Mathematics Mock Test - 2

Concept:

• By the property of the determinant of adj A,

A adj A  = | A |I

⇒ |A adj A|  = || A |I| 

⇒ |A||adj A|  = | A |ⁿ

⇒ |adj A | = | A |^n-1

Where n is the order of the A.

Calculation:

Given:

• If A is a singular matrix

⇒ | A | = 0

⇒ | adj A | = 0 

⇒ adj A is also singular. 

• So, option 1 is correct.

Concept:

Order of a differential equation is the highest order of derivative that occurs in the differential equation.

Degree of a differential equation is the highest power of the highest order derivative that occurs in the equation, after all the derivatives are converted into rational and radical free form.

Calculation:

Getting rid of the radicals by raising both the sides to power 2 will give us:

\(\rm x^2=1+(\frac{d^2y}{dx^2})^1\)

The highest derivative in it is \(\rm \frac{d^2y}{dx^2}\), therefore its order is 2.

∴ The highest power of this derivative in the equation is 1, therefore its degree is 1.

Concept:

Mean = E[X] 

\(E\left[ X \right] = \mathop \smallint \limits_{ - \infty }^{ + \infty } x.{f_x}\left( x \right)dx\)

\(E\left[ {{X^2}} \right] = \mathop \smallint \limits_{ - \infty }^{ + \infty } {x^2}{f_x}\left( x \right)dx\)

Where f_x(x) is the probability density function.

Analysis:

\(Var\left( X \right) = E\left[ {{{\left( {X - {\mu _x}} \right)}^2}} \right]\)

\( = E\left[ {{X^2} - 2{\mu _x}X + \mu _x^2} \right]\)

\( = E\left[ {{X^2}} \right] - 2E\left[ {{\mu _x}X} \right] + E\left[ {\mu _x^2} \right]\)

\( = E\left[ {{X^2}} \right] - 2{\mu _x}E\left[ X \right] + \mu _x^2\)

\( = E\left[ {{X^2}} \right] - 2\mu _x^2 + \mu _x^2\)

\( = E\left[ {{X^2}} \right] - \mu _x^2\)

Since, \( E[X]= \mu _x\)

\(Var\left[ X \right] = E\left[ {{X^2}} \right] - {E^2}\left[ X \right]\)

Concept:

The derivative of a function may be used to determine whether the function is increasing or decreasing on any intervals in its domain. 

If f′(x) > 0, then f is increasing on the interval, and if f′(x) < 0, then f is decreasing on the interval.

Calculations:

Consider the function f(x) = 6x - x2, x > 0

The derivative of a function may be used to determine whether the function is increasing or decreasing on any intervals in its domain. ... If f′(x) > 0, then f is increasing on the interval, and if f′(x) < 0, then f is decreasing on the interval.

⇒f'(x) = 6 - 2x

For increasing function f'(x) > 0

⇒ 6 - 2x > 0

⇒ x < 3

Here, x > 0 and x < 3

⇒ x \(\in (0, 3)\)

Hence, the function f(x) = 6x - x2, x > 0 is increasing in the interval (0, 3).

Concept:

Binomial distribution:

It gives the probability of happening of event 'r' times exactly in 'n' trials.

P(r) = ⁿC_r(p)^r(q)^{n - r}

where n = number of trials, r = number of favourable events, p = probability of happening of an event and q = (1 - p) probability of not happening of an event.

Calculation:

Given:

n = 5 (tossed 5 times), r = 2 (exactly two heads), p = 1/2 (probability of happening event) and q = 1/2 (probability of not happening of an event)

P(r) = nCr(p)r(q)n - r

∴ probability of head occurring exactly 2 times is P(2).

\(P(2)={5_{{C_2}}}\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^3\Rightarrow\frac{5}{16}\)

CONCEPT:

• The patterns and behavior of the data in any time series are based on four components.

1.  Secular trend component (T) –

• ​Also known as the trend series, is the smooth, regular, and long-term variations of the series observed over a long period of time. 
• The figure below shows an upward trend for annual electricity consumption per household in a certain residential locality from the years 1990 – to 2002. In general, trend variations can be either linear or non-linear.

• Seasonal component (S)–
  • When a time series captures the periodic variability in the data, capturing the regular pattern of variability; within one-year periods. The main causes of such fluctuations are usually climate changes, seasons, customs, and habits which people follow at different times.
  • The figure shows seasonal electricity consumption and variations in peak demand in Nepal and India in the year 2018.

• Cyclical component (C)–
  • When a time series shows an oscillatory movement where the period of oscillation is more than a year where one complete period is called a cycle.
•  Irregular component (I)–
  • These kinds of fluctuations are unaccountable, unpredictable, or sometimes caused by unforeseen circumstances like – floods, natural calamities, labor strikes, etc.
  • Such random variations in the time series are caused by short-term, unanticipated, and nonrecurring factors that affect the time series.

EXPLANATION:

• To analyze these four components we use two types of models.
• Additive model: When all the four components of the time series are independent of each other.

⇒ The original value of trend (O) = T + S +C +I

• Multiplication model: When all the four components' time series are dependent on each other.

⇒ The original value of trend (O) = T × S × C × I

So, the correct answer is option 2.

Concept:

For two vectors \(\vec m \ and \ \vec n \) to be collinear,​ \(\vec m\; = \;λ \vec n\) where λ is a scalar.

Calculation:

Given that, the vectors \(4\hat i + \hat j - 3\hat k\) & \(p\hat i + q\hat j - 2\hat k\) are collinear.

Since two vectors \(\vec m \ and \ \vec n \) are collinear then \(\vec m\; = \;λ \vec n\) where λ is a scalar.

⇒ \(4\hat i + \hat j - 3\hat k\;\ = λ × (\;p\hat i + q\hat j - 2\hat k)\)

⇒ \(4\hat i + 1\hat j - 3\hat k\;\ = λ p \hat i + λq \hat j - 2λ \hat k\)

⇒ λp = 4,  λq = 1 and -2λ = -3

⇒  λ = 3/2 

So, by substituting λ = 3/2 in  λp = 4 and λq = 1, we get

⇒ (3/2)p = 4 and (3/2)q = 1

⇒ p = 8/3 and q  = 2/3

∴  \(\frac{8}{3}, \frac{2}{3}\)is the correct answer.

Concept:

• Intersection of Sets: Intersection of two given sets is the largest set which contains all the elements that are common to both the sets.
• The intersection of two sets A and B, denoted by A ∩ B

Calculation:

Given:

aN = {ax : x∈N}

Taking a = 2 and 5, we get

⇒ 2N = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20 ….}

⇒ 5N = {5, 10, 15, 20, 25 ….}

Hence 2N ∩ 5N = {10, 20 ….}

∴ 2N ∩ 5N = 10N

Calculations:

Consider the equation (1 + ex) ydy = ex dx

⇒ y dy = \(\rm \dfrac {e^x}{1+e^x}dx\)

variables are seperated, Integrating w.r.to x, we get

⇒ \(\rm ∫\)y dy = \(\rm ∫ \dfrac {e^x}{1+e^x}dx\)

Let 1 + e^x = t

⇒ e^xdx = dt

⇒ \(\rm \dfrac {y^2}{2} \) = \(\rm ∫ \frac{dt}{t}\)

⇒ \(\rm \dfrac {y^2}{2} \) = log t + log c

⇒ \(\rm \dfrac {y^2}{2} \) = log ct

⇒ \(\rm y^2\) = 2 log ct

⇒ \(\rm y^2\) = log [c²t²]

∴ y2 = ln [c2 (ex + 1)2]

CONCEPT:

• By the definition of the inverse of the matrix,

\( A^{ - 1} = \frac{Adj \ A}{|A|} \)

• Where Adj A will be the transpose of the matrix formed from the cofactors of all the elements A.

CALCULATION:

Given:

∵​ \(I_3= \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right]\)

• The cofactor of all the diagonal elements will be the same as 1 and the cofactors of all the non-diagonal elements will be 0 also.

\(\Rightarrow adj\left( {{I_3}} \right) = \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right]\)

Also, | I3 | = 1

• So, by the definition of the inverse of the matrix,

\(\therefore I_3^{ - 1} = \frac{Adj \ I_3}{|I_3|} = I_3\)

So, option 3 is correct. 

Explanation:

The equations can be re-written as

\(\frac{{{x_1}}}{{500}} + \frac{{{x_2}}}{{\begin{array}{*{20}{c}} {1000}\\ \end{array}}} = 1\)       ---(1)

\(\frac{{{x_1}}}{{800}} + \frac{{{x_2}}}{{800}} = 1\)        ---(2)

x₁ = 400       ---(3)

x₂ = 700       ---(4)

Now, these equations can be plotted on a graph,

The optimum value of the objective function is at one of the extreme points of the feasible region.

We can get the points by solving equations.

 ∴ Optimum values are x₁ = 200, x₂ = 600

Concept:

For maximum -

∘ f'(x) = 0 

∘ f''(x) < 0 at x where f'(x) = 0 (At the critical points)

For minimum -

∘ f'(x) = 0 

∘ f''(x) > 0 at x where f'(x) = 0

Calculation:

Given: f(x) =  x + x-1

⇒ f'(x) = 1 - 1/x²

• For critical points, f'(x) = 0 

⇒ 1 - 1/x2 = 0

⇒ x = ±1

• Now we have got, f''(x) = 2/x³
• At x = 1, f''(x) = 2 > 0 hence minimum
• At x = -1, f''(x) = -2 < 0 hence maximum
• For maximum and minimum value:

⇒  max f(x) < min f(x) 

• ∴ f(x) has both a maximum and minimum and max f(x) < min f(x) 
• So, the correct answer is option 3.

Concept:

Variable Separable Form:

If it is possible to write a differential equation in the form of f(x) dx = g(y) dy where, f(x) is a function of x and g(y) is a function of y, then we say that the variables are separable.

This kind of differential equations can be solved by integrating both sides. The solution is given by \(\int f\left( x \right)~dx=~\int g\left( y \right)~dy+C\)

Calculation:

\(\frac{dx}{dt}~\propto x\)        ---(Given)

\(\Rightarrow \frac{dx}{dt}=k~x\)

As it is given that, proportional constant equal to 2

\(\Rightarrow \frac{dx}{dt}=2x\)

\(\Rightarrow \frac{dx}{2x}=dt\)

Now, by integrating both the sides we get

\(\Rightarrow ~\int \frac{dx}{2x}=~\int dt\)

\(\Rightarrow \ln \sqrt{x}=~t+{{C}_{1}}\)

Now, raising both the side to the power of e, we get

Concept:

Let us consider the system of linear equations:

a₁₁ × x + a₁₂ × y = b₁

a₂₁ × x + a₂₂ × y = b₂

We can write these equations in matrix form as: A X = B, where \(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}\\ {{a_{21}}}&{{a_{22}}} \end{array}} \right],\;X = \left[ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right]\;and\;B = \left[ {\begin{array}{*{20}{c}} {{b_1}}\\ {{b_2}} \end{array}} \right]\)

Case -1: If A is a non-singular matrix then |A| ≠ 0.

Then, X = A^-1 B where A^-1 will exist if and only if |A| ≠ 0 and it is given by: \({A^{ - 1}} = \frac{{adj\;\left( A \right)}}{{\left| A \right|}}\)

Case – 2: If A is a singular matrix then |A| = 0.

In this case, we have to calculate (adj (A)) × B.

If (adj (A)) × B ≠ O, where O is the null matrix then the system of equations is inconsistent and has no solution.

If (adj (A)) × B = O, where O is the null matrix then the system of equations is consistent and has infinitely many solutions.

Calculation:

Given: x + 3y = 5 and 2x + ky = 8

We can write these equations in matrix form as: A X = B, where \(A = \left[ {\begin{array}{*{20}{c}} 1&3\\ 2&k \end{array}} \right],\;X = \left[ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right]\;and\;B = \left[ {\begin{array}{*{20}{c}} 5\\ 8 \end{array}} \right]\)

In order to say that the given system of linear equations is consistent and has unique solution, |A| ≠ 0.

⇒ |A| = k – 6 ≠ 0 ⇒ k ≠ 6

Concept:

\(\rm\vec a.\vec b = |\vec a||\vec b|\cos \theta\)

\(\rm\text{If A = (x, y, z) and B = (x', y', z') then }\\ \vec {AB} = (x' - x)\hat i + (y' - y)\hat j+(z' - z)\hat k\\ |\vec {AB}|=\sqrt{(x' - x)^2 + (y' - y)^2+(z' - z)^2} \)

Calculation:

Here, OA and BC are diagonals of cube 

O (0, 0, 0), A = (a, a, a), B = (0, 0, a), C = (a, a, 0)

\(\rm\vec {OA} = (a - 0)\hat i + (a - 0)\hat j+(a - 0)\hat k\\ =a\hat i + a \hat j+a \hat k\\ |\vec {OA}|=\sqrt{a^2 +a^2 +a^2 }=\sqrt{3} a\\ \rm\vec {BC} = (a - 0)\hat i + (a - 0)\hat j+(0 - a)\hat k\\ =a\hat i + a \hat j-a \hat k\\|\vec {BC}|=\sqrt{a^2 +a^2 +(-a)^2 }=\sqrt{3} a\\\)

Now,

 \(\rm\vec {OA} .\rm\vec {BC} =|\rm\vec {OA}||\rm\vec {BC} | \cos \theta \\ \Rightarrow a^2 +a^2 -a^2=\sqrt 3a\times \sqrt 3a \cos \theta \\ \Rightarrow a^2=3a^2\cos \theta \\ \Rightarrow 3\cos \theta=1 \)

Hence, option (4) is correct.

Concept:

Scalar Triple Product:

A scalar triple product is also called a box product.

It is evident that scalar triple product of vectors means the product of three vectors. It means taking the dot product of one of the vectors with the cross product of the remaining two. It is denoted as [a b c ] = a · [b × c]

The product is cyclic in nature ⇔ [ a b c ] = [ b c a ] = [ c a b ]

Properties of the scalar triple product:

In a scalar triple product, dot and cross can be interchanged without altering the order of occurrences of the vectors ⇔ a · [b × c] = [a × b] ∙ c

Three vectors are coplanar if and only if their Scalar Triple Product is zero.

The Scalar Triple Product of three vectors is zero if any two of them are parallel.

Calculation:

\(\rm [\vec{a} + 2\vec{b} - \vec{c}, \vec{a} - \vec{b}, \vec{a}-\vec{b}- \vec{c}]\\=\rm[\vec {a}\;\vec {b}\;\vec {c}]\begin{vmatrix} 1 & 2 & -1\\ 1 & -1 & 0\\ 1 & -1 & -1 \end{vmatrix}\\=[\vec {a}\;\vec{b}\;\vec {c}][1(1-0)-2(-1-0)-1(-1+1)]\\=3[\vec {a}\;\vec{b}\;\vec {c}]\)

Alternate solution:

\(\rm [\vec{a} + 2\vec{b} - \vec{c}, \vec{a} - \vec{b}, \vec{a}-\vec{b}- \vec{c}]\\=(\vec{a} +2\vec{b} - \vec{c}) \cdot [(\vec{a} - \vec{b}) \times (\vec{a}-\vec{b}- \vec{c})]\)

\(\rm =(\vec{a} - 2\vec{b} - \vec{c}) \cdot [\vec{a} \times \vec{a}-\vec{a} \times \vec{b}-\vec{a} \times \vec{c}-\vec{b} \times \vec{a}+\vec{b} \times \vec{b}+\vec{b} \times \vec{c}]\)

As we know, cross product of parallel vectors is zero.

\(\rm =(\vec{a} + 2\vec{b} - \vec{c}) \cdot [0-\vec{a} \times \vec{b}-\vec{a} \times \vec{c}+\vec{a} \times \vec{b}+0+\vec{b} \times \vec{c}]\)          \((\because \rm \vec{a} \times \vec{b}= -\vec{b} \times \vec{a})\)

\(\rm =(\vec{a} + 2\vec{b} - \vec{c}) \cdot [-\vec{a} \times \vec{c}+\vec{b} \times \vec{c}]\)

\(\rm =-\vec{a}\cdot[\vec{a} \times \vec{c}] - 2\vec{b}\cdot[\vec{a} \times \vec{c}] + \vec{c} \cdot[\vec{a} \times \vec{c}]+\vec{a}\cdot[\vec{b} \times \vec{c}] + 2\vec{b}\cdot[\vec{b} \times \vec{c}] - \vec{c} \cdot[\vec{b} \times \vec{c}]\)

The Scalar Triple Product of three vectors is zero if any two of them are parallel.

\(\rm =0 - 2\vec{b}\cdot[\vec{a} \times \vec{c}] + 0+\vec{a}\cdot[\vec{b} \times \vec{c}] + 0 - 0\)

\(\rm = 2[\vec {a}\;\vec{b}\;\vec {c}] + [\vec {a}\;\vec{b}\;\vec {c}]\\=3[\vec {a}\;\vec{b}\;\vec {c}]\)

Calculation:

f(x) = |x + 1| + |x + 10|

|x + 1| = 0 then x = -1

|x + 10| = 0 then x = -10

f(x) can be zero when x = -1 or x = -10

f(-1) = |-1 + 1| + |-1 + 10| = 9

f(-10) = |-10 + 1| + |-10 + 10| = 9

∴ Minimum value of f(x) = 9 at x = -1 or x = -10

Important Points

If f(x) = |x + a| + |x + b| then

|x + a| and |x + b| = 0

they can be individually zero but can not be zero at same time.

Concept:

The distance of a point (p, q, r) from the plane ax + by + cz + d = 0

D = \(\rm \left|ap+bq+cr + d\over \sqrt{a^2+b^2+c^2}\right|\)

Calculation:

Given the equation of the plane is 

2x - 3y + 6z - 42 = 0

The distance of the plane from the origin (0, 0, 0)

D = \(\rm \left|2\times0-3\times 0+6\times0 -42\over \sqrt{2^2+(-3)^2+6^2}\right|\)

D = \(\rm \left|-42\over \sqrt{4+9+36}\right|\) = \(\left|-42\over7\right|\)

D = 6

Concept:

If l, m, n are direction cosines and a, b, c are direction ratios, then \(\rm l=\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}, m=\frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}} \text { and } n=\frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}\)

Calculation:

A line has direction ratios a + b, b + c, c + a

Let, direction cosines be l, m, and n 

So,

\(\rm l=\frac{a+b}{\sqrt{(a+b)^{2}+(b+c)^{2}+(c+a)^{2}}}\)

\(m=\frac{b+c}{\sqrt{(a+b)^{2}+(b+c)^{2}+(c+a)^{2}}} \)

\(n=\frac{c+a}{\sqrt{(a+b)^{2}+(b+c)^{2}+(c+a)^{2}}}\)

\(\rm l^2+m^2+n^2=\rm (\frac{a+b}{\sqrt{(a+b)^{2}+(b+c)^{2}+(c+a)^{2}}})^2+(\frac{b+c}{\sqrt{(a+b)^{2}+(b+c)^{2}+(c+a)^{2}}})^2 +(\frac{c+a}{\sqrt{(a+b)^{2}+(b+c)^{2}+(c+a)^{2}}})^2\)

\(\\ =(\frac{(a+b)^{2}+(b+c)^{2}+(c+a)^{2}}{(a+b)^{2}+(b+c)^{2}+(c+a)^{2}})\)

= 1

Hence, option (4) is correct.

Calculation:

xe = e\(\rm x^2+y^2\) 

Taking log on both sides, we get

⇒ log xe = llog e\(\rm x^2+y^2\) 

⇒ e log x = (x² + y²) log e             

[∵ log mⁿ = n log m]

⇒ e log x = x2 + y2                       

[∵ log e = 1]

Differentiating w.r.t x, we get

⇒ e(\(\rm 1\over x\)) = \(\rm 2x + 2y{dy\over dx}\)

⇒ \(\rm {e\over x}- 2x = 2y{dy\over dx}\)

∴ \(\rm {dy\over dx}={e- 2x^2\over2xy}\)

Concept:

For a system of equation, the number of possible basic solution is calculated by - \({n_C}_m\)

n = number of variables.

m = number of equations.

Inequalities must be converted into equalities.

Calculation:

Given:

X1 + X2 + X3 ≤ 5

X₁ + X₂ + X₃ + S₁ + 0S₂ = 5     (1)

2X1 + X2 + 3X3 ≤ 10

2X1 + X2 + 3X₃ + 0S₁ + S₂ = 10      (2)

n = number of variables = 5

m = number of equations = 2

∴ number of basic solution = \({n_C}_m ⇒ {5_C}_2\)

∴ \(\frac{5!}{2!\;\times\;(5-2)!}\Rightarrow10\)

Concept:

• Marginal cost is the instantaneous rate of change of cost function with respect to the production output x.

Marginal cost = C'(x) = \( \frac{dC(x)}{dx}\) 

• Marginal revenue is the instantaneous rate of change of revenue function with respect to the production output x.

Marginal revenue = R'(x) = \( \frac{dR(x)}{dx}\) 

• Revenue function R(x) = x p(x)

Calculation:

Given:

Variable cost function, C(x) = x(12 - x3)

Price per unit p(x) = 6x - 2 

Revenue function R(x) = x p(x)

⇒ R(x) = x(6x - 2)

⇒ R(x) = 6x2 - 2x

• Marginal revenue,

⇒ R'(x) = \( \frac{dR(x)}{dx}\) 

⇒ R'(x)= \( \frac{d(6x^2 - 2x)}{dx}=12x - 2\)

⇒ Marginal revenue at (x = 3) = 12 × 3 - 2 = 34 Rs

• Therefore the marginal revenue when 3 units were both produced and in demand will be 34 Rs.

Given:

Amount of sugar and water in the solution = 25% and 75%

The total amount of solution = 100 L

New percentage of sugar in the mixture = 50%

Calculation:

Amount of sugar in the initial solution = 25% × 100 L = 25 L

Amount of water in the initial solution = 75% × 100 L = 75 L

Let x litre of sugar solution is added.

Hence,

(25 + x)/(100 + x) × 100 = 50

⇒ (25 + x)/(100 + x) = 1/2

⇒ 50 + 2x = 100 + x

⇒ x = 50 lt

∴ The amount of sugar added into the mixture to get a 50% sugar solution is 50 L liquid sugar.

Amount of water in the initial solution = 75% × 100 L = 75 L

Since, in new solution, water percent = 50%, liquid sugar percent = 50%

So, 50% of new solution = 75 L

⇒ total quantity of new solution = 100% = 75 × 2 = 150 L

⇒ Quantity of liquid sugar required = (150 - 100)L = 50 L

∴ The amount of sugar added into the mixture to get a 50% sugar solution is 50 L liquid sugar.

CONCEPT:

• Formula used:

P(E) = Favorable events ÷ total events

• In a pack of 52 cards, there will be 4 kings (2 of red and 2 of black color) and 4 ace cards ( 2 of red and 2 of black color) will be there. 
  Note that a card can not be an ace and king simultaneously.
•  The probability that the card drawn is either Ace or a King,

P(K∪ A) =  P(K) +  P(K) - P(K⋂A)       ...(1)

CALCULATION:

• When one card is drawn at random from a pack of 52 cards
  • Probability of choosing 1 king from the pack of cards = P(K) = 4/52 = 1/13
  • Probability of choosing 1 ace card from the pack of cards = P(A) = 4/52 = 1/13
  • The Probability of choosing 1 card which is king and ace both = P(K⋂A) = 0 
• Using the equation 1,

So, the correct answer will be option 3.

Concept:

Expected Value (or mean) of a Discrete Random Variable:
For a discrete random variable, the expected value usually denoted as μ or E(X), is calculated using

μ = E(X) = ∑ x_iP(x_i)

1. The formula means that we multiply each value x, in the support by its respective probability P(x) and then add them all together.

2. It can be seen as an average value but weighted by the likelihood of the value.

Calculation:

We know that, mean

μ = E(X) = ∑ x_iP(x_i)

⇒ E(X) = 0.2 + 0.7 + 0.75 + 0.6 + 0.25

⇒ E(X) = 2.5

Additional Information
The variance of a discrete random variable is given by:

Var(X) = σ²(X) = E(X²) - [E(X)]²

The formula means that first, we sum the square of each value times its probability then subtract the square of the mean.

Concept:

No solution: no feasible region available

Unique solution: a unique point

Unbounded: solution region is open and is not bounded.

Calculation:

z = 5x₁ + 3x₂

\({x_1}--{x_2} \le 2 \Rightarrow \frac{{{x_1}}}{2} - \frac{{{x_2}}}{2} \le 1\)

\({x_1} + {x_2} \ge 3 \Rightarrow \frac{{{x_1}}}{3} + \frac{{{x_2}}}{3} \ge 1\)

First quadrant: x₁ ≥ 0, x₂ ≥ 0

As can be seen, the feasible region

Rolle’s mean value theorem:

Suppose f(x) is a function that satisfies all of the following

1. f(x) is continuous on the closed interval [a, b]

2. f(x) is differentiable on the closed interval (a, b)

3. f(a) = f(b)

Then there is a number c such that a < c < b and f'(c) = 0

Let h(x) = f(x) – 2g(x)

Given that f(x) and g(x) are continuous as well as differentiable

Hence h(x) is also continuous as well as differentiable.

h(0) = f(0) – 2g(0) = 2 – 2(0) = 2

h(1) = f(1) – 2g(1) = 6 – 2(2) = 2

h(0) = h(1)

hence h(x) satisfies all the conditions of Rolle’s theorem

Now, h'(c) = 0

⇒ f'(c) – 2g'(c) = 0

Concept:-

• We know that area under  the probability density function (P.D.F) is equal to 1

\(⇒ 1 = \int_ {0 }^\infty \mathrm{f(x)}^{}\,\mathrm{d}x\)

Calculation:

Given: \(f(x) =​​\frac{e^-{\frac{x}{2}}} {K}\)

\(1 = \int_0^\infty \ { {e^{-x/2}}\,\mathrm\ \over k} {d}x\)

\(⇒ k = \int_0^\infty \ { {e^{-x/2}}\,\mathrm\ } {d}x\)

\(⇒ k = \left.\frac{e^{-x/2}}{{-1/2}}\right|_0^\infty\)

\(⇒ k = -2 [ 0- 1] \)

⇒ k = 2 

• So, the correct answer is  option 1.

Concept:

• As per the given information, the total cost function C(x) is given by, the sum of the fixed cost and the variable cost.
• Total cost function,

C(x)  = F + V(x) 

Where, Fixed cost = F, Variable cost =  V(x)    

Calculation:

Given: F = 1200 Rs, V(x) = 2x(x - 40)

• For minimum value of total cost,

C'(x) = 0

⇒ C'(x) = 4x - 80 = 0

⇒  4x - 80 = 0

⇒  x = 20

Now, 

C''(x) = 4 × 1 - 0 = 4 > 0 for x = 20

• So the cost will be minimum for x = 20
• So the correct answer will be an option 3.

CONCEPT:

Consider the system of m linear equations

a11 x1 + a12 x2 + … + a1n xn = 0

a21 x1 + a22 x2 + … + a2n xn = 0

am1 x1 + am2 x2 + … + amn xn = 0

• The above equations containing the n unknowns x1, x2, …, xn. To determine whether the above system of equations is consistent or not, we need to find the rank of the following matrix.

\(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}& \ldots &{{a_{1n}}}\\ {{a_{21}}}&{{a_{22}}}& \ldots &{{a_{2n}}}\\ \ldots & \ldots & \ldots & \ldots \\ {{a_{m1}}}&{{a_{m2}}}& \ldots &{{a_{mn}}} \end{array}} \right]\)

• A is the coefficient matrix of the given system of equations.

\(x = \frac{|A_x|}{|A|}\), \(y = \frac{|A_y|}{|A|}\), \(z = \frac{|A_z|}{|A|}\)

• Where, A_x, A_y, A_z is the coefficient matrix of the given system of equations after replacing the first, second, and third columns from the constant term column which will be having all the entries as 0.
• In the case of homogeneous equations, the determinants of, Ax, Ay, Az will be 0 definitely.
• So, for the system of homogeneous equations having the the-trivial solution, the determinant of A should be zero.
• The system of homogeneous equations has a unique solution (trivial solution) if and only if the determinant of A is non-zero.

CALCULATION:

For non - trivial solution, the |A| = 0

\(⇒ \;\left| {\begin{array}{*{20}{c}} 1&{ K}&3\\ 4&3&{ k}\\ 2&1&{ 2} \end{array}} \right|\; = \;0\)

⇒ 1(6 - K) - K(8 - 2K) + 3(4 - 6) = 0

⇒ 9K -  2K² = 0

⇒ k = 0 or \(\frac{9}{2}\)

So, the correct answer is option 1.

Concept:

• Formula: for Index number calculated by the simple aggregate method : \(\frac{Σ P_1}{Σ P_0} × 100\;\;\;\;\; \ldots \left( 1 \right)\)

Where,

ΣP₁ = The sum of all the prices of the current year 

ΣP₀ =The sum of all the prices of the current year 

Calculation:

Given:

• Data for the base year and the current year are in the table. So, referring to the table,

ΣP₁ = 90 + 115 + 75 + β 

⇒ΣP1 = 280 + β 

⇒ Σ P₀ = 75 + 85 + α + 100  

⇒ Σ P0 = 260 + α  

• So, referring equation 1,

\(\Rightarrow \frac{\Sigma P_1}{\Sigma P_0}\times 100=\frac{280 + \beta}{260 + \alpha} × 100\)

⇒ \(\frac{125}{100}\) = \(\frac{280 + \beta}{260 + \alpha}\)

⇒ \(\frac{5}{4}\) = \(\frac{280 + \beta}{260 + \alpha}\)

⇒ 4β - 5α = 180​

So, the correct answer is option 3.

Concept:

Let b ∈ Z such that a^b ≡ 1(mod n).

Let us apply division algorithm to b and k then we have,

b = kq + r where 0 ≤ r ≤ k

Consider, a^b = a^{kq + r} = (a^k)^q . a^r

By hypothesis ab ≡ 1(mod n) and ak ≡ 1(mod n).

Hence, ar ≡ 1(mod n) where 0 ≤ r ≤ k

∴ r has to be equal to zero and otherwise the choice of k has the smallest positive integer such that ak ≡ 1(mod n) will be contradicted.

Hence, b = qk

⇒ k | b

⇒ k divides b

Hence, the correct answer is option 2)

Concept:

• A sinking fund is a fund established by a company or business entity by setting aside revenue over a period of time to fund a future capital expense, or repayment of long-term debt.
• It is a fund that is accumulated for the purpose of paying off a financial obligation at some future designated date.

Concept:

If |x| ≤ a then - a ≤ x ≤ a

Calculations:

Given , |3x - 5| ≤ 2 

⇒ - 2 ≤ 3x - 5 ≤ 2

⇒ - 2 + 5 ≤ 3x  ≤ 2 + 5 

⇒ 3 ≤ 3x  ≤ 7

⇒\(\rm 1 \le x \le \dfrac{7}{3}\) 

Hence, if |3x - 5| ≤ 2 then then \(\rm 1 \le x \le \dfrac{7}{3}\)

Concept:

• When we integrate the density function over an interval, we get the probability for that interval, i.e.

P(a ≤ x ≤ b) = \(\mathop \smallint \limits_{ a }^b {f_x}\left( x \right)dx = 1\)

Where fx(x) is any given density function defined for a ≤ x ≤ b.

Explanation:

• In the case of a continuous random variable, the probability at a point = 0
• So, the correct answer will be option 3.

Put x = 1/t

⇒ dx = -1/t² dt    [By using substitution method]

\(I = \mathop \smallint \limits_2^{\frac{1}{2}} t\;sin\left( {\frac{1}{t} - t} \right)\left( { - \frac{1}{{{t^2}}}} \right)dt = \mathop \smallint \limits_{\frac{1}{2}}^2 - \frac{1}{t}sin\left( {t - \frac{1}{t}} \right)dt\)

I = -I

⇒ 2I = 0

Concept:

Reflexive: Each element is related to itself.

• R is reflexive if for all x ∈ A, xRx.

Symmetric: If any one element is related to any other element, then the second element is related to the first.

• R is symmetric if for all x, y ∈ A, if xRy, then yRx.

Transitive: If any one element is related to a second and that second element is related to a third, then the first element is related to the third.

• R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

R is an equivalence relation if A is nonempty and R is reflexive, symmetric and transitive.

Calculation:

Here,  R = {(a, b): a, b ϵ N and a2 = b}

1. Relation R is not reflexive since, 2 ≠ 2²

2. Since, 2² = 4 so,  (2, 4) belong to R

But,  4 ≠ 2 and so, R is not symmetric

3. Since, 4² = 16, so (4, 16) belong to R

Also, 16² = 256,  so (16, 256) belong to R

But, 4² ≠ 256  so R is not transitive.

So, R satisfies none of the reflexivity, symmetry and transitivity.

Hence, option (4) is correct. 

Concept:

\(\rm \int \frac{1}{1+x^2}dx=tan^{-1}x+c\)

\(\rm x^{-1}=\frac1x\)

Calculation:

Let, I = \(\rm \int \frac{1}{e^x+e^{-x}}dx\)

= \(\rm \int \frac{1}{e^x+\frac{1}{e^{x}}}dx\)                (∵ \(\rm x^{-1}=\frac1x\))

= \(\rm \int \frac{e^x}{e^{2x}+{1}}dx\)

Now, let e^x = t

⇒ ex dx = dt

∴ I = \(\rm \int \frac{dt}{t^2+{1}}\)

= \(\rm tan^{-1}t+c\)             (∵ \(\rm \int \frac{1}{1+x^2}dx=tan^{-1}x+c\))

= \(\rm tan^{-1}(e^x)+c\)         (∵ ex = t)

Hence, option (4) is correct. 

Calculation:

Draw the graph of y = cos x in the interval \(\rm 0

Let the required area be A.

Using the formula of the area under the curve, \(\rm A=\left | \int_{a}^{b}f(x)-g(x) dx \right |\)

\(\Rightarrow \rm A=\left | \int_{0}^{\frac{π}{2}}cos x dx \right |=\left | \left [ sin x \right ]_{0}^{\frac{π}{2}} \right |\)

Using the value of sin 0 = 0 and sin π/2 = 1

\(\Rightarrow \rm A=\left | sin {\frac{π}{2}} -sin0 \right |=\left | 1-0 \right |=1\)

Hence, the required answer is option 2.

Concept: 

A function is said to be differentiable at x0,

f'(x0) = lim (h→0⁺) (f(x0+ h) - f(x0))/h = lim (h→0-) (f(x0+ h) - f(x0))/h 

A function is said to be continuous at x0, if

 lim (h→0⁺) f(x0+ h) = f(x0) = lim (h→0^-) f(x0+ h) 

Calculation:

Let a function f(x) attains maxima at x₀,

Then f(x0) ≥  f(x) for all x in |x - x0| < r

Now the derivative at x0 is

f'(x0) = lim _(h→0) (f(x0+ h) - f(x0))/h for all |h|

Since, f(x0 + h) ≤ f(x0) ⇒ f(x0+ h) - f(x0) ≤ 0

Consider, RHD = lim (h→0⁺) (f(x0+ h) - f(x0))/h ≤ 0

And LHD = lim (h→0^-) (f(x0+ h) - f(x0))/h ≥ 0 

Since, LHD = RHD ⇒ f'(x0) = 0

Similarly, at minima, f'(x0) = 0.

So, statement I is true.

Now statement II,

Let a function is differentiable at x₀,

f'(x0) = lim (h→0) (f(x0+ h) - f(x0))/h 

Consider, lim (h→0) (f(x0+ h) - f(x0))

⇒ lim (h→0) (f(x0+ h) - f(x0)) × \(h\over h\)

⇒ lim (h→0) [(f(x0+ h) - f(x0))/h] × h

⇒ lim (h→0) (f(x0+ h) - f(x0)) = lim (h→0) f'(x₀) × h

⇒ lim (h→0) (f(x0+ h) - f(x0)) =  f'(x0) × 0 = 0

⇒ lim (h→0) f(x0+ h) = lim (h→0)f(x0)

⇒ lim (h→0) f(x0+ h) = f(x0)

So, function is continuous when it differentiable.

⇒ Statement II is also true.

∴ The correct answer is option (3).

Concept:

\(A = {\left[ {{a_{ij}}} \right]_{2 × 2}}\) means the given matrix has 2 rows and 2 column as [A]_{m × n} signifies the size of the matrix.

a_ij represents the element's location of a matrix.

\(A=\left[ {\begin{array}{*{20}{c}} {a_{11}}&a_{12}\\ {a_{21}}&a_{22} \end{array}} \right]\)

Calculation:

Given:

\({a_{ij}} = \frac{1}{2}\left( {3i - 2j} \right)\)

\( \Rightarrow {a_{11}} = 1/2,\,{a_{12}} = - 1/2\,and\,{a_{21}} = 2,\,{a_{22}} = 1\)

\(\therefore A = {\left[ {{a_{ij}}} \right]_{2 × 2}} = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}\\ {{a_{21}}}&{{a_{22}}} \end{array}} \right]\)

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

Concept:

Area bounded by function f(x) and g(x) is given as,

Area \(= \left| {\mathop \smallint \limits_{{x_1}}^{{x_2}} \left\{ {f\left( x \right) - g\left( x \right)} \right\}dx} \right|\).

And to find x₁ and x₂ we need to solve for f(x) = g(x).

Calculation:

Given: f(x) = x and g(x) = x² +2

x₁ = -1 and x₂ = 1

Plot a rough graph of given curves and shade the bounded area to be calculated.

Area \(= \left| {\mathop \smallint \limits_{{x_1}}^{{x_2}} \left\{ {f\left( x \right) - g\left( x \right)} \right\}dx} \right|\) 

\(= \left| {\mathop \smallint \limits_{ - 1}^1 \left\{ {x - {x^2} - 2} \right\}dx} \right|\)

\(= \left| {\frac{{{x^2}}}{2} - \frac{{{x^3}}}{3} - 2x} \right|_{ - 1}^1\)

\(= \left| {\frac{1}{2} - \frac{1}{3} - 2 - \left( {\frac{1}{2} + \frac{1}{3} + 2} \right)} \right|\)

= | – 1/3 – 2 – 1/3 – 2|

= | – 2/3 – 4|

= 14/3

CONCEPT:

Addition or subtraction of matrix:

• Two matrices may be added or subtracted only if they have the same dimension; that is, they must have the same number of rows and columns.
• Addition or subtraction is accomplished by adding or subtracting corresponding elements. For example, consider matrix A and matrix B.

Multiplication of matrices:

• Let A be a matrix of m × n, B be a matrix of p × q then AB is a matrix of m × q, and for multiplication to exist n = p.

EXPLANATION:

Given:

A is having order 3 × 3, B is having order 3 × 2, C is having order 1 × 3.

• Definitely, B'B will be possible as both will be having the same order as 3 × 3
• Definitely, C_{1 × 3}A_{3 × 3}B _{3 × 2} will be possible as this is satisfying the condition for multiplication and its order will be 1 × 2.
• A + B' is not satisfying the condition for addition so option 3 is correct.
• A2 + A will be a square matrix of order 3 × 3 as both the added matrix are of the same order.

Additional Information 

Properties of matrix multiplication:

• Associative property of matrix multiplication:
  • A(BC) = (AB)C
• Distributive property of multiplication:
  • A(B + C) = AB + AC
  • (A + B)C = AC + BC
• AIn = InA = A,  In is the appropriate identity matrix
  • c(AB)= (cA)B = A(cB)

Note:  AB ≠ BA in general, i.e. multiplication of matrices is not commutative even if the matrix is square and of the same order.

Properties of Matrix addition and scalar multiplication:

• Commutative Property of addition
  • A + B = B + A
• Associative Property of addition
  • A + (B + C) = (A + B) + C
  • A + O = O + A Where O is the appropriate zero matrix
• Distributive Property of addition
  • c(A + B) = cA + cB
  • (a + b)C = aC + bC

Concept:

• This given parabola is opening upward parabola as k > 0
• Referring to the diagram,
• We have to find the shaded area as shown.
• We can calculate one-half of the shaded area and double it for the complete area.

​\(Area (A) = \rm 2 |\int_0^3 x dy| = 2 |\int_0^3 √{ky} dy|\)

Calculation:

Given: A  =  12 unit2.

Area = \(\rm 2 \int_0^3 x dy = 2 \int_0^3 √{ky} dy =\)

\(Area (A) =| \left[ 2 √ k . \frac{y^{3/2}}{3/2} \right]_0^3|\)

 \(\Rightarrow A = \frac{4 √ k}{3} . 3 √ 3\)

⇒12 = 4√k × √3 

⇒ √k = √3 

∴ k = 3

Concept:

• \(\rm \int \frac{dx}{x^{2}+a^{2}}=\frac{1}{a}tan^{-1}\frac{x}{a}+C\)
• \(\rm \int_{0}^{x}{f(x) dx}=F(x)-F(0)\) where F(x) is the anti-derivative of f(x).

Calculation:

Given: \(\rm f(x)=\int_{0}^{x} \frac{dt}{t^{2}+9}\)

Using the formula, \(\rm \int \frac{dx}{x^{2}+a^{2}}=\frac{1}{a}tan^{-1}\frac{x}{a}+C\)

\(\rm \Rightarrow f(x)=\int_{0}^{x} \frac{dt}{t^{2}+9}=\int_{0}^{x} \frac{dt}{t^{2}+3^{2}}= \frac{1}{3}(tan^{-1}\frac{x}{3}-tan^{-1}\frac{0}{3})=\frac{1}{3}(tan^{-1}\frac{x}{3}-tan^{-1}0)\)

Now, we know that tan^-1 0 = 0

\(\rm \Rightarrow f(x)=\frac{1}{3}tan^{-1}\frac{x}{3}\)

Hence, the correct answer is option 1.

Concept:

\(\rm tan^{-1}x+tan^{-1}y =tan^{-1}\frac{x+y}{1-xy}, \ xy<1\)

\(\rm tan^{-1}x+tan^{-1}y =\pi +tan^{-1}\frac{x+y}{1-xy}, \ xy>1\)

tan(π/4) = 1, tan(π -π/4) = tan(3π/4) = -1

Calculation:

Given: tan-1(2), tan-1(3) are two angles of a triangle

Let, C be the third angle of a triangle.

As we know, sum of angle in triangle is equal to 180° 

So, tan-1(2) + tan-1(3) + C = π 

Here, 2 × 3 = 6 > 1

\(π + \rm tan^{-1}\frac{2+3}{1-(2)(3)}+C=π\\ =π + \rm tan^{-1}\frac{5}{-5}+C=π\)

⇒ π + tan^-1(-1) + C = π 

⇒ π - π/4 + C = π 

⇒ C = π - 3π/4

⇒ π/4 

Hence, option (2) is correct.

Concept:

The domain is the range of inverse trigonometric functions.

Calculation:

Given: f(x) = y = cos sin-1 (x + 1) 

• For the domain of the function the two conditions should be satisfied and these are,

⇒ - ∞ -1 (x + 1) < + ∞      ...(1) and |x + 1| ≤ 1      ...(2)

• Now we have to find the common values of x out of these two conditions.

⇒ -1 ≤  x + 1 ≤ 1 

⇒ -2 ≤  x  ≤ 0  

• Now for these values of x, the first condition will be automatically satisfied so the domain of the given function will be,

⇒ -2 ≤  x  ≤ 0  

∴ x ∈ [-2, 0]

• So, the correct answer is option 1.

Concept:

Area of a triangle with vertices (x1, y1) , (x2, y2), (x3, y3) is given by

​Area = \(\rm \dfrac 12 \rm \begin{vmatrix} \rm x_1&\rm y_1 &1 \\ x_2& \rm y_2&1 \\ \rm\rm x_3 &\rm y_3&1 \end{vmatrix}\)

Calculations:

Given that, vertices of triangle are (-3, 0), (3, 0) and (0, k)

By using the above formula,

​⇒ Area = \(\rm \dfrac 12 \rm \begin{vmatrix} \rm -3&\rm 0 &1 \\ 3& 0&1 \\ 0 &k&1 \end{vmatrix}\)

⇒ Area = \(\dfrac 12\)[-3(0 - k) - 0 + 1(3k)]

⇒ Area = 3k

According to the question, area of triangle is 9 square unit,

⇒ 3k = 9

⇒ k = 3

∴ Required value of k is 3 unit.

Concept:

Derivatives of Trigonometric Functions:

Chain Rule of Derivatives:

• \(\rm \frac{d}{dx}f(g(x))=\frac{d}{d\ g(x)}f(g(x))× \frac{d}{dx}g(x)\).
• \(\rm \frac{dy}{dx}=\frac{dy}{du}× \frac{du}{dx}\).

Calculation:

We have y = cos (sin2 x).

Differentiating w.r.t. x, we get:

\(\rm\frac{dy}{dx}\) = - [sin (sin² x)] × (2 sin x) (cos x)

At x = \(\rm\frac{\pi}{2}\), we get:

\(\rm\frac{dy}{dx}\) = - [sin (sin2 \(\rm\frac{\pi}{2}\))] × (2 sin \(\rm\frac{\pi}{2}\)) (cos \(\rm\frac{\pi}{2}\)) = - [sin (1)] × (2 × 1) (0) = 0.