Quantitative Aptitude PYQs

Concept:

• A function f(x) is said to be continuous at a point x = a, in its domain if \(\mathop {\lim }\limits_{x \to {\rm{a}}} f\left( x \right) = f\left( a \right)\) exists or its graph is a single unbroken curve.
• f(x) is Continuous at x = a ⇔ \(\mathop {\lim }\limits_{x \to {{\rm{a}}^ + }} f\left( x \right){\rm{}} = {\rm{}}\mathop {\lim }\limits_{x \to {{\rm{a}}^ - }} f\left( x \right){\rm{}} = {\rm{}}\mathop {\lim }\limits_{x \to {\rm{a}}} f\left( x \right)\)

 

Calculation:

Given: f(x) = |x| + |x – 1|

The possible points of discontinuity are x = 0, 1 since the sign of the modulus changes at these points.

For continuity at x = 0

\({\rm{LHL}} = \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \;\mathop {\lim }\limits_{x \to {0^ - }} \left| {\rm{x}} \right|{\rm{\;}} + {\rm{\;}}\left| {{\rm{x\;}}-{\rm{\;}}1} \right|\)

\( = \;\mathop {\lim }\limits_{h \to 0} \left| {0 - {\rm{h}}\left| {{\rm{\;}} + {\rm{\;}}} \right|0 - {\rm{h\;}}-{\rm{\;}}1} \right|\)     ...(∵ x → 0^- ⇒ x = 0 – h)

\( = \;\mathop {\lim }\limits_{h \to 0} \left| {\rm{h}} \right|{\rm{\;}} + {\rm{\;}}\left| {{\rm{h}} + {\rm{\;}}1} \right|\)     ...(∵ |-x| = x)

= 0 + 1 = 1

\({\rm{RHL}} = \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \;\mathop {\lim }\limits_{x \to {0^ + }} \left| {\rm{x}} \right|{\rm{\;}} + {\rm{\;}}\left| {{\rm{x\;}}-{\rm{\;}}1} \right|\)

\( = \;\mathop {\lim }\limits_{h \to 0} \left| {0 + {\rm{\;h}}\left| {{\rm{\;}} + {\rm{\;}}} \right|0 + {\rm{\;h\;}}-{\rm{\;}}1} \right|\)    ...(∵ x → 0⁺ ⇒ x = 0 + h)

\( = \;\mathop {\lim }\limits_{h \to 0} \left| {\rm{h}} \right|{\rm{\;}} + {\rm{\;}}\left| {{\rm{h}} - {\rm{\;}}1} \right|\)

= 0 + 1 = 1     ...(∵ |-x| = x)

Value of f(x) at x = 0

⇒ f(0) = |0| + |0 – 1| = 1

Since LHL = RHL = f(0)

So, Function is continuous at x = 0

For continuity at x = 1

\({\rm{LHL}} = \mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \;\mathop {\lim }\limits_{x \to {1^ - }} \left| {\rm{x}} \right|{\rm{\;}} + {\rm{\;}}\left| {{\rm{x\;}}-{\rm{\;}}1} \right|\)

\(= \mathop {\lim }\limits_{h \to 0} \left| {1 - {\rm{h}}\left| {{\rm{\;}} + {\rm{\;}}} \right|1 - {\rm{h\;}}-{\rm{\;}}1} \right|\)     ...(∵ x → 1^- ⇒ x = 1 – h)

\( = \;\mathop {\lim }\limits_{h \to 0} \left| {1 - {\rm{\;h}}\left| {{\rm{\; + }}} \right| - {\rm{h}}} \right|\)

\( = \;\mathop {\lim }\limits_{h \to 0} \left| {1 - {\rm{\;h}}} \right|{\rm{\;}} + {\rm{\;}}\left| {\rm{h}} \right|\)     ...(∵ |-x| = x)

= 1 + 0 = 1

\({\rm{RHL}} = \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \;\mathop {\lim }\limits_{x \to {1^ + }} \left| {\rm{x}} \right| + \left| {{\rm{x}}-1} \right|\)

\( = \;\mathop {\lim }\limits_{h \to 0} \left| {1 + {\rm{\;h}}\left| {{\rm{\;}} + {\rm{\;}}} \right|1 + {\rm{\;h\;}}-{\rm{\;}}1} \right|\)     ...(∵ x → 1⁺ ⇒ x = 1 + h)

\( = \;\mathop {\lim }\limits_{h \to 0} \left| {1 + {\rm{\;h}}} \right|{\rm{\;}} + {\rm{\;}}\left| {\rm{h}} \right|\)

= 1 + 0 = 1

Value of f(x) at x = 1

⇒ f(1) = |1| + |1 – 1| = 1

Since LHL = RHL = f(1)

So, Function is continuous at x = 1

∴ f(x) is continuous at x = 0 and x = 1

Concept:

1. Integration by parts: Integration by parts is a method to find integrals of products.

• The formula for integrating by parts is given by;

\(\Rightarrow \;\smallint u\;vdx = u\;\smallint vdx - \;\smallint \left( {\frac{{du}}{{dx}}\;\smallint vdx} \right)dx\)

Where u is the function u(x) and v is the function v(x)

2. ILATE Rule: Usually, the preference order of this rule is based on some functions such as Inverse, Logarithm, Algebraic, Trigonometric and Exponent. 3. \(\smallint {x^n}dx = \;\frac{{{x^{n\; + \;1}}}}{{n + 1}} + c\) 4. \(\mathop \smallint \nolimits_a^b f'\left( x \right)dx = \;\left[ {f\left( x \right)} \right]_a^b = \left[ {f\left( b \right) - f\left( a \right)} \right]\)

Calculation:

We have to find the value of \(\mathop \smallint \nolimits_1^e x\ln x\;dx\)

\(\Rightarrow \mathop \smallint \nolimits_1^e (\ln x)\;x\;dx = \left[ {\ln x\;\left( {\frac{{{x^2}}}{2}} \right)} \right]_1^e - \;\mathop \smallint \nolimits_1^e \frac{1}{x}\left( {\frac{{{x^2}}}{2}} \right)dx\)

\(\Rightarrow \left[ {\ln x\;\left( {\frac{{{x^2}}}{2}} \right)} \right]_1^e - \frac{1}{2}\mathop \smallint \nolimits_1^e xdx\)

\(\Rightarrow \left[ {\ln x\;\left( {\frac{{{x^2}}}{2}} \right)} \right]_1^e - \frac{1}{2}\left[ {\frac{{{x^2}}}{2}} \right]_1^e\)

\(\Rightarrow \left[ {\ln e\;\frac{{{e^2}}}{2} - \;\ln 1\;\frac{{{1^2}}}{2}} \right] - \frac{1}{4}\left[ {{e^2} - 1} \right]\)           (∵ loge = 1)

\(\Rightarrow \left[ {\frac{{{e^2}}}{2}\; - \;0} \right] - \frac{1}{4}\left[ {{e^2} - 1} \right]\)

\(\Rightarrow \frac{{{e^2} + 1}}{4}\)

∴ Option 2 is correct.

Concept:

• \(\mathop \smallint \nolimits_a^b f\left( x \right)dx = \;\mathop \smallint \nolimits_a^c f\left( x \right)dx + \;\mathop \smallint \nolimits_c^b f\left( x \right)dx\)
• \(\mathop \smallint \nolimits_a^b f'\left( x \right)dx = \;\left[ {f\left( x \right)} \right]_a^b = \left[ {f\left( b \right) - f\left( a \right)} \right]\)
• \(\smallint {x^n}dx = \;\frac{{{x^{n + 1}}}}{{n + 1}} + c\)

Calculation:

We have to find the value of \(\mathop \smallint \nolimits_0^{\sqrt 2 } \left[ {{x^2}} \right]\;dx\)

Now, f(x) = \(\left[ {{x^2}} \right]\; \Rightarrow \;f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {0,\;\;0 \le x < 1}\\ {1,\;\;1 \le x < \sqrt 2 } \end{array}} \right.\)

\(\mathop \smallint \nolimits_0^{\sqrt 2 } \left[ {{x^2}} \right]\;dx = \;\mathop \smallint \nolimits_0^1 0\;dx + \mathop \smallint \nolimits_1^{\sqrt 2 } 1\;dx\)

\( = 0 + \left[ x \right]_1^{\sqrt 2 }\)

= √2 – 1

Concept:

Following steps to finding maxima and minima using derivatives.

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

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

Calculation:

We have to calculate the maximum value of 16 sin θ – 12 sin² θ

Let f(θ) = 16 sin θ – 12 sin² θ

Differentiating, both sides, we get

⇒ f’(θ) = 16 cos θ – 12 × 2 × sin θ × cos θ = 16 cos θ – 24 × sin θ × cos θ

For maxima and minima, f’(θ) = 0

⇒ 16 cos θ – 24 × sin θ × cos θ = 0

⇒ 8 cos θ (2 – 3 sin θ) = 0

⇒ 8 cos θ = 0 and (2 – 3 sin θ) = 0

∴ θ = 90° and sin θ = 2/3 or θ = sin^-1 (2/3)

Now we have to find second derivative,

⇒ f”(θ) = -16 sin θ – 24 [cos² θ – sin² θ]

⇒ f”(θ) = -16 sin θ – 24 [1 - sin² θ - sin² θ] = -16 sin θ – 24 [1 – 2 sin² θ]

When θ = 90° ⇒ \({\left. {{\rm{f}}\left( {\rm{\theta }} \right)} \right|_{{\rm{\theta \;}} = {\rm{\;}}90^\circ }}\) = -16 sin 90° – 24 [1 – 2 sin² 90°] = -16 + 24 = 8 > 0

So, function gives minimum value at θ = 90°

When θ = sin^-1 (2/3) ⇒ \({\left. {{\rm{f''}}\left( {\rm{\theta }} \right)} \right|_{{\rm{\theta \;}} = {\rm{\;}}{{\sin }^{ - 1}}\left( {\frac{2}{3}} \right)}}\) = -16 sin sin^-1 (2/3) – 24 [1 – 2 sin² sin^-1 (2/3)]

⇒ f”(θ) = -16 (2/3) – 24 [1 – 2 (2/3)²] = -(32/3) – (24/9) < 0

So, function gives maximum value at θ = sin^-1 (2/3) or sin θ = 2/3

⇒ f(θ)_max = 16(2/3) – 12 (2/3)² = (32/3) – (16/3) = 16/3

Concept:

Period of a Function:

• If a function repeats over at a constant period we say that is a periodic function.
• It is represented like f(x) = f(x + T), T is the real number and this is the period of the function.

Calculation:

We have to find the period of the function f(x) = sin x

Now,

f(x + 2π) = sin (x + 2π) = sin x

⇒ f(x + 2π) = f(x)

∴ Period of sin x is 2π

Concept:

• \(\frac{{d\left( {{a^x}} \right)}}{{dx}} = \;{a^x}\;\ln a\)
• \(\smallint \frac{1}{{\rm{x}}}{\rm{\;dx}} = {\rm{}}\ln {\rm{x\;}} + {\rm{c}}\)

 

Calculation:

We have to find the value of \(\smallint \frac{{dx}}{{{2^x} - 1}}\)

Let I = \(\smallint \frac{{dx}}{{{2^x} - 1\;}} = \;\smallint \frac{{dx}}{{{2^x}\;\left( {1\; - \;\frac{1}{{{2^x}}}} \right)\;}}\)

\({\rm{I}} = \smallint \frac{{{2^{ - {\rm{x}}}}{\rm{\;dx}}}}{{{\rm{\;}}\left( {1{\rm{\;}} - {\rm{\;}}{2^{ - {\rm{x}}}}} \right){\rm{\;}}}}\)

Let (1 – 2^-x) = t

Now, Differentiating, Both sides, we get

\( \Rightarrow {\rm{\;}}\left( {0{\rm{\;}} + {\rm{\;}}{2^{ - {\rm{x}}}}\ln 2} \right){\rm{}} = {\rm{}}\frac{{{\rm{dt}}}}{{{\rm{dx}}}}\)

\(\Rightarrow \;{2^{ - x}}\;dx = \frac{{dt}}{{\ln 2}}\)

Now, \({\rm{I}} = \frac{1}{{\ln 2}}\smallint \frac{{{\rm{dt}}}}{{{\rm{\;}}t{\rm{\;}}}}\)

\( \Rightarrow I = \frac{{\ln t}}{{\ln 2}} + c = \frac{{\ln \left( {1 - \;{2^{ - x}}} \right)}}{{\ln 2}} + c\)

∴ Option 2 is correct.

Concept:

• Order: The order of a differential equation is the order of the highest derivative appearing in it.
• Degree: The degree of a differential equation is the power of the highest derivative occurring in it, after the Equation has been expressed in a form free from radicals as far as the derivatives are concerned.

Calculation:

y² = 4a (x – a)     ...(1)

Differentiating, both sides, we get

\( \Rightarrow 2y\;\frac{{dy}}{{dx}} = 4a\)

\(\Rightarrow \left( {\frac{y}{2}} \right)\frac{{dy}}{{dx}} = a\)

Put the value of a in equation 1^st

\(\Rightarrow {y^2} = 4\left( {\frac{y}{2}} \right)\frac{{dy}}{{dx}}\;\left( {(x - \frac{y}{2}\;\frac{{dy}}{{dx}}} \right)\)

\(\Rightarrow {y^2} = 2y\;\frac{{dy}}{{dx}}\;\left( {(x - \frac{y}{2}\;\frac{{dy}}{{dx}}} \right)\)

\(\Rightarrow {y^2} = 2xy\;\frac{{dy}}{{dx}} - {y^2}{\left( {\frac{{dy}}{{dx}}} \right)^2}\)

Hence order is 1 and degree is 2.

Concept:

1. Let f(x) and g(x) be two functions such that their limits \(\mathop {\lim }\limits_{x \to a} {\rm{f}}\left( {\rm{x}} \right)\) and \(\mathop {\lim }\limits_{x \to a} {\rm{g}}\left( {\rm{x}} \right)\) exist.

• Limit of the sum of two functions is the sum of the limits of the functions.

\( \Rightarrow \mathop {\lim }\limits_{x \to a} \left[ {{\rm{f}}\left( {\rm{x}} \right) + {\rm{\;g}}\left( {\rm{x}} \right)} \right] = {\rm{\;}}\mathop {\lim }\limits_{x \to a} {\rm{f}}\left( {\rm{x}} \right){\rm{\;}} + {\rm{\;}}\mathop {\lim }\limits_{x \to a} {\rm{g}}\left( {\rm{x}} \right)\)

• Limit of the difference of two functions is the difference of the limits of the functions.

\( \Rightarrow \mathop {\lim }\limits_{x \to a} \left[ {{\rm{f}}\left( {\rm{x}} \right) - {\rm{\;g}}\left( {\rm{x}} \right)} \right] = {\rm{\;}}\mathop {\lim }\limits_{x \to a} {\rm{f}}\left( {\rm{x}} \right){\rm{}} - {\rm{}}\mathop {\lim }\limits_{x \to a} {\rm{g}}\left( {\rm{x}} \right)\)

• Limit of product of two functions is the product of the limits of the functions.

\(\Rightarrow \mathop {\lim }\limits_{x \to a} \left[ {{\rm{f}}\left( {\rm{x}} \right){\rm{\;}} \times {\rm{\;g}}\left( {\rm{x}} \right)} \right] = {\rm{\;}}\mathop {\lim }\limits_{x \to a} {\rm{f}}\left( {\rm{x}} \right){\rm{\;}} \times {\rm{\;}}\mathop {\lim }\limits_{x \to a} {\rm{g}}\left( {\rm{x}} \right)\)

• Limit of quotient of two functions is the quotient of the limits of the functions.

\( \Rightarrow \mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{\mathop {\lim }\limits_{x \to a} {\rm{f}}\left( {\rm{x}} \right)}}{{\mathop {\lim }\limits_{x \to a} {\rm{g}}\left( {\rm{x}} \right)}}\)

2. \(\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1\) 3. \(\mathop {\lim }\limits_{x \to 0} \frac{{\tan x}}{x} = 1\)

Calculation:

We have to find the value of \(\mathop {\lim }\limits_{x \to 0} \frac{{\tan x}}{{\sin 2x}}\)

\(\Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{\tan x}}{{\sin 2x}} = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\frac{{\tan x}}{x}} \right)\; \times x}}{{\left( {\frac{{\sin 2x}}{{2x}}} \right)\; \times 2x}} = \frac{1}{2}\mathop {\lim }\limits_{x \to 0} \frac{{\left( {\frac{{\tan x}}{x}} \right)}}{{\left( {\frac{{\sin 2x}}{{2x}}} \right)}}\)

We know that \(\Rightarrow \mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{\mathop {\lim }\limits_{x \to a} {\rm{f}}\left( {\rm{x}} \right)}}{{\mathop {\lim }\limits_{x \to a} {\rm{g}}\left( {\rm{x}} \right)}}\)

So, \(\frac{1}{2}\mathop {\lim }\limits_{x \to 0} \frac{{\left( {\frac{{\tan x}}{x}} \right)}}{{\left( {\frac{{\sin 2x}}{{2x}}} \right)}} = \frac{1}{2}\frac{{\mathop {\lim }\limits_{x \to 0} \frac{{\tan x}}{x}}}{{\mathop {\lim }\limits_{x \to 0} \frac{{\sin 2x}}{{2x}}}} = \frac{1}{2} \times 1 = \frac{1}{2}\)

∴ \(\mathop {\lim }\limits_{x \to 0} \frac{{\tan x}}{{\sin 2x}} = \frac{1}{2}\)

Concept:

• \({\rm{Coefficient\;of\;variation}} = {\rm{}}\frac{{{\rm{standard\;deviaiton}}}}{{{\rm{Mean}}}} \times 100\)
• Adding a constant to each value: The median, mean, and quartiles will be changed by adding a constant to each value. However, the range, interquartile range, standard deviation and variance will remain the same.
• Multiplying every value by a constant: however, will multiply the mean, median, quartiles, range, interquartile range, and standard deviation by that constant, and multiply the variance by the square of that constant.

 

Calculation:

Given:

Mean = 5 and Standard deviation = 2

If 5 is added to each value,

We know that adding a constant to each value Mean will be changed by adding a constant value and standard deviation remain the same.

Therefore, Mean = 5 + 5 = 10

Standard deviation will not change.

Now,

\(\Rightarrow {\rm{Coefficient\;of\;variation}} = {\rm{}}\frac{{{\rm{standard\;deviaiton}}}}{{{\rm{Mean}}}} \times 100 = \frac{2}{{10}} \times 100\)

∴ Coefficient of variation = 20

Concept:

• \({\rm{Speed}} = \frac{{{\rm{Distance}}}}{{{\rm{Time}}}}\)
• \({\rm{Average\;Speed}} = \frac{{{\rm{Total\;Distance}}}}{{{\rm{Total\;Time}}}}\)

 

Calculation:

Given:

Total distance = 5 + 15 = 20 km

Time taken for first part of the journey = distance/speed = 5/30 = 1/6 hr

Time taken to cover the second part = distance/speed = 15/45 = 1/3 hr

Total time taken = 1/6 + 1/3

= 1/2 hr = 0.5 hr

Now, \({\rm{Average\;Speed}} = \frac{{{\rm{Total\;Distance}}}}{{{\rm{Total\;Time}}}}\)

\(\Rightarrow {\rm{Average\;Speed}} = {\rm{}}\frac{{20}}{{0.5}} = 40\;{\rm{km}}/{\rm{hr}}\)

∴ Option 4 is correct.

Concept:

Law of Total Probability:

• If E₁, E₂, E₃, ⋯ is a partition of the sample space S, then for any event A we have

\(\Rightarrow P\;\left( A \right) = \mathop \sum \limits_{i\; = \;1} P\left( {{E_i}} \right)\;P\left( {\frac{A}{{{E_i}}}} \right)\)

Calculation:

Given:

Bag one contain 3 white and 2 black balls and bag two contains 5 white and 3 black balls

Let E₁, E₂ and A be the three events as defined below:

E₁ = Selecting bag one

E₂ = Selecting bag two

A = Drawing a white ball

Now, Bag is chosen randomly;

∴ P (E₁) = P (E₂) = 1/2

Now,

Probability of getting white ball from bag 1 = P (A/ E₁) = 3/5

Probability of getting white ball from bag 2 = P (A/ E₂) = 5/8

Using the law of total probability, we get

Required probability P (A) = P (E₁) P (A/ E₁) + P (E₂) P (A/ E₂) = (1/2) × (3/5) + (1/2) × (5/8)

\(= \frac{3}{{10}} + \frac{5}{{16}} = \frac{{24 + 25}}{{80}} = \frac{{49}}{{80}}\)

Concept:

1) Combination: Selecting r objects from given n objects.

• The number of selections of r objects from the given n objects is denoted by \(^n{C_r}\)
• \(^n{C_r} = \frac{{n!}}{{r!\left( {n - r} \right)!}}\)

2. Probability of an event happening = \(\frac{{{\rm{Number\;of\;ways\;it\;can\;happen}}}}{{{\rm{Total\;number\;of\;outcomes}}}}\)

Note: Use combinations if a problem calls for the number of ways of selecting objects.

Calculation:

We have to find the probability that exactly two members are men

Favourable outcomes: 2 men selected from 4 men and 1 more extra person has to be chosen from 5 women

⇒ n (E) = ⁴C₂ × ⁵C₁ = 6 × 5 = 30

Total sample space n (s) = ⁹C₃ = 84

⇒ required probability P (E) = \(\frac{{{\rm{n\;}}\left( {\rm{E}} \right)}}{{{\rm{n\;}}\left( {\rm{s}} \right)}} = \frac{{30}}{{84}} = \frac{5}{{14}}\)

Concept:

Formula used:

• \({\log _a}b = \frac{1}{{{{\log }_b}a}}\)
• log_a M + log_a N = log_a (MN)
• log_a a = 1

where a ≠ 1, a > 0 and b ≠ 1, b > 0 and M, N are arbitrary positive numbers and p is any real number.

Factorial:

• n! = 1 × 2 × 3 × ⋯ × (n – 1) × n

Calculation:

Given: n = (2017)!           

To find: \(\frac{1}{{{{\log }_2}n}} + \frac{1}{{{{\log }_3}n}} + \frac{1}{{{{\log }_4}n}} + \ldots + \frac{1}{{{{\log }_{2017}}n}}\)

Using \({\log _a}b = \frac{1}{{{{\log }_b}a}}\)

\(\Rightarrow \frac{1}{{{{\log }_2}n}} + \frac{1}{{{{\log }_3}n}} + \ldots + \frac{1}{{{{\log }_{2017}}n}} = {\log _n}2 + {\log _n}3 + \ldots + {\log _n}2017\)

= log_n (2017!)

= log_n (n)

= 1

Concept:

General term: General term in the expansion of (x + y) ⁿ is given by

• \({{\rm{T}}_{\left( {{\rm{r\;}} + {\rm{\;}}1} \right)}} = {\rm{\;}}{{\rm{\;}}^{\rm{n}}}{{\rm{C}}_{\rm{r}}} \times {{\rm{x}}^{{\rm{n}} - {\rm{r}}}} \times {{\rm{y}}^{\rm{r}}}\)

Formula used:

• ⁿC_r = ⁿC_{(n – r)}
• If ⁿC_x = ⁿC_y then x = y

 

Calculation:

Given:

Coefficients of T_(2r+1) = Coefficients of  T_(r+2)

⇒ ⁴³C_2r = ⁴³C_{(r + 1)}

So, 2r = r + 1

⇒ r = 1

But r ≠ 1.

Using ⁿC_r = ⁿC_{(n – r)}

⁴³C_{(r + 1)} = ⁴³C_{(43 – r – 1)}

So, ⁴³C_2r = ⁴³C_{(43 – r – 1)}

We know that If ⁿC_x = ⁿC_y then x = y

⇒ 2r = 42 – r

⇒ 3r = 42

⇒ r = 14

Concept:

Let z = x + iy be a complex number.

Argument:

• \(\arg {\rm{z}} = \arg \left( {{\rm{x}} + {\rm{iy}}} \right) = {\rm{\theta }} = {\tan ^{ - 1}}\left( {\frac{{\rm{y}}}{{\rm{x}}}} \right)\)

Note:

• Principal argument (θ) ranges –π < θ ≤ π

Calculation:

Given: z = -1 – i

As z lies in third quadrant so arg (z) = -π + θ

arg (–1 – i) = -π + tan^-1 (-1/-1)

= -π + tan^-1 (1)

= -π + π/4

= -3π/4

Concept:

If z is a complex number and a quadratic is in terms of z i.e., az² + bz + c = 0, then

• Roots are imaginary number i.e., z = α ± iβ.
• Discriminant should be less than zero i.e., D < 0, where D = b² – 4ac
• Sum of roots is a real number.

 

Calculation:

As z is a complex number so, roots of the equation will be z = a ± ib

But Re(z) = 1. So, a = 1

And, z = 1 ± ib      ----(1)

Roots of the given quadratic equation z² + αz + β = 0 will be,

\({\rm{z}} = \frac{{{\rm{\alpha }} \pm \sqrt {{{\rm{\alpha }}^2} - 4{\rm{\beta }}} }}{2}\)      ----(2)

On comparing (1) and (2),

α/2 = 1

⇒ α = 2

For roots to be non-real and distinct, D < 0

So, \(\sqrt {{{\rm{\alpha }}^2} - 4{\rm{\beta }}} < 0\)

Squaring both sides

⇒ α² – 4β < 0

⇒ 4 – 4β < 0

⇒ 4 < 4β

⇒ 1 < β

∴ β ∈ (1, ∞)

Concept:

Let A and B be the subsets of X, then then Venn Diagram to represent that is:

• Union of sets (A ∪ B):

 

• Intersection of subsets (A ∩ B):

 

• Complement of subset (A’):

 

 

• Difference of subsets (A – B):

 

Calculation:

Let figure given below denotes A and B as subsets of X.

Now, (A ∩ B’) is A intersection not B i.e., (A ∩ B’) = A – (A ∩ B) as shown below.

Similarly, (A’ ∩ B) is B intersection not A i.e., (A’ ∩ B) = B – (A ∩ B) as shown below.

Now, C = (A ∩ B’) ∪ (A’ ∩ B) means (A ∪ B) – (A ∩ B) as shown below.

Concept:

General term: General term in the expansion of (x + y)ⁿ is given by

• \({{\rm{T}}_{\left( {{\rm{r\;}} + {\rm{\;}}1} \right)}} = {\rm{\;}}{{\rm{\;}}^{\rm{n}}}{{\rm{C}}_{\rm{r}}} \times {{\rm{x}}^{{\rm{n}} - {\rm{r}}}} \times {{\rm{y}}^{\rm{r}}}\)

Formula used:

• ⁿC_r = ⁿC_{(n – r)}

Calculation:

General term in expansion of (1 + a)^{m + n}

T_{(r + 1)} = ⁿC_r a^r

For coefficient of a^m, r = m

α = ^{(m + n)}C_m

For coefficient of aⁿ, r = n

β = ^{(m + n)}C_n

⇒ β = ^{(m + n)}C_{(m + n - n)}

⇒ β = ^{(m + n)}C_m

∴  β = α

Concept:

Logarithmic properties:

• Log_a (MN) = log_a M + log_b N
• Log_a a^p = p
• Log_a M^p = p × log_a M

where a ≠ 1, a > 0 and b ≠ 1, b > 0 and M, N are arbitrary positive numbers and p is any real number.

Calculation:

To find: x

x + log₁₅ (1 + 3^x) = x log₁₅ 5 + log₁₅ 12

Using log₁₅ 15^x = x

⇒ log₁₅ 15^x + log₁₅ (1 + 3^x) = xlog₁₅ 5 + log₁₅ 12

⇒ log₁₅ 15^x + log₁₅ (1 + 3^x) = log₁₅ 5^x + log₁₅ 12      (Using xlog_a b = log_ab^x)

⇒ log₁₅ {15^x(1 + 3^x)} = log₁₅ (5^x × 12)      [Using log_a M + log_a N = log_a (M × N)]

⇒ 15^x(1 + 3^x) = 5^x × 12

⇒ (15/5)^x × (1 + 3^x) = 12

⇒ 3^x × (1 + 3^x) = 3 × 4

⇒ 3^x-1 (1 + 3^x) = (1 + 3)

⇒ 3^x-1 (1 + 3^x) = 3⁰ × (1 + 3)

On comparing powers

⇒ x = 1

Concept:

Permutation:

Arrangement of n different things taken r at a time = ⁿP_r \(= \frac{{n!}}{{\left( {n - r} \right)!}}\)

Calculation:

For any number to be divisible by 10, the last digit of the number should be 0.

Now,

 

So, we have 3 places and 4 choices to be arranged in that places.

Numbers can be formed = ⁴P₃

\(= \frac{{4!}}{{\left( {4 - 3} \right)!}}\)

= 4!

= 4 × 3 × 2 × 1

= 24

Alternate method:

For 1^st place we have 4 choices (1, 5, 6, 7), for 2^nd place we will have only 3 choices as repetition of digit is not allowed and for 3^rd place only 2 choices.

∴ Number of digits that can be formed = 4 × 3 × 2

= 24