{20}{c}} 1&1&2\ z&x&{z + x}\ y&z&{x + y} \end{array}} \right|

Apply C₂ → C₂ - C₁ and C₃ → C₃ - 2C₁

= \left( {x + ;y; + z} \right)\left| {\begin{array}{*{20}{c}} 1&0&0\ z&{x - z}&{z + x - 2z}\ y&{z - y}&{x + y - 2y} \end{array}} \right|

= \left( {x + ;y; + z} \right)\left| {\begin{array}{*{20}{c}} 1&0&0\ z&{x - z}&{x - z}\ y&{z - y}&{x - y} \end{array}} \right|

Now, Expanding along R₁

= (x + y + z) × [1 × {(x - z) (x - y) - (x - z) (z - y)} - 0 + 0]

= (x + y + z) (x - z) (x - y - z + y)

= (x + y + z) (x - z)²

= (z - x)² (x + y + z)

∴ Option 4 is correct.

Concept:

1. Elementary row or column transformations do not change the value of the determinant of a matrix.
2. If A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}}\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}}\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}} \end{array}} \right] then determinant of A is given by:

|A| = a₁₁ × {(a₂₂ × a₃₃) – (a₂₃ × a₃₂)} - a₁₂ × {(a₂₁ × a₃₃) – (a₂₃ × a₃₁)} + a₁₃ × {(a₂₁ × a₃₂) – (a₂₂ × a₃₁)}

3. sin² θ + cos² θ = 1

Calculation:

Given:

\left| {\begin{array}{*{20}{c}} 1&1&1\ {1 + {\rm{sinA}}}&{1 + {\rm{sinB}}}&{1 + {\rm{sinC}}}\ {{\rm{sinA}} + {{\sin }^2}{\rm{A}}}&{\sin {\rm{B}} + {{\sin }^2}{\rm{B}}}&{\sin {\rm{C}} + {{\sin }^2}{\rm{C}}} \end{array}} \right| = 0

Now, Applying R₁ → R₁ - R₂, R₃ → R₃ – R₂

{\rm{;}}\left| {\begin{array}{*{20}{c}} {1 - \left( {{\rm{;}}1 + {\rm{sinA}}} \right)}&{1 - {\rm{;}}\left( {1 + {\rm{sinB}}} \right)}&{1 - {\rm{;}}\left( {1 + {\rm{sinC}}} \right)}\ {1 + {\rm{sinA}}}&{1 + {\rm{sinB}}}&{1 + {\rm{sinC}}}\ {{\rm{sinA}} + {{\sin }^2}{\rm{A;}} - {\rm{;}}\left( {{\rm{;}}1 + {\rm{sinA}}} \right)}&{\sin {\rm{B}} + {{\sin }^2}{\rm{B}}; - {\rm{;}}\left( {1 + {\rm{sinB}}} \right)}&{\sin {\rm{C}} + {{\sin }^2}{\rm{C;}} - {\rm{;}}\left( {1 + {\rm{sinC}}} \right)} \end{array}} \right| = 0

\Rightarrow ;\left| {\begin{array}{*{20}{c}} { - {\rm{;sinA}}}&{ - {\rm{;sinB}}}&{ - {\rm{;sinC}}}\ {1 + {\rm{sinA}}}&{1 + {\rm{sinB}}}&{1 + {\rm{sinC}}}\ {{{\sin }^2}{\rm{A}} - 1}&{{{\sin }^2}{\rm{B}} - 1}&{{{\sin }^2}{\rm{C}} - 1} \end{array}} \right| = 0

\Rightarrow {\rm{;}}\left| {\begin{array}{*{20}{c}} { - {\rm{;sinA}}}&{ - {\rm{;sinB}}}&{ - {\rm{;sinC}}}\ {1 + {\rm{sinA}}}&{1 + {\rm{sinB}}}&{1 + {\rm{sinC}}}\ { - {\rm{;co}}{{\rm{s}}^2}{\rm{A}}}&{ - {\rm{;}}{{\cos }^2}{\rm{B}}}&{ - {\rm{;co}}{{\rm{s}}^2}{\rm{C}}} \end{array}} \right| = 0

Now, Applying R₂ → R₂ + R₁

\Rightarrow {\rm{;}}\left| {\begin{array}{*{20}{c}} { - {\rm{;sinA}}}&{ - {\rm{;sinB}}}&{ - {\rm{;sinC}}}\ 1&1&1\ { - {\rm{;co}}{{\rm{s}}^2}{\rm{A}}}&{ - {\rm{;}}{{\cos }^2}{\rm{B}}}&{ - {\rm{;co}}{{\rm{s}}^2}{\rm{C}}} \end{array}} \right| = 0

Now, Applying R₃ → R₃ + R₂

\Rightarrow {\rm{;}}\left| {\begin{array}{*{20}{c}} { - {\rm{;sinA}}}&{ - {\rm{;sinB}}}&{ - {\rm{;sinC}}}\ 1&1&1\ { - {\rm{;co}}{{\rm{s}}^2}{\rm{A}} + 1}&{ - {\rm{;}}{{\cos }^2}{\rm{B}} + 1}&{ - {\rm{;co}}{{\rm{s}}^2}{\rm{C}} + 1} \end{array}} \right| = 0

\Rightarrow ;\left| {\begin{array}{*{20}{c}} { - {\rm{;sinA}}}&{ - {\rm{;sinB}}}&{ - {\rm{;sinC}}}\ 1&1&1\ {{{\sin }^2}{\rm{A}}}&{{{\sin }^2}{\rm{B}}}&{{{\sin }^2}{\rm{C}}} \end{array}} \right| = 0

Now, Applying C₃ → C₃ – C₂ and C₂ → C₂ – C₁

\Rightarrow ;\left| {\begin{array}{*{20}{c}} { - {\rm{;sinA}}}&{ - {\rm{;sinB}} - \left( { - {\rm{;sinA}}} \right)}&{ - {\rm{;sinC}} - \left( { - {\rm{;sinB}}} \right)}\ 1&0&0\ {{{\sin }^2}{\rm{A}}}&{{{\sin }^2}{\rm{B}} - {\rm{;}}{{\sin }^2}{\rm{A}}}&{{{\sin }^2}{\rm{C}} - {\rm{;}}{{\sin }^2}{\rm{B}}} \end{array}} \right| = 0

\Rightarrow ;\left| {\begin{array}{*{20}{c}} { - {\rm{;sinA}}}&{{\rm{sinA;}} - {\rm{;sinB}}}&{{\rm{sinB;}} - {\rm{;sinC}}}\ 1&0&0\ {{{\sin }^2}{\rm{A}}}&{{{\sin }^2}{\rm{B}} - {\rm{;}}{{\sin }^2}{\rm{A}}}&{{{\sin }^2}{\rm{C}} - {\rm{;}}{{\sin }^2}{\rm{B}}} \end{array}} \right| = 0

Now, expanding R₁

⇒ - 1 × [{(sin A – sin B) (sin² C – sin² B) – (sin B – sin C) (sin² B – sin² A)} + 0 – 0] = 0

⇒ -1 × [(sin A – sin B) (sin C –sin B) (sin C + sin B) – (sin C –sin B) (sin A – sin B) (sin A + sin B)] = 0

⇒ -1 × (sin A – sin B) (sin C –sin B) [sin C + sin B - sin A - sin B] = 0

⇒ (sin A – sin B) (sin C –sin B) (sin C – sin A) = 0

Therefore, sin A – sin B = 0, sin C –sin B = 0 and sin C – sin A = 0

So, sin A = sin B = sin c

Hence, A = B = C

∴ The triangle ABC is equilateral.

Shortcut Trick:

Take A = B = C

\(\left| {\begin{array}{{20}{c}} 1&1&1\ {1 + {\rm{sinA}}}&{1 + {\rm{sinB}}}&{1 + {\rm{sinC}}}\ {{\rm{sinA}} + {{\sin }^2}{\rm{A}}}&{\sin {\rm{B}} + {{\sin }^2}{\rm{B}}}&{\sin {\rm{C}} + {{\sin }^2}{\rm{C}}} \end{array}} \right| = {\rm{;}}\left| {\begin{array}{{20}{c}} 1&1&1\ {1 + {\rm{sinA}}}&{1 + \sin {\rm{A}}}&{1 + \sin {\rm{A}}}\ {{\rm{sinA}} + {{\sin }^2}{\rm{A}}}&{\sin {\rm{B}} + {{\sin }^2}{\rm{B}}}&{\sin {\rm{C}} + {{\sin }^2}{\rm{C}}} \end{array}} \right|\)

= \left( {1 + {\rm{sinA}}} \right){\rm{;}}\left| {\begin{array}{*{20}{c}} 1&1&1\ 1&1&1\ {{\rm{sinA}} + {{\sin }^2}{\rm{A}}}&{\sin {\rm{B}} + {{\sin }^2}{\rm{B}}}&{\sin {\rm{C}} + {{\sin }^2}{\rm{C}}} \end{array}} \right|

Here two rows are identical,

= 0

If we take A = B = C then determinant of a matrix is zero so triangle ABC is equilateral.

Note: If, we have any matrix with two identical rows or columns then its determinant is equal to zero.

Concept:

• Matrix product is commutative if both are diagonal matrices of same order.

           If A and B are diagonal matrices of same order then;

• The product of any square matrix and the appropriate identity matrix is always the original matrix.

            AI = IA = A

• I² = I, Here I is identity matrix.

 

Calculation:

1. Statement 1 is wrong because A and B are not diagonal matrices of same order.
2. (A – I) (I + A) = O

⇒ AI – I² + A² – IA = 0

⇒ - I² + A² = 0

⇒ A² = I² =I

So, statement 2 is correct.

∴ Option 2 is correct.

Concept:

• Sin² θ + cos² θ = 1
• 1 + tan² θ = sec² θ
• 1 +cot² θ = cosec² θ
• 2 sin θ cos θ = sin 2θ

Calculation:

We have to find the value of  $\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}$

$\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }} = {\rm{;}}\frac{{2\tan {\rm{\theta }}}}{{{{\sec }^2}{\rm{\theta }}}} = 2 \times \frac{{\sin {\rm{\theta }}}}{{\cos {\rm{\theta }}}} \times {\cos ^2}{\rm{\theta }} = 2 \times {\rm{;}}\sin {\rm{\theta }} \times {\rm{;}}\cos {\rm{\theta ;}} = {\rm{;}}\sin 2{\rm{\theta }}$

∴ Option 3 is correct.

Concept:

• If A, B and C are in AP then 2B =A + C
• 2 × Middle term = sum of 1^st term and last term
• cos (A + B) + cos (A – B) = 2 cos A cos B
• cos (A + B) cos (A – B) = cos² B - sin² A

 

Calculation:

Given: sec (θ – α), sec θ and sec (θ + α) are in AP.

So, 2 sec θ = sec (θ – α) + sec (θ + α)

$\Rightarrow {\rm{;}}\frac{2}{{\cos {\rm{\theta }}}} = {\rm{;}}\frac{1}{{\cos \left( {{\rm{\theta }} - {\rm{;\alpha }}} \right)}} + {\rm{;}}\frac{1}{{\cos \left( {{\rm{\theta }} + {\rm{;\alpha }}} \right)}}$

$\Rightarrow {\rm{;}}\frac{2}{{\cos {\rm{\theta }}}} = {\rm{;}}\frac{{\cos \left( {{\rm{\theta }} + {\rm{;\alpha }}} \right) + {\rm{;}}\cos \left( {{\rm{\theta }} - {\rm{;\alpha }}} \right)}}{{\cos \left( {{\rm{\theta }} + {\rm{;\alpha }}} \right)\cos \left( {{\rm{\theta }} - {\rm{;\alpha }}} \right)}}$

$\Rightarrow {\rm{;}}\frac{2}{{\cos {\rm{\theta }}}} = {\rm{;}}\frac{{\cos \left( {{\rm{\theta }} + {\rm{;\alpha }}} \right) + {\rm{;}}\cos \left( {{\rm{\theta }} - {\rm{;\alpha }}} \right)}}{{\cos \left( {{\rm{\theta }} + {\rm{;\alpha }}} \right)\cos \left( {{\rm{\theta }} - {\rm{;\alpha }}} \right)}}$

$\Rightarrow {\rm{;}}\frac{2}{{\cos {\rm{\theta }}}} = {\rm{;}}\frac{{2\cos {\rm{\theta }}\cos {\rm{\alpha }}}}{{{{\cos }^2}{\rm{\alpha ;}} - {\rm{;}}{{\sin }^2}{\rm{\theta }}}}$

⇒ cos² α - sin² θ = cos² θ cos α

⇒ cos² α = sin² θ + cos² θ cos α

⇒ cos² α = 1 - cos² θ + cos² θ cos α

⇒ cos² θ – cos² θ cos α = 1- cos² α

⇒ cos² θ (1 – cos α ) = (1 – cos α )(1 + cos α )

⇒ 1 – sin² θ = 1 + cos α

⇒ sin² θ + cos α =0

∴ Option 1 is correct.

Concept:

• $\sin {\rm{A;}} + {\rm{;}}\sin {\rm{B}} = 2{\rm{;}}\sin \left( {\frac{{{\rm{A;}} + {\rm{;B}}}}{2}} \right){\rm{;}}\cos \left( {\frac{{{\rm{A;}} - {\rm{;B}}}}{2}} \right)$
• $\sin {\rm{A;}} - {\rm{;}}\sin {\rm{B}} = 2{\rm{;}}\sin \left( {\frac{{{\rm{A;}} - {\rm{;B}}}}{2}} \right){\rm{;}}\cos \left( {\frac{{{\rm{A;}} + {\rm{;B}}}}{2}} \right)$
• sin (A + B) + sin (A – B) = 2 sin A cos B
• 2 sin A cos A = sin 2A

Calculation:

Given: A + B + C = 180° ⇒ C = 180° - A – B or A + B = 180° - C

We have to find the value of sin 2A – sin 2B – sin 2C

⇒ sin 2A – sin 2B – sin 2C

$= 2{\rm{;}}\cos \left( {\frac{{2{\rm{A}} + 2{\rm{B}}}}{2}} \right).{\rm{;}}\sin \left( {\frac{{2{\rm{A;}} - {\rm{;}}2{\rm{B}}}}{2}} \right) - {\rm{;}}\sin 2(180^\circ {\rm{;}} - {\rm{;A;}}-{\rm{;B}})$

= 2 cos (A + B).sin (A – B) – sin (360° - 2A – 2B)

= 2 cos (A + B).sin (A – B) + sin (2A + 2B)

= 2 cos (A + B).sin (A – B) + sin 2(A + B)

= 2 cos (A + B).sin (A – B) + 2 sin (A + B).cos (A + B)

= 2 cos (A + B). [sin (A – B) + sin (A + B)]

= 2 cos (180° - C). (2 sin A cos B)

= 4 × - cos C × sin A × cos B

= - 4 sin A cos B cos C

Concept:

• Angle of depression: The angle of depression is formed when the observer is higher than the object he/she is looking at. When an observer looks at an object that is situated at a distance lower than the observer, an angle is formed below the horizontal line drawn with the level of the eye of the observer and line joining object with the observer’s eye.

 

 

• tan (90 – θ) = cot θ
• cot (90 – θ) = tan θ

Calculation:

Let AC be the height of the balloon above the bridge = 122 meters

Since, the angle of depression of the other end of the bridge from the balloon is 48°.

Let the length of the bridge = AB = 'x' meters

So, Alternate angles ∠ CBA =48°

In triangle ABC,

tan 48° = AC/AB

⇒ AB = AC/ tan 48°

⇒ AB = AC cot 48°

⇒ x = 122 cot (90 – 42)°

⇒ x = 122 tan 42°

∴ Option 2 is correct.

Concept:

• tan (270 + θ) = - cot θ
• tan 300° = tan (270 + 30)° = - cot 30° = -  √3

Calculation

Given:

3 (3 – tan² A - cot A)² = 1

⇒ (3 – tan² A - cot A)² = 1/3

⇒ 3 – tan² A - cot A = 1/ √3

⇒ 3√3 - √3 tan² A – √3 (1/tan A) = 1

⇒ 3√3 tan A - √3 tan³ A - √3 = tan A

⇒ √3 tan³ A + tan A (1 – 3√3) + √3 = 0

Assume, tan A = - √3

So, √3 (- √3)³ + (- √3) (1 – 3√3) + √3 = - 9 - √3 + 9 + √3 = 0

So   tan A = - √3 is zeroes of polynomial.

⇒ tan A = - √3

⇒ tan A = tan 300°

∴ A = 300°

Shortcut Trick:

A = 300°

⇒ tan A = tan 300° = tan (270 + 30)° = - cot 30° = -  √3

⇒ tan A = - √3

⇒ tan A = tan (- 60° )

∴ A = - 60° 

 L.H.S = 3 (3 – tan² A - cot A)²

= 3 (3 – tan² (- 60°)   - cot (- 60°) )²

= 3 (3 – 3 + 1/√3 )² = 3 × 1/3 = 1

So, option A is correct.

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.

Calculation:

Given: height of the building = h

Let BC be the height of the hill = H

And distance between hill and building = x

In triangle ABC,

tan π/3 = BC/AB = H/x

√3 = H/x

⇒ x = H/√3       …. (1)

  Now, In triangle DEC,

AB = DE = x

⇒ tan π/6 = CE/DE = (H – h)/x

⇒ 1/√3 = (H – h)/x

Put the value of x from equation 1^st

$\frac{1}{{\sqrt 3 }} = {\rm{;}}\frac{{{\rm{H}} - {\rm{h}}}}{{\left( {\frac{{\rm{H}}}{{\sqrt 3 }}} \right)}}{\rm{;}}$

$\Rightarrow \frac{1}{{\sqrt 3 }}{\rm{}} \times {\rm{}}\left( {\frac{{\rm{H}}}{{\sqrt 3 }}} \right){\rm{}} = {\rm{;H}} - {\rm{h}}$

⇒ H /3 = H – h

⇒ H = 3H – 3h

⇒ 3h = 2H

∴ H = 3h/2

Concept:

• cosec² x – cot² x = 1
• a² – b² = (a – b) (a + b)

Calculation:

Given:

cosec x + cot x = √3       …. (1)

We know that, cosec² x – cot² x = 1

⇒ (cosec x - cot x) (cosec x + cot x) = 1

⇒ (cosec x - cot x) × √3 = 1

⇒ cosec x - cot x = 1/√3       …. (2)

Adding equation 1 and equation 2;

2 cosec x = √3 + 1/√3  

⇒ 2 cosec x = 4/√3

⇒ cosec x = 2/√3

⇒ sin x = √3/2

∴ x = π/3

Concept:

• ${\cos ^2}\frac{\pi }{8} = ;\frac{{2 + ;\sqrt 2 }}{4};$
• $\cos \frac{\pi }{4} = ;\frac{1}{{\sqrt 2 }}$

Calculation:

We have to find the value of (2cos θ + 1)¹⁰ (2 cos 2θ – 1)¹⁰ (2 cos θ – 1)¹⁰ (2 cos 4θ – 1)¹⁰

(2cos θ + 1)¹⁰ (2 cos 2θ – 1)¹⁰ (2 cos θ – 1)¹⁰ (2 cos 4θ – 1)¹⁰

= [(2cos θ + 1) (2 cos θ – 1)]¹⁰ (2 cos 2θ – 1)¹⁰ (2 cos 4θ – 1)¹⁰

= (4 cos² θ – 1)¹⁰ (2 cos 2θ – 1)¹⁰ (2 cos 4θ – 1)¹⁰

Putting, θ = π /8

2 θ = π/4 and 4 θ = π /2

= (4 cos² π /8 – 1)¹⁰ (2 cos π/4 – 1)¹⁰ (2 cos π /2 – 1)¹⁰

$= {\left( {4; \times ;\frac{{2 + ;\sqrt 2 }}{4} - 1} \right)^{10}}{\left( {2; \times ;\frac{1}{{\sqrt 2 }} - 1} \right)^{10}}{\left( {0; - ;1} \right)^{10}}$

= (√2 + 1)¹⁰ (√2 – 1)¹⁰ × 1 

= [(√2 + 1) (√2 – 1)]¹⁰

= (2 – 1)¹⁰

= 1