NDA Mathematics 17 Nov 2019 Official Paper (Previous Year Paper)

Concept:

For any quadratic equation, ax² + bx + c = 0. We have discriminant, D = b² - 4ac then the given quadratic equation has:

1. Distinct and real roots if D > 0.
2. Real and repeated roots, if D = 0.
3. Complex roots and conjugate of each other, D < 0.

Calculation:

Given:

Both p and q belong to the set {1, 2, 3, 4} and the quadratic equation of the form, px² + qx + 1 = 0 has real roots.

By comparing the quadratic equation of the form, px² + qx + 1 = 0 with the standard quadratic equation ax² + bx + c = 0. We get, a = p, b = q and c = 1.

D = q² - 4p

∵ The roots are real.

⇒ D ≥ 0 ⇒ q2 - 4p ≥ 0

⇒ q² ≥ 4p

∵ p and q belong to the set {1, 2, 3, 4}

Case -1:

• If p = 1 then q² ≥ 4 ⇒ q ∈ {2, 3, 4}
• Hence, for p = 1, q can take 3 values
• So, 3 quadratic equation of the form, px² + qx + 1 = 0 can be formed for p = 1.

Case -2:

• If p = 2 then q² ≥ 8 ⇒ q ∈ {3, 4}
• Hence, for p = 2, q can take 2 values
• So, 2 quadratic equation of the form, px² + qx + 1 = 0 can be formed for p = 2.

Case -3:

• If p = 3 then q² ≥ 12 ⇒ q ∈ {4}
• Hence, for p = 1, q can take 1 value
• So, only 1 quadratic equation of the form, px² + qx + 1 = 0 can be formed for p = 3.

Case -4:

• If p = 4 then q² ≥ 16 ⇒ q ∈ {4}
• Hence, for p = 4 , q can take 1 value
• So, only 1 quadratic equation of the form, px² + qx + 1 = 0 can be formed for p = 4.

So, in total we get, 3 + 2 + 1 + 1 = 7 possible quadratic equations of the form px² + qx + 1 = 0 that can be formed such that both p and q belong to the set {1, 2, 3, 4} and have real roots.

Concept:

If a₁, a₂, …., a_n be an AP then ${S_n} = {a_1} + {a_2} + \ldots + {a_n} = \frac{n}{2} \times \left( {2a + \left( {n - 1} \right)d} \right)$ where a is the 1^st term and d is the common difference.

If a₁, a₂, …, a_n be an AP then the general term is given by: a_n = a + (n - 1) × d where a is the 1^st term and d is the common difference.

Calculation:

Here, we have to find the value of 1 - 2 + 3 - 4 + 5 - ______ + 101.

⇒ 1 - 2 + 3 - 4 + 5 - ______ + 101 = (1 + 3 + …… + 101) - (2 + 4 + ….. + 100)

As, we can see that (1, 3, ……, 101) is an AP with a = 1 and d = 2.

⇒ a_n = 101 = 1 + (n - 1) × 2

⇒ n = 51

$\Rightarrow {S_{51}} = 1 + 3 + \ldots + 101 = \frac{{51}}{2} \times \left( {2 + 50 \times 2} \right) = 2601$

Similarly, (2, 4, …, 100) is an AP with a = 2 and d = 2

⇒ a_n = 100 = 2 + (n - 1) × 2

⇒ n = 50

$\Rightarrow {S_{50}} = 2 + 4 + \ldots + 100 = \frac{{50}}{2} \times \left( {4 + 49 \times 2} \right) = 2550$

⇒ 1 - 2 + 3 - 4 + 5 - ______ + 101 = (1 + 3 + …… + 101) - (2 + 4 + ….. + 100) = 2601 - 2550 = 51.

Concept:

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

I. A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

II. A ∪ A = A, A ∩ (A ∪ B) = A, A ∪ (A ∩ B) = A and A ∩ A = A

III. (A ∩ B) ∪ C = (A ∪ C) ∩ (B ∪ C)

IV. (A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C)

Calculation:

⇒ A ∪ (A ∩ B) = (A ∪ A) ∩ (A ∪ B) = A ∩ (A ∪ B) = A       --- (Using property I and II)

So, option 1 is not correct.

⇒ A ∩ (A ∪ B) = (A ∩ A) ∪ (A ∩ B) = A ∪ (A ∩ B) = A       --- (Using property I and II)

So, option 2 is correct.

⇒ (A ∩ B) ∪ C = (A ∪ C) ∩ (B ∪ C)       --- (Using property III)

So, option 3 is correct.

⇒ (A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C)       --- (Using property IV)

So, option 4 is correct.

Concept:

S_n denotes the sum of the first n terms of the sequence.

T_n = a_n = S_n - S_n-1

Calculation:

Given: S_n = n + 12

By substituting n = 2 in the given equation, we get

⇒ S₂ = 2 + 12 = 14

Similarly by substituting n = 3 in the given equation, we get

⇒ S₃ = 3 + 12 = 15

⇒ a₃ = S₃ - S₂ = 15 - 14 = 1.

Concept:

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

$\alpha + \beta = - \frac{b}{a};and;\alpha × \beta = \frac{c}{a}$

Calculation:

Given: 2x² - 2(k - 2) x - (k + 1) = 0

By comparing the given equation with the quadratic equation, ax² + bx + c = 0.

We get, a = 2, b = - 2(k - 2) and c = - (k + 1).

Let α and β be the roots of the equation 2x² - 2(k - 2) x - (k + 1) = 0.

$\Rightarrow \alpha + \beta = - \frac{{ - ;2\left( {k - 2} \right)}}{2} = k - 2$

$\Rightarrow \alpha × \beta = \frac{{ - \left( {k + 1} \right)}}{2}$

As we know that, α² + β² = (α + β) ² - 2αβ

⇒ α² + β² = (k - 2) ² + (1 + k)

⇒ α2 + β2 = k² + 4 - 4k + k + 1

⇒ α2 + β2 = k2 - 3k + 5

⇒ α2 + β2 = k2 - (3 × 2/2)k + 9/4 - 9/4 + 5

⇒ α² + β² = (k - 3/2)² + 11/4

So, the minimum value of α² + β² is found when k - 3/2 = 0 ⇒ k = 3/2.

Concept:

For any quadratic equation, ax² + bx + c = 0. We have discriminant, D = b² - 4ac, then the given quadratic equation has:

I. Distinct and real roots if D > 0.

II. Real and repeated roots, if D = 0.

III. Complex roots and conjugate of each other, D < 0.

If a, b and c are in HP, then the harmonic mean, $b = \frac{{2ac}}{{a + c}}$

Calculation:

Given: a (b - c) x² + b (c - a) x + c (a - b) = 0 has equal roots ⇒ Discriminant, D = 0.

By comparing the given equation with the quadratic equation, ax² + bx + c = 0. We get, a’ = a (b - c) , b’ = b (c - a) and c’ = c (a - b)

As, Discriminant, D = 0.

⇒ D = b² - 4ac = b² × (c - a) ² - 4 × a (b - c) × c (a - b) = 0

⇒ D = (bc + ab - 2ac) ² = 0

⇒ bc + ab - 2ac = 0

⇒ b × (a + c) = 2ac

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

Hence a, b and c are in HP.

Concept:

If f (x) = |x|, then \(f\left( x \right) = ;\left{ {\begin{array}{*{20}{c}} {x;,;x \ge 0}\ { - x,;x < 0} \end{array}} \right.\)

Calculation:

Given: |x² - 3x + 2| > x² - 3x + 2

Case-1: If x ≥ 0

⇒ x² - 3x + 2 > x² - 3x + 2

∴ No real value of x satisfies the above equation.

Case-2:- If x < 0

⇒ - (x² - 3x + 2) > x² - 3x + 2

⇒ x² - 3x + 2 < 0

⇒ (x - 1) (x - 2) < 0

⇒ 1 < x < 2.

Concept:

If a₁, a₂, ….., a_n are in GP then common ratio is given by: r = a_{i + 1} / a_i ∀ i =1, 2, …., n - 1.

The n^th term in a GP is given by: a_n = ar^{n - 1}, where a is the first term and r is the common ratio.

Calculation:

Given: The sum of odd terms of the GP is m and sum of even terms of the GP is n.

$\Rightarrow \frac{{Sum;of;even;terms;of;GP}}{{Sum;of;odd;terms;of;GP}} = \frac{{\left( {ar + a{r^3} + \ldots + a{r^{199}}} \right)}}{{\left( {a + a{r^2} + \ldots + a{r^{198}}} \right)}} = \frac{n}{m}$

$\Rightarrow \frac{{Sum;of;even;terms;of;GP}}{{Sum;of;odd;terms;of;GP}} = \frac{{ar \times \left( {1 + {r^2} + \ldots + {r^{198}}} \right)}}{{a \times \left( {1 + {r^2} + \ldots + {r^{198}}} \right)}} = r = \frac{n}{m}$

⇒ r = n / m.

Concept:

If A and B are two sets then, n (A ∪ B) = n (A) + n (B) - n (A ∩ B)

Calculation:

Given: n (A) = 3 and n (B) = 6.

As we know that, if A and B are two sets then, n (A ∪ B) = n (A) + n (B) - n (A ∩ B)

⇒ n (A ∪ B) = 3 + 6 - n (A ∩ B)

In order to minimize n (A ∪ B) we have to maximize n (A ∩ B) .

If A is a subset of B, then A ∩ B = A ⇒ n (A ∩ B) = n (A) = 3

⇒ n (A ∪ B) = 3 + 6 - 3 = 6.

Hence, the minimum number of elements that (A∪B) can have is 6.

Concept:

The no. of diagonals that can be drawn by joining the angular points of a polygon having n sides is given by: ${;^n}{C_2} - n = \frac{{n \times \left( {n - 3} \right)}}{2}$

Calculation:

Given: The polygon is an octagon ⇒ The no. of sides of the given polygon n = 8.

As we know that, no. of diagonals that can be drawn by joining the angular points of a polygon having n sides is given by: ${;^n}{C_2} - n = \frac{{n \times \left( {n - 3} \right)}}{2}$

⇒ ${;^8}{C_2} - 8 = \frac{{8 \times 5}}{2} = 20$

Concept:

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) .

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:

\(\Rightarrow \left| {\begin{array}{*{20}{c}} {1!}&{2!}&{3!}\ {2!}&{3!}&{4!}\ {3!}&{4!}&{5!} \end{array}} \right| \)

$= 1! \times \left( {3! \times 5! - 4! \times 4!} \right) - 2! \times \left( {2! \times 5! - 4! \times 3!} \right) + 3! \times \left( {2! \times 4! - 3! \times 3!} \right);$

= 144 - 192 + 72 = 24

Concept:

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) .

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:

\(\Rightarrow \left| {\begin{array}{*{20}{c}} x&0&2\ {2x}&2&1\ 1&1&1 \end{array}} \right| = x \times \left( {2 - 1} \right) - 0 \times \left( {2x - 1} \right) + 2 \times \left( {2x - 2} \right) = 5x - 4\)

\(\Rightarrow \left| {\begin{array}{*{20}{c}} {3x}&0&2\ {{x^2}}&2&1\ 0&1&1 \end{array}} \right| = 3x \times \left( {2 - 1} \right) - 0 \times \left( {{x^2} - 0} \right) + 2 \times \left( {{x^2} - 0} \right) = 2{x^2} + 3x\)

\(\Rightarrow \left| {\begin{array}{{20}{c}} x&0&2\ {2x}&2&1\ 1&1&1 \end{array}} \right| + \left| {\begin{array}{{20}{c}} {3x}&0&2\ {{x^2}}&2&1\ 0&1&1 \end{array}} \right| = \left( {5x - 4} \right) + \left( {2{x^2} + 3x} \right) = 2{x^2} + 8x - 4 = 0\)

⇒ 2x² + 8x - 4 = 0

By comparing the equation 2x² + 8x - 4 = 0 with ax² + bx + c = 0, we get a = 2, b = 8 and c = - 4.

$\Rightarrow x = \frac{{ - b \pm \sqrt {{b^2} - 4ac;} }}{{2a}} = \frac{{ - 8 \pm \sqrt {64 + 32} }}{4} = ; - 2 \pm \sqrt 6 ;$

Concept:

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

Calculation:

Given: x + a + b + c = 0

By applying R₁ ↔ R₁ - R₂ we get

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

By applying R₂ ↔ R₂ - R₃ we get

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

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

\(\Rightarrow \left| {\begin{array}{*{20}{c}} x&{ - x}&0\ 0&x&{ - x}\ a&b&{x + c} \end{array}} \right| = x \times \left( {{x^2} + cx + bx} \right) + x \times \left( {ax} \right) = x \times \left( {{x^2} + cx + bx + ax} \right)\)

As we know it is given that, x + a + b + c = 0

\(\Rightarrow \left| {\begin{array}{*{20}{c}} x&{ - x}&0\ 0&x&{ - x}\ a&b&{x + c} \end{array}} \right| = x \times \left( {{x^2} + cx + bx + ax} \right) = {x^2} \times \left( {x + a + b + c} \right) = 0\)

Concept:

Let A and B be any two matrices of order m × n and p × q respectively. Then the product AB exists if and only if n = p. Similarly, the product BA will exist if and only if q = m.

Zero Matrix or Null Matrix: 

• Any matrix in which all the elements are zero is called zero matrix. It is also known as null matrix and is denoted by O.

i.e \(O = ;{\left[ {\begin{array}{*{20}{c}} 0& \cdots &0\ \vdots & \ddots & \vdots \ 0& \cdots &0 \end{array}} \right]_{m \times ;n}}\) 

Identity Matrix:

• A diagonal matrix in which all the principal diagonal elements are equal to 1 is called an identity matrix. It is also known as unit matrix whereas an identity matrix of order n is denoted by I or In

If A and B are two matrices such that the no. of columns of A is equal to the no. of rows of B. If A = [a_ij] is a m × n matrix and B = [b_ij] be a n × p matrix, then the product AB is the resultant matrix of order m × p and is defined as:

\({\left( {AB} \right){ij}} = \mathop \sum \limits{k = 1}^n {a_{ik}} \times {b_{kj}} = {a_{i1}};{b_{1j}} + ;{a_{i2}};{b_{2j}} + \ldots + {a_{in}};{b_{nj}};\forall ;i = 1,;2, \ldots ,m;and;j = 1,;2,; \ldots .,;p\)

Calculation:

Given: \(A = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}\ { - 1}&1 \end{array}} \right]\)

\(\Rightarrow {A^3} = \left[ {\begin{array}{{20}{c}} 4&{ - 4}\ { - 4}&4\end{array}} \right];and;2{A^2} = 2 \times \left[ {\begin{array}{{20}{c}} 2&{ - 2}\ { - 2}&2 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 4&{ - 4}\ { - 4}&4 \end{array}} \right]\)

\(\Rightarrow ;{A^3} - 2{A^2} = ;\left[ {\begin{array}{{20}{c}} 4&{ - 4}\ { - 4}&4 \end{array}} \right] - ;\left[ {\begin{array}{{20}{c}} 4&{ - 4}\ { - 4}&4 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { 0}&0\ 0&{ 0} \end{array}} \right] \)

Calculation:

Given:

m and n (m < n) be the roots of the equation x² - 16x + 39 = 0

⇒ x² - 13x - 3x + 39 = 0

⇒ x (x - 13) - 3 (x - 13) = (x - 13) × (x - 3) = 0

⇒ x = 13 or 3

⇒ m = 3 and n = 13

∵ m, p, q, r, s and n are in AP

⇒ p + q + r + s = (m + d) + (m + 2d) + (m + 3d) + (m + 4d)

= 4m + 10d = 2 × (2m + 5d)

= 2 × [m + (m + 5d)]

= 2 × (m + n) ----------∵ (n = m + 5d)

⇒ p + q + r + s = 2 × (m + n)

∵ m + n = 16

⇒ p + q + r + s = 2 × (16) = 32

Concept:

For any quadratic equation, ax² + bx + c = 0. We have discriminant, D = b² - 4ac, then the given quadratic equation has:

I. Distinct and real roots if D > 0.

II. Real and repeated roots, if D = 0.

III. Complex roots and conjugate of each other, D < 0.

Calculation:

Given: x² + mx + 2 = 0

By comparing the given equation with the standard quadratic equation ax² + bx + c = 0, we get a = 1, b = m and c = 2.

⇒ D = b² - 4ac = m² - 8

For real roots, D ≥ 0

⇒ D = m² - 8 ≥ 0 ⇒ m² ≥ 8

∴ m2 ≥ 8.

Concept:

• i² = -1, i³ = -i, i⁴ⁿ = 1, where n = natural number
• The cube roots of unity are: 1, , ω and ω2
• $ω = ;\frac{{ - 1 + i\sqrt 3 ;}}{2};and;{ω ^2} = ;\frac{{ - 1 - i\sqrt 3 ;}}{2};$
• ω³ = 1

Calculation:

We know that,​

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

and i2 = -1

⇒ $ω = ;\frac{{{i^2} + i\sqrt 3 ;}}{2};and;{ω ^2} = ;\frac{{{i^2} - i\sqrt 3 ;}}{2};$

⇒ $- ;iω = \frac{{i + \sqrt 3 }}{2};and - i{ω ^2} = \frac{{i - \sqrt 3 }}{2}$

$\Rightarrow {\left[ {\frac{{i + \sqrt 3 }}{2}} \right]^{2019}} + {\left[ {\frac{{i - \sqrt 3 }}{2}} \right]^{2019}} = {\left( { - ;iω } \right)^{2019}} + {\left( { - i{ω ^2}} \right)^{2019}}$

⇒ -(i)^{2016 +3} (ω)²⁰¹⁹ - (i)^{2016 +3} (ω)4038

⇒ -(i)2016. (i)³ × 1 - (i)2016. (i)3 × 1

⇒ - 1 × (-i) × 1 - 1 × (-i) × 1 

⇒ (i) + (i) = 2i

∴ The required value is 2i.

Concept:

The cube roots of unity are: 1, ω, ω²

1 + ω + ω² = 0 and ω³ = 1

Calculation:

Given: x² + x + 1 = 0, here a = 1, b = 1 and c = 1.

$\Rightarrow x = \frac{{ - 1 \pm \sqrt {1 - 4} }}{2} = \frac{{ - 1 \pm \sqrt 3 i}}{2}$

⇒ The roots of the given equation are ω and ω² .

Let α = ω and β = ω²

\(\Rightarrow \mathop \sum \limits_{j = 0}^3 \left( {{α ^j} + {β ^j}} \right) = 2 + ω + {ω ^2} + {ω ^2} + {\left( {{ω ^2}} \right)^2} + {ω ^3} + {\left( {{ω ^2}} \right)^3} = 2\)

Calculation:

Let A = Students who play cricket, B = Students who play football

⇒ n (A) = 50% and n (B) = 40%

⇒ n (A ∩ B) = 10%

⇒ n (A ∪ B) = n (A) + n (B) - n (A ∩ B) = 50 + 40 - 10 = 80%

As we know that (A ∪ B) ’ = A’ ∩ B’

⇒ n ( (A ∪ B) ’) =n (U) - n (A ∪ B) = 100 - 80 = 20% = n (A’ ∩ B’)

Concept:

The prime numbers = A number that can be divided only by itself and 1

Calculation:

Given: A = (x : 0 ≤ x ≤ 2) and B = {y; y is a prime number}

A = (0, 1, 2) and B = (2, 3, 5, 7, 11, 13, 17, 19, .....)

So, from the above, The common number between A and B is 2.

⇒ A ∩ B = {2}

Concept: 

i² = - 1, i3 = - i

Calculation:

Given: x = 1 + i    ----(i)

On squaring both side, we get 

⇒ x² = 1 + i² + 2i

⇒ x2 = 1 - 1 + 2i

⇒ x2 = 2i    ----(ii)

Again On squaring both side, we get

⇒ x⁴ = 4i²

⇒ x4 = -4    ----(iii)

Now, we have to find the value of x6 + x4 + x2 + 1

⇒ x⁶ + x⁴ + x² + 1

⇒ (x2)³ + x4 + x2 + 1

Now from (i), (ii) and (iii) we get

⇒ (2i) ³ + (- 4) + (2i) + 1

⇒ -8i - 4 + 2i + 1 = -6i - 3

∴ The value of  x6 + x4 + x2 + 1 is (-6i - 3).

Calculation:

Let, $y = 2 + \frac{1}{{2 + \frac{1}{{2 + \frac{1}{{2 + \ldots \infty }}}}}}$

$\Rightarrow y = 2 + \frac{1}{y}$

⇒ y² - 2y - 1 = 0

By comparing the equation y² - 2y - 1 = 0 with the standard quadratic equation ax² + bx + c = 0. We get, a = 1, b = -2 and c = - 1.

$\Rightarrow y = \frac{{2 \pm \sqrt {4 + 4;} }}{2} = 1 \pm \sqrt 2$

 $\because y = 2 + \frac{1}{y}$ ⇒ y > 2.

Hence, y = 1 + √2

Concept:

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

$\frac{{{;^n}{P_r}}}{{{;^n}{C_r}}} = r!$

Calculation:

Given: P (n, r) = 2520 and C (n, r) = 21

As we know that, $\frac{{{;^n}{P_r}}}{{{;^n}{C_r}}} = r!$

⇒ P (n, r) / C (n, r) = 2520 / 21 = 120 = r!

⇒ r = 5.

As we know that, ${;^n}{P_r} = \frac{{n!}}{{\left( {n - r} \right)!}}$ 

$⇒ {;^n}{P_5} = \frac{{n!}}{{\left( {n - 5} \right)!}} = 2520$

$⇒ \frac{{n(n-1)(n-2)(n-3)(n-4)(n-5)!}}{{\left( {n - 5} \right)!}} = 2520$

⇒ n(n - 1)(n - 2)(n - 3)(n - 4) = 7 × 6 × 5 × 4 × 3 

On comparing both side, 

⇒ n = 7

⇒ C (n + 1, r + 1) = C (8, 6)

$⇒ {;^8}{C_6} = ;\frac{{8!}}{{6! × \left( {8 - 6} \right)!}} = 28$

Concept:

In the binomial expansion of (a + b) ⁿ, there are total n + 1 terms.

Calculation:

⇒ (1 + 2x + x²) ⁵ + (1 + 4y + 4y²) ⁵ = (1 + x) ^2 × 5 

• (1 + 2y)2 × 5

 = (1 + x) ¹⁰ + (1 + 2y)¹⁰

As we know that, in the binomial expansion of (a + b) ⁿ, there are total n + 1 terms.

⇒ The no. of terms in the binomial expansion of (1 + x) ¹⁰ and (1 + 2y)¹⁰ is 11

∴ The no. of terms in the given binomial expansion is 11 + 11 - 1 = 21

Mistake Points

We have subtracted 1 from (11 + 11) just because there is only one term common the expression of the binomial which is 1.

Concept:

The general term in the binomial expansion of (a + b) ⁿ is given by:

T_(r-1) = $^nC_r \times a^{n-r}\times b^r$

In the expansion of (a + b) ⁿ :

I. The middle term is ^{$\left( {\frac{{\rm{n}}}{2} + 1} \right)th$} term if n is even.

II. If n is odd the two middle terms are: $\left( {\frac{{{\rm{n}} + 1}}{2}} \right){\rm{th;and}};\left( {\frac{{n + 1}}{2} + 1} \right)th$ terms.

Calculation:

Given:

The middle term in the expansion of ^{${\left( {{x^2} + \frac{1}{x}} \right)^{2n}}$} is 184756x¹⁰.

As 2n is even

⇒ The middle term is $\left( {\frac{{2{\rm{n}}}}{2} + 1} \right) = ;\left( {n + 1} \right);th$ term

As we know that, The general term in the binomial expansion of (a + b) ⁿ is given by: ${T_{r + 1}} = {;^n}{C_r} \times {a^{n - r}} \times {b^r}$

\(\Rightarrow {T_{n + 1}} = {;^{2n}}{C_n} \times {x^{4n - 2n}} \times {\left( {\frac{1}{x}} \right)^n} \= ;{;^{2n}}{C_n} \times {x^n} = 184756 \times {x^{10}}\)

⇒ n = 10

Concept:

Let A and B be any two matrices of order m × n and p × q respectively. Then the product AB exists if and only if n = p. Similarly, the product BA will exist if and only if q = m.

Calculation:

Given: \(A = \left( {\begin{array}{{20}{c}} 1&2\ 2&3\ 3&4 \end{array}} \right)\) and \(B = \left( {\begin{array}{{20}{c}} 1&2\ 2&1 \end{array}} \right)\)

So, order of the matrix A is 3 × 2 whereas the order of the matrix A is 2 × 2.

As we know that, if A and B be any two matrices of order m × n and p × q respectively. Then the product AB exists if and only if n = p

Here, n = 2 = p ⇒ The product AB exists.

As we know that, if A and B be any two matrices of order m × n and p × q respectively. Then the product BA exists if and only if q = m.

Here, q = 2 and m = 3 ⇒ m ≠ q ⇒ The product BA does not exists.

Concept:

The no. of zeros present in a n! = The highest power of 5 in $\left[ {\frac{n}{5}} \right] + \left[ {\frac{n}{{25}}} \right] + \left[ {\frac{n}{{125}}} \right]$

Calculation:

Given: n! has 17 zeros

As we know that, the no. of zeros present in a n! = the highest power of 5 in $\left[ {\frac{n}{5}} \right] + \left[ {\frac{n}{{25}}} \right] + \left[ {\frac{n}{{125}}} \right]$

If n = 95, then $\left[ {\frac{{95}}{5}} \right] + \left[ {\frac{{95}}{{25}}} \right] + \left[ {\frac{{95}}{{125}}} \right] = 23 \ne 17$

Hence, n ≠ 95

If n = 85, then $\left[ {\frac{{85}}{5}} \right] + \left[ {\frac{{85}}{{25}}} \right] = 20 \ne 17$

Hence, n ≠ 85

If n = 80, then $\left[ {\frac{{80}}{5}} \right] + \left[ {\frac{{80}}{{25}}} \right] = 19 \ne 17$

Hence, n ≠ 80

Calculation:

Given: A ∪ B = {x|(x - a)(x - b) > 0, where a < b}

Here, if we take any value greater than the b 

Let say (c > b)

Now, if we put x = c in the above inequalities we get 

(c - a) (c - b)

Since, c > b > a 

So, (c - a) (c - b) > 0 

So, x > b satisfy the given inequalities 

Now, if we take a value between a and b 

Let say d: a < d < b 

Now, on putting x = d, we get 

(d - a) (d - b) = -ve × +ve = -ve

According to the condition we get 

(d - a) (d - b) < 0 

So, a < x < b doesn't satisfy the above inequalities 

Now, if we take a value between less than a 

Let say e: e < a

Now, on putting x = e in the inequality we get 

(e - a) (e - b) = -ve × -ve = +ve

So, x < a satisfy the given inequalities 

Hence, The correct relation is A = {x|x < a} and B = {x|x > b}.

Concept:

In the binomial expansion of (a + b)ⁿ, the term which does not involve any variable is said to be an independent term.

The general term in the binomial expansion of (a + b)ⁿ is given by: ${T_{r + 1}} = {;^n}{C_r} \times {a^{n - r}} \times {b^r}$

Calculation:

Given: The constant term in the expansion of ${\left( {\sqrt x - \frac{k}{{{x^2}}}} \right)^{10}}$ is 405

i.e the independent term of ${\left( {\sqrt x - \frac{k}{{{x^2}}}} \right)^{10}}$ is 405.

Let (r + 1)^th term be the independent term.

$\Rightarrow {T_{r + 1}} = {;^{10}}{C_r} \times {\left( {\sqrt x } \right)^{10 - r}} \times {\left( {\frac{{ - k}}{{{x^2}}}} \right)^r} = ;{;^{10}}{C_r} \times {\left( { - k} \right)^r} \times {\left( x \right)^{\frac{{10 - 5r}}{2}}};$

∵ (r + 1)th term is the independent term

$\Rightarrow \frac{{10 - 5r}}{2} = 0 \Rightarrow r = 2$

$\Rightarrow {;^{10}}{C_2} \times {\left( { - k} \right)^2} = 405 \Rightarrow 45{k^2} = 405$

⇒ k = ± 3

Concept:

C (n, r - 1) + C (n, r) = C (n + 1, r)

Calculation:

⇒ C(47, 4) + C(51, 3) + C(50, 3) + C(49, 3) + C(48, 3) + C(47, 3)

= {C(47, 4) + C(47, 3)} + C(51, 3) + C(50, 3) + C(49, 3) + C(48, 3)

As we know that,

C (n, r - 1) + C (n, r) = C (n + 1, r)

⇒ {C(47, 4) + C(47, 3)} + C(51, 3) + C(50, 3) + C(49, 3) + C(48, 3)

= {C (48, 4) + C(48, 3)} + C(51, 3) + C(50, 3) + C(49, 3)

⇒ {C (48, 4) + C(48, 3)} + C(51, 3) + C(50, 3) + C(49, 3)

= { C(49, 4) + C(49, 3)} + C(51, 3) + C(50, 3)

⇒ {C(49, 4) + C(49, 3)} + C(51, 3) + C(50, 3)

= {C(50, 4) + C(50, 3)} + C(51, 3)

⇒ {C(50, 4) + C(50, 3)} + C(51, 3)

= { C(51, 4) + C(51, 3)}

= C(52, 4)

Concept:

1. If a, b and c are in A.P 

• Then ka, kb, kc are in AP
• eg. 2, 4, 6 are in A.P then (2 × 2, 4 × 2, 6 × 2) = (4, 8, 12) are also in A.P

2. If a, b and c are in A.P 

• Then k - a, k - b, k - c are in AP
• eg. 2, 4, 6 are in A.P then (8 - 2, 8 - 4, 8 - 6) = (6, 4, 2) are also in A.P

3. If a, b and c are in A.P 

• Then $\frac{a}{k},\frac{b}{k},\frac{c}{k}$ are in AP
• eg. 2, 4, 6 are in A.P then (2/2, 4/2, 6/2) = (1, 2, 3) are also in A.P
• Hence, All the options are correct.

Concept:

If a₁, a₂, ….,a_n forms an AP then the nth term of an AP is given by : a_n = a + (n - 1) × d, where a is the first term and d is the common difference.

Calculation:

Two digit numbers which are divisible by 4 are:

12, 16, 20, ..., 96 forms an AP with,

First term (a) = 12,

Common difference (d) = 4 and,

n^th terms (a_n) = 96.

⇒ a_n = a + (n - 1) × d = 96

⇒ 12 + (n - 1) × 4 = 96

∴ n = 22

Concept:

If a and d are the first term and common difference of arithmetic progression. Then sum of first n terms of an arithmetic progression is given by: ${S_n} = \frac{n}{2} \times \left( {2a + \left( {n - 1} \right)d} \right)$

Calculation:

Given: S_2n = 3n + 14n²

If n = 1, then S₂ = 17

⇒ S₂ = 2a + d = 17     ---(1)

If n = 2, then S₄ = 62

⇒ S₄ = 4a + 6d = 62

⇒ 2a + 3d = 31     ---(2)

Subtracting (1) from (2), we get

⇒ 2d = 14 ⇒ d = 7.

Concept:

If a is the first term and r is the common ratio of a GP then the nth term of GP is given by: a_n = ar^{n - 1}

Calculation:

Given: 3rd, 8th and 13th terms of a GP are p, q and r respectively.

Let a is the first term and R is the common ratio.

⇒ a₃ = aR² = p     ---(1)

⇒ a₈ = aR⁷ = q    ----(2)

⇒ a₁₃ = aR¹² = r     ---(3)

Dividing (2) by (1), we get

⇒ q / p = R⁵          ---(4)

Similarly, by dividing (3) by (2), we get

⇒ r / q = R⁵          ---(5)

From (4) and (5), we get

⇒ q² = p × r

When x ≤ 4, y ≥ 0, it represents the region R₁

When x ≤ -4, y ≤ 0, it represents the region R₂

The solution set is represented by R₁ ∩ R₂ $\Rightarrow {R_1} \cap ;{R_2};:x \le - ;4;and;y = 0$

Concept:

${{\log }_{a}}b=\frac{\log b}{\log a}$

Calculation:

${{x}^{{{\log }_{7}}x}}>7$

By taking log both the sides we get,

\(⇒ \left( {{\log }_{7}}x \right)\times \log x>\log 7\)

As we know that, ${{\log }_{a}}b=\frac{\log b}{\log a}$

$⇒ \frac{{{\left( \log x \right)}^{2}}}{\log 7}>\log 7⇒ {{\left( \log x \right)}^{2}}>{{\left( \log 7 \right)}^{2}}$

$⇒ {{\left( \log x \right)}^{2}}-{{\left( \log 7 \right)}^{2}}>0$

$⇒ \left( \log x+\log 7 \right)\times \left( \log x-\log 7 \right)>0$

⇒ logx + log 7 < 0, logx - log 7 > 0

⇒ logx < -log 7 , logx > log 7

⇒ logx < log 7^-1 , logx > log 7

⇒ x < 1/7, x > 7

∴ $x\epsilon \left( 0,\frac{1}{7} \right)\cup \left( 7,~\infty \right)$

Concept:

If \(f\left( x \right)=\left| x \right|thenf\left( x \right)=~\left{ \begin{matrix} x,~x\ge 0 \ -x,~x<0 \ \end{matrix} \right.\)

Calculation:

Given: x² + 3|x| + 2 = 0

Case - 1: when x ≥ 0

⇒ x² + 3x + 2 = 0

⇒ x² + x + 2x + 2 = 0

⇒ (x + 1) (x + 2) = 0

⇒ x = - 1 or - 2

∵ x ≥ 0 ⇒ The given quadratic equation has no real roots for x ≥ 0.

Case - 2: when x < 0

⇒ x² - 3x + 2 = 0

⇒ x² - x - 2x + 2 = 0

⇒ (x - 1) (x - 2) = 0

⇒ x = 1 or 2

∵ x < 0 ⇒ The given quadratic equation has no real roots for x < 0.

Concept:

For any quadratic equation, ax² + bx + c = 0. We have discriminant, D = b² - 4ac then the given quadratic equation has:

1. Distinct and real roots if D > 0.
2. Real and repeated roots, if D = 0.
3. Complex roots and conjugate of each other, D < 0.

Calculation:

Given: 4(x - p)(x - q) - r² = 0

⇒ x² - (p + q) x + pq - r² /4 = 0

By comparing the quadratic equation x2 - (p + q) x + (pq - (r /2)2) = 0 with the standard quadratic equation ax² + bx + c = 0. We get, a = 1, b = - (p + q) and c =  pq - r2 /4

⇒ D = b² - 4ac = (p + q)² - 4pq + r² = (p - q)² + r² ≥ 0

Hence statement I is true

The given quadratic equation 4(x - p)(x - q) - r² = 0 will have repeated roots if D = 0.

⇒ D = (p - q)2 + r2 = 0 if and only if p = q and r = 0.

Hence statement II is true.

Concept:

1. Cardinality of a set A is defined as the no. of elements present in the set A, denoted by n(A).
2. Power set: Let A be a non-empty set. Then, \(P\left( A \right)=\left{ B~|B\subseteq A \right}\) is called as the power set of A.
3. For any non-empty set A with n(A) = x. Then the total no. of subsets of set A is given by: 2^x.

i.e n(P(A)) = 2^x.

Calculation:

Given: S = {2, 4, 6, 8, ______ 20}

⇒ The no. of elements in set S, n (S) = 10.

As we know that, for any non-empty set A with n(A) = x. Then the total no. of subsets of set A is given by: 2^x.

⇒ The maximum number of subsets of S = 2¹⁰ = 1024.

Calculation:

A binary number is a number expressed in the base-2 numeral system or binary numeral system, a method of mathematical expression that uses only two symbols: typically "0" (zero) and "1" (one).

Here we have,

c > d

Hence, c must be 1 and d must be 0

Hence, (cdccddcccddd) can be written as,

⇒ (cdccddcccddd)₂ = (101100111000)₂

Using concept of binary to decimal conversion,

⇒ (101100111000)₂ = (1 × 2¹¹) + (1 × 2⁹) + (1 × 2⁸) + (1 × 2⁵) + (1 × 2⁴) + (1 × 2³) + 0 = (2872)₁₀

Concept:

\(\sec \theta =\frac{1}{\cos \theta }and\tan \theta =~\frac{\sin \theta }{\cos \theta }\)

sin² x + cos² x = 1

Calculation:

Given: $cosec~\theta =\frac{29}{21}$

As we know that, sin² x + cos² x = 1

\(\Rightarrow \sin \theta =\frac{21}{29}and\cos \theta =\frac{20}{29}\)

\(\Rightarrow 4\sec \theta +4\tan \theta =4\times \frac{29}{20}+4\times \frac{21}{20}=10\)

Concept:

If A, G and H is the arithmetic, geometric and harmonic mean respectively between any two numbers a and b.

Then A ≥ G ≥ H

where,

A = (a + b)/2,

G = √(ab) and

H = 2ab/(a + b)

Calculation:

Let a = cos θ and b = sec θ

$\Rightarrow A=\frac{\cos \theta +\sec \theta }{2}$

and, $G=\sqrt{\cos \theta \times \sec \theta }$

According to the concept used

$\Rightarrow A\ge G$

$\Rightarrow ~\frac{\cos \theta +\sec \theta }{2}\ge 1$

$\Rightarrow \cos \theta +~\sec \theta ~\ge 2$

Hence, the statement I is true.

Similarly, Let a = tan θ and b = cot θ

$\Rightarrow A=\frac{\tan \theta +\cot \theta }{2}$

and, $G=\sqrt{\tan \theta \times \cot \theta }$

According to the concept used

\(\Rightarrow A\ge G\Rightarrow \frac{\tan \theta +\cot \theta }{2}\ge 1\Rightarrow \tan \theta +\cot \theta ~\ge 2\)

Hence, statement II is also true.

Calculation:

Let CD be the vertical flagstaff and AB is the ladder as shown in the figure below

Given: AB(a) = 9 m, CD = h m, BC = (h - 9) and ∠ DAC = 60°.

In Δ ACD, ∠ ADC = 180° - (150°) = 30°

In Δ ABD

As, AB = BD

⇒ ∠ DAB = ∠ ADB = 30°

⇒ ∠ DAC = 60° = ∠ BAC + ∠ DAB = ∠ BAC + 30°

⇒ ∠ BAC = 30°

In Δ ABC,

\(\Rightarrow \sin 30{}^\circ =\frac{h-9}{9}=\frac{1}{2}\Rightarrow h=13.5m\)

Concept:

The length of a chord subtending angle θ at the centre of a circle with radius r is given by: $2r× \sin \left( \frac{\theta }{2} \right)$

Calculation:

Given: r = 1 unit

⇒ The length of a chord subtending angle θ at the centre of a circle with radius 1 unit = $2× \sin \left( \frac{\theta }{2} \right)$

Alternate Solution:

In Triangle AOB,

Apply cosine rule

$\cos {\rm{\theta }} = {\rm{;}}\frac{{{\rm{A}}{{\rm{O}}^2} + {\rm{;B}}{{\rm{O}}^2}{\rm{;}} - {\rm{;A}}{{\rm{B}}^2}}}{{2{\rm{;}} × {\rm{AO;}} × {\rm{BO}}}}$

Where, AO = BO = Radius = 1

$\cos {\rm{\theta }} = {\rm{;}}\frac{{1 + {\rm{;}}1{\rm{;}} - {\rm{;A}}{{\rm{B}}^2}}}{{2{\rm{;}} × 1{\rm{;}} × 1}}$

⇒ 2 cos θ = 2 – AB²

⇒ AB² = 2 - 2 cos θ = 2 (1 - cos θ) = 2 × 2 sin² (θ/2)

∴ AB = 2 × sin (θ/2) = Length of chord

Concept:

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

Calculation:

Given: \(\tan \left{ 2{{\tan }^{-1}}\left( \frac{1}{3} \right) \right}\) ,

here x = 1/3 and - 1 < x < 1

$\Rightarrow 2\times {{\tan }^{-1}}\left( \frac{1}{3} \right)={{\tan }^{-1}}\left( \frac{2\times \frac{1}{3}}{1-{{\left( \frac{1}{3} \right)}^{2}}} \right)~$

$\Rightarrow {{\tan }^{-1}}\left( \frac{3}{4} \right)$

\(\Rightarrow \tan \left{ 2{{\tan }^{-1}}\left( \frac{1}{3} \right) \right}=\tan \left( {{\tan }^{-1}}\left( \frac{3}{4} \right) \right)=\frac{3}{4}\)

∴ The value of \(\tan \left{ 2{{\tan }^{-1}}\left( \frac{1}{3} \right) \right}\)is 3/4.

Concept:

Let \(\vec{u}and\vec{v}\) be two vectors. Then the scalar projection of vector \(\vec{u}on\vec{v}\) is given by: $\frac{\vec{u}\cdot ~\vec{v}}{\left| {\vec{v}} \right|}$

Calculation:

Given: $\vec{a}=\hat{i}-2\hat{j}+\hat{k}$ and $\vec{b}=4\hat{i}-4\hat{j}+7\hat{k}$

$\Rightarrow ~\vec{a}\cdot ~\vec{b}=\left( \hat{i}-2\hat{j}+\hat{k} \right)\cdot \left( 4\hat{i}-4\hat{j}+7\hat{k} \right)=4+8+7=~19$

$\Rightarrow \left| {\vec{b}} \right|=\sqrt{{{4}^{2}}+{{\left( -~4 \right)}^{2}}+{{7}^{2}}}=9$

∴ The scalar projection of vector \(\vec{a}on\vec{b}=~\frac{\vec{a}\cdot ~\vec{b}}{\left| {\vec{b}} \right|}=\frac{19}{9}\)

Concept:

Let $\vec{u}$ be a vector then ${{\left| {\vec{u}} \right|}^{2}}=~\vec{u}\cdot ~\vec{u}$

Calculation:

Let \(\vec{u}and\vec{v}\) be two non-zero vectors.

Given: The magnitude of the sum of two non-zero vectors is equal to the magnitude of their difference.

\(\Rightarrow \left| \vec{u}+\vec{v} \right|=\left| \vec{u}-~\vec{v} \right|\)

\(\Rightarrow {{\left| \vec{u}+\vec{v} \right|}^{2}}={{\left| \vec{u}-~\vec{v} \right|}^{2}}\)

As we know that, if $\vec{u}$ be a vector then ${{\left| {\vec{u}} \right|}^{2}}=~\vec{u}\cdot ~\vec{u}$

\(\Rightarrow \left( \vec{u}+\vec{v} \right)\cdot \left( \vec{u}+\vec{v} \right)=\left( \vec{u}-\vec{v} \right)\cdot \left( \vec{u}-\vec{v} \right)\)

\(\Rightarrow {{\left| {\vec{u}} \right|}^{2}}+{{\left| {\vec{v}} \right|}^{2}}+2~\left| {\vec{u}} \right|\times \left| {\vec{v}} \right|\times \cos \theta ={{\left| {\vec{u}} \right|}^{2}}+{{\left| {\vec{v}} \right|}^{2}}-2\left| {\vec{u}} \right|\times \left| {\vec{v}} \right|\times \cos \theta \)

$\Rightarrow 4\times ~\left| {\vec{u}} \right|\times \left| {\vec{v}} \right|\times \cos \theta =0$

∵ \(\vec{u}and\vec{v}\) be two non-zero vectors ⇒ $\left| {\vec{u}} \right|\times \left| {\vec{v}} \right|\ne 0$

⇒ cos θ = 0 ⇒ θ = 90°

Hence, \(\vec{u}and\vec{v}\) are perpendicular to each other.

Concept:

Let $\vec{u}$ be a vector then ${{\left| {\vec{u}} \right|}^{2}}=~\vec{u}\cdot ~\vec{u}$

Calculation:

\(\left( \vec{a}+\vec{b} \right)\cdot \left( \vec{a}-\vec{b} \right)=\{{\left| {\vec{a}} \right|}^{2}}-\left| {\vec{a}} \right|\times \left| {\vec{b}} \right|\times \cos \theta +\left| {\vec{a}} \right|\times \left| {\vec{b}} \right|\times \cos \theta -{{\left| {\vec{b}} \right|}^{2}}=~{{\left| {\vec{a}} \right|}^{2}}-{{\left| {\vec{b}} \right|}^{2}}\)

Hence, statement 1 is true.

\(\left( \left| \vec{a}+\vec{b} \right| \right)\left( \left| \vec{a}-\vec{b} \right| \right)=\\sqrt{{{\left| {\vec{a}} \right|}^{2}}+{{\left| {\vec{b}} \right|}^{2}}+2\times \left| {\vec{a}} \right|\times \left| {\vec{b}} \right|\times \cos \theta }~\times \sqrt{{{\left| {\vec{a}} \right|}^{2}}+{{\left| {\vec{b}} \right|}^{2}}-2\times \left| {\vec{a}} \right|\times \left| {\vec{b}} \right|\times \cos \theta }\)

If cos θ = 1

$\Rightarrow \left( \left| \vec{a}+\vec{b} \right| \right)\left( \left| \vec{a}-\vec{b} \right| \right)={{\left| {\vec{a}} \right|}^{2}}-{{\left| {\vec{b}} \right|}^{2}}$

But if cos θ ≠ 1

$\Rightarrow \left( \left| \vec{a}+\vec{b} \right| \right)\left( \left| \vec{a}-\vec{b} \right| \right)\ne {{\left| {\vec{a}} \right|}^{2}}-{{\left| {\vec{b}} \right|}^{2}}$

Hence, statement 2 is not true .

\({{\left| \vec{a}\cdot \vec{b} \right|}^{2}}+{{\left| \vec{a}\times \vec{b} \right|}^{2}}=\{{\left| {\vec{a}} \right|}^{2}}\times {{\left| {\vec{b}} \right|}^{2}}\times {{\cos }^{2}}\theta +{{\left| {\vec{a}} \right|}^{2}}\times {{\left| {\vec{b}} \right|}^{2}}\times {{\sin }^{2}}\theta ={{\left| {\vec{a}} \right|}^{2}}{{\left| {\vec{b}} \right|}^{2}}\)

Hence, statement 3 is true.

Concept:

If  $\vec{a}$ and $\vec{b}$ be two vectors, then $\left| \vec{a}\times ~\vec{b} \right|$ represents the area of a parallelogram with sides $\vec{a}$ and $\vec{b}$

Hence, statement 1 is false.

If  $\vec{a}\times \vec{b}=\vec{0}$ and it is given that $\vec{a}\ne \vec{0},~\vec{b}\ne \vec{0},$ ⇒  $\vec{a}$ and $\vec{b}$ are parallel to each other i.e $\vec{a}=\lambda \vec{b}$

Hence, statement 2 is true.

Concept:

If $\vec{a}$ is a unit vector then $\left| {\vec{a}} \right|=1$

If $\vec{u}$ be a vector then ${{\left| {\vec{u}} \right|}^{2}}=~\vec{u}\cdot ~\vec{u}$

Calculation:

Given: If $\vec{a}:and:\vec{b}$ are unit vectors and θ is the angle between them

As we know that, if $\vec{u}$ be a vector then ${{\left| {\vec{u}} \right|}^{2}}=\vec{u}\cdot ~\vec{u}$

\(\Rightarrow {{\left| \vec{a}-\vec{b} \right|}^{2}}=\left( \vec{a}-\vec{b} \right)\cdot \left( \vec{a}-~\vec{b} \right)={{\left| {\vec{a}} \right|}^{2}}+{{\left| {\vec{b}} \right|}^{2}}-2\times \left| {\vec{a}} \right|\times ~\left| {\vec{b}} \right|\times \cos \theta \)

∵ $\vec{a}:and:\vec{b}$ are unit vectors 

$\Rightarrow {{\left| \vec{a}-~\vec{b} \right|}^{2}}=2-2\times \cos \theta$