NDA Mathematics 10 Sept 2017 Official Paper (Previous Year Paper)

CONCEPT:

Formulas and Properties of Logarithms:

1. ${\log _a}a = 1$
2. ${\log _a}\left( {x.y} \right) = ;{\log _a}x + {\log _a}y$
3. ${\log _a}\left( {\frac{x}{y}} \right) = ;{\log _a}x - {\log _a}y$
4. ${\log _a}(\frac{1}{x}) = - {\log _a}x$
5. ${\log _a}{x^p} = p{\log _a}x$

CALCULATION:

Given:

\({\rm{x}} + {\rm{;lo}}{{\rm{g}}{10}}\left( {1 + {2^{\rm{x}}}} \right) = {\rm{xlo}}{{\rm{g}}{10}}5 + {\rm{lo}}{{\rm{g}}_{10}}6\)

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$\begin{array}{l} \Rightarrow {\rm{x}} - {\rm{xlog}}5 = {\rm{log}}6 - \log \left( {1 + {2^{\rm{x}}}} \right)\ \Rightarrow {\rm{x}}\left( {1 - {\rm{log}}5} \right) = {\rm{log}}6 - \log \left( {1 + {2^{\rm{x}}}} \right)\ \Rightarrow {\rm{x}}\left( {{\rm{log}}10 - {\rm{log}}5} \right) = {\rm{log}}6 - \log \left( {1 + {2^{\rm{x}}}} \right){\rm{;}}\left( {{\rm{;log}}_{\rm{a}}^{\rm{a}} = 1} \right) \end{array}$

\(\Rightarrow {\rm{xlog}}\left( {\frac{{10}}{5}} \right) = \log \left( {\frac{6}{{1 + {2^{\rm{x}}}}}} \right){\rm{;}}\left( {{\rm{lo}}{{\rm{g}}{\rm{a}}}\left( {\frac{{\rm{x}}}{{\rm{y}}}} \right) = {\rm{;lo}}{{\rm{g}}{\rm{a}}}{\rm{x}} - {\rm{;lo}}{{\rm{g}}_{\rm{a}}}{\rm{y}}} \right)\)

$\Rightarrow {\rm{xlog}}\left( 2 \right) = \log \left( {\frac{6}{{1 + {2^{\rm{x}}}}}} \right)$

\(\Rightarrow \log \left( {{2^{\rm{x}}}} \right) = \log \left( {\frac{6}{{1 + {2^{\rm{x}}}}}} \right){\rm{;}}\left( {{\rm{lo}}{{\rm{g}}{\rm{a}}}{{\rm{x}}^{\rm{p}}} = {\rm{plo}}{{\rm{g}}{\rm{a}}}{\rm{x}}} \right)\)

Taking “ANTILOG” on both sides

$\Rightarrow {2^{\rm{x}}} = \frac{6}{{1 + {2^{\rm{x}}}}}$

Now simplify this equation

⇒ 2^x (1 + 2^x) = 6

Let’s take 2^x = u

⇒ u(1 + u) = 6

⇒ u² + u = 6

⇒ u² + u - 6 = 0

⇒ u² + 3u - 2u - 6 = 0

⇒ u(u + 3) - 2(u + 3) = 0

⇒ (u + 3) (u - 2) = 0

⇒ u = -3, 2

Now replace “u” with “2^x”

⇒ 2^x = 2, 2^x = -3

Observe both equations.

Clearly, 2^x = -3 is impossible

So, 2^x = 2

⇒ x = 1

CONCEPT:

BINARY TO DECIMAL CONVERSION:

For Example, to convert 10011011 into decimal the following process must be followed.

DECIMAL TO BINARY CONVERSION:

To convert 57 from decimal to binary, the following two methods can be used.

CALCULATION:

The easiest way to solve this problem is by converting the given binary numbers into decimal numbers.

(101110)₂ = (1 × 2⁵) + (0 × 2⁴) + (1 × 2³) + (1 × 2²) + (1 × 2¹) + (1 × 2⁰)

= 32 + 0 + 8 + 4 + 2 + 0

= (46)₁₀

(110)₂ = (1 × 2²) + (1 × 2¹) + (0 × 2⁰)

= 4 + 2 + 0

= (6)₁₀

Now divide the decimal numbers,

46/6 ⇒ Quotient =7, Remainder = 4

Now again convert 7, 4 into binary numbers

⇒ (7)₁₀ = (1 × 2²) + (1 × 2¹) + (1 × 2⁰)

= (111)₂

(4)₁₀ = (1 × 2²) + (0 × 2¹) + (1 × 2⁰)

= (100)₂

Hence, Remainder = (100)₂ and Quotient = (111)₂

CONCEPT:

If A, B are two matrices, then matrix AB exist when “No. of columns of Matrix A = No. of rows of Matrix B “

Similarly, matrix BA exist when “No. of columns of Matrix B = No. of rows of Matrix A “

CALCULATION:

For Matrix AB to exist, No. of columns of Matrix A = No. of rows of Matrix B)

⇒ x + 5 = y

For Matrix BA to exist, No. of columns of Matrix B = No. of rows of Matrix A

⇒ 11 - y = x

Solving both the equations, we get: x = 3, y = 8.

CONCEPT:

Arithmetic Progression:

The standard form of arithmetic progression (AP) is given by: a, a + d, a + 2d, a + 3d……. so on.

The nth term of an AP series is T_n = a + (n-1) d,

Sum of n terms of an A.P is S_n = n/2 [2a + (n − 1) d] 

where ‘a’= first term and ‘d’= common difference.

CALCULATION:

Given, ${{\rm{S}}_{\rm{n}}} = {\rm{nP}} + \frac{{{\rm{n}}\left( {{\rm{n}} - 1} \right){\rm{Q}}}}{2}$

Comparing the given equation with the formula of S_n of an A.P

$\Rightarrow {\rm{nP}} + \frac{{{\rm{n}}\left( {{\rm{n}} - 1} \right){\rm{Q}}}}{2} = \frac{{\rm{n}}}{2}\left[ {2{\rm{a}} + \left( {{\rm{n}} - 1} \right){\rm{d}}} \right]$

RHS equation can be written as,

$\Rightarrow \frac{{\rm{n}}}{2}\left[ {2{\rm{P}} + \left( {{\rm{n}} - 1} \right){\rm{Q}}\left] { = \frac{{\rm{n}}}{2}} \right[2{\rm{a}} + \left( {{\rm{n}} - 1} \right){\rm{d}}} \right]$

By comparing LHS and RHS, we get: a = P and d = Q

So, common difference d = Q.

CONCEPT:

Quadratic Equation:

The standard quadratic equation is given by: f(x) = ax² + bx + c = 0

The roots of a quadratic equation can be calculated by making f(x) = 0

Thus, the roots of a quadratic function are given by: $x = ;\frac{{ - b; \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

CALCULATION:

Given: (q - r)x² + (r - p)x + (p - q) = 0

By Comparing the given equation with the standard quadratic equation, ax² + bx + c = 0 , we get: a = (q − r), b = (r − p) and c = (p − q)

So, Discriminant b²- 4ac = (r - p)² - {4 × (p - q) × (q - r)}

⇒ b² - 4ac = r² +p² - 2rp - 4[pq - rp - q² + rq]

⇒ b² - 4ac = r² +p² - 2rp - 4pq + 4rp - 4q² - 4rq

⇒ b² - 4ac = r² +p² - 4pq + 2rp - 4q² - 4rq

⇒ b² - 4ac = (r + p - 2q)²

Roots of the given Quadratic Equation is given by: ${\rm{x}} = \frac{{ - \left( {{\rm{r}} - {\rm{p}}} \right){\rm{;}} \pm \sqrt {{{\left( {{\rm{r}} + {\rm{p}} - 2{\rm{q;}}} \right)}^2}} }}{{2\left( {{\rm{q}} - {\rm{r}}} \right)}}$

$\Rightarrow {\rm{x}} = \frac{{ - \left( {{\rm{r}} - {\rm{p}}} \right) + \sqrt {{{\left( {{\rm{r}} + {\rm{p}} - 2{\rm{q;}}} \right)}^2}} }}{{2\left( {{\rm{q}} - {\rm{r}}} \right)}}{\rm{;}},{\rm{;x}} = {\rm{;}}\frac{{ - \left( {{\rm{r}} - {\rm{p}}} \right) - \sqrt {{{\left( {{\rm{r}} + {\rm{p}} - 2{\rm{q;}}} \right)}^2}} }}{{2\left( {{\rm{q}} - {\rm{r}}} \right)}}$

$\Rightarrow {\rm{x}} = \frac{{ - \left( {{\rm{r}} - {\rm{p}}} \right) + \left( {{\rm{r}} + {\rm{p}} - 2{\rm{q}}} \right)}}{{2\left( {{\rm{q}} - {\rm{r}}} \right)}}{\rm{;}},{\rm{;x}} = {\rm{;}}\frac{{ - \left( {{\rm{r}} - {\rm{p}}} \right) - \left( {{\rm{r}} + {\rm{p}} - 2{\rm{q}}} \right)}}{{2\left( {{\rm{q}} - {\rm{r}}} \right)}}$

$\Rightarrow {\rm{x}} = \frac{{2{\rm{p}} - 2{\rm{q}}}}{{2\left( {{\rm{q}} - {\rm{r}}} \right)}}{\rm{;}},{\rm{;x}} = {\rm{;}}\frac{{2{\rm{q}} - 2{\rm{r}}}}{{2\left( {{\rm{q}} - {\rm{r}}} \right)}}$

Hence, ${\rm{x}} = \frac{{{\rm{p}} - {\rm{q}}}}{{{\rm{q}} - {\rm{r}}}},{\rm{;}}1$

CONCEPT:

If there are two sets P and Q, then

1. n(P U Q) represents the number of elements present in both the sets P or Q.
2. n(P ⋂ Q) represents the number of elements which are common in both the sets P & Q.
3. n(P U Q) = n(P) + (n(Q) - n (P ⋂ Q)

 

If U is a Universal set, A is a subset, then

1. U - A = A’ (A’ is the complement of A).
2. U - A’ = A

 

CALCULATION:

Given that, E is a universal set and A is a set.

⇒ Here, E - A = A’ called as (A complement)

⇒ Similarly, E - A’ = A

⇒ Hence, E-(E-(E-(E-(E-A)))) = E-(E-(E-(E-A’))) (∵ E - A = A’)

= E-(E-(E-A)) (∵ E - A’ = A)

= E-(E-A’)

= E-A

= A’

= (B∪ C)’ = B’∩ C’ (∵ we were given that A = B∪ C)

CONCEPT:

The intersection of two or more sets is the set of elements that are common to each of the sets.

i.e If A and B are any two sets, then A ∩ B = {x : x ∈ A and B}

CALCULATION:

Given,

A = {x : x is a multiple of 2},

B = {x : x is a multiple of 5},

C = {x : x is a multiple of 10}

From the given information, it is clear that multiples of 2 include multiples of 10. So, C is subset of A.

Similarly, multiples of 5 include multiples of 10. So, C is subset of B.

Also, C = A ∩ B

But, B = A ∩ C, B ∩ c = B

Hence, (A ∩ B ∩ C) = A ∩ B = C

CONCEPT:

Quadratic Equation Roots:

The Standard Quadratic Equation is given by: f(x) = ax² + bx + c

The roots of a Quadratic Equation f(x) = 0, Satisfies the Given Quadratic Equation.

If α, β are roots of a Quadratic Equation f(x) = 0, then f(α) = 0 and f(β) = 0.

Sum of the roots ${\rm{\alpha }} + {\rm{\beta }} = \frac{{ - {\rm{b}}}}{{\rm{a}}}$, Product of the roots $\left( {{\rm{\alpha }} \times {\rm{\beta }}} \right) = \frac{{\rm{c}}}{{\rm{a}}}$

Matrix Multiplication:

If \({\rm{A}} = \left[ {\begin{array}{{20}{c}} {\rm{a}}&{\rm{b}}\ {\rm{c}}&{\rm{d}} \end{array}} \right],{\rm{;B}} = \left[ {\begin{array}{{20}{c}} {\rm{e}}&{\rm{f}}\ {\rm{g}}&{\rm{h}} \end{array}} \right],\) then AB is given by:

 \({\rm{AB}} = \left[ {\begin{array}{*{20}{c}} {ae + bg}&{af + bh}\ {ce + dg}&{cf + dh} \end{array}} \right]\)

CALCULATION:

Given that α , β are the roots of the Quadratic Equation 1+ x + x² = 0.

Hence, 1 + α + α² = 0 and 1+ β + β² = 0

⇒ Sum of the roots ${\rm{\alpha }} + {\rm{\beta }} = \frac{{ - {\rm{b}}}}{{\rm{a}}} = {\rm{;}} - 1$

⇒ Product of the roots $\left( {{\rm{\alpha }} \times {\rm{\beta }}} \right) = \frac{{\rm{c}}}{{\rm{a}}} = 1{\rm{;}}$

\(\left[ {\begin{array}{{20}{c}} 1&{\rm{\beta }}\ {\rm{\alpha }}&{\rm{\alpha }} \end{array}} \right]\left[ {\begin{array}{{20}{c}} {\rm{\alpha }}&{\rm{\beta }}\ 1&{\rm{\beta }} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {{\rm{\alpha }} + {\rm{\beta }}}&{{\rm{\beta }} + {{\rm{\beta }}^2}}\ {{\rm{\alpha }} + {{\rm{\alpha }}^2}}&{{\rm{\alpha \beta }} + {\rm{\alpha \beta }}} \end{array}} \right]\)

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

CONCEPT:

PROPERTIES OF MODULUS FUNCTION:

1. |ab| = |a||b|
2. |a + b| ≤ |a| + |b|
3. |a - b| ≥ ||a| - |b||

 

EXPLANATION:

Given that,

Statement 1: |ab| = |a||b|

⇒ From the properties of Modulus Function, it is clear that Statement 1 is true.

Statement 2: |a + b|

⇒ From the properties of Modulus Function, it is clear that Statement 2 is true.

Statement 3: |a - b|

⇒ From the properties of Modulus Function, it is clear that Statement 3 is true.

Hence, All the given statements are true.

CONCEPT:

PERMUTATION:

Number of Permutations of ‘n’ things taken ‘r’ at a time:

${\rm{p}}\left( {{\rm{n}},{\rm{;r}}} \right) = \frac{{{\rm{n}}!}}{{\left( {{\rm{n}} - {\rm{r}}} \right)!}}$

Number of Permutations of ‘n’ objects where there are n₁ repeated items, n₂ repeated items, ….. n_k repeated items taken ‘r’ at a time:

\({\rm{p}}\left( {{\rm{n}},{\rm{;r}}} \right) = \frac{{{\rm{n}}!}}{{{{\rm{n}}_1}!{{\rm{n}}_2}!{{\rm{n}}3}! \ldots .{{\rm{n}}{\rm{k}}}!}}\)

CALCULATION:

For the given word ‘PERMUTATION’

The number of letters in the given word is 11.

The letter ‘t’ is repeated 2 times and there is no repetition of any other letter.

So, different permutations that can be made out of the letters of the word ‘PERMUTATION’ is: $\frac{{11!}}{{2!}}$

CONCEPT:

Scalar multiplication:

If A = (a_ij) is a matrix and k is any scalar then the matrix kA is obtained by multiplying every element of matrix A by a scalar k.

CALCULATION:

Given: \(A = \left[ {\begin{array}{*{20}{c}} {4i - 6}&{10i}\ {14i}&{6 + 4i} \end{array}} \right]\) and $k = \frac{1}{{2i}}$

\(\Rightarrow kA = \frac{1}{{2i}} \times \left[ {\begin{array}{*{20}{c}} {4i - 6}&{10i}\ {14i}&{6 + 4i} \end{array}} \right]\)

\(\Rightarrow kA = \left[ {\begin{array}{*{20}{c}} {\frac{1}{{2i}} \times \left( {4i - 6} \right)}&{\frac{1}{{2i}} \times \left( {10i} \right)}\ {\frac{1}{{2i}} \times \left( {14i} \right)}&{\frac{1}{{2i}} \times \left( {6 + 4i} \right)} \end{array}} \right]\)

\(\Rightarrow kA = \left[ {\begin{array}{*{20}{c}} {2 - \frac{3}{i}}&5\ 7&{\frac{3}{i} + 2} \end{array}} \right]\)

\(\Rightarrow kA = \left[ {\begin{array}{*{20}{c}} {2 - \frac{3}{i}\left( {\frac{i}{i}} \right)}&5\ 7&{\frac{3}{i}\left( {\frac{i}{i}} \right) + 2} \end{array}} \right]\)     …(by multiplying and dividing by “i”)

\(\Rightarrow kA = \left[ {\begin{array}{*{20}{c}} {2 + 3i}&5\ 7&{2 - 3i} \end{array}} \right]\)

Hence, \(kA = \left[ {\begin{array}{*{20}{c}} {2 + 3i}&5\ 7&{2 - 3i} \end{array}} \right]\)

CONCEPT:

Modulus function:

Modulus of a real number x, denoted by |x| is the non-negative value of x without regard to its sign. Namely, |x| = x if x is positive, and |x| = -x if x is negative.

The modulus function is defined as: \({\rm{f}}\left( {\rm{x}} \right) = \left| {\rm{x}} \right| = \left{ {\begin{array}{*{20}{c}} { - x,;;x < 0}\ {x,;;x \ge 0} \end{array}} \right.\)

CALCULATION:

Given equation is |x – 3|² + |x – 3| - 2 = 0

Let us assume x – 3 = t     …equation (1)

So, |t|² + |t| - 2 = 0

We know that*,* \(\left| {\rm{t}} \right| = \left{ {\begin{array}{*{20}{c}} { - t,;;t < 0}\ {t,;;t \ge 0} \end{array}} \right.\)

Case 1: (t ≥ 0)

t² + t - 2 = 0

⇒ t² + 2t - t - 2 = 0

⇒ t(t + 2) - 1(t + 2) = 0

⇒ (t - 1) (t + 2) = 0

⇒ t = 1, -2

As we have taken in the above case that (t > 0) so, t = 1 is the possible solution.

Now substitute t = 1 in equation (1)

⇒ x – 3 = 1

⇒ x = 4.

CASE 2: (t < 0)

(-t)² + (-t) - 2 = 0

⇒ t² - t - 2 = 0

⇒ t² – 2t + t - 2 = 0

⇒ t (t +1) - 2(t + 1) = 0

⇒ (t – 2) (t +1) = 0

⇒ t = -1, 2

As we have taken in the above case that (t < 0) so, t = -1 is the possible solution.

Now substitute t = -1 in equation (1)

⇒ x – 3 = -1

⇒ x = 2.

Hence, the sum of the real roots of the given equation is 6.

CONCEPT:

Nature of roots of a quadratic equation:

The roots of a quadratic function are given by, $x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

We call the term b² – 4ac as discriminant (Δ).

Case 1: If Δ = 0, then the roots are real and equal.

Case 2: If Δ > 0, then the roots are real and unequal.

Case 3: If Δ < 0, then the roots are imaginary.

CALCULATION:

Given that roots of the equation x² – 4x – log₃ P = 0 are real.

Comparing the given equation with the standard form of quadratic equation: f(x) = ax² + bx + c = 0

⇒ a = 1

⇒ b = -4

⇒ c = -log­₃ P

For roots to be real, Δ ≥ 0

⇒ Δ = b² – 4ac = [(- 4)]² – (4× 1× (- log₃ P)

⇒ Δ = 16 + (4 × log₃ P)

⇒ 16 + (4 × log₃ P) ≥ 0

⇒ 4log₃ P ≥ -16

⇒log₃ P ≥ -4

Applying “ANTILOG” on both sides

⇒ P ≥ 3^-4

⇒ P ≥ 1/81

The minimum value of P for roots to be real is 1/81.

CONCEPT:

Adjoint Matrix:

If B_{n× n} is a cofactor matrix of matrix A_{n× n} then the adjoint matrix of A_{n× n} is denoted by adj(A) and is defined as B^T. So, adj(A) = B^T.

Property of adjoint matrix:

If A is a square matrix of order ‘n’ then adj(A^T) = (adj(A))^T

CALCULATION:

As we know that  adj (A^T) = (adj (A))^T

Hence, adj (A^T) - (adj (A))^T = adj(A^T) – adj(A^T) = O (Null Matrix whose order is same as the order of A)

CONCEPT:

Infinite Geometric series:

Standard infinite geometric series is given by: a₁ + a₁r + a₁r² + a₁r³ + …… +a₁r^n-1 + ….

Sum of infinite geometric series is given by ${{\rm{s}}_{\rm{n}}} = \frac{{\rm{a}}}{{1 - {\rm{r}}}}$ as long as -1 < r < 1

CALCULATION:

Given that: ${6^{\frac{1}{2}}} \times {6^{\frac{1}{4}}} \times {6^{\frac{1}{8}}} \times {6^{\frac{1}{{16}}}} \times \cdots$ upto infinite terms

⇒ ${6^{\left( {\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots } \right)}}$     ...(∵ a^m × aⁿ = a^m+n)

The series: $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots$is in form of infinite geometric series.

Comparing it with standard infinite G.P series, we get a = 1/2 and r = 1/2

Sum of infinite terms of a G.P, ${s_n} = \frac{a}{{1 - r}}$

$\Rightarrow {{\rm{s}}_{\rm{n}}} = \left( {\frac{{\frac{1}{2}}}{{1 - \frac{1}{2}}}} \right)$

$\Rightarrow {{\rm{s}}_{\rm{n}}} = \left( {\frac{{\frac{1}{2}}}{{\frac{1}{2}}}} \right) = 1$

$\Rightarrow {6^{\left( {\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots } \right)}} = {6^1} = 6$

Hence, the value of product of ${6^{\frac{1}{2}}} \times {6^{\frac{1}{4}}} \times {6^{\frac{1}{8}}} \times {6^{\frac{1}{{16}}}} \times \cdots$ upto infinite terms = 6

CONCEPT:

Determinant of a square matrix of order two:

If \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}\ {{a_{21}}}&{{a_{22}}} \end{array}} \right]\) is a square matrix of order 2 then determinant is given by:

\(\det \left( A \right) = \left| A \right| = {\rm{\Delta }} = \left| {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}\ {{a_{21}}}&{{a_{22}}} \end{array}} \right| = {a_{11}}{a_{22}} - ;{a_{12}}{a_{21}}\)

CALCULATION:

Given that: \(\left| {\begin{array}{*{20}{c}} {{{\cos }^2}\frac{\theta }{2}}&{{{\sin }^2}\frac{\theta }{2}}\ {{{\sin }^2}\frac{\theta }{2}}&{{{\cos }^2}\frac{\theta }{2}} \end{array}} \right|\)

Determinant of the given matrix is given by: ${\rm{\Delta }} = co{s^4}\left( {\frac{\theta }{2}} \right) - si{n^4}\left( {\frac{\theta }{2}} \right)$

$\Rightarrow {\rm{\Delta }} = \left[ {co{s^2}\left( {\frac{\theta }{2}} \right) - si{n^2}\left( {\frac{\theta }{2}} \right)} \right]\left[ {co{s^2}\left( {\frac{\theta }{2}} \right) + si{n^2}\left( {\frac{\theta }{2}} \right)} \right]$     …{∵ A⁴ – B⁴ = (A² – B²)(A² + B²)}

$\Rightarrow {\rm{\Delta }} = \left[ {co{s^2}\left( {\frac{\theta }{2}} \right) - ;si{n^2}\left( {\frac{\theta }{2}} \right)} \right]\left[ 1 \right]$     …{∵ cos²A – sin²A = cos(2A)}

$\Rightarrow {\rm{\Delta }} = [\cos \left( {\frac{{\theta \times 2}}{2}} \right)]\left[ 1 \right]$

⇒ Δ = cos θ

CONCEPT:

Binomial Theorem:

According to the theorem, it is possible to expand any non-negative power of x + y into a sum of the form:

(x + y)ⁿ = ⁿC₀ xⁿy⁰ + ⁿC₁x^n-1y­_­¹+ ⁿC₂x^n-2y­_­²+ ⁿC₃x^n-3y­_­³+ …………. + ⁿC_nx⁰y­_­ⁿ where n≥ 0 is an integer.

CALCULATION:

(x + a)¹⁰⁰ = ¹⁰⁰C₀ x¹⁰⁰ + ¹⁰⁰C₁x⁹⁹a­_­¹ + ¹⁰⁰C₂x⁹⁸a­_­² + ¹⁰⁰C₃x⁹⁸y­_­² +     ………… + ¹⁰⁰C₁₀₀a­_­¹⁰⁰     …equation (1)

(x – a)¹⁰⁰ = ¹⁰⁰C₀ x¹⁰⁰ - ¹⁰⁰C₁x⁹⁹a­_­¹+ ¹⁰⁰C₂x⁹⁸a­_­² - ¹⁰⁰C₃x⁹⁸y­_­²+     ………… + ¹⁰⁰C₁₀₀a­_­¹⁰⁰     …equation (2)

The above two equations are written with the help of “BINOMIAL THEOREM”

By adding equation (1) and equation (2), we get

(x + a)¹⁰⁰ = 2× [¹⁰⁰C₀ x¹⁰⁰ + ¹⁰⁰C₂x⁹⁸a­_­² + ………… + ¹⁰⁰C₁₀₀a­_­¹⁰⁰]     …equation (3)

Total no. of terms in equation (1) = 101 (odd terms = 50, even terms = 51)

By adding equation (1) and equation (2), odd terms will get cancelled

Hence, the total no. of terms in the expansion (x + a)¹⁰⁰ + (x - a)¹⁰⁰ = 51

Formula used:

(1 + x)n  = [nC0 + nC1 x + nC2 x2 + … +nCn xn]

• C₀ + C₁ + C₂ + … + C_n = 2ⁿ
• C₀ + C₂ + C₄ + … =  2n-1
• C₁ + C₃ + C₅ + … = 2^n-1

 

Calculation:

(1 + x)⁵⁰  = [⁵⁰C0 + ⁵⁰C1 x + ⁵⁰C2 x2 + … +⁵⁰Cn x⁵⁰]    ----(1)

Here, n = 50

Using the above formula, the sum of odd terms of the coefficient is

S = (50C1 + 50C3­ + 50C5 + ……. + 50C49)

⇒ S = 2^50-1 = 2⁴⁹

∴ Sum of odd terms of the coefficient = 249

CONCEPT:

Inverse of a matrix: The Inverse of a n × n matrix is given by:

${A^{ - 1}} = \frac{{adj\left( A \right)}}{{\left| A \right|}}$ where adj(A) is called adjoint matrix.

Adjoint Matrix: If B_{n× n} is a cofactor matrix of matrix A_{n× n} then the adjoint matrix of A_{n× n} is denoted by adj(A) and is defined as B^T. So, adj(A) = B^T.

CALCULATION:

Given: \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} a&0&0\ 0&b&0\ 0&0&c \end{array}} \right]\)

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

⇒ |A| = [a (bc - 0) - 0(0) - 0(0)]

⇒ |A| = abc

\(\Rightarrow {\rm{adj}}\left( {\rm{A}} \right) = {\left[ {\begin{array}{*{20}{c}} {bc}&0&0\ 0&{ac}&0\ 0&0&{ab} \end{array}} \right]^T}\)

\(\Rightarrow {\rm{adj}}\left( {\rm{A}} \right) = \left[ {\begin{array}{*{20}{c}} {bc}&0&0\ 0&{ac}&0\ 0&0&{ab} \end{array}} \right]\)

\(\Rightarrow {A^{ - 1}} = \frac{{\left[ {\begin{array}{*{20}{c}} {bc}&0&0\ 0&{ac}&0\ 0&0&{ab} \end{array}} \right]}}{{abc}}\)

\(\Rightarrow {A^{ - 1}} = \left[ {\begin{array}{*{20}{c}} {{a^{ - 1}}}&0&0\ 0&{{b^{ - 1}}}&0\ 0&0&{{c^{ - 1}}} \end{array}} \right]\)

CONCEPT:

Arithmetic progression:

The Standard form of an arithmetic progression (AP) is given by: a, a+ d, a + 2d, a + 3d ……. so on.

⇒ The nth term of an AP series is T_n = a+ (n - 1) d,

⇒ Sum of n terms of an A.P is S_n = n/2 [2a + (n − 1) d]

‘a’= first term and

‘d’= common difference.

CALCULATION:

Given that: a₁ = a₂ = a₃ = ……. = a₁₀ = 150 and a₁₀, a₁₁, a₁₂, … are in AP.

So, the person counts 1500 notes in 10 minutes. The remaining notes which he need to count are

⇒ 4500 – 1500 = 3000

It is given that a₁₁, a₁₂, a₁₃ are in AP with a common difference of -2.

Hence, a₁₀ = 150, a₁₁ = 148, a₁₂ = 146, a₁₃ = 144 … and so on.

So, the sum of all the notes must be 3000.

Hence, s_n = 3000, a = 148 (as he already counted 1500 notes in 10 minutes, so we need to start from

a₁₁ = 148, d = -2

$\Rightarrow {{\rm{s}}_{\rm{n}}} = \frac{{\rm{n}}}{2}[\left( {2{\rm{a}} + \left( {{\rm{n}} - 1} \right){\rm{d}}} \right]$

$\Rightarrow 3000 = \frac{{\rm{n}}}{2}\left[ {\left( {2 \times 148} \right) + \left( {{\rm{n}} - 1} \right)\left( { - 2} \right)} \right]$

$\Rightarrow 3000 = \frac{{\rm{n}}}{2}\left[ {296 + 2 - 2{\rm{n}}} \right]$

⇒ 3000 = n (149 – n)

⇒ 3000 = 149n – n²

⇒ n² – 149n + 3000 = 0

⇒ n² – 125n – 24n + 3000 = 0

⇒ n (n – 125) – 24(n – 125) = 0

⇒ (n – 24) (n – 125) = 0

⇒ n = 24, 125

So, the time taken by the person to count 3000 notes is 24 minutes.

Now, the count will take place until a₃₅ term because the count of 3000 notes started from a₁₁ but not a₁.

Hence, the total time required to count 4500 notes = 10 + 24 = 34 minutes

CONCEPT:

• i² = -1
• i⁴ⁿ = 1

 

CALCULATION:

Given:

By rationalising with (1+i)

$\Rightarrow {\left( {\frac{{\left( {1 + i} \right)\left( {1 + i} \right)}}{{\left( {1 - i} \right)\left( {1 + i} \right)}}} \right)^n} = {\left( {\frac{{1 + 2i + {i^2}}}{{1 - {i^2}}}} \right)^n}$ {∵ (A+ B) (A – B) = A² – B²)}

$\Rightarrow {\left( {\frac{{1 + 2i - 1}}{{1 + 1}}} \right)^n} = {\left( {\frac{{2i}}{2}} \right)^n}$ {∵ i² = -1}

⇒ iⁿ = 1

The smallest value of ‘n’ for which iⁿ = 1 is 4.

CONCEPT:

Equivalence Relation:

Any relation R on a non-empty set A which is reflexive, symmetric and transitive is called an equivalence relation on A.

Reflexive Relation:

For all a ∈ A, if (a, a) ∈ R, then R is called a reflexive relation on A.

Symmetric Relation:

For all a, b ∈ A, if (a, b) ∈ R⇒ (b, a) ∈ R then R is called a symmetric relation on A.

Transitive Relation:

For any a, b, c ∈ A if (a, b) ∈ R, (b, c) ∈ R ⇒ (a, c) ∈ R, then R is called a transitive relation on A.

 

CALCULATION:

Given that: (a, b) R (c, d) ⇔ a + d = b + c

Let (a, b) ∈ N × N

⇒ (a + b) = (b + a)

⇒ (a, b) R (a, b)

Hence, R is reflexive.

Let (a, b) R (c, d), where (a, b), (c, d) ∈ N × N

⇒ a + d = b + c

⇒ b + c = a + d

⇒ c + b = a + d

⇒ (c, d) R (a, b)

Hence, R is symmetric.

Given that: (a, b) R (c, d)

⇒ a + d = b + c

⇒ a – b = c – d … equation (1)

Let us suppose (c, d) R (e, f)

⇒ c + f = d + e

⇒ c – d = e – f …equation (2)

From equations (1) and (2)

⇒ a – b = e – f

⇒ a + f = b + e

⇒ (a, b) R (e, f)

Hence, R is transitive.

Relation R satisfies reflexive, symmetric and transitive.

Hence, R is an equivalence relation.

CONCEPT:

Infinite Geometric series:

Standard infinite geometric series is given by: a , ar , ar² , ar³ ,……,ar^n-1 , ….up to infinite terms

Sum of infinite terms of Geometric Progression is given by: ${{\rm{s}}_{\rm{n}}} = \frac{{\rm{a}}}{{1 - {\rm{r}}}}$

<br>

CALCULATION:

Given that: y = x + x² + x³ + …… up to infinite terms

Comparing it with standard infinite G.P series, we get a = x and r = x

Sum of infinite terms of geometric progression is given by: ${{\rm{s}}_{\rm{n}}} = \frac{{\rm{a}}}{{1 - {\rm{r}}}}$

$\Rightarrow ;{{\rm{s}}_{\rm{n}}} = \left( {\frac{{\rm{x}}}{{1 - {\rm{x}}}}} \right)$

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

⇒ y (1 – x) = x

⇒ y – xy = x

⇒ y = x + xy

⇒ y = x(1+y)

$\Rightarrow x = \left( {\frac{y}{{1 + y}}} \right)$

CONCEPT:

If α and β are the roots of ax² + bx + c = 0, then the equation can be written as: a (x² – (α +β) x + α β) = 0 i.e., (x - α) (x - β) = 0 

CALCULATION:

Given that: α and β are the roots of the equation 3x² + 2x + 1 = 0

$\Rightarrow {\rm{\alpha }} + {\rm{\beta }} = \frac{{ - 2}}{3}$ and ${\rm{\alpha }} \times {\rm{\beta }} = \frac{1}{3}$

We know that, the standard quadratic equation is given by: x² - (sum of roots) x + (product of roots) = 0

Sum of roots is given by:

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

$\Rightarrow \left( {{\rm{\alpha }} + {{\rm{\beta }}^{ - 1}}} \right) + \left( {{\rm{\beta }} + {{\rm{\alpha }}^{ - 1}}} \right) = \frac{{ - 2}}{3} + \frac{{\frac{{ - 2}}{3}}}{{\frac{1}{3}}} = \frac{{ - 2}}{3} + \left( { - 2} \right) = \frac{{ - 8}}{3}$

Product of the roots is given by:

$\Rightarrow \left( {{\rm{\alpha }} + {{\rm{\beta }}^{ - 1}}} \right)\left( {{\rm{\beta }} + {{\rm{\alpha }}^{ - 1}}} \right) = \left( {{\rm{\alpha \beta }} + 1 + 1 + \frac{1}{{{\rm{\alpha \beta }}}}} \right) = \frac{1}{3} + 1 + 1 + 3 = \frac{{16}}{3}$

The quadratic equation whose roots are α + β^-1 and β + α^-1 is given by:

$\Rightarrow {x^2} - \left( {\frac{{ - 8}}{3}} \right)x + \frac{{16}}{3} = 0$

$\Rightarrow {x^2} + \frac{8}{3}x + \frac{{16}}{3} = 0$

⇒ 3x²+ 8x + 16 = 0

CONCEPT:

Formulas and Properties of Logarithms:

1. \({\log {\rm{a}}}{{\rm{m}}^{\rm{n}}} = {\rm{;nlo}}{{\rm{g}}{\rm{a}}}{\rm{m}}\)
2. $\frac{1}{{{{\log }_a}m}} = ;{\log _m}a$

 

CALCULATION:

Given that: $\frac{1}{{{{\log }_3}e}} + \frac{1}{{{{\log }_3}{e^2}}} + \frac{1}{{{{\log }_3}{e^4}}} + \ldots$ up to infinite terms

$\Rightarrow \frac{1}{{{{\log }_3}{\rm{e}}}} + \frac{1}{{{{\log }_3}{{\rm{e}}^2}}} + \frac{1}{{{{\log }_3}{{\rm{e}}^4}}} + \ldots = \frac{1}{{{{\log }_3}{\rm{e}}}} + \frac{1}{{2{{\log }_3}{\rm{e}}}} + \frac{1}{{4{{\log }_3}{\rm{e}}}} + ......$

$\Rightarrow \frac{1}{{{{\log }_3}{\rm{e}}}} + \frac{1}{{{{\log }_3}{{\rm{e}}^2}}} + \frac{1}{{{{\log }_3}{{\rm{e}}^4}}} + \ldots = \frac{1}{{{{\log }_3}{\rm{e}}}}\left[ {1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots } \right]$

\(\left{\because {{{\log }{\rm{a}}}{{\rm{m}}^{\rm{n}}} = {\rm{;nlo}}{{\rm{g}}{\rm{a}}}{\rm{m}}} \right}\)

The series: $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + {\rm{;}} \cdots {\rm{;}}$is in the form of infinite geometric series.

Comparing it with standard infinite G.P series, we get a = 1 and r = 1 / 2

Sum of infinite terms of Geometric Progression is given by: ${{\rm{s}}_{\rm{n}}} = \frac{{\rm{a}}}{{1 - {\rm{r}}}}$

$\Rightarrow {{\rm{s}}_{\rm{n}}} = {\rm{;}}\frac{1}{{1 - \frac{1}{2}}} = 2$

As we know that, \(\frac{1}{{{{\log }a}{\rm{m}}}} = {\rm{;log}}{\rm{m}}{\rm{a}}\), by using this property we get

\(\Rightarrow \frac{1}{{{\rm{log}}3{\rm{e}}}}\left[ {1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots } \right] = \frac{2}{{{\rm{log}}3{\rm{e}}}} = 2{\rm{log}}{\rm{e}}3 = ;{\rm{log}}{\rm{e}}9\)

CONCEPT:

Permutations:

The number of permutations of n distinct things taken r at a time is denoted by ⁿP_r = n! /(n-r)!

CALCULATION:

Let us consider the below diagram as a long table which contain two sides named as “A” side, “B” side.

Given that: 4 particular men wish to sit on one particular side and 2 particular men on the other side.

So, let us suppose that 4 particular men sat on “A” side and the other 2 sat on “B” side.

The number of ways in which 4 people can sit in 8 chairs on “A” side are ⁸P₄ ways and the number of ways in which 2 people can sit in 8 chairs on “B” side are ⁸P₂ ways.

Now, we are left with 10 people and there are 10 chairs remaining.

Hence, the total number of ways of arranging 10 people in 10 chairs are 10! Ways.

Hence, the required answer is ⁸P₄ × ⁸P_2­ × 10!  = 210 × 8! × 8! Ways

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, the 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: 

kx + y + z = 1, x + ky + z = k and x + y + kz = k²

\( \Rightarrow {\rm{A}} = {\rm{}}\left[ {\begin{array}{{20}{c}} {\rm{k}}&1&1\ 1&{\rm{k}}&1\ 1&1&{\rm{k}} \end{array}} \right],{\rm{B}} = {\rm{}}\left[ {\begin{array}{{20}{c}} {\rm{x}}\ {\rm{y}}\ {\rm{z}} \end{array}} \right]{\rm{and;C}} = \left[ {\begin{array}{*{20}{c}} 1\ {\rm{k}}\ {{{\rm{k}}^2}} \end{array}} \right]\)

⇒ For the given equations to have no solution, |A| = 0

\(\Rightarrow \left| {\begin{array}{*{20}{c}} {\rm{k}}&1&1\ 1&{\rm{k}}&1\ 1&1&{\rm{k}} \end{array}} \right| = 0\)

⇒ k(k² – 1) - 1(k – 1) + 1(1 – k) = 0

⇒ k³ – k – k + 1 + 1 – k = 0

⇒ k³ -3k +2 = 0

⇒ (k – 1) (k – 1) (k + 2) = 0

⇒ k = 1, -2

If we put k = 1 in the above given equations, then all the equations will become the same.

Hence, the given equations have no solution if k = - 2.

CONCEPT:

Geometric Progression:

The standard form of geometric progression is given by: a, ar, ar², ar³, …., arⁿ

The sum of n terms of a Geometric progression is given by: ${{\rm{s}}_{\rm{n}}} = \frac{{{\rm{a}}\left( {{{\rm{r}}^{\rm{n}}} - 1} \right)}}{{{\rm{r}} - 1}}$ if r≠ 1

CALCULATION:

Given that: $1.3 + {2.3^2} + \ldots + {\rm{n}}{.3^{\rm{n}}} = \frac{{\left( {\left( {2{\rm{n}} - 1} \right){3^{\rm{a}}} + {\rm{b}}} \right)}}{4}{\rm{;}}$

Let L.H.S = $1.3 + {2.3^2} + \ldots + {\rm{n}}{.3^{\rm{n}}}$ = S

Let us put x = 3

⇒ L.H.S = s = 1.x + 2.x² + ……+ n.xⁿ … equation (1)

Multiply equation (1) with ‘x’

⇒ xs = 1.x² + 2.x³ + …… +n.xⁿ⁺¹….equation(2)

Equation (1) – equation (2) ⇒ s – xs = (1.x + 2.x² + ……+ n.xⁿ) – (1.x² + 2.x³ + …… +n.xⁿ⁺¹)

⇒ s (1 – x) = x + x² + x³ + ……+xⁿ – n.xⁿ⁺¹

$\Rightarrow {\rm{s}}\left( {1 - {\rm{x}}} \right) = \frac{{{\rm{x}}\left( {1 - {{\rm{x}}^{\rm{n}}}} \right)}}{{1 - {\rm{x}}}} - {\rm{n}}.{{\rm{x}}^{{\rm{n}} + 1}}$   {∵ ${{\rm{s}}_{\rm{n}}} = \frac{{{\rm{a}}\left( {{{\rm{r}}^{\rm{n}}} - 1} \right)}}{{{\rm{r}} - 1}}$  }

$\Rightarrow {\rm{s}} = \frac{1}{{\left( {{\rm{x}} - 1} \right)}}\left[ {\frac{{ - {\rm{x}}\left( {{{\rm{x}}^{\rm{n}}} - 1} \right) + {\rm{n}}.{{\rm{x}}^{{\rm{n}} + 1}}\left( {{\rm{x}} - 1} \right)}}{{{\rm{x}} - 1}}} \right]$

Put x = 3

$\Rightarrow {\rm{s}} = \frac{1}{2}\left( {\frac{{ - 3\left( {{3^{\rm{n}}} - 1} \right) + {\rm{n}}{{.3}^{{\rm{n}} + 1}}\left( 2 \right)}}{2}} \right) = {\rm{;}}\frac{{{3^{{\rm{n}} + 1}}\left( {2{\rm{n}} - 1} \right) + 3}}{4}$

Comparing L.H.S with R.H.S we get,

a = n+1, b = 3.

CONCEPT:

Trigonometric Identity:

$\tan \left( {{\rm{\alpha }} \pm {\rm{\beta }}} \right) = {\rm{;}}\frac{{{\rm{tan\alpha }} \pm {\rm{tan\beta }}}}{{1 \mp {\rm{tan\alpha tan\beta }}}}$

CALCULATION:

Given that: $\tan \left( {\frac{P}{2}} \right)$ and $\tan \left( {\frac{Q}{2}} \right)$ are the roots of the equation ax² + bx + c = 0,$\angle R = \frac{\pi }{2}$

$\Rightarrow \tan \left( {\frac{{\rm{P}}}{2}} \right) + \tan \left( {\frac{{\rm{Q}}}{2}} \right) = \frac{{ - {\rm{b}}}}{{\rm{a}}}$

$\Rightarrow \tan \left( {\frac{{\rm{P}}}{2}} \right) \times \tan \left( {\frac{{\rm{Q}}}{2}} \right) = \frac{{\rm{c}}}{{\rm{a}}}$

We know that, Sum of interior angles of a triangle = π

⇒ ∠ P + ∠ Q + ∠ R = π

⇒ ∠ P + ∠ Q = π /2

We know that, $\tan \left( {{\rm{\alpha }} \pm {\rm{\beta }}} \right) = {\rm{;}}\frac{{{\rm{tan\alpha }} \pm {\rm{tan\beta }}}}{{1 \mp {\rm{tan\alpha tan\beta }}}}$

Now replace α = P/2 and β = Q/2

$\Rightarrow \tan \left( {\frac{{\rm{P}}}{2} + \frac{{\rm{Q}}}{2}} \right) = \frac{{\tan \left( {\frac{{\rm{P}}}{2}} \right) + \tan \left( {\frac{{\rm{Q}}}{2}} \right)}}{{1 - {\rm{;}}\tan \left( {\frac{{\rm{P}}}{2}} \right)\tan \left( {\frac{{\rm{Q}}}{2}} \right)}}$

\(\Rightarrow \tan \left( {\frac{{{\rm{P}} + {\rm{Q}}}}{2}} \right) = ;\frac{{\frac{{ - b}}{a}}}{{1 - \frac{c}{a}}};\left{ \because {\tan \left( {\frac{{\rm{P}}}{2}} \right) + \tan \left( {\frac{{\rm{Q}}}{2}} \right) = \frac{{ - {\rm{b}}}}{{\rm{a}}};and{\rm{;tan}}\left( {\frac{{\rm{P}}}{2}} \right) \times \tan \left( {\frac{{\rm{Q}}}{2}} \right) = \frac{{\rm{c}}}{{\rm{a}}}} \right}\)

$\Rightarrow \tan \left( {\frac{\pi }{4}} \right) = ;\frac{{ - b}}{{a - c}}$  {∵ ∠ P + ∠ Q = π /2}

\(\Rightarrow 1 = ;\frac{{ - b}}{{a - c}}\left{\because {\tan \left( {\frac{\pi }{4}} \right) = 1} \right}\)

⇒ a – c = (-b)

⇒ a + b = c

CONCEPT:

Properties of Modulus Function:

1. |ab| = |a||b|
2. |a + b| ≤ |a| + |b|
3. |a - b| ≤ |a| + |b|
4. |a - b| ≥ ||a| - |b||

Quadratic Equation:

The standard quadratic equation is given by: f(x) = ax² + bx + c = 0

The roots of a quadratic equation can be calculated by making f(x) = 0

Thus, the roots of a quadratic function are given by: ${\rm{x}} = \frac{{ - {\rm{b;}} \pm {\rm{;}}\sqrt {{{\rm{b}}^2} - 4{\rm{ac}}} }}{{2{\rm{a}}}}{\rm{;}}$

CALCULATION:

Given that $\left| {z - \frac{4}{z}} \right| = 2$

$\left| Z \right| = \left| {z - \frac{4}{z} + \frac{4}{z}} \right|$

Let us consider $z - \frac{4}{z}$ as ‘a’ and

 as ‘b’

We know that, |a + b| ≤ |a| + |b|

$\Rightarrow \left| {z - \frac{4}{z} + \frac{4}{z}} \right| \le \left| {z; - \frac{4}{z}} \right| + \left| {\frac{4}{z}} \right|$

$\Rightarrow \left| z \right| \le 2 + \left| {\frac{4}{z}} \right|$   {∵ $\left| {z - \frac{4}{z}} \right| = 2$ }

$\Rightarrow \left| z \right| - ;2 - \left| {\frac{4}{z}} \right| \le 0$

$\Rightarrow \frac{{{{\left| z \right|}^2} - 2\left| z \right| - 4}}{{\left| z \right|}} \le 0$

$\Rightarrow {\left| {\rm{z}} \right|^2} - 2\left| {\rm{z}} \right| - 4 \le 0$ (∵ |z| > 0)

Finding Roots of |z|² - 2|z| - 4 = 0

⇒ |z| = $\frac{{ - \left( { - 2} \right); \pm ;\sqrt {{{\left( { - 2} \right)}^{2;}} - ;4; \times ;1; \times ; - 4} }}{{2 \times 1}} = \frac{{2; \pm ;\sqrt {20} }}{2} = 1 \pm \sqrt 5$

The roots of the equation $\frac{{{{\left| z \right|}^2} - 2\left| z \right| - 4}}{{\left| z \right|}} \le 0$ are given by:|Z| = 1± √5

Hence, the maximum value of |Z| is given by: 1 + √5

CONCEPT:

Trigonometric Identities:

• ${\rm{sin\theta }} = \frac{{{\rm{opposite;side;}}}}{{{\rm{hypotenuse;}}}}$
• ${\rm{cos\theta }} = \frac{{{\rm{adjacent;side;}}}}{{{\rm{hypotenuse;}}}}$
• ${\rm{tan\theta }} = \frac{{{\rm{opposite;side;}}}}{{{\rm{;adjacent;side;}}}}$
• $\tan \left( {{\rm{\alpha }} \pm {\rm{\beta }}} \right) = \frac{{{\rm{tan\alpha }} \pm {\rm{tan\beta }}}}{{1 \mp {\rm{tan\alpha tan\beta }}}}$

 

CALCULATION:

AB = DE = 25 m

AE = PE = x m

As we know that, tan θ = opposite side / adjacent side

In Δ AEC, $\tan \left( {15^\circ } \right) = \frac{{h - 25}}{x}$

As we know that,

$\Rightarrow \tan \left( {15^\circ } \right) = \tan \left( {45^\circ - 30^\circ } \right) = ;\frac{{1 - \frac{1}{{\sqrt 3 }}}}{{1 + \frac{1}{{\sqrt 3 }}}} = \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}$

$\Rightarrow \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}} = {\rm{;}}\frac{{{\rm{h}} - 25}}{{\rm{x}}}$        …equation (1)

In Δ APE, tan (45°) = AP/PE

$\Rightarrow \frac{{25 + {\rm{h;}}}}{{\rm{x}}} = 1$

$\Rightarrow {\rm{x}} = 25 + {\rm{h;}}$       …equation (2)

Substituting equation (2) in equation (1)

$\Rightarrow \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}} = \frac{{{\rm{h}} - 25}}{{{\rm{h}} + 25}}$

⇒ (√3 – 1) (h + 25) = (√3 +1) (h – 25)

⇒ 50√3 = 2h

⇒ h = 25√3 m

CONCEPT:

Trigonometric Identities:

• Sin (90° - θ) = cos θ
• Cos (90° - θ) = sin θ
• tan (90° - θ) = cot θ
• ${\rm{tan\theta }} = {\rm{;}}\frac{{{\rm{sin\theta }}}}{{{\rm{cos\theta }}}}$
• $\cot {\rm{\theta }} = {\rm{;}}\frac{{{\rm{cos\theta }}}}{{{\rm{sin\theta }}}}$
• sin² θ + cos² θ = 1
• 2sinθcosθ = sin2θ
• ${\rm{sin}}18^\circ = {\rm{;}}\frac{{\sqrt 5 - 1}}{4}$
• ${\rm{sin}}54^\circ = \frac{{\sqrt 5 + 1}}{4}$

 

CALCULATION:

Given that, tan 9° - tan 27° - tan 63° + tan 81°

$\Rightarrow \tan 9^\circ - \tan 27^\circ - \tan \left( {90 - 27} \right)^\circ + \tan \left( {{\rm{;}}90 - 9} \right)^\circ$

As we know that,

$\begin{array}{l} \Rightarrow \tan 9^\circ - \tan 27^\circ - \cot 27^\circ + \cot 9^\circ \ \Rightarrow \tan 9^\circ + \cot 9^\circ - (\tan 27^\circ + \cot 27^\circ ) \end{array}$

As we know that, ${\rm{tan\theta }} = {\rm{;}}\frac{{{\rm{sin\theta }}}}{{{\rm{cos\theta }}}}{\rm{;and}}\cot {\rm{\theta }} = {\rm{;}}\frac{{{\rm{cos\theta }}}}{{{\rm{sin\theta }}}}$

$\Rightarrow \frac{{\sin 9;^\circ }}{{\cos 9;^\circ }} + \frac{{\cos 9^\circ }}{{\sin 9^\circ }} - \left( {\frac{{\sin 27^\circ }}{{\cos 27^\circ }} + \frac{{\cos 27^\circ }}{{\sin 27^\circ }}} \right)$

$\Rightarrow \frac{{{\rm{si}}{{\rm{n}}^2}9^\circ + {\rm{co}}{{\rm{s}}^2}9^\circ }}{{\cos 9{\rm{;}}^\circ \sin 9^\circ }} - \left( {\frac{{{\rm{si}}{{\rm{n}}^2}27^\circ + {\rm{co}}{{\rm{s}}^2}27^\circ }}{{\cos 27^\circ \sin 27^\circ }}} \right)$

As we know that, sin² θ + cos² θ = 1

$\Rightarrow \frac{1}{{\cos 9{\rm{;}}^\circ \sin 9^\circ }} - \left( {\frac{1}{{\cos 27^\circ \sin 27^\circ }}} \right)$

$\Rightarrow \frac{2}{{2{\rm{;cos}}9^\circ \sin 9^\circ }} - \left( {\frac{2}{{2{\rm{;cos}}27^\circ \sin 27^\circ }}} \right)$

As we know that, 2sinθcosθ = sin2θ

$\Rightarrow \frac{2}{{\sin 18^\circ }} - \left( {\frac{2}{{\sin 54^\circ }}} \right)$

As we know that, ${\rm{sin}}18^\circ = {\rm{;}}\frac{{\sqrt 5 - 1}}{4}{\rm{;and;sin}}54^\circ = \frac{{\sqrt 5 + 1}}{4}$

$\Rightarrow \frac{2}{{\frac{{\sqrt 5 - 1}}{4}}} - ;\frac{2}{{\frac{{\sqrt 5 + 1}}{4}}}$

$\Rightarrow \frac{8}{{\sqrt 5 - 1}} - \left( {\frac{8}{{\sqrt 5 + 1}}} \right)$

$\Rightarrow \frac{{8\left( {\sqrt 5 + 1} \right) - 8\left( {\sqrt 5 - 1} \right)}}{{\left( {\sqrt 5 - 1} \right)\left( {\sqrt 5 + 1} \right)}}$

As we know that, (A + B) (A – B) = A² – B²

⇒ 16/4 = 4

CONCEPT:

Trigonometric Identities:

• ${\rm{cosec\theta }} = \frac{1}{{{\rm{sin\theta }}}}$
• ${\rm{sec\theta }} = \frac{1}{{{\rm{cos\theta }}}}$
• ${\rm{cos;}}30^\circ = {\rm{sin;}}60^\circ = \frac{{\sqrt 3 }}{2}$
• ${\rm{cos;}}60^\circ = {\rm{sin;}}30^\circ = \frac{1}{2}$
• (cos A× cos B) – (sin A × sin B) = cos(A + B)
• 2 × sin A × cos A = sin (2A)
• sin (90 - θ) = cos θ

 

CALCULATION:

Given that: √3 cosec 20° – sec 20°

As we know that, cosec θ = 1/ sin θ and sec θ = 1/ cos θ

$\Rightarrow \frac{{\sqrt 3 }}{{\sin 20^\circ }} - \frac{1}{{\cos 20^\circ }}$

\(\Rightarrow 2\left{ {\frac{{\frac{{\sqrt 3 }}{2}}}{{\sin 20^\circ }} - \frac{{\frac{1}{2}}}{{\cos 20^\circ }}} \right}\)

As we know that, cos 30° = √3 / 2, sin 30° = ½

\(\Rightarrow 2\left{ {\frac{{\cos 30^\circ }}{{\sin 20^\circ }} - \frac{{\sin 30^\circ ;}}{{\cos 20^\circ }}} \right}\)

\(\Rightarrow 2\left{ {\frac{{\cos 30^\circ \cos 20^\circ - \sin 30^\circ \sin 20^\circ }}{{\sin 20^\circ \cos 20^\circ }}} \right}\)

As we know that, (cos A × cos B) – (sin A × sin B) = cos(A + B)

\(\Rightarrow 2\left{ {\frac{{\cos \left( {50^\circ } \right)}}{{\sin 20^\circ \cos 20^\circ }}} \right}\)

\(\Rightarrow 2 \times 2;\left{ {\frac{{\cos \left( {50^\circ } \right)}}{{2 \times \sin 20^\circ \cos 20^\circ }}} \right}\)

As we know that, 2 × sin A × cos A = sin (2A)

\(\Rightarrow 4\left{ {\frac{{\cos \left( {50^\circ } \right)}}{{\sin 40^\circ }}} \right}\)

\(\Rightarrow 4\left{ {\frac{{\cos \left( {50^\circ } \right)}}{{{\rm{sin}}\left( {90^\circ - 50^\circ } \right)}}} \right}\)

As we know that, sin (90 - θ) = cos θ

\(\Rightarrow 4\left{ {\frac{{\cos \left( {50^\circ } \right)}}{{\cos 50^\circ }}} \right}\)

⇒ 4