NDA Mathematics 18 Sept 2016 Official Paper (Previous Year Paper)

Calculation:

Given: We have to form all distinct numbers of the form $\frac{{\rm{p}}}{{\rm{q}}}$

As 1/2, 2/4, 3/6 are same and 1/1, 2/2, 3/3...6/6 are same.

We will find only co-prime numbers.

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∴ Cardinality of set S = 6 + 3 + 4 + 3 + 5 + 2 = 23

Calculation:

Given: c > 0 and 4a + c < 2b

Let f(x) = ax² – bx + c

Putting x = 0, we get 

f(x = 0) = 0 – 0 + c

⇒ f(0) = c

∴ f(x) is positive at x = 0                        (∵ c > 0)

Now Putting x = 2, we get

f(x) = 4a − 2b + c  

⇒ f(x) = (4a + c) − (2b)

As we know that 4a + c < 2b

∴ f(x) is negative at x = 2

f(x) changes it sign between 0 & 2

So a root must lie between x = 0 and x = 2. 

In Other Options f(x) is not changing its sign in between the given values.

∴ Option 1^st is correct.

Calculation:

Given:

A = {x ∈ R : x² + 6x – 7 < 0}

⇒ x² + 6x – 7 < 0

⇒ x² + 7x - x – 7 < 0

⇒ x (x + 7) – 1 (x + 7) < 0

⇒ (x – 1) (x + 7) < 0

∴ A ∈ (-7, 1)

Now, B = {x ∈ R : x² + 9x + 14 > 0}

⇒ x² + 9x + 14 > 0

⇒ x² + 7x + 2x + 14 > 0

⇒ x (x + 7) + 2(x + 7) > 0

⇒ (x + 7) (x + 2) > 0

∴ B ∈ (-∞, -7) ∪ (-2, ∞)

Now, A ∩ B

∴ A ∩ B = (-2, 1) ⇔ A ∩ B = {x ∈ R : - 2 < x < 1}

So, statement 1 is true,

Now, A ∪ B = R – {-7}

So, statement 2 is wrong,

Concept:

• $\det \left( {{{\rm{A}}^{ - 1}}} \right) = \frac{1}{{\det ({\rm{A}})}}$
• If A is an n × n matrix and k is any scalar, then det (kA) = kⁿ det (A)

Calculation:

Given: A is a square matrix of order 3 and det A = 5

Now, we have to find the value of det [(2A)^-1]

$\Rightarrow \det \left( {2{{\rm{A}}^{ - 1}}} \right) = \frac{1}{{{\rm{det}}\left( {2{\rm{A}}} \right)}};$               $\because \left( {\det \left( {{{\rm{A}}^{ - 1}}} \right) = \frac{1}{{\det ({\rm{A}})}}} \right)$

$\Rightarrow \det \left( {2{{\rm{A}}^{ - 1}}} \right) = \frac{1}{{{2^3}{\rm{det}}\left( {\rm{A}} \right)}}$               (∵ det (kA) = kⁿ det (A))

$\Rightarrow \det \left( {2{{\rm{A}}^{ - 1}}} \right) = \frac{1}{{8; \times 5}} = ;\frac{1}{{40}}$

∴ Option 4 is correct.

Concept:

Cube Roots of unity are 1, ω and ω²

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

Property of cube roots of unity

• ω³ = 1
• 1 + ω + ω² = 0
• ω = 1 / ω ² and ω² = 1 / ω
• ω³ⁿ = 1

<br>

Calculation:

We have to find the value of ω¹⁰⁰ + ω²⁰⁰ + ω³⁰⁰

⇒ ω¹⁰⁰ + ω²⁰⁰ + ω³⁰⁰

= ω⁹⁹ ω + ω¹⁹⁸ ω² + (ω³)¹⁰⁰

= (ω³)³³ ω + (ω³)⁶⁶ ω² + 1     (∵ ω³ = 1)

= ω + ω² + 1

= 0                       (∵ 1 + ω + ω² = 0)

Concept:

1. Equality of complex numbers.

• Two complex numbers z₁ = a + ib and z₂ = x + iy are equal if and only if a = x and b = y
• Or Re (z₁) = Re (z₂) and Im (z₁) = Im (z₂).

2. Let z = x + iy

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

 

Calculation:

Given:

${\rm{Re}}\left( {\frac{{{\rm{z}} - 1}}{{{\rm{z}} + 1}}} \right) = 0$

$\Rightarrow {\rm{Re}}\left( {\frac{{{\rm{x;}} + {\rm{;iy;}} - 1}}{{{\rm{x;}} + {\rm{;iy;}} + 1}}} \right) = 0$

$\Rightarrow {\rm{Re}}\left[ {\frac{{\left( {{\rm{x;;}} - 1} \right){\rm{;}} + {\rm{;iy}}}}{{\left( {{\rm{x;;}} + 1} \right){\rm{;}} + {\rm{;iy}}}}} \right] = 0$

$\Rightarrow {\rm{Re}}\left[ {\frac{{\left( {{\rm{x;;}} - 1} \right){\rm{;}} + {\rm{;iy}}}}{{\left( {{\rm{x;;}} + 1} \right){\rm{;}} + {\rm{;iy}}}}; \times ;\frac{{\left( {{\rm{x;;}} + 1} \right) - {\rm{;iy}}}}{{\left( {{\rm{x;;}} + 1} \right) - {\rm{;iy}}}}} \right] = 0$

$\Rightarrow {\rm{Re}}\left[ {\frac{{\left( {{\rm{x;;}} - 1} \right)\left( {{\rm{x;;}} + 1} \right) - {\rm{;}}{{\left( {{\rm{iy}}} \right)}^2} + i;\left[ { - y\left( {{\rm{x;;}} - 1} \right) + y\left( {{\rm{x;;}} + 1} \right)} \right]}}{{{{\left( {{\rm{x;;}} + 1} \right)}^2} - {{\left( {{\rm{;iy}}} \right)}^2}}};} \right] = 0$

$\Rightarrow {\rm{Re}}\left[ {\frac{{\left( {{x^2} - 1 + {\rm{;}}{{\rm{y}}^2}} \right) + i;\left[ { - y\left( {{\rm{x;;}} - 1} \right) + y\left( {{\rm{x;;}} + 1} \right)} \right]}}{{{{\left( {{\rm{x;;}} + 1} \right)}^2} + {\rm{;}}{{\rm{y}}^2}}};} \right] = 0$                             (∵ i² = -1)

$\Rightarrow {\rm{Re}}\left[ {\frac{{\left( {{x^2} + {\rm{;}}{{\rm{y}}^2}; - {\rm{;}}1} \right)}}{{{{\left( {{\rm{x;;}} + 1} \right)}^2} + {\rm{;}}{{\rm{y}}^2}}}; + \frac{{i;\left[ { - y\left( {{\rm{x;;}} - 1} \right) + y\left( {{\rm{x;;}} + 1} \right)} \right]}}{{{{\left( {{\rm{x;;}} + 1} \right)}^2} + {\rm{;}}{{\rm{y}}^2}}}} \right] = 0$

$\Rightarrow {\rm{;}}\frac{{\left( {{x^2} + {\rm{;}}{{\rm{y}}^2}; - {\rm{;}}1} \right)}}{{{{\left( {{\rm{x;;}} + 1} \right)}^2} + {\rm{;}}{{\rm{y}}^2}}} = ;0$

⇒ x² + y² – 1 = 0

⇒ x² + y² = 1

∴ |z| = 1

Calculation:

We have to find the value of \(\left[ {{\rm{x;;y;;z}}} \right]\left[ {\begin{array}{*{20}{c}} {\rm{a}}&{\rm{h}}&{\rm{g}}\ {\rm{h}}&{\rm{b}}&{\rm{f}}\ {\rm{g}}&{\rm{f}}&{\rm{c}} \end{array}} \right]\)

\( \Rightarrow \left[ {{\rm{x;;y;;z}}} \right]\left[ {\begin{array}{*{20}{c}} {\rm{a}}&{\rm{h}}&{\rm{g}}\ {\rm{h}}&{\rm{b}}&{\rm{f}}\ {\rm{g}}&{\rm{f}}&{\rm{c}} \end{array}} \right]\)

= [ax + hy + gz    hx + by + fz    gx + fy + cz]

Concept:

Number of ways to select 3 points out of the n collinear points = ${;^n}{C_3}$

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

 

Calculation:

Number of triangles that can be formed is equal to the number of ways to select 3 non-collinear points.

⇒ Number of ways to select 3 points from 15 points = ¹⁵c₃

Let n points be collinear.

⇒ Number of ways to select 3 points out of the n collinear points = ⁿc₃

So, Number of ways to select 3 non-collinear points = (Number of ways to select 3 points using all the points - Number of ways to select 3 points using the collinear points)

⇒ Number of ways to select 3 non-collinear points = ¹⁵c₃ - ⁿc₃

⇒ Number of triangles that can be formed = ¹⁵c₃ - ⁿc₃

⇒ 445 = ¹⁵c₃ - ⁿc₃

⇒ ⁿc₃ = ¹⁵c₃ – 445 = 455 – 445 = 10

$\Rightarrow \frac{{n!}}{{\left( {n - 3} \right)!; \times 3!}} = 10$

$\Rightarrow \frac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{6} = 10$

⇒ n (n – 1) (n – 2) = 60

∴ n = 5

Concept:

Cube Roots of unity are 1, ω and ω²

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

Property of cube roots of unity

• ω³ = 1
• 1 + ω + ω² = 0
• ω = 1 / ω ² and ω² = 1 / ω
• ω³ⁿ = 1

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Calculation:

Given: ${\rm{z}} = {\left( {\frac{{\sqrt 3 }}{2} + \frac{{\rm{i}}}{2}} \right)^{107}} + {\left( {\frac{{\sqrt 3 }}{2} - \frac{{\rm{i}}}{2}} \right)^{107}}$

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

$\Rightarrow {\rm{ω }} = {\rm{;}}\left( {\frac{{ - 1 + {\rm{i}}\sqrt 3 }}{2}} \right) × \frac{{\rm{i}}}{{\rm{i}}}{\rm{;and;;}}{{\rm{ω }}^2} = {\rm{;}}\left( {\frac{{ - 1 - {\rm{i}}\sqrt 3 }}{2}} \right) × \frac{{\rm{i}}}{{\rm{i}}}$

$\Rightarrow {\rm{ω }} = {\rm{;}}\left( {\frac{{ - {\rm{i}} + {{\rm{i}}^2}\sqrt 3 }}{{2{\rm{i}}}}} \right){\rm{;and;;}}{{\rm{ω }}^2} = {\rm{;}}\left( {\frac{{ - {\rm{i}} - {{\rm{i}}^2}\sqrt 3 }}{{2{\rm{i}}}}} \right){\rm{;}}$

$\Rightarrow {\rm{ω }} = {\rm{;}}\left( {\frac{{ - {\rm{i}} - \sqrt 3 }}{{2{\rm{i}}}}} \right){\rm{;and;;}}{{\rm{ω }}^2} = {\rm{;}}\left( {\frac{{ - {\rm{i;}} + \sqrt 3 }}{{2{\rm{i}}}}} \right)$

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

Now,

${\rm{z}} = {\left( {\frac{{\sqrt 3 }}{2} + \frac{{\rm{i}}}{2}} \right)^{107}} + {\left( {\frac{{\sqrt 3 }}{2} - \frac{{\rm{i}}}{2}} \right)^{107}} = {\rm{;}}{\left( { - {\rm{iω }}} \right)^{107}} + {\left( { {\rm{i}}{{\rm{ω }}^2}} \right)^{107}}$

⇒ z = - (i⁴)²⁶ × i³ × [(ω³)³⁵ × ω² ] + (i4)26 × i3 × [(ω3)71 × ω]        

⇒ z = i × (ω² - ω) = $\sqrt 3$

So, the Imaginary part of z is 0.

Calculation:

Given quadratic equation x² – 2kx + k² - 4 = 0

⇒ (x – k) ² – 2² = 0

⇒ (x – k – 2) (x – k + 2) = 0         [∵ a² – b² = (a – b) (a + b)]

∴ x = k + 2 and k – 2

Therefore Larger root = k + 2 and Smaller root = k – 2

Given Roots lie in the interval (−3, 5)

 

So, larger root < 5

⇒ k + 2 < 5

⇒ k < 3

Also, smaller root > −3

⇒ k – 2 > −3

⇒ k > −1

∴ −1 < k < 3

Concept:

Let z = x + iy be a complex number.

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

Equality of complex numbers: Two complex numbers z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂ are equal

• If and only if x₁ = x₂ and y₁ = y₂ Or Re (z₁) = Re (z₂) and Im (z₁) = Im (z₂).

 

Calculation:

Given:

z² + |z| = 0

$\Rightarrow {\rm{;}}{\left( {{\rm{x;}} + {\rm{;iy}}} \right)^2} + {\rm{;}}\sqrt {{{\rm{x}}^2} + {{\rm{y}}^2}} = 0{\rm{;}}$

$\Rightarrow {\rm{;}}{{\rm{x}}^2} + {\rm{i}}2{\rm{xy}} + {\left( {{\rm{iy}}} \right)^2} + {\rm{;}}\sqrt {{{\rm{x}}^2} + {{\rm{y}}^2}} = 0$

$\Rightarrow \left( {{{\rm{x}}^2} - {{\rm{y}}^2} + {\rm{;}}\sqrt {{{\rm{x}}^2} + {{\rm{y}}^2}} } \right) + {\rm{i}}2{\rm{xy}} = 0 + {\rm{i}}0$

Equate real and imaginary part, we get

$\Rightarrow \left( {{{\rm{x}}^2} - {{\rm{y}}^2} + {\rm{;}}\sqrt {{{\rm{x}}^2} + {{\rm{y}}^2}} } \right) = 0$            …. (1)

⇒ 2xy = 0

∴ x = 0 or y = 0

Case-1: put x = 0 in equation 1^st, we get

⇒ 0 – y² + √(0 + y²) = 0

⇒ |y| - y² = 0                              (∵ √(y²) = |y|)

⇒ ± y - y² = 0

⇒ y (±1 – y) = 0

∴ y = 0 and ±1

Solutions are (0, 0), (0, 1) and (0, -1)

Case-2: put y = 0 in equation 1^st, we get

⇒ x² – 0 + √(x² + 0) = 0

⇒ x² + |x| = 0 

We know that x² ≥ 0 and |x| ≥ 0

∴ x = 0   

Solution is (0, 0)

Number of distinct solutions are (0, 0), (0, 1) and (0, -1)

Concept:

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

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

<br>

Calculation:

Suppose 27, 8 and 12 are the pth, qth and rth terms of a G.P with first term A and common ratio is R.

T_p = 27

⇒ AR^p−1 = 27                     …. (1)

T_q = 8

⇒ AR^q−1 = 8                        …. (2)

T_r = 12

⇒ AR^r−1 = 12                      …. (3)

Dividing (1) & (3), we get

$\Rightarrow \frac{{{\rm{;}}27}}{{12}}{\rm{;}} = {\rm{;}}\frac{{{\rm{A}}{{\rm{R}}^{{\rm{p}} - 1}}}}{{{\rm{A}}{{\rm{R}}^{{\rm{r}} - 1}}}} = {{\rm{R}}^{{\rm{p}} - {\rm{r}}}}$

$\therefore \frac{{9}}{4} = {{\rm{R}}^{{\rm{p}} - {\rm{r}}}}$

$\Rightarrow {\rm{;}}{\left( {\frac{3}{2}} \right)^2} = {{\rm{R}}^{{\rm{p}} - {\rm{r}}}}$         …. (4)

Dividing (3) by (2), we get

$\Rightarrow \frac{{12}}{8}{\rm{;}} = {\rm{;}}\frac{{{\rm{A}}{{\rm{R}}^{{\rm{r}} - 1}}}}{{{\rm{A}}{{\rm{R}}^{{\rm{q}} - 1}}}} = {{\rm{R}}^{{\rm{r}} - {\rm{q}}}}$

$\therefore \frac{{3}}{2} = {{\rm{R}}^{{\rm{r}} - {\rm{q}}}}$              …. (5)

from (4) and (5),

(R^{r − q}) ² = R^{p − r}

⇒ R^{2r − 2q} = R^{p – r}

Base is same so power,

⇒ 2r − 2q = p − r

⇒ p + 2q = 3r             …. (6)

As there can be infinite natural numbers for p, q, and r to satisfy equation (6)

∴ There exist an infinite number of G.P having 27, 8 and 12 as their terms.

Given:

A = {1, 2, 3, 4} to B = {1, 3, 5}

R = {(a, b) : a < b, where a ∈ A and b ∈ B}.

Calculation:

R = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5))

∴ R⁻¹ = {(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)}

To compute RoR^-1, we pick 1 element from R^-1 and its corresponding relation from R.

For example (3, 1) ∈ R^-1 and (1, 3) ∈ R

⇒ (3, 3) belongs to RoR-1

Similarly, by doing same optration, we have 

ROR⁻¹ = {(3, 3) (3, 5) (5, 3)(5, 5)}

Concept:

• For a number to be divisible by 3 the sum of the digits must be divisible by 3.

<br>

Calculation:

We have six digits: 0, 1, 2, 3, 4, and 5.

We should determine which 5 digits number divisible by 3.

For a number to be divisible by 3 the sum of the digits must be divisible by 3.

Possible cases:

Case-1: {1, 2, 3, 4, 5}

Their sum = 15

 

Total possible ways for case-1 = 5 × 4 × 3 × 2 × 1 = 120

Case-2: {0, 1, 2, 4, 5}

Their sum = 12

Here we cannot use 0 as the first digit, otherwise number won't be any more 5 digit and become 4 digit.

∴ First digit can be filled by 4 ways (1, 2, 4, and 5)

Reaming 4 digit can be filled by 4! Ways

Total possible ways for case-2 = 4 × 4! = 96

∴ Total ways = 120 + 96 = 216

Concept:

• ⁿC_r + ⁿC_r+1 = ⁿ⁺¹C_r+1

<br>

Calculation:

We have to find the value of ⁴⁷C₄ + ⁵¹C₃ + \(\mathop \sum \limits_{{\rm{j}} = 2}^5\) ^52-jC₃

⁴⁷C₄ + ⁵¹C₃ + \(\mathop \sum \limits_{{\rm{j}} = 2}^5\) ^52-jC₃

= ⁴⁷C₄ + ⁵¹C₃ + ⁵⁰C₃ + ⁴⁹C₃ + ⁴⁸C₃ + ⁴⁷C₃

= (⁴⁷C₃ + ⁴⁷C₄) + ⁵¹C₃ + ⁵⁰C₃ + ⁴⁹C₃ + ⁴⁸C₃

= (⁴⁸C₃ + ⁴⁸C₄) + ⁵¹C₃ + ⁵⁰C₃ + ⁴⁹C₃

= (⁴⁹C₃ + ⁴⁹C₄) + ⁵¹C₃ + ⁵⁰C₃

= (⁵⁰C₃ + ⁵⁰C₄) + ⁵¹C₃

= ⁵¹C₃ + ⁵¹C₄

= ⁵²C₄

Concept:

• If a, b, c are in A.P ⇔ 2b = a + c ⇔ First term + Last term = 2 × Middle term
• If a, b, c are in G.P ⇔  b² = ac
• If a, b, c are in H.P ⇔ ${\rm{b}} = {\rm{;}}\frac{{2{\rm{ac}}}}{{{\rm{a;}} + {\rm{;c}}}}$

<br>

Calculation:

Given: a, x, y, z, b be in AP

⇒ x – a = y – x = z – y = b – z = common difference

⇒ y – x = z – y

∴ 2y = x + z

Now,

x + y + z = 15

⇒ 2y + y = 15

∴ y = 5

In AP, First term + Last term = 2 × Middle term

⇒ a + b = 2y

∴ a + b = 2 × 5 = 10                  …. (1)

Again,

Given: a, p, q, r, b are in HP

$\therefore \frac{1}{{\rm{a}}},\frac{1}{{\rm{p}}},\frac{1}{{\rm{q}}},\frac{1}{{\rm{r}}},\frac{1}{{\rm{b}}}{\rm{;are;in;AP}}$

$\Rightarrow \frac{1}{{\rm{p}}} - \frac{1}{{\rm{a}}} = \frac{1}{{\rm{q}}} - \frac{1}{{\rm{p}}} = \frac{1}{{\rm{r}}} - \frac{1}{{\rm{q}}} = \frac{1}{{\rm{b}}} - \frac{1}{{\rm{r}}} = {\rm{common;difference}}$

$\Rightarrow \frac{1}{{\rm{q}}} - \frac{1}{{\rm{p}}} = \frac{1}{{\rm{r}}} - \frac{1}{{\rm{q}}} \Rightarrow \frac{2}{{\rm{q}}} = \frac{1}{{\rm{p}}} + \frac{1}{{\rm{r}}}$

⇒ p^-1 + r^-1 = 2q^-1

Given: p^-1 + q^-1 + r^-1 = 5/3

⇒ 2q^-1+ q^-1 = 5/3

∴ q^-1 = 5/9

In AP, First term + Last term = 2 × Middle term

⇒ a^-1 + b^-1 = 2q^-1 = 10/9

$\Rightarrow \frac{{{\rm{b}} + {\rm{a}}}}{{{\rm{ab}}}} = \frac{{10}}{9}$

From equation 1^st,

$\Rightarrow \frac{{10}}{{{\rm{ab}}}} = \frac{{10}}{9}$

∴ ab = 9

Calculation:

From previous solution we get, ab = 9

⇒ b = 9/a             …. (2)

Put the value of b in equation 1^st,

⇒ a + (9/a) = 10

⇒ a² − 10a + 9 = 0

⇒ (a − 1) (a − 9) = 0

∴ a = 1, 9

Put the value ‘a’ in equation 2^nd,

⇒ b = 9, 1

Given: a, x, y, z, b be in AP

⇒ Last term = First term + (n – 1) × common difference

⇒ b = a + (5 – 1) × d

⇒ 9 = 1 + 4d

∴ d = 2

Now,

x = a + d = 3

y = a + 2d = 5

z = a + 3d = 7

So, xyz = 105

Calculation:

We have value of a and b,

a = 1, 9 and b = 9, 1 ⇒ (a, b) = (1, 9) and (9, 1)

$\frac{1}{{\rm{a}}},\frac{1}{{\rm{p}}},\frac{1}{{\rm{q}}},\frac{1}{{\rm{r}}},\frac{1}{{\rm{b}}}{\rm{;are;in;AP}}$

⇒ Last term = First term + (n – 1) × common difference

⇒ b^-1 = a^-1 + (5 – 1) × d

Taking (a, b) = (9, 1)

⇒ 1^-1 = 9^-1 + 4d

⇒ 4d = 1 – 1/9 = 8/9

∴ d = 2/9

Now,

p^-1 = a^-1 + d = (1/9) + (2/9) = 1/3

∴ p = 3

q^-1 = a^-1 + 2d and r^-1 = a^-1 + 3d = 7

After solving we get q = 9/5 and r = 9/7

So, pqr = 243/35

Concept:

1. Let us consider a1, a2, a3 …. an is an A.P.

• Common difference “d”= a2 – a1 = a3 – a2 = …. = an – an – 1
• nth term of the A.P. is given by an = a + (n – 1) d

2. Following steps to finding maxima and minima using derivatives.

• Find the derivative of the function.
• Set the derivative equal to 0 and solve. This gives the values of the maximum and minimum points.
• Now we have find second derivative.

1. f``(x) is less than 0 then the given function is said to be maxima
2. If f``(x) Is greater than 0 then the function is said to be minima

 

Calculation:

Given:

Sixth term of an A.P is 2

⇒ a + 5d = 2      (∵an = a + (n – 1) d)

⇒ a = 2 – 5d     …. (1)

Let, p = a1​a4​a5​

⇒ p = a (a + 3d) (a + 4d)

Now substitute a = 2 − 5d in the above equation, we get;

⇒ p = (2 – 5d) (2 − 5d + 3d) (2 − 5d + 4d)

⇒ p = (2 − 5d) (2 – 2d) (2 – d)

⇒ p = 2 (-5d3 + 17d2 -16d + 4)

We have to maximize this,

Let f (d) = 2 (-5d3 + 17d2 -16d + 4)

⇒ f’(d) = 2 (-15d2 + 34d - 16)

⇒ f”(d) = 2 (-30d  + 34)

For maxima and minima, f’ (d) = 0

⇒ 2 (-15d2 + 34d - 16) = 0

⇒ 15d2 - 34d + 16 = 0

⇒ 15d2 – 10d – 24d + 16 = 0

⇒ 5d (3d – 2) – 8 (3d – 2) = 0

⇒ (5d – 8) (3d – 2) = 0

∴ d = 8/5 and 2/3

Now check for second derivative test,

At d = 8/5

⇒ f”(8/5) = 2 (-30 × 8/5  + 34) = 2 (-48 + 34) = -28 < 0

Therefore, d = 8/5 ​gives maximum value.

Calculation:

From the previous solution, we see that at d = 8/5 ​ product of the first, fourth and fifth terms gives maximum value.

From equation 1st,

a = 2 - 5d

⇒ a = 2 - 5 × (8/5)

⇒ a = 2 - 8

∴ a = -6

Calculation:

\({\rm{Let;A}} = \left| {\begin{array}{*{20}{c}} {{\rm{x}} + 1}&{2{\rm{x}}}&{3{\rm{x}}}\ {2{\rm{x}} + 3}&{{\rm{x}} + 1}&{\rm{x}}\ {2 - {\rm{x}}}&{3{\rm{x}} + 4}&{5{\rm{x}} - 1} \end{array}} \right|\)

Apply R₂ → R₂ – 2R₁ and R₃ → R₃ + R₁, we get

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

Apply R₃ → R₃ – 3R₂, we get

\( = \left| {\begin{array}{*{20}{c}} {{\rm{x}} + 1}&{2{\rm{x}}}&{3{\rm{x}}}\ 1&{ - 3{\rm{x}} + 1}&{ - 5{\rm{x}}}\ 0&{14{\rm{x}} + 1}&{23{\rm{x}} - 1} \end{array}} \right|\)

Expanding along C₁,

A = (x + 1) [(-3x + 1) (23x – 1) + 5x (14x + 1)] – 1 [2x (23x – 1) – 3x (14x + 1)] + 0

= (x + 1) [-69x² + 3x + 23x -1 + 70x² + 5x] – x (46x – 2 - 42x – 3)

= (x + 1) (x² + 31x - 1) – x (4x – 5)

= (x³ + 31x² – x + x² + 31x – 1 – 4x² + 5x)

= x³ + 28x² + 35x – 1

Given A = ax³ + bx² + cx + d

Compare with above result,

∴ c = 35

Calculation:

We get,

A = ax³ + bx² + cx + d = x³ + 28x² + 35x – 1

Put x = 1, we get

⇒ a + b + c + d = 1 + 28 + 35 – 1 = 63

Concept:

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

• Common difference “d”= a₂ – a₁ = a₃ – a₂ = …. = a_n – a_{n – 1}
• n^th term of the A.P. is given by a_n = a + (n – 1) d
• n^th term from the last is given by a_n = l – (n – 1) d
• sum of the first n terms = S = n/2[2a + (n − 1) × d] = n/2(a + l)

Where, a = First term, d = Common difference, n = number of terms, a_n = n^th term

 

Calculation:

Let there be in n sides in the polygon.

Sum of all n interior angles of polygon = (n – 2) × 180°

The angles are in A. P. with the smallest angle = 120° and common difference = 5°

A P is 120, 125, 130 ….

∴ Sum of all interior angles of polygon = n/2[2 × 120 + (n – 1) × 5]

⇒ n/2[2 × 120 + (n – 1) × 5] = (n – 2) × 180

⇒ n/2 [5n + 235] = (n – 2) × 180

⇒ 5n² + 235n = 360n – 720

⇒ 5n² – 125n + 720 = 0

⇒ n² – 25n + 144 = 0

⇒ (n – 16) (n – 9) = 0

⇒ n = 16, 9

Also if n = 16 then 16th angle = 120 + 15 × 5 = 195° > 180°

∴This is not possible as interior angle cannot be greater than 180°

Hence n = 9.

Calculation:

Let there be in n sides in the polygon.

Sum of all n interior angles of polygon = (n – 2) × 180°

The angles are in A. P with the smallest angle = 120° and common difference = 5°

A P is 120, 125, 130 ….

∴ Sum of all interior angles of polygon = n/2[2 × 120 + (n – 1) × 5]

⇒ n/2[2 × 120 + (n – 1) × 5] = (n – 2) × 180

⇒ n/2 [5n + 235] = (n – 2) × 180

⇒ 5n2 + 235n = 360n – 720

⇒ 5n2 – 125n + 720 = 0

⇒ n2 – 25n + 144 = 0

⇒ (n – 16) (n – 9) = 0

⇒ n = 16, 9

Now, Largest interior angle of the polygon will be nth tem of AP

Nth term of AP = a + (n – 1) d

⇒ Tn = 120 + (n – 1) × 5

For n = 9

⇒ T9 = 120 + (9 – 1) × 5 = 120 + 40 = 160

For n = 16

⇒ T16 = 120 + (16 – 1) × 5 = 120 + 75 = 195

So, Answer is 160° only

Concept:

Scalar Multiplication of Matrices

• If A = [a_ij] _{m × n} is a matrix and k is a scalar, then kA is another matrix which is obtained by multiplying each element of A by the scalar k ⇔ [a_ij]_m×n = [k (a_ij)]_m×n
• 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).
• sin² x + cos² x = 1

 

Calculation:

Given:

\(m = \left[ {\begin{array}{{20}{c}} 1&0\ 0&1 \end{array}} \right]\) And \(n = \left[ {\begin{array}{{20}{c}} 0&1\ { - 1}&0 \end{array}} \right]\)

Now,

\({\rm{m;cos;\theta }} = \left[ {\begin{array}{{20}{c}} {{\rm{cos;\theta }}}&0\ 0&{{\rm{cos;\theta }}} \end{array}} \right]\) And \({\rm{n;sin;\theta }} = \left[ {\begin{array}{{20}{c}} 0&{{\rm{sin;\theta }}}\ { - {\rm{sin;\theta }}}&0 \end{array}} \right]\)

Now find the value of (m cos θ – n sin θ)

\(\Rightarrow {\rm{m;cos;\theta ;}}-{\rm{;n;sin;\theta ;}} = {\rm{;}}\left[ {\begin{array}{{20}{c}} {{\rm{cos;\theta }}}&0\ 0&{{\rm{cos;\theta }}} \end{array}} \right] - ;\left[ {\begin{array}{{20}{c}} 0&{{\rm{sin;\theta }}}\ { - {\rm{sin;\theta }}}&0 \end{array}} \right]\)

\(= \left[ {\begin{array}{*{20}{c}} {{\rm{cos;\theta }} - {\rm{;}}0}&{0 - {\rm{;sin;\theta }}}\ {0 - \left( { - {\rm{sin;\theta }}} \right)}&{{\rm{cos;\theta }} - 0} \end{array}} \right]\)

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

Determinant of m cos θ – n sin θ = (cos θ × cos θ) – (-sin θ × sin θ)

= cos² θ + sin² θ = 1

∴ Option 3 is correct.

Concept:

• For Even function: f(-x) = f(x)
• cos (x + y) = cos x cos y – sin x sin y
• sin (x + y) = sin x cos y + cos x sin y

<br>

Calculation:

Given:

\({\rm{f}}\left( {\rm{x}} \right) = \left[ {\begin{array}{*{20}{c}} {\cos {\rm{x}}}&{ - \sin {\rm{x}}}&0\ {\sin {\rm{x}}}&{\cos {\rm{x}}}&0\ 0&0&1 \end{array}} \right]\)

Statement 1:

f(θ) × f(ϕ) = f(θ + ϕ)

LHS = f(θ) × f(ϕ) = \(\left[ {\begin{array}{{20}{c}} {\cos {\rm{\theta }}}&{ - \sin {\rm{\theta }}}&0\ {\sin {\rm{\theta }}}&{\cos {\rm{\theta }}}&0\ 0&0&1 \end{array}} \right] \times \left[ {\begin{array}{{20}{c}} {\cos {\rm{\varphi }}}&{ - \sin {\rm{\varphi }}}&0\ {\sin {\rm{\varphi }}}&{\cos {\rm{\varphi }}}&0\ 0&0&1 \end{array}} \right]\)

\(= \left[ {\begin{array}{*{20}{c}} {\cos {\rm{\theta }}\cos {\rm{\varphi }} - \sin {\rm{\theta }}\sin {\rm{\varphi ;}}}&{ - \cos {\rm{\theta }}\sin {\rm{\varphi }} - {\rm{;}}\sin {\rm{\theta }}\cos {\rm{\varphi }}}&0\ {\sin {\rm{\theta }}\cos {\rm{\varphi }} + {\rm{;}}\cos {\rm{\theta }}\sin {\rm{\varphi }}}&{ - \sin {\rm{\theta }}\sin {\rm{\varphi }} + {\rm{;}}\cos {\rm{\theta }}\cos {\rm{\varphi }}}&0\ 0&0&1 \end{array}} \right]\)

\(= \left[ {\begin{array}{*{20}{c}} {\cos {\rm{\theta }}\cos {\rm{\varphi }} - \sin {\rm{\theta }}\sin {\rm{\varphi ;}}}&{ - \left( {\sin {\rm{\theta }}\cos {\rm{\varphi }} + {\rm{;}}\cos {\rm{\theta }}\sin {\rm{\varphi }}} \right)}&0\ {\sin {\rm{\theta }}\cos {\rm{\varphi }} + {\rm{;}}\cos {\rm{\theta }}\sin {\rm{\varphi }}}&{\cos {\rm{\theta }}\cos {\rm{\varphi }} - \sin {\rm{\theta }}\sin {\rm{\varphi }}}&0\ 0&0&1 \end{array}} \right]\) 

\( = {\rm{;}}\left[ {\begin{array}{*{20}{c}} {\cos \left( {{\rm{\theta }} + {\rm{;\varphi }}} \right)}&{ - \sin \left( {{\rm{\theta }} + {\rm{;\varphi }}} \right)}&0\ {\sin \left( {{\rm{\theta }} + {\rm{;\varphi }}} \right)}&{\cos \left( {{\rm{\theta }} + {\rm{;\varphi }}} \right)}&0\ 0&0&1 \end{array}} \right]{\rm{;}}\)

= f(θ + ϕ) = RHS

So, Statement 1 is correct.

Statement 2:

Determinant of the matrix f(θ) × f(ϕ) = Determinant of the matrix f(θ + ϕ)

\(= {\rm{;}}\left| {\begin{array}{*{20}{c}} {\cos \left( {{\rm{\theta }} + {\rm{;\varphi }}} \right)}&{ - \sin \left( {{\rm{\theta }} + {\rm{;\varphi }}} \right)}&0\ {\sin \left( {{\rm{\theta }} + {\rm{;\varphi }}} \right)}&{\cos \left( {{\rm{\theta }} + {\rm{;\varphi }}} \right)}&0\ 0&0&1 \end{array}} \right|\)

= cos² (θ + ϕ) + sin² (θ + ϕ) = 1

So, Statement 2 is also correct.

Statement 3:

\({\rm{f}}\left( {\rm{x}} \right) = \left[ {\begin{array}{*{20}{c}} {\cos {\rm{x}}}&{ - \sin {\rm{x}}}&0\ {\sin {\rm{x}}}&{\cos {\rm{x}}}&0\ 0&0&1 \end{array}} \right]\)

Determinant of f(x) = cos² x + sin² x

Replace, x → -x

⇒ f(-x) = cos² (-x) + sin² (-x) = cos² x + sin² x = f(x)

So determinant of f(x) is an even function.

So, Statement 3 is also correct.

Concept

Let the system of equations be,

a₁x + b₁y + c₁z = d₁

a₂x + b₂y + c₂z = d₂

a₃x + b₃y + c₃z = d₃

\({\rm{;}} ⇒ \left[ {\begin{array}{{20}{c}} {{{\rm{a}}_1}}&{{{\rm{b}}_1}}&{{{\rm{c}}_1}}\ {{{\rm{a}}_2}}&{{{\rm{b}}_2}}&{{{\rm{c}}_2}}\ {{{\rm{a}}_3}}&{{{\rm{b}}_3}}&{{{\rm{c}}_3}} \end{array}} \right]{\rm{;}}\left[ {\begin{array}{{20}{c}} {\rm{x}}\ {\rm{y}}\ {\rm{z}} \end{array}} \right] = {\rm{;}}\left[ {\begin{array}{*{20}{c}} {{{\rm{d}}_1}}\ {{{\rm{d}}_2}}\ {{{\rm{d}}_3}} \end{array}} \right]\)

⇒ AX = B

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

⇒ 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)

Calculation:

Given system of equation x + y + z = 8, x – y + 2z = 6 and 3x – y + 5z = k

\(⇒ \left[ {\begin{array}{{20}{c}} 1&1&1\ 1&{ - 1}&2\ 3&{ - 1}&5 \end{array}} \right]{\rm{;}}\left[ {\begin{array}{{20}{c}} {\rm{x}}\ {\rm{y}}\ {\rm{z}} \end{array}} \right] = {\rm{;}}\left[ {\begin{array}{*{20}{c}} 8\ 6\ {\rm{k}} \end{array}} \right]\)

⇒ AX = B

Determinant of A = |A| = 1 (-5 + 2) – 1 (5 – 6) + 1 (-1 + 3) = -3 + 1 + 2 = 0

So we can say that equations have either an infinite solution or no solution.

A unique solution is not possible.

 ∴ Statement 3 is wrong.

We have adj A = \(\left[ {\begin{array}{*{20}{c}} -3&-6&3\ 1&{2}&-1\ 2&{ 4}&-2 \end{array}} \right]\)

If k = 15,

B = \(\left[ {\begin{array}{*{20}{c}} 8\ 6\ {\rm{15}} \end{array}} \right]\)

Now (adj A). B will be 

\(\left[ {\begin{array}{{20}{c}} -3&-6&3\ 1&{2}&-1\ 2&{ 4}&-2 \end{array}} \right] \cdot \left[ {\begin{array}{{20}{c}} 8\ 6\ {\rm{15}} \end{array}} \right] \ne 0 \)

⇒ no solution

If K = 20,

B = \(\left[ {\begin{array}{*{20}{c}} 8\ 6\ {\rm{20}} \end{array}} \right]\)

Now (adj A). B will be 

\(\left[ {\begin{array}{{20}{c}} -3&-6&3\ 1&{2}&-1\ 2&{ 4}&-2 \end{array}} \right] \cdot \left[ {\begin{array}{{20}{c}} 8\ 6\ {\rm{20}} \end{array}} \right] = 0 \)

⇒ infinitely many solutions

Hence Option 1 is correct.

Concept:

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

Steps for obtaining Inverse of the matrix:

1. We need to calculate cofactors of all elements

⇒ \({{\rm{C}}{{\rm{ij}}}}{\rm{;or;}}{{\rm{A}}{{\rm{ij}}}} = {\left( { - 1} \right)^{{\rm{i}} + {\rm{j}}}}{\rm{;}}{{\rm{M}}_{{\rm{ij}}}}\)

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

2. Transpose of cofactor is called adjoint of the matrix

⇒ Adj (A) = C^T

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

Shortcut Method to finding adjoint:

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

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

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

Calculation:

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

\({\rm{AB}} = \left[ {\begin{array}{{20}{c}} 1&{ - 1}\ 2&3 \end{array}} \right]\left[ {\begin{array}{{20}{c}} 2&3\ { - 1}&{ - 2} \end{array}} \right] = {\rm{;}}\left[ {\begin{array}{*{20}{c}} 3&5\ 1&0 \end{array}} \right]\)

Determinant of A = |A| = 3 + 2 = 5

Determinant of B = |B| = -4 + 3 = -1

\({\rm{Adj;}}\left( {\rm{A}} \right) = \left[ {\begin{array}{{20}{c}} 3&1\ { - 2}&1 \end{array}} \right]{\rm{;;and;Adj;}}\left( {\rm{B}} \right) = \left[ {\begin{array}{{20}{c}} { - 2}&{ - 3}\ 1&2 \end{array}} \right]\)

Now, Inverse of the matrix,

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

\({{\rm{A}}^{ - 1}}{{\rm{B}}^{ - 1}} = \frac{1}{5}\left[ {\begin{array}{{20}{c}} 3&1\ { - 2}&1 \end{array}} \right]\left[ {\begin{array}{{20}{c}} 2&3\ { - 1}&{ - 2} \end{array}} \right] = \frac{1}{5}\left[ {\begin{array}{*{20}{c}} 5&7\ { - 5}&{ - 8} \end{array}} \right]{\rm{;}}\)

Statement 1:

AB (A^-1B^-1) = \(\left[ {\begin{array}{{20}{c}} 3&5\ 1&0 \end{array}} \right] \times \frac{1}{5}\left[ {\begin{array}{{20}{c}} 5&7\ { - 5}&{ - 8} \end{array}} \right] = \frac{1}{5}\left[ {\begin{array}{*{20}{c}} { - 10}&{ - 19}\ 5&7 \end{array}} \right] \ne {{\rm{I}}_2}\) 

So, statement 1 is wrong.

Statement 2:

\({\rm{AB}} = \left[ {\begin{array}{*{20}{c}} 3&5\ 1&0 \end{array}} \right]\)

Determinant of AB = |AB| = 0 – 5 = -5

\({\rm{Adj;}}\left( {{\rm{AB}}} \right) = \left[ {\begin{array}{*{20}{c}} 0&{ - 5}\ { - 1}&3 \end{array}} \right]\)

\({\left( {{\rm{AB}}} \right)^{ - 1}}{\rm{;}} = \frac{{\left[ {\begin{array}{{20}{c}} 0&{ - 5}\ { - 1}&3 \end{array}} \right]}}{{ - 5}} = \frac{1}{5}\left[ {\begin{array}{{20}{c}} 0&5\ 1&{ - 3} \end{array}} \right] \ne {{\rm{A}}^{ - 1}}{{\rm{B}}^{ - 1}}\)

So, statement 2 is also wrong.

Concept:

Logarithm properties

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

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

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

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

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

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

Calculation:

${{\rm{x}}^{{\rm{In}}\left( {\frac{{\rm{y}}}{{\rm{z}}}} \right)}} \cdot {{\rm{y}}^{{\rm{In}}{{\left( {{\rm{xz}}} \right)}^2}}} \cdot {{\rm{z}}^{{\rm{In}}\left( {\frac{{\rm{x}}}{{\rm{y}}}} \right)}} = {{\rm{y}}^{4{\rm{;In;y}}}}$

Taking log both sides, we get

$⇒ \ln \left( {{{\rm{x}}^{{\rm{In}}\left( {\frac{{\rm{y}}}{{\rm{z}}}} \right)}} \cdot {{\rm{y}}^{{\rm{In}}{{\left( {{\rm{xz}}} \right)}^2}}} \cdot {{\rm{z}}^{{\rm{In}}\left( {\frac{{\rm{x}}}{{\rm{y}}}} \right)}}} \right) = {\rm{;}}\ln \left( {{{\rm{y}}^{4{\rm{;In;y}}}}} \right)$

$⇒ \ln \left( {{{\rm{x}}^{{\rm{In}}\left( {\frac{{\rm{y}}}{{\rm{z}}}} \right)}}} \right) + \ln \left( {{{\rm{y}}^{{\rm{In}}{{\left( {{\rm{xz}}} \right)}^2}}}} \right) + {\rm{;}}\ln \left( {{{\rm{z}}^{{\rm{In}}\left( {\frac{{\rm{x}}}{{\rm{y}}}} \right)}}} \right){\rm{;}} = 4{\rm{;In;y}}\ln \left( {\rm{y}} \right)$

$\% MathType!MTEF!2!1!+- \% feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn \% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr \% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrpepu0de9LqFHe9Lq \% pepeea0xd9q8as0-LqLs-Jirpepeea0-as0Fb9pgea0lrP0xe9Fve9 \% Fve9qapdbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 \% qacqGHshI3caqGjbGaaeOBamaabmaabaWaaSaaaeaacaWG5baabaGa \% amOEaaaaaiaawIcacaGLPaaaciGGSbGaaiOBamaabmaapaqaa8qaca \% qG4baacaGLOaGaayzkaaGaey4kaSIaaeysaiaab6gadaqadaWdaeaa \% peGaaeiEaiaabQhaaiaawIcacaGLPaaapaWaaWbaaSqabeaapeGaaG \% OmaaaakiGacYgacaGGUbWaaeWaa8aabaWdbiaabMhaaiaawIcacaGL \% PaaacqGHRaWkpaGaaGjbV-qacaqGjbGaaeOBamaabmaabaWaaSaaae \% aacaWG4baabaGaamyEaaaaaiaawIcacaGLPaaaciGGSbGaaiOBamaa \% bmaapaqaa8qacaqG6baacaGLOaGaayzkaaWdaiaaysW7peGaeyypa0 \% JaaGina8aacaaMe8+dbmaabmaapaqaa8qacaqGjbGaaeOBa8aacaaM \% e8+dbiaabMhaaiaawIcacaGLPaaapaWaaWbaaSqabeaapeGaaGOmaa \% aaaaa!68E6! \Rightarrow {\rm{In}}\left( {\frac{y}{z}} \right)\ln \left( {\rm{x}} \right) + {\rm{In}}{\left( {{\rm{xz}}} \right)^2}\ln \left( {\rm{y}} \right) + ;{\rm{In}}\left( {\frac{x}{y}} \right)\ln \left( {\rm{z}} \right); = 4;{\left( {{\rm{In}};{\rm{y}}} \right)^2}$

⇒ [In (y) - In (z)] In (x) + 2 [In (x) + In (z)] + In (y) + [In (x) - In (y)] In (z) = 4 (In y)²  

⇒ In (y) In (x) - In (z) In (x) + 2 In (x) In (y) + 2 In (z) In (y) + In (x) In (z) - In (y) In (z) = 4 (In y)²

⇒ 3In (y) In (x) + In (z) In (y) = 4 (In y)²

⇒ 3ln (x) + ln (z) = 4ln (y)

⇒ ln (x) + ln (x) + ln (x) + ln (z) = 4ln (y)

∴ In y is the AM of In x, In x, In x, In z

Concept:

Convert decimal to binary

Conversion steps:

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

<br>

Calculation:

 

∴ 235₁₀ = 11101011₂

Concept:

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

Discriminant = D = b² – 4ac

For Real roots: Discriminant (D) > 0      

Calculation:

x² - (1 – 2a²) x + (1 – 2a²) = 0

For real roots,

D ≥ 0

⇒ (- (1 – 2a²))² – 4 × 1 × (1 – 2a²) ≥ 0

⇒ (1 – 2a²)² – 4 × (1 – 2a²) ≥ 0

⇒ (1 – 2a²) [(1 – 2a²) – 4] ≥ 0

⇒ (1 – 2a²) (-2a² – 3) ≥ 0

⇒ (1 – 2a²) (2a² + 3) ≤ 0

∴ (1 – 2a²) ≤ 0               (∵ 2a² + 3 > 0, Square can’t be negative)

⇒ 1 ≤ 2a²

⇒ a² ≥ 1/2

Hence Option 4 is correct.

Concept:

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

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

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

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

• The product of the roots is given by:

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

Calculation:

Let α and β be the two roots of the equation x² - (1 – 2a²) x + (1 – 2a²) = 0

⇒ α + β = (1 – 2a²)

⇒ α × β = (1 – 2a²)

Given: $\frac{1}{{{{\rm{\alpha }}^2}}} + \frac{1}{{{{\rm{\beta }}^2}}} < 1$

$\Rightarrow \frac{{{{\rm{\alpha }}^2} + {{\rm{\beta }}^2}{\rm{;}}}}{{{{\rm{\alpha }}^2}{{\rm{\beta }}^2}}} < 1$

$\Rightarrow \frac{{{{\left( {{\rm{\alpha }} + {\rm{\beta }}} \right)}^2} - 2{\rm{\alpha \beta ;}}}}{{{{\rm{\alpha }}^2}{{\rm{\beta }}^2}}} < 1$

$\Rightarrow \frac{{{{\left( {1{\rm{;}}-{\rm{;}}2{{\rm{a}}^2}} \right)}^2} - 2\left( {1{\rm{;}}-{\rm{;}}2{{\rm{a}}^2}} \right){\rm{;}}}}{{{{\left( {1{\rm{;}}-{\rm{;}}2{{\rm{a}}^2}} \right)}^2}}} < 1$

$\Rightarrow \frac{{\left( {1{\rm{;}}-{\rm{;}}2{{\rm{a}}^2}} \right) \times \left[ {\left( {1{\rm{;}}-{\rm{;}}2{{\rm{a}}^2}} \right) - 2} \right]{\rm{;}}}}{{{{\left( {1{\rm{;}}-{\rm{;}}2{{\rm{a}}^2}} \right)}^2}}} < 1$

$\Rightarrow \frac{{\left( {1{\rm{;}}-{\rm{;}}2{{\rm{a}}^2}} \right) - 2{\rm{;}}}}{{\left( {1{\rm{;}}-{\rm{;}}2{{\rm{a}}^2}} \right)}} < 1$

$\Rightarrow \frac{{-{\rm{;}}2{{\rm{a}}^2} - 1{\rm{;}}}}{{\left( {1{\rm{;}}-{\rm{;}}2{{\rm{a}}^2}} \right)}} - 1 < 0$

$\Rightarrow \frac{{-{\rm{;}}2{{\rm{a}}^2} - 1 - 1 + 2{{\rm{a}}^2}{\rm{;}}}}{{\left( {1{\rm{;}}-{\rm{;}}2{{\rm{a}}^2}} \right)}} < 0$

$\Rightarrow \frac{{-{\rm{;}}2{\rm{;}}}}{{\left( {1{\rm{;}}-{\rm{;}}2{{\rm{a}}^2}} \right)}} < 0$

$\Rightarrow \frac{{1{\rm{;}}}}{{\left( {1{\rm{;}}-{\rm{;}}2{{\rm{a}}^2}} \right)}} > 0$

∴ 1 – 2a² > 0

⇒ 1 > 2a²

⇒ a² < ½

Concept:

Cube Roots of unity are 1, ω and ω²

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

Property of cube roots of unity

• ω³ = 1
• 1 + ω + ω² = 0
• ω = 1 / ω ² and ω² = 1 / ωsum
• ω³ⁿ = 1

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Calculation:

We have to find the value of $\sqrt {\frac{{1 + {{\rm{\omega }}^2}}}{{1 + {\rm{\omega }}}}}$

As we know that 1 + ω + ω² = 0

⇒ 1 + ω² = - ω and 1 + ω = - ω²

Now,

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

Calculation:

We have to find a minimum number of students failing in all four subjects.

70% students passed in physics that means 30% students failed in Physics.

⇒ n (Fail in physics) = 0.3

80% student passed in Chemistry that means 20% student failed in chemistry.

⇒ n (Fail in Chemistry) = 0.2

75% students passed in Mathematics that means 25% student failed in maths.

⇒ n (Fail Mathematics) = 0.25

85% students passed in Biology that means 15% student failed in biology.

⇒ n (Fail in Biology) = 0.15

Therefore minimum number of students failing in all four subjects will be = 0.3 × 0.2 × 0.25 × 0.15

⇒ n (Fail in all 4 subjects) = 0.00225

Percent of students failing in all 4 subjects = 0.00225 × 100% = 0.225%

So, the minimum value of x is 0.225

∴ Option 4 is correct.

Concept

Let the system of equations be,

a₁x + b₁y + c₁z = d₁

a₂x + b₂y + c₂z = d₂

a₃x + b₃y + c₃z = d₃

\(\Rightarrow \left[ {\begin{array}{{20}{c}} {{{\rm{a}}_1}}&{{{\rm{b}}_1}}&{{{\rm{c}}_1}}\ {{{\rm{a}}_2}}&{{{\rm{b}}_2}}&{{{\rm{c}}_2}}\ {{{\rm{a}}_3}}&{{{\rm{b}}_3}}&{{{\rm{c}}_3}} \end{array}} \right]{\rm{;}}\left[ {\begin{array}{{20}{c}} {\rm{x}}\ {\rm{y}}\ {\rm{z}} \end{array}} \right] = {\rm{;}}\left[ {\begin{array}{*{20}{c}} {{{\rm{d}}_1}}\ {{{\rm{d}}_2}}\ {{{\rm{d}}_3}} \end{array}} \right]\)

⇒ AX = B

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

⇒ 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)

Calculation:

Given linear equations 2x + 3y + 5z = 9, 7x + 3y – 2z = 8 and 2x + 3y + λz = μ 

\(\Rightarrow \left[ {\begin{array}{{20}{c}} 2&3&5\ 7&3&{ - 2}\ 2&3&{\rm{\lambda }} \end{array}} \right]{\rm{;}}\left[ {\begin{array}{{20}{c}} {\rm{x}}\ {\rm{y}}\ {\rm{z}} \end{array}} \right] = {\rm{;}}\left[ {\begin{array}{*{20}{c}} 9\ 8\ {\rm{\mu }} \end{array}} \right]\)

⇒ AX = B

\(\therefore {\rm{A}} = \left[ {\begin{array}{*{20}{c}} 2&3&5\ 7&3&{ - 2}\ 2&3&{\rm{\lambda }} \end{array}} \right]\)

Minor of A = \(\left[ {\begin{array}{*{20}{c}} {3{\rm{\lambda }} + 6}&{7{\rm{\lambda }} + 4}&{15}\ {3{\rm{\lambda }} - 15}&{2{\rm{\lambda }} - 10}&0\ { - 21}&{ - 39}&{ - 15} \end{array}} \right]\)

Cofactor of A = \(\left[ {\begin{array}{*{20}{c}} {3{\rm{\lambda }} + 6}&{ - 7{\rm{\lambda }} - 4}&{15}\ { - 3{\rm{\lambda }} + 15}&{2{\rm{\lambda }} - 10}&0\ { - 21}&{ + 39}&{ - 15} \end{array}} \right]\)

Now, Adj(A) = (Cofactor of A)^T = \(\left[ {\begin{array}{*{20}{c}} {3{\rm{\lambda }} + 6}&{ - 3{\rm{\lambda }} + 15}&{ - 21}\ { - 7{\rm{\lambda }} - 4}&{2{\rm{\lambda }} - 10}&{39}\ {15}&0&{ - 15} \end{array}} \right]\)

For infinitely many solutions:

(adj A). B = 0

\(\Rightarrow \left[ {\begin{array}{{20}{c}} {3{\rm{\lambda }} + 6}&{ - 3{\rm{\lambda }} + 15}&{ - 21}\ { - 7{\rm{\lambda }} - 4}&{2{\rm{\lambda }} - 10}&{39}\ {15}&0&{ - 15} \end{array}} \right] \times \left[ {\begin{array}{{20}{c}} 9\ 8\ {\rm{\mu }} \end{array}} \right] = 0\)

\(\Rightarrow \left[ {\begin{array}{*{20}{c}} {27{\rm{\lambda }} + 54 + 120 - 24{\rm{\lambda }} - 21{\rm{\mu }}}\ { - 63{\rm{\lambda }} - 36 + 16{\rm{\lambda }} - 80 + 39{\rm{\mu }}}\ {135 + 0 - 15{\rm{\mu }}} \end{array}} \right] = 0\)

So, 27λ + 54 + 120 − 24λ − 21μ = 0 and 135 + 0 − 15μ = 0

135 + 0 − 15μ = 0

∴ μ = 9

Again, 27λ + 54 + 120 − 24λ − 21μ = 0

⇒ 3λ + 174 – 21 × 9 = 0

∴ λ = 5

Calculation:

\({\rm{A}} = \left[ {\begin{array}{*{20}{c}} 2&3&5\ 7&3&{ - 2}\ 2&3&{\rm{\lambda }} \end{array}} \right]\)

For unique solution:

Determinant of A = |A|≠ 0

⇒ 2 (3λ + 6) – 3 (7λ + 4) + 5(21 – 6) ≠ 0

⇒ 6λ + 12 – 21λ -12 + 75 ≠ 0

⇒ -15λ + 75 ≠ 0

∴ λ ≠ 5

And μ has any real value

Calculation:

We have to form odd integers between 1000 and 9999 with no digits repeated. 

For odd integers unit digit can be 1, 3, 5, 7, 9

Case-1: Let us suppose all the digits of the 4-digit odd integer are non-zero:

No. of such integers = 6 × 7 × 8 × 5 = 1680

Case-2: Let us suppose hundredth digit of the 4-digit odd integer is zero.

No. of such integers = 7 × 1 × 8 × 5 = 280

Case-3: Let us suppose units digit of the 4-digit odd integer is zero. 

No. of such integers = 7 × 8 × 1 × 5 = 280

Total No. of integers = 1680 + 280 + 280 = 2240 

Hence, Option 3 is correct.