Quantitative Aptitude PYQs

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 \(\frac{{{\rm{m}} + {\rm{n}}}}{2}\)

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

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