NDA Mathematics 23 April 2017 Official Paper (Previous Year Paper)

Concept:

Arithmetic progression (AP) is a mathematical sequence in which the difference between two consecutive terms is always a constant.

Common difference (d) in AP:

• Let’s suppose, \({{\rm{a}}{\rm{1}}},{{\rm{a}}{{\rm{2;}}}},{{\rm{a}}{{\rm{3;}}}}, \ldots .,{{\rm{a}}{\rm{n}}}\) is an AP, the **common difference ‘ d ’**can be obtained as:
  

d = \({{\rm{a}}{\rm{2}}} - {{\rm{a}}{{\rm{1;}}}} = {{\rm{a}}{\rm{3}}} - {{\rm{a}}{{\rm{2;}}}} = ; \ldots \ldots = {{\rm{a}}{\rm{n}}} - {{\rm{a}}{{\rm{n - 1}}}}\)

 

To find nth term of an AP:

• T_n = a + (n − 1) × d
• T_n = S_n – S_n-1

 

To find sum of ‘n’ terms of an AP:

• ${{\rm{S}}_{\rm{n}}} = {\rm{}}\frac{{\rm{n}}}{2}\left[ {2{\rm{a;}} + {\rm{;}}\left( {{\rm{n;}} - {\rm{;}}1} \right){\rm{;}} \times {\rm{;d}}} \right]$ also,
• ${{\rm{S}}_{\rm{n}}} = \frac{{\rm{n}}}{2}\left[ {{\rm{first;term}} + {\rm{last;term}}} \right]$

 

Where,

a = First term

d = Common difference

n = number of terms

T_n = nth term

S_n = sum of ‘n’ terms

Calculation:

Given: Sum of n terms of an AP = n² – 2n.

To find: The fifth term = ?

Sum of ‘n’ terms of an AP = n² – 2n

∴ Sum of 1^st 5 terms

$\Rightarrow {{\rm{S}}_{\rm{5}}}{\rm{; = ;}}{{\rm{5}}^{\rm{2}}};{\rm{ - ;2}} \cdot {\rm{(5)}}$

= 25 – 10 = 15

Similarly,

Now, Sum of first 4 terms

${{\rm{S}}_{\rm{4}}}{\rm{; = ;}}{{\rm{4}}^{\rm{2}}};{\rm{ - ;2}} \cdot {\rm{(4)}}$

= 16 – 8 = 8

∴"The 5^th term of an AP"

⇒ T₅ = S₅ – S₄     ...\({\rm{(Using;}}{{\rm{T}}{\rm{n}}}; = ;{{\bf{S}}{\bf{n}}} - {{\bf{S}}_{{\bf{n}} - 1}})\)

= 15 – 8

= 7

So, option 2 is correct.

Concept:

Series in general is the sum of all terms in the sequence, where sequence is just an arrangement of items/numbers.

• For example, if \({{\rm{a}}{\rm{1}}},{{\rm{a}}{{\rm{2;}}}},{{\rm{a}}_{{\rm{3;}}}}, \ldots\) is a sequence
• Then series is simply, \({{\rm{S}}{\rm{N}}} = ;{{\rm{a}}{\rm{1}}} + {{\rm{a}}{{\rm{2;}}}} + {{\rm{a}}{{\rm{3;}}}} + \ldots\)

 

Geometric progression (GP) is a mathematical sequence in which the common ratio (r) is always constant.

• Let’s suppose, \({{\rm{a}}{\rm{1}}},{{\rm{a}}{{\rm{2;}}}},{{\rm{a}}{{\rm{3;}}}}, \ldots .,{{\rm{a}}{\rm{n}}}\) is GP, then the common ratio (r) is obtained by dividing any term by the preceding term,
  

i.e., r = \(\frac{{{{\rm{a}}{{\rm{2;}}}}}}{{{{\rm{a}}{{\rm{1}}}}}} = \frac{{{{\rm{a}}{{\rm{3;}}}}}}{{{{\rm{a}}{\rm{2}}}}} = \ldots = \frac{{{{\rm{a}}{{\rm{n - 1;}}}}}}{{{{\rm{a}}{{\rm{n;}}}}}}\)

• To find the nth term of a geometric sequence we use: ⇔ ${{\bf{a}}_{\bf{n}}} = {{\bf{a}}_1}{{\bf{r}}^{{\bf{n}} - 1}}$
• Sum of n terms in GP: ⇔ ${{\bf{S}}_{\bf{n}}} = \frac{{{{\bf{a}}_1}\left( {1 - {{\bf{r}}^{\bf{n}}}} \right)}}{{1 - {\bf{r}}}},;{\bf{r}} \ne 1$

Where,

r = common ratio, n = number of terms, a₁ = First term

a_n = The nth term, S_n = sum of GP with n terms

Calculation:

The sum of the first ‘n’ terms of the series

${{\rm{S}}_{\rm{N}}} = \frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{{15}}{{16}} + \ldots + {\rm{nth;term}}$

$\Rightarrow \left( {1 - \frac{1}{2}} \right) + \left( {1 - \frac{1}{4}} \right) + \left( {1 - \frac{1}{8}} \right) + \left( {1 - \frac{1}{{16}}} \right) + \ldots + {\rm{;nth;term}}$

$\Rightarrow \left( {1 - \frac{1}{2}} \right) + \left( {1 - \frac{1}{{{2^2}}}} \right) + \left( {1 - \frac{1}{{{2^3}}}} \right) + \left( {1 - \frac{1}{{{2^4}}}} \right) + \ldots + \left( {1 - \frac{1}{{{2^{\rm{n}}}}}} \right)$

 

$\Rightarrow \left( {1 + 1 + 1 + ; \ldots + {\rm{n}}} \right) + \left[ {\left( { - \frac{1}{2}} \right) + \left( { - \frac{1}{{{2^2}}}} \right) + \left( { - \frac{1}{{{2^3}}}} \right) + \left( { - \frac{1}{{{2^4}}}} \right) + \ldots + \left( { - \frac{1}{{{2^{\rm{n}}}}}} \right)} \right]$

$\Rightarrow \left( {1 + 1 + 1 + ; \ldots + {\rm{n}}} \right) - \left[ {\left( {\frac{1}{2}} \right) + \left( {\frac{1}{{{2^2}}}} \right) + \left( {\frac{1}{{{2^3}}}} \right) + \left( {\frac{1}{{{2^4}}}} \right) + \ldots + \left( {\frac{1}{{{2^{\rm{n}}}}}} \right)} \right]$

Now, $\left( {1 + 1 + 1 + ; \ldots + {\rm{n}}} \right)$ = n        ...(1)

And $\left[ {\left( {\frac{1}{2}} \right) + \left( {\frac{1}{{{2^2}}}} \right) + \left( {\frac{1}{{{2^3}}}} \right) + \left( {\frac{1}{{{2^4}}}} \right) + \ldots + \left( {\frac{1}{{{2^{\rm{n}}}}}} \right)} \right]$ is clearly sum of GP

As the common ratio (r) = $\frac{{\left( {\frac{1}{{{2^2}}}} \right)}}{{\left( {\frac{1}{2}} \right)}}$ = $\frac{{\left( {\frac{1}{{{2^3}}}} \right)}}{{\left( {\frac{1}{{{2^2}}}} \right)}}$ = …. = $\frac{{\left( {\frac{1}{{{2^{\rm{n}}}}}} \right)}}{{\left( {\frac{1}{{{2^{{\rm{n}} - 1}}}}} \right)}}$ = $\frac{1}{2}$ which is constant.

We have ${{\rm{a}}_1}$ = 1/2, r = 1/2

Sum of n terms of GP: ${{\bf{S}}_{\bf{n}}} = \frac{{{{\bf{a}}_1}\left( {1 - {{\bf{r}}^{\bf{n}}}} \right)}}{{1 - {\bf{r}}}},;{\bf{r}} \ne 1$

$= \frac{{\left( {\frac{1}{2}} \right)\left( {1 - {{\left( {\frac{1}{2}} \right)}^{\rm{n}}}} \right)}}{{1 - \left( {\frac{1}{2}} \right)}}$

$= \frac{{\left( {\frac{1}{2}} \right)\left( {1 - {{\left( {\frac{1}{2}} \right)}^{\rm{n}}}} \right)}}{{\left( {\frac{1}{2}} \right)}}$

$= 1 - {\left( {\frac{1}{2}} \right)^{\rm{n}}}$

= 1 – 2^-n     ...2)

Now from (1) and (2),

S_n = n – (1 – 2^-n)

= 2^-n + n - 1

Hence, option 3 is correct.

Concept:

Venn diagrams are the simplest way to express the relationships between various sets.

The following example can help you understand the concepts relating to Venn diagrams:

• A ∪ B : It’s called “A union B”

Venn diagram:

• A ∩ B : It’s called “A intersection B”

We said “A intersection B” if there is a common region occupied by sets.

Venn diagram:

• A – B : It’s called “Difference of two sets”

Venn diagram:

• ${\rm{A}} \subseteq {\rm{B}}$ : It’s called “A is subset of B”

Set A is called subset of another set B, if all elements of the set A are elements of the set B.

In other words, the set A is contained inside the set B.

Venn diagram:

Calculation:

We know the use and importance of Venn diagram, let’s check which relationships are correct.

1. (A - B) ∪ B = A

Here if we take union of (A – B) and B we will get union of A and B

It means, this statement is incorrect.

2. (A - B) ∪ A = A

From the above Venn diagram, we can easily say that this statement is correct.

3. (A - B) ∩ B = ϕ 

There is no common region between (A – B) and B so the result will be null (ϕ)

This statement is correct.

4. A ⊆ B ⇒ A ∪ B = B

Here A is subset of B that means set A is contained inside set B,

If we take union of both A and B, the obvious result will be B

This statement is also correct.

From above analysis we can say statement (2), (3), and (4) are correct.

Hence option 2 is correct.

Concept:

• In square matrix number of columns equal to the number of rows.
• A matrix is singular matrix if its det = 0
• If the determinant of a matrix is not equal to zero, then the matrix is called a non-singular matrix. i.e., det ≠ 0.
• det(AB) = det(A) × det(B)
• det(A) × det(B) = det(B) × det(A)
• ${\rm{det}}\left( {{{\rm{A}}^{ - 1}}} \right) = {\rm{;}}\frac{1}{{{\rm{detA;}}}} \Rightarrow ;{\rm{det}}\left( {{{\rm{A}}^{ - 1}}} \right) \times {\rm{;det}}\left( {\rm{A}} \right) = 1$

 

Calculation:

Given: B is a non-singular matrix and A is a square matrix

To find: det(B-1 AB) = ?

Now,

${\rm{det}}\left( {{{\rm{B}}^{{\rm{ - 1}}}}{\rm{AB}}} \right) = {\rm{det}}\left( {{{\rm{B}}^{ - 1}}} \right) \times {\rm{det}}\left( {\rm{A}} \right) \times {\rm{det}}\left( {\rm{B}} \right)$     ...(∵ det(AB) = det(A) × det(B))

${\rm{ = det}}\left( {{{\rm{B}}^{ - 1}}} \right) \times {\rm{det}}\left( {\rm{B}} \right) \times {\rm{det}}\left( {\rm{A}} \right)$     ...(∵ det(A) × det(B) = det(B) × det(A))

= 1 × det (A)     ...$\left( \because \text{det}\left( {{\text{A}}^{-1}} \right)\times \text{ }!!~!!\text{ det}\left( \text{A} \right)=1 \right)$

= det (A)

Hence, option 2 is correct.

Concept:

Suppose there are ‘n’ observations {\({{\text{x}}{\text{1}}},{{\text{x}}{\text{2 }!!!!\text{ }}},{{\text{x}}{\text{3 }!!~!!\text{ }}},\ldots ,{{\text{x}}{\text{n }!!!!\text{ }}}\)}

• Mean \(\left( {\bar x} \right) = \frac{{{\rm{;}}({{\rm{x}}1} + {{\rm{x}}{2{\rm{;}}}} + {{\rm{x}}{3{\rm{;}}}} + \ldots + {{\rm{x}}{{\rm{n;}}}})}}{{\rm{n}}} = ;\frac{{\mathop \sum \nolimits_{{\rm{i = 1}}}^{\rm{n}} {{\rm{x}}_{\rm{i}}}}}{{\rm{n}}}\)
• Variance \(({\sigma ^2}) = {\rm{;}}\frac{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{n}} {{\left( {{{\rm{x}}_{\rm{i}}} - {\rm{\bar x}}} \right)}^2}}}{{\rm{n}}}\)

 

Calculation:

Given:

Variance = 5,

Number of observations (n) = 20

To find: New variance (If each observation is multiplied by 3) = ?

Here number of observations (n) = 20

Suppose, the observations are \({{\rm{x}}{\rm{1}}},{{\rm{x}}{{\rm{2;}}}},{{\rm{x}}{{\rm{3;}}}}, \ldots ,{{\rm{x}}{{\rm{20;}}}}\)

We have, variance = \(\frac{{\mathop \sum \nolimits_{{\rm{i = 1}}}^{\rm{n}} {{\left( {{{\rm{x}}_{\rm{i}}} - {\rm{\bar x}}} \right)}^2}}}{{\rm{n}}}\)

\( \Rightarrow 5 =\frac{{\mathop \sum \nolimits_{{\rm{i = 1}}}^{\rm{n}} {{\left( {{{\rm{x}}_{\rm{i}}} - {\rm{\bar x}}} \right)}^2}}}{{{\rm{20}}}}\)

\(\Rightarrow \mathop \sum \limits_{{\rm{i = 1}}}^{\rm{n}} {\left( {{{\rm{x}}_{\rm{i}}} - {\rm{\bar x}}} \right)^2} = 100\)     ...1)

We need to find relation between old and new observations

If Old observations are multiplied by 3, we get new observations,

Suppose, the new observations are \({{\rm{y}}{\rm{1}}},{{\rm{y}}{{\rm{2;}}}},{{\rm{y}}{{\rm{3;}}}}, \ldots ,{{\rm{y}}{{\rm{20}}}}\)

\(\therefore {{\rm{y}}{\rm{i}}} = 3{{\rm{x}}{\rm{i}}}\)     ...2) 

New mean \(\left( {{\rm{\bar y}}} \right){\rm{}} = {\rm{}}\frac{{\mathop \sum \nolimits_{{\rm{i = 1}}}^{\rm{n}} {{\rm{y}}_{\rm{i}}}}}{{\rm{n}}}\)

\(\Rightarrow \left( {{\rm{\bar y}}} \right) = {\rm{}}\frac{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{n}} 3{{\rm{x}}_{\rm{i}}}}}{{20}}\)     ...(From (2))

\(\Rightarrow \left( {{\rm{\bar y}}} \right) = \frac{{3\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{n}} {{\rm{x}}_{\rm{i}}}}}{{20}}\)

$\Rightarrow \left( {{\rm{\bar y}}} \right) = 3{\rm{\bar x}}$     ...3)

Now new variance = \(\frac{{\mathop \sum \nolimits_{{\rm{i = 1}}}^{\rm{n}} {{\left( {{{\rm{y}}_{\rm{i}}} - {\rm{\bar y}}} \right)}^2}}}{{\rm{n}}}\)

\(\Rightarrow \frac{{\mathop \sum \nolimits_{{\rm{i = 1}}}^{\rm{n}} {{\left( {3{{\rm{x}}_i} - 3{\rm{\bar x}}} \right)}^2}}}{{{\rm{20}}}}\)     ...(From (2) and (3))

\(\Rightarrow \frac{{\left( {{3^2}} \right)\mathop \sum \nolimits_{{\rm{i = 1}}}^{\rm{n}} {{\left( {{{\rm{x}}_i} - {\rm{\bar x}}} \right)}^2}}}{{{\rm{20}}}}\)

$\Rightarrow \frac{{9;{\rm{ \times 100}}}}{{{\rm{20}}}}$     ...(From (1))

⇒ 9 × 5

⇒ variance = 45

Short cut:

• If variance and number of observations (n) are known, observations are now multiplied with ‘k’ (k = integer) and we asked to find out new variance. Hence, New variance = k² × old variance

Concept:

Direction cosine: It is the cosines of each angles that the directed line makes with the x-axis, y-axis, and z-axis i.e. α, β and γ respectively.

Represented as: {l, m, n}

Where,

 l = cos α, m = cos β and n = cos γ

Also, l² + m² + n² = 1

Calculation:

Given:

Direction cosines of a line {0, 1, 0}

To find: Orientation with coordinate axes.

l = cos α = 0

⇒ α = 90°

m = cos β = 1

⇒ β = 0°

n = cos γ = 0

⇒ n = 90°

So, Cosine angle along y- axis is zero hence the line is parallel to y-axis.

Concept:

Coplanar Vectors: All the vectors that lie on same plane or parallel to same plane.

Condition for coplanar of vectors: Let three vectors $\vec a,{\rm{;}}\vec b{\rm{;and;}}\vec c$ which are coplanar then:

• Their scalar triple product is zero, i.e. $\left[ {\vec a;\vec b;\vec c} \right] = 0$ or $\vec a.\left( {\vec b \times \vec c} \right) = 0$
• They are linearly independent.
• In case of n vectors, if no more than two vectors are linearly independent, then all are coplanar.
• Linearly independent vectors: A set vectors are said to be linearly independent if, no vector can be defined as a linear combination of the others.
• Scalar triple product is zero, because the volume of the parallelepiped becomes zero, as all three vectors lie on same plane.

 

Formulas used:

• Scalar Triple Product:

\(\vec a.\left( {\vec b \times \vec c} \right) = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}&{{a_3}}\ {{b_1}}&{{b_2}}&{{b_3}}\ {{c_1}}&{{c_2}}&{{c_3}} \end{array}} \right|\)

\(\Rightarrow \vec a.\left( {\vec b \times \vec c} \right) = {a_1}\left{ {\left( {{b_2}{c_3}} \right) - \left( {{b_3}{c_2}} \right)} \right} - {a_2}\left{ {\left( {{b_1}{c_3}} \right) - \left( {{b_3}{c_1}} \right)} \right} + {a_3}\left{ {\left( {{b_1}{c_2}} \right) - \left( {{b_2}{c_1}} \right)} \right}\)

• Modulus of a vector:

$\left| {\vec a} \right| = \sqrt {{a_1}^2 + {a_2}^2 + {a_3}^2}$

Calculation:

Given:

${\rm{\vec a}} = {\rm{\hat i}} - {\rm{\hat j}} + {\rm{\hat k}}$

${\rm{\vec b}} = 2{\rm{\hat i}} + 3{\rm{\hat j}} + 2{\rm{\hat k}}$

${\rm{\vec c}} = {\rm{\hat i}} + {\rm{m\hat j}} + {\rm{n\hat k}}$

$\left| {{\rm{\vec c}}} \right| = \sqrt 6$

Find: Value of m and n

Calculating scalar triple product

\({\rm{\vec a}}.\left( {{\rm{\vec b}} \times {\rm{\vec c}}} \right) = \left| {\begin{array}{*{20}{c}} 1&-1&1\ 2&3&2\ 1&{\rm{m}}&{\rm{n}} \end{array}} \right|\)

$\Rightarrow {\rm{\vec a}}.\left( {{\rm{\vec b}} \times {\rm{\vec c}}} \right) = 1\left( {3 \times {\rm{n}} - 2 \times {\rm{m}}} \right) - \left( -1 \right)\left( {2 \times {\rm{n}} - 2 \times 1} \right) + 1\left( {2 \times {\rm{m}} - 3 \times 1} \right)$

$\Rightarrow {\rm{\vec a}}.\left( {{\rm{\vec b}} \times {\rm{\vec c}}} \right) = \left( {{3\rm{n}} - 2{\rm{m}}} \right) + \left( {2{\rm{n}} - 2} \right) + \left( {2{\rm{m}} - 3} \right)$

$\Rightarrow {\rm{\vec a}}.\left( {{\rm{\vec b}} \times {\rm{\vec c}}} \right) = 5{\rm{n}} - 5$

We know that, for vectors to be coplanar ${\rm{\vec a}}.\left( {{\rm{\vec b}} \times {\rm{\vec c}}} \right) = 0$.

${\rm{\vec a}}.\left( {{\rm{\vec b}} \times {\rm{\vec c}}} \right) = 5{\rm{n}} - 5 = 0$

$\Rightarrow 5{\rm{n}} = 5$

$\Rightarrow {\rm{n}} = 1$

Now,

$\left| {{\rm{\vec c}}} \right| = \sqrt 6$

$\Rightarrow \sqrt {{1^2} + {{\rm{m}}^2} + {{\rm{n}}^2}} = \sqrt 6$

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

⇒ m² = 4

⇒ m = ± 2

Concept:

1. Combination of r objects out of n objects:

• Number of ways selecting r objects out of n objects is given by:

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

2. Addition principle:

• Number of ways of choosing n objects or m objects is done in n + m ways.

 

Calculation:

Number of elements in the given set is n (A) = 10.

Number of subset containing two elements is same as choosing two elements out of the given 10 elements. It is given by ¹⁰C₂ = 45.

Similarly, number of subset containing three elements is same as choosing three elements out of the given 10 elements. It is given by ¹⁰C₃ = 120

Therefore, number of subsets of A containing two or three elements is given by ¹⁰C₂ + ¹⁰C₃ = 45 + 120 = 165

Therefore, number of subsets of A containing two or three elements is 165.

Concept:

1. Sum of the roots of a quadratic equation

• For a general quadratic equation ax² + bx + c = 0 where a, b, c are real numbers, if α and β are the roots of the equation then sum of the roots is given by: ${\bf{\alpha }} + {\bf{\beta }} = - \frac{{\bf{b}}}{{\bf{a}}}$

2. Product of the roots of a quadratic equation

• For a general quadratic equation ax² + bx + c = 0 where a, b, c are real numbers, if α and β are the roots of the equation then sum of the roots is given by: ${\bf{\alpha \beta }} = \frac{{\bf{c}}}{{\bf{a}}}$

Calculation:

Comparing given quadratic equation with standard form ax² + bx + c = 0 we get a = 1, b = k, and c = 1.

Let α and β denote the roots of the given equation.

Therefore, sum of the roots is α + β = -k

Similarly, product of the roots is αβ = 1.

From the given condition we have ${\rm{\alpha }} - {\rm{\beta }} < \sqrt 5$ and |k| ≥ 2.

Let us focus on the first condition. Express the difference of roots in terms of sum and product of roots:

${\left( {{\rm{\alpha }} - {\rm{\beta }}} \right)^2} = {\left( {{\rm{\alpha }} + {\rm{\beta }}} \right)^2} - 4{\rm{\alpha \beta }}$

Combine this with the given condition and write:

${\rm{\alpha }} - {\rm{\beta }} < \sqrt 5$

${\left( {{\rm{\alpha }} - {\rm{;\beta }}} \right)^2} < 5$     ...squaring both sides

${\left( {{\rm{\alpha }} + {\rm{\beta }}} \right)^2} - 4{\rm{\alpha \beta }} < 5$

Substitute the values from the obtained equations we write:

${\left( { - {\rm{k}}} \right)^2} - 4 < 5$

${{\rm{k}}^2} - {\rm{}}4 < 5$

${{\rm{k}}^2} < 9$

k² - 9 < 0

(k - 3) (k + 3) < 0

• 3 < k < 3

|k| < 3 

Therefore, $k \in \left( { - 3,3} \right)$.

Now, from the second condition we have $\left| {\rm{k}} \right| \ge 2 \Rightarrow {\rm{k}} \in \left( { - \infty , - 2\left] \cup \right[2,\infty } \right)$.

Combining both the conditions we get ${\rm{k}} \in \left( { - 3, - 2\left] \cup \right[2,3} \right)$.

Concept:

1. Triangle inequality:

• For any two complex numbers α, β the following inequality always holds:

$\left| {{\rm{\alpha }} + {\rm{\beta }}} \right| \le \left| {\rm{\alpha }} \right| + \left| {\rm{\beta }} \right|$

Calculation:

Express |z + 1| in terms of |z + 4| as follows:

|z + 1| = |z + 4 + (-3)|

It is given that $\left| {z + 4} \right| \le 3$.

Use the above condition along with triangle inequality as follows:

$\left| {{\rm{z}} + 1} \right| = \left| {{\rm{z}} + 4 + \left( { - 3} \right)} \right|$

$\le \left| {{\rm{z}} + 4| + |\left( { - 3} \right)} \right|$     ...Using triangle inequality

$\le 3 + 3$

$\le 6$

Therefore, maximum possible value for $\left| {z + 1} \right|$ is 6.

Concept:

1. Representation of a complex number and its conjugate:

• A complex number z is represented by its real and complex component as follows:

z = x + iy

Where, x and y are real numbers and I = √(-1).

• The conjugate of $z,\bar z$ is given by:

$\bar z = x - iy$

2. Comparison of two complex numbers:

Two complex numbers ${{\rm{z}}_1} = {{\rm{x}}_1} + {\rm{i}}{{\rm{y}}_1}{\rm{;and;}}{{\rm{z}}_2} = {{\rm{x}}_2} + {\rm{i}}{{\rm{y}}_2}$ are said to be equal if there real and complex components are equal. Symbolically:                ${{\rm{z}}_1} = {{\rm{z}}_2} \Rightarrow {{\rm{x}}_1} = {{\rm{x}}_2}{\rm{; and;}}{{\rm{y}}_1} = {{\rm{y}}_2}$

Calculation:

It is given that ${z^2} = 2\bar z$.

Let $z = x + iy \Rightarrow \bar z = x - iy$.

${{\rm{z}}^2} = 2{\rm{\bar z}}$

${\left( {{\rm{x}} + {\rm{iy}}} \right)^2} = 2\left( {{\rm{x}} - {\rm{iy}}} \right)$

$\left( {{{\rm{x}}^2} - {\rm{;}}{{\rm{y}}^2}} \right) + {\rm{;}}2{\rm{xyi}} = 2{\rm{x}} - 2{\rm{yi}}$

Comparing LHS and RHS we get x² – y² = 2x and 2xy = -2y.

Consider $2xy = - 2y$ first and proceed as follows:

2xy = -2y

xy + y = 0

y(x + 1) = 0

Therefore, y = 0 or x = -1.

If y = 0 then the first equation gets reduced to:

${{\rm{x}}^2} - {\rm{;}}{\left( 0 \right)^2} = 2{\rm{x}}$

${{\rm{x}}^2} - {\rm{;}}2{\rm{x}} = 0$

x(x - 2) = 0 

Therefore, if y = 0 then x = 0 or x = 2.

Thus, in this case the solutions are z = 0 or z = 2.

Now, let x = -1 then the first equation gets reduced to:

${\left( { - 1} \right)^2} - {\rm{;}}{{\rm{y}}^2} = 2\left( { - 1} \right)$

$1 - {{\rm{y}}^2} = - 2$

${{\rm{y}}^2} = 3$

${\rm{y}} = \pm {\rm{;}}\sqrt 3$

Thus, in this case the solutions are $z = - 1 + \sqrt 3 i$ and $z = - 1 - \sqrt 3 i$.

Hence in general there are 4 different solutions to the given equation and the solution set is given by \(\left{ {0,2, - 1 + \sqrt 3 i, - 1 - \sqrt 3 i} \right}\).

Concept:

1. For a general quadratic equation ${\rm{a}}{{\rm{x}}^2} + {\rm{bx}} + {\rm{c}} = 0$ where $\text{a},\text{b},\text{c}\in \mathbb{R}$, if α and β are the roots of the equation

• Sum of the roots of a quadratic equation is given by $\mathbf{\alpha }+\mathbf{\beta }=-\frac{\mathbf{b}}{\mathbf{a}}$
• Product of the roots of a quadratic equation is given by $\mathbf{\alpha \beta }=\frac{\mathbf{c}}{\mathbf{a}}$

2. Three terms in AP:

• Three terms a, b, c are in AP if the following condition is satisfied:

$b=\frac{a+c}{2}$ Or 2b = a + c

3. Three terms in HP:

• Three terms a, b, c are in HP if the following condition is satisfied:

$b=\frac{2ac}{a+c}$ Or b(a + c) = 2ac

4. Three terms in GP:

• Three terms a, b, c are in GP if the following condition is satisfied:

$\frac{b}{a} = \frac{c}{b}$ Or ${{\rm{b}}^2} = {\rm{ac}}$

Calculation:

The given quadratic equation is x² + bx + c = 0 where b, c are non-zero.

Let α and β be the roots of the given quadratic equation then sum of the roots is α + β = -b and product of the roots is αβ = c.

The terms $\frac{1}{c},b,\frac{c}{b}$ will be in AP,

$2{\rm{b}} = {\rm{;}}\frac{1}{{\rm{c}}} + {\rm{;}}\frac{{\rm{c}}}{{\rm{b}}}$

$⇒ 2{{\rm{b}}^2}{\rm{c}} = {\rm{b}} + {{\rm{c}}^2}$

The terms $\frac{1}{c},b,\frac{c}{b}$ will be in HP,

${\rm{b}} = \frac{{2\left( {\frac{1}{{\rm{c}}}} \right)\left( {\frac{{\rm{c}}}{{\rm{b}}}} \right)}}{{\frac{1}{{\rm{c}}} + {\rm{;}}\frac{{\rm{c}}}{{\rm{b}}}}}$

$⇒ {\rm{b}} = {\rm{}}\frac{{2{\rm{c}}}}{{{\rm{b}} + {\rm{c^2}}}}$

$⇒ {{\rm{b}}^2} + {\rm{b}}{{\rm{c}}^2} = 2{\rm{c}}$

The terms $\frac{1}{c},b,\frac{c}{b}$ will be in GP,

$⇒ {{\rm{b}}^2} = \left( {\frac{1}{{\rm{c}}}} \right)\left( {\frac{{\rm{c}}}{{\rm{b}}}} \right)$

⇒ b³ = 1 

From the given condition we have:

${\rm{\alpha }} + {\rm{\beta }} = {\rm{}}\frac{1}{{{{\rm{\alpha }}^2}}} + {\rm{}}\frac{1}{{{{\rm{\beta }}^2}}}$

${\rm{\alpha }} + {\rm{\beta }} = \frac{{{{\rm{\alpha }}^2} + {{\rm{\beta }}^2}}}{{{\rm{\alpha^2 \beta^2 }}}}{\rm{}}$

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

Substitute the values of sum and product for the given equation in the above equation we get:

$\left( { - {\rm{b}}} \right) = {\rm{}}\frac{{{{\left( { - {\rm{b}}} \right)}^2} - 2{\rm{c}}}}{{\rm{c^2}}}$

$- {\rm{bc^2}} = {{\rm{b}}^2} - {\rm{;}}2{\rm{c}}$

⇒ 2c = b² + bc² 

⇒ 2/b = b/c + c 

⇒ b/c ,b, c are in AP

$⇒\frac{1}{c},b,\frac{c}{b}$ are in HP

Concept:

Definition of Equivalence relation

• A relation is an equivalence relation if & only if it is:
• Reflexive property: aRa
• Symmetric property: if a R b, then b R a
• Transitive property: if a R b and b R c, then a R c

 

Calculation:

Given that,

S = persons living in Delhi x, y ∈ S.

x R y if x & y born on same day.

⇒ First we check reflexive property so,

x R x or y R y both will be reflective.

So reflexive property true for Relation.

⇒ Now symmetric property,

Given that x R y born on same day so it’s reverse y R x also true y and x also born on same day.

Symmetric property also true for relation.

⇒ For transitive property,

Given that x R y born on same day

Let’s take z who also born on same day of y so y R z ∈ S

So we can say that x R z ∈ S because given that x R y and y R z ∈ S.

Transitive property also true for relation.

So this is an equivalence relation.

Concept:

• i³ = - i
• i² = - 1

 

Calculation:

Given that,

⇒ i²ⁿ + i²ⁿ⁺¹ + i²ⁿ⁺² + i²ⁿ⁺³ where ${\rm{i}} = \sqrt { - 1}$

Take i²ⁿ common from all terms

⇒ i²ⁿ (1 + i¹ + i² + i³)

⇒ i²ⁿ (1 + i – 1 – i) [${{\rm{i}}^2} = - 1,{\rm{;}}{{\rm{i}}^3} = - {\rm{i}}$]

⇒ i²ⁿ (0)

⇒ 0

Concept:

1. Let us consider the standard form of a quadratic equation, ax² + bx + c = 0 Where a, b and c are constants with a ≠ 0
2. Let α and β be the two roots of the above quadratic equation.

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

\(\text{ }!!\alpha!!\text{ }+\text{ }!!\beta!!\text{ }=-\frac{\text{b}}{\text{a}}=-\frac{\text{coefficient }!!!!\text{ of }!!!!\text{ x}}{\text{coefficient }!!!!\text{ of }!!!!\text{ }{{\text{x}}^{2}}}\)

• The product of the roots is given by:

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

1. (a + b)² = a² + b² + 2ab

 

Calculation:

Let’s take α, β roots of x² + px + q = 0 so,

${\rm{\alpha }} + {\rm{\beta }} = - \frac{{\rm{p}}}{1} = - {\rm{p}}$     ...1)

${\rm{\alpha }}.{\rm{\beta }} = \frac{{\rm{q}}}{1} = {\rm{q}}$      ...2)

Same as let’s assume that γ, δ roots of x² + lx + m = 0 so,

${\rm{\gamma }} + {\rm{\delta }} = - \frac{{\rm{l}}}{1} = - {\rm{l}}$     ...3)

${\rm{\gamma }}.{\rm{\delta }} = \frac{{\rm{m}}}{1} = {\rm{m}}$     ...4)

Given that,

$\frac{{\rm{\alpha }}}{{\rm{\beta }}} = \frac{{\rm{\gamma }}}{{\rm{\delta }}}$ …. (5)

$\frac{{\rm{\beta }}}{{\rm{\alpha }}} = \frac{{\rm{\delta }}}{{\rm{\gamma }}}$ …. (6)

Add both equation 5 and 6

$\Rightarrow \frac{{({{\rm{\alpha }}^2} + {{\rm{\beta }}^2})}}{{{\rm{\alpha }}.{\rm{\beta }}}} = \frac{{({{\rm{\gamma }}^2} + {{\rm{\delta }}^2})}}{{{\rm{\gamma }}.{\rm{\delta }}}}$

Now, add both side 2

$\Rightarrow \frac{{({{\rm{\alpha }}^2} + {{\rm{\beta }}^2})}}{{{\rm{\alpha }}.{\rm{\beta }}}} + 2 = \frac{{({{\rm{\gamma }}^2} + {{\rm{\delta }}^2})}}{{{\rm{\gamma }}.{\rm{\delta }}}}$ + 2

$\Rightarrow \frac{{({{\rm{\alpha }}^2} + {{\rm{\beta }}^2}) + 2{\rm{\alpha \beta }}}}{{{\rm{\alpha }}.{\rm{\beta }}}} = \frac{{({{\rm{\gamma }}^2} + {{\rm{\delta }}^2}) + 2{\rm{\gamma \delta }}}}{{{\rm{\gamma }}.{\rm{\delta }}}}{\rm{;}}[{\left( {{\rm{a}} + {\rm{b}}} \right)^2} = {{\rm{a}}^2} + {{\rm{b}}^2} + 2{\rm{ab}}$]

$\Rightarrow \frac{{{{\left( {{\rm{\alpha }} + {\rm{\beta }}} \right)}^2}}}{{{\rm{\alpha }}.{\rm{\beta }}}} = \frac{{{{\left( {{\rm{\gamma }} + {\rm{\delta }}} \right)}^2}}}{{{\rm{\gamma }}.{\rm{\delta }}}}$

Put all values from equation 1, 2 and 3, 4

$\Rightarrow \frac{{{{\left( { - {\rm{p}}} \right)}^2}}}{{\rm{q}}} = \frac{{{{\left( { - {\rm{l}}} \right)}^2}}}{{\rm{m}}}$

$\Rightarrow \frac{{{{\rm{p}}^2}}}{{\rm{q}}} = \frac{{{{\rm{l}}^2}}}{{\rm{m}}}$

⇒ p²m = l²q

Concept:

• ω = $\frac{{ - 1 + {\rm{i}}\sqrt 3 }}{2}$ and ω² = $\frac{{ - 1 - {\rm{i}}\sqrt 3 }}{2}$ where ω is cube root of unity.
• ${{\rm{\omega }}^{3{\rm{k}}}} = 1,{\rm{;}}{{\rm{\omega }}^{\left( {3{\rm{k}} + 1} \right)}} = {\rm{;\omega ;and;}}{{\rm{\omega }}^{\left( {3{\rm{k}} + 2} \right)}} = {\rm{;}}{{\rm{\omega }}^2}$
• $1 + {\rm{\omega }} + {{\rm{\omega }}^2} = 0$

 

Calculation:

Given that,

${\left( {\frac{{ - 1 + {\rm{i}}\sqrt 3 }}{2}} \right)^{\rm{n}}} + {\left( {\frac{{ - 1 - {\rm{i}}\sqrt 3 }}{2}} \right)^{\rm{n}}}$ where n is not a multiple of 3.

$\Rightarrow {{\rm{\omega }}^{\rm{n}}} + {\rm{}}{{\rm{\omega }}^{2{\rm{n}}}}{\rm{}}\left[ {{\rm{}}{\rm{\omega }} = {\rm{}}\frac{{ - 1 + {\rm{i}}\sqrt {3,} }}{2},{\rm{;}}{{\rm{\omega }}^2}{\rm{}} = {\rm{}}\frac{{ - 1 - {\rm{i}}\sqrt 3 }}{2}{\rm{}}} \right]$

Here n is not a multiple of 3 so,

Put n = 3k + 1 where k is integer k =1, 2, 3 …..

$\Rightarrow {{\rm{\omega }}^{3{\rm{k}} + 1}} + {{\rm{\omega }}^{6{\rm{k}} + 2}}$

$\Rightarrow {{\rm{\omega }}^{3{\rm{k}}}}.{{\rm{\omega }}^1} + {{\rm{\omega }}^{6{\rm{k}}}}.{{\rm{\omega }}^2}\left[ {{{\because\rm{\omega }}^{3{\rm{k}}}} = 1} \right]$

$\Rightarrow {\rm{\omega }} + {{\rm{\omega }}^2}\left[ {\because 1 + {\rm{\omega }} + {{\rm{\omega }}^2} = 0} \right]$

⇒ -1

Concept:

Sum of formed n digit numbers = (sum of given numbers). (n – 1)! (10⁰ + 10¹ + 10² ………..10^n-1)

Calculation:

Given that,

n = 3 and digits are 1, 2 and 3

Put all values in given formula

Sum of formed numbers = (sum of given numbers). (n – 1)! (10⁰ + 10¹ + 10² ………..10^n-1)

Sum of all three digit number = (1 + 2 + 3). (3 – 1)! (10⁰ + 10¹ + 10²)

Sum of all three digit number = 6 × 2! × (1 + 10 + 100)

Sum of all three digit number = 12 × 111

Sum of all three digit number = 1332

Concept:

• Sum of n terms of GP = $\frac{{{\rm{a}}\left( {1 - {{\rm{r}}^{\rm{n}}}} \right)}}{{1 - {\rm{r}}}}$ where a is first letter of GP and r is the common ratio.

 

Calculation:

⇒ 0.3 + 0.33 + 0.333 + …n terms

Take 3 common from all terms

⇒ 3 (0.1 + 0.11 + 0.111 + ….n terms)

Divide and multiply by 9

⇒ 3/9 (0.9 + 0.99 + 0.999 + ….n terms)

$\Rightarrow \frac{3}{9}\left( {\frac{9}{{10}} + \frac{{99}}{{100}} + \frac{{999}}{{1000}} + \ldots {\rm{n;terms}}} \right)$

$\Rightarrow \frac{3}{9}\left( {1 - \frac{1}{{10}} + 1 - \frac{1}{{100}} + 1 - \frac{1}{{1000}} + \ldots {\rm{n;terms}}} \right)$

⇒ $\frac{3}{9}\left[ {\left( {1 + 1 + 1 + \ldots {\rm{n;times}}} \right) - \left( {\frac{1}{{10}} + \frac{1}{{100}} + \frac{1}{{1000}} + \ldots {\rm{n;terms}}} \right)} \right]$     ...1)

GP series,

$\Rightarrow \frac{1}{{10}} + \frac{1}{{100}} + \frac{1}{{1000}} + \ldots {\rm{n;terms}}$

Where a = 1/10 and r = $\frac{{\frac{1}{{100}}}}{{\frac{1}{{10}}}} = \frac{1}{{10}}$

Sum of GP = $\frac{{{\rm{a}}\left( {1 - {{\rm{r}}^{\rm{n}}}} \right)}}{{1 - {\rm{r}}}}$ 

⇒ S_n = $\frac{{\frac{1}{{10}}\left( {1 - {{\left( {\frac{1}{{10}}} \right)}^{\rm{n}}}} \right)}}{{1 - \frac{1}{{10}}}}$ 

⇒ S_n = $\frac{{\frac{1}{{10}}\left( {1 - {{\left( {\frac{1}{{10}}} \right)}^{\rm{n}}}} \right)}}{{1 - \frac{1}{{10}}}}$ 

⇒ S_n = $\frac{1}{9}{\rm{;}}\left( {1 - \frac{1}{{{{10}^{\rm{n}}}}}} \right)$     ...2)

From equation 1 and 2

$\Rightarrow \frac{1}{3}\left[ {\left( {\rm{n}} \right) - \frac{1}{9}{\rm{;}}\left( {1 - \frac{1}{{{{10}^{\rm{n}}}}}} \right)} \right]$

Short trick:

Put n = 1 in all option and check which option satisfy the first term of series.

Concept:

• ω = $\frac{{ - 1 + {\rm{i}}\sqrt 3 }}{2}$ and ω² = $\frac{{ - 1 - {\rm{i}}\sqrt 3 }}{2}$ where ω is cube root of unity.
• ${{\rm{\omega }}^{3{\rm{k}}}} = 1,{\rm{;}}{{\rm{\omega }}^{\left( {3{\rm{k}} + 1} \right)}} = {\rm{;\omega ;and;}}{{\rm{\omega }}^{\left( {3{\rm{k}} + 2} \right)}} = {\rm{;}}{{\rm{\omega }}^2}$
• $1 + {\rm{\omega }} + {{\rm{\omega }}^2} = 0$

 

Calculation:

Given that,

(1 + ω) (1 + ω²) (1 + ω³) (1 + ω + ω²) where 1, ω, ω² are the cube roots of unity

As we know,

$\Rightarrow 1 + {\rm{\omega }} + {{\rm{\omega }}^2} = 0$

So,

⇒ (1 + ω) (1 + ω2) (1 + ω3) (1 + ω + ω2) = (1 + ω) (1 + ω²) (1 + ω³) (0) = 0

Concept:

Sum of n AP terms = ${{\rm{S}}_{\rm{n}}} = \frac{{\rm{n}}}{2}\left[ {2{\rm{a}} + \left( {{\rm{n}} - 1} \right){\rm{d}}} \right]$ 

Where a is first term of AP and d is difference between two consecutive terms of AP.

Calculation:

Given that,

Sum of m terms of an AP is n so,

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

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

⇒ 2n = 2am + m (m - 1)d     ...1)

Same as given that

The sum of n terms is m so,

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

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

⇒ 2m = 2an + n(n - 1)d     ...2)

Subtracting 2 from 1

$\Rightarrow 2\left( {{\rm{n}} - {\rm{m}}} \right) = 2{\rm{a}}\left( {{\rm{m}} - {\rm{n}}} \right) + \left[ {\left( {{{\rm{m}}^2} - {{\rm{n}}^2}} \right) - \left( {{\rm{m}} - {\rm{n}}} \right)} \right]{\rm{d}}$

$\Rightarrow - 2\left( {{\rm{m}} - {\rm{n}}} \right) = 2{\rm{a}}\left( {{\rm{m}} - {\rm{n}}} \right) + \left[ {\left( {{\rm{m}} - {\rm{n}}} \right)\left( {{\rm{m}} + {\rm{n}}} \right) - \left( {{\rm{m}} - {\rm{n}}} \right)} \right]{\rm{d}}$     ...[$\because{{\rm{a}}^2} - {{\rm{b}}^2} = \left( {{\rm{a}} - {\rm{b}}} \right)\left( {{\rm{a}} + {\rm{b}}} \right)$]

Divide both side by (m - n)

⇒ -2 = 2a + [(m + n) -1]d

Multiply both side by $\frac{{{\rm{m}} + {\rm{n}}}}{2}$

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

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

Concept:

Let z = x + iy be a complex number.

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

 

Calculation:

Given that,

$\frac{{1 + 2{\rm{i}}}}{{1 - {{\left( {1 - {\rm{i}}} \right)}^2}}}$

⇒ (1 - i)² = 1² + i² - 2.1.i      ...[∵ (a- b)² = a² + b²- 2ab]

⇒ (1 - i)² = 1 – 1 - 2i

⇒ (1 - i)² = - 2i

So,

⇒ 1 - (1 - i)² = 1 + 2i

Put this value in main equation

$\Rightarrow \frac{{1 + 2{\rm{i}}}}{{1 + 2{\rm{i}}}} = 1$ = 1 + 0 i

So,

Modulus = $\sqrt {{1^2} + {0^2}}$ = 1

Argument = ${\tan ^{ - 1}}\left( {\frac{0}{1}} \right) = 0$

Concept:

• Find the coordinates of the points of intersection with the x axis, we must solve the equation f(x) = 0 or we can say that to finding roots of quadratic equation we must calculate the intersection point with x axis.
• The points where the graph of a polynomial cuts the x-axis are the zeros of the polynomial
• If the graph of a polynomial doesn't cut or touches the x- axis it means that the polynomial has no real root i.e it has complex roots.

 

Calculation:

Given that the graph of a quadratic polynomial lies entirely above x-axis so it never cuts the x axis so quadratic equation does not have any real roots or we can say it does have complex roots.

Concept:

Let us consider the standard form of a quadratic equation, ax² + bx + c =0 Where a, b and c are constants with a ≠ 0

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

The sum of the roots of a quadratic equation are: ${\rm{\alpha }} + {\rm{\beta }} = - \frac{{\rm{b}}}{{\rm{a}}} = - \frac{{{\rm{coefficient;of;x}}}}{{{\rm{coefficient;of;}}{{\rm{x}}^2}}}$

The product of the roots is given by: ${\rm{\alpha \beta }} = \frac{{\rm{c}}}{{\rm{a}}} = \frac{{{\rm{constant;term}}}}{{{\rm{coefficient;of;}}{{\rm{x}}^2}}}$

• $\cot \left( {{\rm{a}} + {\rm{b}}} \right) = \frac{{\cot {\rm{;a}}.\cot {\rm{;b;}} - 1}}{{\cot {\rm{;a}} + \cot {\rm{;b}}}}$

 

Calculation:

Given that, cot α and cot β are the roots of the equation x² + bx + c = 0 so,

⇒ cot α + cot β = - b/1 = - b     ...1)

⇒ cot α. cot β = c/1 = c     ...2)

value of cot (α + β)

$\Rightarrow \cot \left( {{\rm{\alpha }} + {\rm{\beta }}} \right) = \frac{{\cot {\rm{;\alpha }}.\cot {\rm{;\beta }} - 1}}{{\cot {\rm{;\alpha }} + \cot {\rm{;\beta }}}}$     ...3) [$\because\cot \left( {{\rm{a}} + {\rm{b}}} \right) = \frac{{\cot {\rm{a}}.\cot {\rm{b}} - 1}}{{\cot {\rm{a}} + \cot {\rm{b}}}}$]

From equation 1, 2 and 3

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

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

Concept:

1. Let us consider the standard form of a quadratic equation, ax² + bx + c = 0 Where a, b and c are constants with a ≠ 0
   

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

 

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

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

• The product of the roots is given by:

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

1. If a, b, c are in A.P ⇔ b = $\frac{{{\rm{a;}} + {\rm{;c}}}}{2}$

 

Calculation:

Given that,

Let’s take α, β are roots of ax² + x + c = 0 so,

$\Rightarrow {\rm{\alpha }} + {\rm{\beta }} = {\rm{}} - \frac{1}{{\rm{a}}}$     ...1)

$\Rightarrow {\rm{\alpha }}.{\rm{\beta }} = \frac{{\rm{c}}}{{\rm{a}}}$     ...2)

Given

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

$\Rightarrow {\rm{\alpha }} + {\rm{\beta }} = \frac{{{{\rm{\alpha }}^2} + {{\rm{\beta }}^2}}}{{{{\rm{\alpha }}^2}.{{\rm{\beta }}^2}}}$

$\Rightarrow {\rm{\alpha }} + {\rm{\beta }} = \frac{{{{\left( {{\rm{\alpha }} + {\rm{\beta }}} \right)}^2} - 2{\rm{\alpha \beta }}}}{{{{\left( {{\rm{\alpha \beta }}} \right)}^2}}}$     ...[∵ (a+ b)² = a² + b²+ 2ab]

Put all values,

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

$\Rightarrow - \frac{1}{{\rm{a}}} = {\rm{;}}\frac{{1 - 2{\rm{ac}}}}{{{{\rm{c}}^2}}}$

Now, Cross multiply

${{\rm{c}}^2} = {\rm{}} - {\rm{a}} + 2{{\rm{a}}^2}{\rm{c}}$

$\Rightarrow {{\rm{c}}^2} + {\rm{a}} = 2{{\rm{a}}^2}{\rm{c}}$

$\Rightarrow {{\rm{a}}^2}{\rm{c}} = \frac{{{{\rm{c}}^2} + {\rm{a}}}}{2}$     ...[∵ a, b, c are in A.P ⇔ b = (a + c)/2]

So we can say that a, ca², c² are in AP.

Concept:

• C(n, 0) + C(n, 1) + C(n, 2) + C(n, 3) ………….. C(n, n) = 2ⁿ
• C(n, 0) = C(n, n) = 1
• C(n, r) = $^{\rm{n}}{{\rm{C}}_{\rm{r}}} = \frac{{{\rm{n}}!}}{{{\rm{r}}!\left( {{\rm{n}} - {\rm{r}}} \right)!}}$

 

Calculation:

Given that,

⇒

C(7, 0) + C(7, 1) + C(7, 1) + C(7, 2) + C(7, 2) … + C(7, 6) + C(7, 7)

⇒ C(7, 0) + 2.[ C(7, 1) + C(7, 2) + … + C(7, 6)] + C(7, 7)

Add and subtract C(7, 0),C(7, 7) so,

⇒

C(7, 0) + 2.[ C(7, 1) + C(7, 2) + … + C(7, 6)] + C(7, 0) + C(7, 7) + C(7, 0) + C(7, 7) - C(7, 0) - C(7, 7)

⇒ 2[ C(7, 0) + C(7, 1) + C(7, 2) + … + C(7, 6) + C(7, 7)] – C(7,0) – C(7, 7)

⇒

2[2⁷] - C(7,0) – C(7, 7)     ...[∵ C(n, 0) + C(n, 1) + C(n, 2) + C(n, 3) ………….. C(n, n) = 2ⁿ]

⇒

2 × 128 – 1 – 1

⇒ 254