Quantitative Aptitude PYQs

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.